Gate by Rk Kanodia
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GATE EC BY RK Kanodia MULTIPLE CHOICE QUESTION Electronics & Communication Engineering Fifth Edition R. K. Kanodia...
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GATE EC BY RK Kanodia
MULTIPLE CHOICE QUESTION
Electronics & Communication Engineering Fifth Edition R. K. Kanodia B.Tech.
NODIA & COMAPNY JAIPUR
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GATE EC BY RK Kanodia
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GATE EC BY RK Kanodia
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GATE EC BY RK Kanodia
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GATE EC BY RK Kanodia
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GATE EC BY RK Kanodia
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GATE EC BY RK Kanodia
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GATE EC BY RK Kanodia
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GATE EC BY RK Kanodia
CHAPTER
1.1 BASIC CONCEPTS
1. A solid copper sphere, 10 cm in diameter is deprived of 10
20
electrons by a charging scheme. The charge on
the sphere is (A) 160.2 C
(B) -160.2 C
(C) 16.02 C
(D) -16.02 C
(A) 1 A
(B) 2 A
(C) 3 A
(D) 4 A
6. In the circuit of fig P1.1.6 a charge of 600 C is delivered to the 100 V source in a 1 minute. The value of v1 must be v1
2. A lightning bolt carrying 15,000 A lasts for 100 ms. If the lightning strikes an airplane flying at 2 km, the charge deposited on the plane is (A) 13.33 mC
(B) 75 C
(C) 1500 mC
(D) 1.5 C
60 V
20 W
100 V
3. If 120 C of charge passes through an electric
Fig. P.1.1.6
conductor in 60 sec, the current in the conductor is (A) 0.5 A
(B) 2 A
(C) 3.33 mA
(D) 0.3 mA
(A) 240 V
(B) 120 V
(C) 60 V
(D) 30 V
4. The energy required to move 120 coulomb through 7. In the circuit of the fig P1.1.7, the value of the
3 V is (B) 360 J
(C) 40 J
(D) 2.78 mJ
voltage source E is 0V +
+
(A) 25 mJ
–
–
2V
5. i = ?
+ – +
5V
–
–
4V
i
E
+
1A
1V
10 V
2A
5A
Fig. P.1.1.7 3A
4A
Fig. P.1.1.5
(A) -16 V
(b) 4 V
(C) -6 V
(D) 16 V
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GATE EC BY RK Kanodia
UNIT 1
8. Consider the circuit graph shown in fig. P1.1.8. Each
Networks
12. v1 = ?
branch of circuit graph represent a circuit element. The
v1 –
+
value of voltage v1 is
1 kW
2
7V
+ 105 V –
– 15 V +
+
+
55 V + v1 –
–
8V
35 V
–
–
5V
3
6V
kW
+
–
100 V
– 10 V + +
65 V
kW
V
30
V
30
4 kW
+
–
Fig. P1.1.12 Fig. P.1.1.8
(A) -30 V
(B) 25 V
(C) -20 V
(D) 15 V
9. For the circuit shown in fig P.1.1.9 the value of
(A) -11 V
(B) 5 V
(C) 8 V
(D) 18 V
13. The voltage vo in fig. P1.1.11 is always equal to 1A
voltage vo is 5W + vo
4W
+ vo –
1A
15 V
5V
Fig. P1.1.11
–
Fig. P.1.1.9
(A) 10 V
(B) 15 V
(C) 20 V
(D) None of the above
(A) 1 V
(B) 5 V
(C) 9 V
(D) None of the above
14. Req = ? 5W
10 W
10 W
10 W
10. R1 = ? 60 W
Req
10 W
+
10 W
10 W
up to ¥
R1 100 V R2
+ 20 V –
70 V
Fig. P1.1.14
–
Fig. P.1.1.10
(A) 11.86 W
(B) 10 W
(C) 25 W
(D) 11.18 W
15. vs = ? (A) 25 W
(B) 50 W
(C) 100 W
(D) 2000 W
180 W + 60 W
11. Twelve 6 W resistor are used as edge to form a cube.
vs
The resistance between two diagonally opposite corner of the cube is 5 (A) W 6 (C) 5 W
Page 4
(B)
90 W
6 W 5
(D) 6 W
40 W
20 V –
Fig. P.1.1.15
(A) 320 V
(B) 280 V
(C) 240 V
(D) 200 V
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180 W
GATE EC BY RK Kanodia
UNIT 1
24. Let i( t) = 3te-100 t A and v( t) = 0.6(0.01 - t) e -100 t V for
Networks
28. vab = ?
a
the network of fig. P.1.1.24. The power being absorbed
2 8
–
b 0.3i1
A 2
W
W
N
0.2i1
i1
A
6
+
W
R
i v
2
by the network element at t = 5 ms is
Fig. P.1.1.24
(A) 18.4 mW
(B) 9.2 mW
(C) 16.6 mW
(D) 8.3 mW
Fig. P.1.1.28
25. In the circuit of fig. P.1.1.25 bulb A uses 36 W when lit, bulb B uses 24 W when lit, and bulb C uses 14.4 W
(A) 15.4 V
(B) 2.6 V
(C) -2.6 V
(D) 15.4 V
29. In the circuit of fig. P.1.1.29 power is delivered by
when lit. The additional A bulbs in parallel to this
500 W
circuit, that would be required to blow the fuse is
400 W
ix
20 A
200 W
2ix
40 V 12 V C
B
A
Fig. P.1.1.29
(A) dependent source of 192 W
Fig. P.1.1.25
(A) 4
(B) 5
(B) dependent source of 368 W
(C) 6
(D) 7
(C) independent source of 16 W
26. In the circuit of fig. P.1.1.26, the power absorbed by
(D) independent source of 40 W
the load RL is
30. The dependent source in fig. P.1.1.30
i1
5W RL = 2 W
2i1
1W
1V
v1
20 V
v1 5
5W
Fig. P.1.1.26
(A) 2 W
(B) 4 W
(C) 6 W
(D) 8 W
Fig. P.1.1.30
27. vo = ?
(A) delivers 80 W
(B) delivers 40 W
(C) absorbs 40 W
(D) absorbs 80 W
31. In the circuit of fig. P.1.1.31 dependent source 0.2 A
5W
+ v1 0.3v1 –
8W
+ v2 –
5v2
18 W
+ vo –
+ 8V – ix
Page 6
(A) 6 V
(B) -6 V
(C) -12 V
(D) 12 V
2ix
4A
Fig. P.1.1.27
Fig. P.1.1.31
(A) supplies 16 W
(B) absorbs 16 W
(C) supplies 32 W
(D) absorbs 32 W
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Basic Concepts
Chap 1.1
32. A capacitor is charged by a constant current of 2 mA
36. The waveform for the current in a 200 mF capacitor
and results in a voltage increase of 12 V in a 10 sec
is shown in fig. P.1.1.36 The waveform for the capacitor
interval. The value of capacitance is
voltage is
(A) 0.75 mF
(B) 1.33 mF
(C) 0.6 mF
(D) 1.67 mF
i(mA) 5
33. The energy required to charge a 10 mF capacitor to 100 V is
Fig. P. 1.1.36
(A) 0.10 J (C) 5 ´ 10
t(ms)
4
(B) 0.05 J
-9
v
(D) 10 ´ 10
J
-9
J
v 50m
5
34. The current in a 100 mF capacitor is shown in fig.
t(ms)
4
P.1.1.34. If capacitor is initially uncharged, then the
4
(A)
(B)
v
waveform for the voltage across it is
t(ms)
v 50m
250m
i(mA) t(ms)
4
2
4
(C)
t(ms)
t(ms)
(D)
Fig. P. 1.1.34 v
37. Ceq = ?
v 10
10
2
4
t(ms)
2
(A)
v
2.5 mF
4
t(ms) 1.5 mF
(B)
v
2
4
1 mF
Ceq
0.2
0.2 t(ms)
2
(C)
4
2 mF
t(ms)
(D) Fig. P.1.1.37
35. The voltage across a 100 mF capacitor is shown in fig. P.1.1.35. The waveform for the current in the
(A) 3.5 mF
(B) 1.2 mF
capacitor is
(C) 2.4 mF
(D) 2.6 mF
v 6
38. In the circuit shown in fig. P.1.1.38 1
2
3
t(ms)
iin ( t) = 300 sin 20 t mA, for t ³ 0.
Fig. P.1.1.35 i(mA) 6
C2
C2
C2
C2
+
i(mA) 600
vin C1
iin 1
2
3
t(ms)
1
2
3
C1
C1
C1
60 mF
t(ms)
–
(A)
(B)
i(mA) 6
Fig. P. 1.1.38
i(mA) 600 2 1
(C)
3
t(ms)
2 1
3
t(ms)
Let C1 = 40 mF and C2 = 30 mF. All capacitors are initially uncharged. The vin ( t) would be
(D)
(A) -0.25cos 20t V
(B) 0.25cos 20t V
(C) -36cos 20t mV
(D) 36cos 20t mV
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GATE EC BY RK Kanodia
Basic Concepts
Chap 1.1
v3 - 30 = v2
SOLUTIONS
Þ
v3 = 65 V
105 + v4 - v3 - 65 = 0 Þ
v4 = 25 V
v4 + 15 - 55 + v1 = 0 Þ v1 = 15 V
1. (C) n = 10 20 , Q = ne = e10 20 = 16.02 C Charge on sphere will be positive.
9. (B) Voltage is constant because of 15 V source.
2. (D) DQ = i ´ D t = 15000 ´ 100m = 15 . C
10. (C) Voltage across 60 W resistor = 30 V 30 Current = = 0.5 A 60
3. (B) i =
dQ 120 = =2 A dt 60
4. (B) W = Qv = 360 J
Voltage across R1 is = 70 - 20 = 50 V 50 = 100 W R1 = 0.5
6. (A)
11. (C) The current i will be distributed in the cube branches symmetrically 1A
i=1A
a
5A
2A 3A
4A
i 3
i i 3
6A
1A
i 3
2A
i 6
Fig. S 1.1.5
6. (A) In order for 600 C charge to be delivered to the 100 V source, the current must be anticlockwise. dQ 600 i= = = 10 A dt 60 Applying KVL we get
0V
Fig. S. 1.1.11
+
+
1V
–
–
+ –
12. (C) If we go from +side of 1 kW through 7 V, 6 V and 5 V, we get v1 = 7 + 6 - 5 = 8 V
E
13. (D) It is not possible to determine the voltage across
–
–
+
+
5V
1 A source.
10 V
Fig. S 1.1.7
14. (D) Req = 5 +
10 + 5 + E + 1 = 0 or E = -16 V 8. (D) 100 = 65 + v2 Þ
v2 = 35 V
+ 105 V –
+
30 +
Req
Fig. S 1.1.14
–
–
10 W
–
V
v2
Req
55 V + v1 –
5W
– 10 V +
30
V
–
+ v3 –
10 + 5 + Req 5W
+
v4
10 ( Req + 5)
+
– +
– 15 V +
+
65 V 100 V
b
6i 6i 6i + + = 5 i, 3 6 3 v Req = ab = 5 W i
7. (A) Going from 10 V to 0 V
4V
i
vab =
v1 + 60 - 100 = 10 ´ 20 or v1 = 240 V
2V
i 3
Þ Req2 + 15 Req = 5 Req + 75 + 10 Req + 50 Fig. S 1.1.8
Þ
Req = 125 = 1118 . W
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GATE EC BY RK Kanodia
Basic Concepts
vo - 20 vo 20 + = 5 5 5 Power is P = vo ´
Þ
We can say Cd = 20 mF, Ceq = 20 + 40 = 60 mF 1 1 æ 300 ö cos 20 t ÷ ´ 10 -3 = - 0.25 cos 20 t V vC = ò idt= ç60m è 20 C ø
vo = 20 V
20 v1 = 20 ´ = 80 W 5 5
39. (C) iC1 =
31. (D) Power P = vi = 2 ix ´ ix = 2 ix2 ix = 4 A, P = 32 W (absorb) 32. (D) vt 2 - vt1 = Þ
1 C
t2
ò idt Þ
iin C1 = 0.8 sin 600 t mA C1 + C2
At t = 2 ms, iC1 = 0.75 mA
Þ
12 =
t1
12 C = 2m ´ 10
Chap 1.1
1 2m ( t2 - t1 ) C
40. (B) vC1 =
4 vin vin C2 = C1 + C2 6 + 4
Þ
vc1 = 0.4 vin
C = 1.67 mF 41. (D) V = 2 + 3 + 5 = 10, Q = 1 C, C =
1 33. (B) E = Cv 2 = 5 ´ 10 -6 ´ 100 2 = 0.05 J 2
42. (A) vL = L
di Þ dt
This 0.2 V increases linearly from 0 to 0.2 V. Then
43. (B) vL = L
di = 0.01 ´ 2( 377 cos 377 t) V dt
current is zero. So capacitor hold this voltage.
= 7.54 cos 377 t V
1 34. (D) vc = c
35. (D) i = C
-3
10 ´ 10 -3 ò0 idt = 100 ´ 10 -6 (2 ´ 10 ) = 0.2 V
2m
dv dt
For 0 < t < 1 , C
dv 6 -0 = 100 ´ 10 -6 ´ = 600 mA dt 10 -3 - 0
1 1 12000 120 cos 3t dt = vdt = sin 377 t 0.01 ò Lò 377 12000 ´ 120 P = vi = (sin 377 t)(cos 377 t) 377
45. (D) vL = L
1 1 idt = ò 200 ´ 10 -6 C
5m
ò 4m tdt = 3125 t
vC = 3vL Þ
iC = 3 LC
46. (B) vL = L
It will be parabolic path. at t = 0 t-axis will be tangent.
combination is in series with 1.5 mF. 15 . (2 + 1) = 1mF, C1 is in parallel with 2.5 mF C1 = 15 . +2+1 . mF Ceq = 1 + 2.5 = 35
diL dt æ -100 - 0 ö vL = (0.05)ç ÷ = - 2.5 V 2 è ø
For 4 < t £ 8,
æ 100 + 100 ö vL = (0.05)ç ÷ = 2.5 V 4 è ø
For 8 < t £ 10,
æ 0 - 100 ö vL = (0.05)ç ÷ = - 2.5 V 2 è ø
Thus (B) is correct option.
30 ´ 60 30(20 + 40) 38. (A) Ca = = 20 mF, Cb = = 20 mF 30 + 60 30 + 20 + 0 C2
Cc
C2
Cb
C2
Ca
47. (C) Algebraic sum of the current entering or leaving a cutset is equal to 0. 6 16 i2 + i4 + i3 = 0 Þ + + i3 = 0 2 4
C2
+
i3 = - 7 A, C1
d 2 iL = - 9.6 sin 4 t A dt
For 2 < t £ 4,
37. (A) 2 mF is in parallel with 1 mF and this
vin
diL dv , iC = C C dt dt
2
At t = 4 ms, vc = 0.05 V
iin
L = 2 mH
= 1910 sin 754 t W
36. (B) For 0 £ t £ 4,
Cd
Þ
44. (A) i =
For 1 ms < t < 2 ms, dv 0-6 C = 100 ´ 10 -6 ´ = - 600 mA ( 3 - 2)m dt
vC =
100m = L
200m 4m
Q = 0.1 F V
C1
C1
C1
v3 = - 7 ´ 3 = - 21 V
60 mF
–
Fig. S 1.1.38
*********
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GATE EC BY RK Kanodia
CHAPTER
1.2 GRAPH THEORY
1. Consider the following circuits :
Non-planner graphs are (A) 1 and 3
(B) 4 only
(C) 3 only
(D) 3 and 4
3. A graph of an electrical network has 4 nodes and 7 branches. The number of links l, with respect to the (1)
(2)
chosen tree, would be (A) 2
(B) 3
(C) 4
(D) 5
4. For the graph shown in fig. P.1.1.4 correct set is
(3)
(4) Fig. P.1.1.4
The planner circuits are (A) 1 and 2
(B) 2 and 3
(C) 3 and 4
(D) 4 and 1
2. Consider the following graphs
Node
Branch
Twigs
Link
(A)
4
6
4
2
(B)
4
6
3
3
(C)
5
6
4
2
(D)
5
5
4
1
5. A tree of the graph shown in fig. P.1.2.5 is c
(1)
(2)
b
3 g
e
d
4
f
2
a
1
5
h
Fig. P.1.2.5
(3) Page 12
(4)
(A) a d e h
(B) a c f h
(C) a f h g
(D) a e f g
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GATE EC BY RK Kanodia
UNIT 1
é 1 -1 0 ù (A) ê-1 0 1ú ê ú 1 -1úû êë 0 é-1 -1 0 ù (C) ê 0 1 1ú ê ú êë 1 0 -1ûú
é 1 0 -1ù (B) ê-1 -1 0 ú ê ú 1 1úû êë 0 1ù é-1 0 ê (D) 0 1 -1ú ê ú êë 1 -1 0 úû
2
2
4
1
3
4
1
5
13. The incidence matrix of a graph is as given below
3
5
(C)
é-1 1 1 0 0 0 ù ê 0 0 -1 1 1 0 ú ú A =ê ê 0 -1 0 -1 0 -1ú ê 1 0 0 0 -1 -1ú ë û
(D)
15. The incidence matrix of a graph is as given below é-1 1 1 0 0 0 ù ê 0 0 -1 1 1 0 ú ú A =ê ê 0 -1 0 -1 0 0 ú ê 1 0 0 0 -1 -1ú ë û
The graph is 2
Networks
2
The graph is 4
4 3
1
3
1
(A)
1
2
3
1
2
3
(B)
2
2
4
4
(A) 4
(B)
4 3
1
3
1
(C)
1
2
3
1
2
3
(D)
14. The incidence matrix of a graph is as given below é- 1 0 0 - 1 1 0 0 ù ê 0 0 0 1 0 1 1ú ê ú A =ê 0 0 1 0 0 0 -1ú ê 0 1 0 0 -1 -1 0 ú ê ú ëê 1 -1 -1 0 0 0 0 úû
4
4
(C)
(D)
16. The graph of a network is shown in fig. P.1.1.16. The number of possible tree are
The graph is 2
2
4
1
3
4
1
3
Fig. P.1.1.16
5
(A) Page 14
5
(A) 8
(B) 12
(C) 16
(D) 20
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GATE EC BY RK Kanodia
UNIT 1
Networks
22. The fundamental cut-set matrix of a graph is é1 -1 0 0 0 ê0 -1 1 0 1 QF = ê 0 0 0 0 1 ê ê0 0 0 1 -1 ë
0ù 0ú ú 1ú 0 úû
The oriented graph of the network is
2
1
(C)
24. A graph is shown in fig. P.1.2.24 in which twigs are
2
1
3
3
4
solid line and links are dotted line. For this tree
4
5
5
6
fundamental loop matrix is given as below é1 1 1 0 ù BF = ê ú ë1 0 1 1û
6
(A)
(D)
(B)
1
2
1
2
1
3
3
2
4
3
4
4
5
5 6
6
Fig. P.1.2.24
(C)
(D)
The oriented graph will be
23. A graph is shown in fig. P.1.2.23 in which twigs are solid line and links are dotted line. For this chosen tree fundamental set matrix is given below. é 1 1 BF = ê 0 -1 ê êë 0 0
0 1 0
0 1 1
1 0 1
0ù 0ú ú 1úû
(A)
(B)
3 2
1
4
5
6
(C)
Fig. P. 1.2.23
The oriented graph will be
(D)
25. Consider the graph shown in fig. P.1.2.25 in which twigs are solid line and links are dotted line.
1
5
6
(A)
2
4
3
(B)
Fig. P. 1.2.25
Page 16
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GATE EC BY RK Kanodia
Graph Theory
A fundamental loop matrix for this tree is given as below é 1 BF = ê 0 ê êë 0
0 1 0
1ù 0 -1 -1 0 ú ú 1 0 1 -1úû 0
1
0
é 1 ê 0 (D) ê ê 0 ê-1 ë
Chap 1.2
1
0
0
0
0
1 -1
1
0
0 0
1 0
1 1
0 1
0
1
0
1
0
1 0
0 0
0 0
0ù 0ú ú 0ú 1úû
27. Branch current and loop current relation are
The oriented graph will be
expressed in matrix form as
(A)
é i1 ù é 0 êi ú ê 0 ê 2ú ê ê i3 ú ê 1 êi ú ê-1 ê 4ú = ê ê i5 ú ê 1 êi ú ê 0 6 ê ú ê i ê 7ú ê 0 êë i8 úû êë 0
(B)
1 -1 0 -1 0 0 1
0
0 1
0 0
0 0
1 0
0ù 1ú ú -1ú é I1 ù 0 ú êI 2 ú úê ú 0 ú êI 3 ú 0 ú êëI 4 úû ú 0ú 1úû
where i j represent branch current and I k loop current. The number of independent node equation are
(C)
(D)
(A) 4
(B) 5
(C) 6
(D) 7
28. If the number of branch in a network is b, the
26. In the graph shown in fig. P.1.2.26 solid lines are twigs and dotted line are link. The fundamental loop
number of nodes is n and the number of dependent loop is l, then the number of independent node equations will be
matrix is i c
a
e
(A) n + l - 1
(B) b - 1
(C) b - n + 1
(D) n - 1
Statement for Q.29–30: h
b
Branch current and loop current relation are
f
d
expressed in matrix form as 1 0ù é i1 ù é 0 0 êi ú ê-1 -1 -1 0 ú ê 2ú ê ú 1 0 0 ú é I1 ù ê i3 ú ê 0 êi ú ê 1 0 0 0 ú ê I ú ê 4ú = ê ú ê 2ú i 0 0 1 1 ê 5ú ê ú êI 3 ú ê i ú ê 1 1 0 -1ú êI ú ë 4û 6 ê ú ê ú ê i7 ú ê 1 0 0 0 ú êë i8 úû êë 0 0 0 1úû
g
Fig. P.1.2.26
é 1 ê 0 (A) ê ê 0 ê-1 ë
é-1 1 0 0 ê 0 -1 -1 -1 (B) ê ê 0 0 0 -1 ê 1 0 1 0 ë é ê (C) ê ê ê ë
1 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
0ù 0ú ú 0ú 1úû
0 0 0 0 1 -1
0 1 0
1 0 0
1 0 0 0 1 -1 -1 0 0 0 1 -1 0 -1 0 -1
1 1 0 0
0 1 0 1
1
0
0
0
0ù 0ú ú 0ú 1úû
0 0 1 0 1 -1 0 -1
0 0 1 0
0 -1 1 0 0 0 0 0
0ù 0ú ú 0ú 1úû
where i j represent branch current and I k loop current. 29. The rank of incidence matrix is (A) 4
(B) 5
(C) 6
(D) 8
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GATE EC BY RK Kanodia
UNIT 1
30. The directed graph will be
Networks
33. The oriented graph for this network is
8
8
5
5
6
2
3
1
2
7
1
6
3
1
4
3
2
7
4
1
5
8 5
6
3
2
4
1
2
6 5
5
(C) 2
3
1
4
3
4
5
8
5
3
4 1
2
7
3
1
4
(D)
7
4
(C)
************
(D)
31. A network has 8 nodes and 5 independent loops. The number of branches in the network is (A) 11
(B) 12
(C) 8
(D) 6
32. A branch has 6 node and 9 branch. The independent loops are (A) 3
(B) 4
(C) 5
(D) 6
Statement for Q.33–34: For a network branch voltage and node voltage relation are expressed in matrix form as follows: 1ù é v1 ù é 1 0 0 êv ú ê 0 1 0 0ú ê 2ú ê ú 1 0 ú é V1 ù ê v3 ú ê 0 0 êv ú ê 0 0 0 1ú êV2 ú ê 4ú = ê úê ú ê v5 ú ê 1 -1 0 0 ú ê V3 ú êv ú ê 0 1 -1 0 ú êëV4 úû 6 ê ú ê ú 1 -1ú ê v7 ú ê 0 0 êë v8 úû êë 1 0 -1 0 úû where vi is the branch voltage and Vk is the node voltage with respect to datum node. 33. The independent mesh equation for this network are (A) 4
(B) 5
(C) 6
(D 7
Page 18
2
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GATE EC BY RK Kanodia
Graph Theory
Chap 1.2
7. (D) D is not a tree
SOLUTIONS 1. (A) The circuit 1 and 2 are redrawn as below. 3 and 4 can not be redrawn on a plane without crossing other branch.
(A) (1)
(B)
(2) Fig. S1.2.1
2. (B) Other three circuits can be drawn on plane without crossing
(C) (1)
(D)
(2)
Fig. S .1.2.7
8. (D) it is obvious from the following figure that 1, 3, and 4 are tree 2
2
(3) a
Fig. S1.2.1
c
1
3. (C) l = b - ( n - 1) = 4.
a
b
e
(1) 2
a c
1 g
e
a
b
f
a
(2)
2
c
d
4
4
5. (C) From fig. it can be seen that a f h g is a tree of
b
f
f
l=b-n+1=3
given graph
3 e
d
4. (B) There are 4 node and 6 branches. t = n - 1 = 3,
c
1
3
d
b
c
1
3 e
d
b 3 e
d
f
f 4
4
h
(3)
Fig. S 1.2.5
(4) 2
6. (B) From fig. it can be seen that a d f is a tree. c b
a
a e
d
b c
1
e
d
f
3
f 4
(5) Fig. S. 1.2.6
Fig. S. 1.2.8
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GATE EC BY RK Kanodia
UNIT 1
So independent mesh equation =Number of link.
i l1
a
34. (D) We know that [ vb ] = ArT [ Vn ]
e
c
So reduced é1 0 ê0 1 Ar = ê ê0 0 ê0 0 ë
l4
l2 h
b
l3
Networks
d
f
g
Fig. S 1.2.26
incidence matrix is 0 0 1 0 0 1ù 0 0 -1 1 0 0 ú ú 1 0 0 -1 1 -1ú 0 1 0 0 -1 0 úû
At node-1, three branch leaves so the only option is (D).
This in similar to matrix in (A). Only place of rows has been changed. 27. (A) Number of branch =8
***********
Number of link =4 Number of twigs =8 - 4 = 4 Number of twigs =number of independent node equation. 28. (D) The number of independent node equation are n - 1. 29. (A) Number of branch b = 8 Number of link l = 4 Number of twigs t = b - l = 4 rank of matrix = n - 1 = t = 4 30. (B) We know the branch current and loop current are related as [ ib ] = [ B T ] [ I L ] So fundamental loop matrix is 1 1 0ù é0 -1 0 1 0 ê0 -1 1 0 0 1 0 0ú ú Bf = ê ê1 -1 0 0 -1 0 0 0 ú ê0 0 0 0 -1 -1 0 1ú û ë f-loop 1 include branch (2, 4, 6, 7) and direction of branch–2 is opposite to other (B only). 31. (B) Independent loops =link l = b - ( n - 1) Þ
5 = b - 7, b = 12
32. (B) Independent loop =link l = b - ( n - 1) = 4 33. (A) There are 8 branches and 4 + 1 = 5 node Number of link = 8 - 5 + 1 = 4 Page 22
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GATE EC BY RK Kanodia
CHAPTER
1.3 METHODS OF ANALYSIS
1. v1 = ?
6R
6R
4vs
+ v1 –
(B) -120 V (D) -90 V
(A) 120 V (C) 90 V
3R
vs
4. va = ?
10 W 4W
12 V
10 V
Fig. P1.3.1
(A) 0.4vs
(B) 1.5vs
(C) 0.67vs
(D) 2.5vs
va 4A
1W
2. va = ?
2W
Fig. P1.3.4
3A
(A) 4.33 V (C) 8.67 V 2W
5. v2 = ?
va
3W
(B) 4.09 V (D) 8.18 V
20 W
1A
+ 30 W
60 W
v2 – 10 V
0.5 A
30 W
Fig. P1.3.2
(A) -11 V
(B) 11 V
(C) 3 V
(D) -3 V
Fig. P1.3.5
(A) 0.5 V (C) 1.5 V
3. v1 = ? 10 W
6. ib = ?
30 V
3A
(B) 1.0 V (D) 2.0 V
64 W
30 W
37 W ib
20 W
0.5 A
10 V
60 W
69 W
36 W
– v1 +
9A
6A
60 W
Fig. P1.3.6
Fig. P1.3.3
(A) 0.6 A (C) 0.4 A www.gatehelp.com
(B) 0.5 A (D) 0.3 A Page 23
GATE EC BY RK Kanodia
UNIT 1
7. i1 = ? 6A
8W
(A) 20 mA
(B) 15 mA
(C) 10 mA
(D) 5 mA
11. i1 = ?
2W
i1
Networks
50 W
10 W 5W
3W
4W
75 V
100 W i1
6.6 V
Fig. P1.3.7
(A) 3.3 A
(B) 2.1 A
(C) 1.7 A
(D) 1.1 A
0.1 A
40 W
0.06 A
60W
8. i1 = ? 0.1A i2
90 kW 7.5mA
i1
10 kW
Fig. P1.3.11
10 kW
75 V
(A) 0.01 A
(B) -0.01 A
(C) 0.03 A
(D) 0.02 A
90 kW
12. The value of the current measured by the ammeter in Fig. P1.3.12 is
Fig. P1.3.8
(A) 1 mA
(B) 1.5 mA
(C) 2 mA
(D) 2.5 mA
2A
Ammeter
4W
7W
9. i1 = ? 2A
6W
3A 5W
3V
4W
2W
Fig. P1.3.12 3W
4A
2W i1
(B) 3 A
(C) 6 A
(D) 5 A
2 A 3
(C) -
Fig. P1.3.9
(A) 4 A
(A)
5 A 6
(B)
5 A 3
(D)
2 A 9
13. i1 = ? 200 W
10. i1 = ? 2 kW
45 V
i1
40 mA
500 W
Fig. P1.3.10
Page 24
15 mA
100 W
50 W
i1
Fig. 1.3.13
(A) 10 mA
(B) -10 mA
(C) 0.4 mA
(D) -0.4 mA
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10 mA
GATE EC BY RK Kanodia
Methods of Analysis
Chap 1.3 5W
8W
4W
14. The values of node voltage are va = 12 V, vb = 9.88 V and vc = 5.29 V. The power supplied by the voltage
i1
12 V
source is
2W
i3
2W
i2
20 V
8V 6W
va
4W
3W
vb
12 V
1A
vc
Fig. 1.3.17
0 ù éi1 ù é12 ù é 4 -2 (A) ê-2 8 -2 ú êi2 ú = ê-8 ú ê úê ú ê ú 5 úû êëi3 úû êë20 úû êë 0 -2
2W
-2 0 ù éi1 ù é12 ù é6 (B) ê2 -12 2 ú êi2 ú = ê 8 ú ê úê ú ê ú 2 -7 úû êëi3 úû êë20 úû êë0
Fig. 1.3.14
(A) 19.8 W
(B) 27.3 W
(C) 46.9 W
(D) 54.6 W
0 ù éi1 ù é12 ù é 6 -2 (C) ê-2 12 -2 ú êi2 ú = ê 8 ú ê úê ú ê ú 7 úû êëi3 úû êë20 úû êë 0 -2
15. i1 , i2 , i3 = ? 2W
3W
15 V
9W
i1
6W
i2
0 ù éi1 ù é12 ù é 4 -2 ê (D) 2 -8 2 ú êi2 ú = ê 8 ú ê úê ú ê ú 2 -5 úû êëi3 úû êë20 úû êë 0 18. For the circuit shown in Fig. P1.3.18 the mesh
21 V
i3
equation are 6 kW
Fig. P1.3.15 6 kW
(A) 3 A, 2 A, and 4 A
(B) 3 A, 3 A, and 8 A
(C) 1 A, 3 A, and 4 A
(D) 1 A, 2 A, and 8 A
i3
6 kW
i1 6V
i2 5 mA
6 kW
16. vo = ? Fig. 1.3.18 4 mA
2 kW
2 mA
é 6 k -12 k -12 k ù éi1 ù é-6 ù ê ú ê ú (A) -6 k 6 k -18 k i2 = ê 0 ú ê úê ú ê ú -1k 0 k úû ëêi3 úû êë -1k êë 5 úû
1 kW
1 kW
+ 1 kW
2 kW
1 mA
vo
é 6 k 12 k -12 k ù éi1 ù é-6 ù (B) ê-6 k -6 k 18 k ú êi2 ú = ê 0 ú ê úê ú ê ú 1k 0 k úû êëi3 úû êë -1k êë 5 úû
–
Fig. P1.3.16
(A)
6 V 5
(B)
8 V 5
(C)
6 V 7
(D)
5 V 7
é-6 k -12 k 12 k ù éi1 ù é-6 ù (C) ê 6 k -6 k 18 k ú êi2 ú = ê 0 ú ê úê ú ê ú 1k 0 k úû êëi3 úû êë 1k êë 5 úû
17. The mesh current equation for the circuit in Fig. P1.3.17 are
é-6 k 12 k -12 k ù éi1 ù é-6 ù (D) ê-6 k 6 k -18 k ú êi2 ú = ê 0 ú ê úê ú ê ú 1k 0 k úû êëi3 úû êë -1k êë 5 úû
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GATE EC BY RK Kanodia
Methods of Analysis
R1
i3
R2
Chap 1.3
(A) 66.67 mA
(B) 46.24 mA
(C) 23.12 mA
(D) 33.33 mA
R3
29. va = ?
i1
R4
i2
10 W
v2
v1 25i2
4A
50 W
va
40 W
Fig. P1.3.25 200 W
10 A
The value of R4 is (A) 40
(B) 15
(C) 5
(D) 20
2.5 kW 10 kW
20 V
10 kW
(A) 342 V
(B) 171 V
(C) 198 V
(D) 396 V
30. ia = ?
va
5 kW
20 A
Fig. P1.3.29
26. va = ?
10 kW
20 W
100 W
5A
50 W
150 W
4 mA
225 W
100 W
200 W
ia
Fig. P1.3.26
2V
(A) 26 V
(B) 19 V
(C) 13 V
(D) 18 V
75 W
2A
50 W
(A) 14 mA
(B) -6.5 mA
(C) 7 mA
(D) -21 mA
31. v2 = ?
20 W
v
8V
Fig. P1.3.30
27. v = ?
10 W
4V
50 W – v2 +
4A
100 W
10 V
15 W
0.04v2
5W
Fig. P.3.1.27
Fig. P1.3.31
(A) 60 V
(B) -60 V
(A) 5 V
(B) 75 V
(C) 30 V
(D) -30 V
(C) 3 V
(D) 10 V
32. i1 = ?
28. i1 = ?
2W
300 W
40 V
i1
500 W
8V
0.4i1
0.5i1
4A
4W
6V
i1
Fig. P1.3.32
Fig. P1.3.28
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GATE EC BY RK Kanodia
UNIT 1
(A) -1.636 A
(B) -3.273 A
(C) -2.314 A
(D) -4.628 A
Networks
37. va = ? 0.8va
16 A
33. vx = ?
+ 100 W
1.6 A
2.5 W
5W
10 A
vx
50 W
0.02vx
2W
va
–
Fig. P1.3.37 Fig. P1.3.33
(A) 32 V
(B) -32 V
(C) 12 V
(D) -12 V
(A) 25.91 V
(B) -25.91 V
(C) 51.82 V
(D) -51.82 V
38. For the circuit of Fig. P1.3.38 the value of vs , that
34. ib = ?
will result in v1 = 0, is 1 kW
va
3A
3 kW
2A
ib
0.1v1 6V
4va
2 kW
10 W
20 W +
vs
40 W
Fig. P1.3.34
(A) 4 mA
(B) -4 mA
(C) 12 mA
(D) -12 mA
4 kW
vb
(A) 28 V
(B) -28 V
(C) 14 V
(D) -14 V
39. i1 , i2 = ?
2ix
4W 2V
48 V
Fig. P1.3.38
35. vb = ? ia
v1 –
2W
5ia
2 kW
ix 15 V
6W
i1
i2
18 V
Fig. P1.3.35
(A) 1 V
(B) 1.5 V
(C) 4 V
(D) 6 V
Fig. P1.3.39
36. vx = ?
(A) 2.6 A, 1.4 A
(B) 2.6 A, -1.4 A
(C) 1.6 A, 1.35 A
(D) 1.2 A, -1.35 A
40. v1 = ?
50 W
3W
iy + 2A
100 W
25iy
50 W
vx
+ vy –
0.2vx
–
3W
2A 6W
14 V + v1 –
Fig. P1.3.36
Page 28
(A) -3 V
(B) 3 V
(C) 10 V
(D) -10 V
7A
Fig. P1.3.40
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2W
2vy
GATE EC BY RK Kanodia
Methods of Analysis
(A) 10 V
(B) -10 V
(C) 7 V
(D) -7 V
SOLUTIONS 1. (B) Applying the nodal analysis v 4 vs + s R R 6 3 v1 = = 15 . vs 1 1 1 + + 6 R 3R 6 R
41. vx = ? – vx +
500 W
0.5vx
500 W 900 W
600 W
0.6 A
Chap 1.3
2. (C) va = 2( 3 + 1) + 3 (1) = 11 V
0.3 A
3. (D) -
v1 -v + 1 + 6 =9 60 60
Þ
v1 = - 90 V
Fig. P1.3.41
(A) 9 V
(B) -9 V
(C) 10 V
(D) -10 V
42. The power being dissipated in the 2 W resistor in the
4. (C)
va - 10 va + =4 4 2
5. (D)
v2 v2 + 10 + = 0.5 20 30
Þ
va = 8.67 V
Þ
v2 = 2 V
circuit of Fig. P1.3.42 is 5W
6. (B) Using Thevenin equivalent and source transform
ia
3W
8W 3
2W
i1
2W
10 W
va
60 V
2A 2.5 A
4W
6ia
3W
25 V
30 V
Fig. P1.3.42
Fig. S.1.3.6
(A) 76.4 W
(B) 305.6 W
(C) 52.5 W
(D) 210.0 W
43. i1 = ? 500 W + vx – 100 W
180 V 400 W
5W
25 60 + + 2 15 va = = 15.23 V 3 1 1 + + 14 3 15 25 - 15.23 = 2.09 A i1 = 14 3 8 3
0.6 A
+ vy –
7. (A) ib = 0.001vy
100 W
10 + 0.5 = 0.6 A 64 + 36
8. (B) 75 = 90 ki1 + 10 k( i1 - 7.5m) 150 = 100 ki1
Þ
i1 = 15 . mA
9. (B) 3 = 2 i1 + 3( i1 - 4)
0.005vy
Þ
i1 = 3 A
Fig. P1.3.43
10. (B) 45 = 2 ki1 + 500 ( i1 + 15m) (A) 0.12 A
(B) 0.24 A
(C) 0.36 A
(D) 0.48 A
Þ
i1 = 15 mA
11. (D) 6.6 = 50 i1 + 100( i1 + 0.1) + 40( i1 - 0.06) + 60( i1 - 0.1)
*****************
i1 = 0.02 A www.gatehelp.com
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GATE EC BY RK Kanodia
UNIT 1
38. (D) If v1 = 0, the dependent source is a short circuit v1 v - vs v1 - 48 + 1 + =2 - 3 Þ 40 10 20 48 v - s = - 1 Þ vs = - 14 V 10 20 3A
v1 = 0
ia =
Networks
30 + 7.5 + 2 = 329 . A 12
Þ
6 ia = 19.75 V
voltage across 2 W resistor 30 - 19.75 = 10.25 V, P=
2A
(10.25) 2 = 52.53 W 2
43. (A) vx = 500 i1 10 W 40 W
vs
v y = 400( i1 - 0.001vx ) = 400 ( i1 - 0.5 i1 ) = 200 i1
20 W + v1 –
180 = 500 i1 + 100( i1 - 0.6) + 200 i1 + 100( i1 + 0.005 v y)
48 V
180 = 900 i1 - 60 + 100 ´ 0.005 ´ 200 i1 i1 = 0.12 A
Fig. S1.3.38 ************
39. (D) ix = i1 - i2 15 = 4 i1 - 2( i1 - i2 ) + 6( i1 - i2 ) 8 i1 - 4 i2 = 15
Þ
K(i)
-18 = 2 i2 + 6( i2 - i1 ) 3i1 - 4 i2 = 9
Þ
K(ii)
. A, i2 = -1.35 A i1 = 12 40. (B) 14 = 3i1 + v y + 6( i1 - 2 - 7) + 2 v y + 2( i1 - 7) v y = 3( i1 - 2) 14 = 3i1 + 9( i1 - 2) + 6( i1 - 9) + 2( i1 - 7) 14 = 20 i1 - 18 - 54 - 14
Þ
i1 = 5 A
v1 = 6(5 - 2 - 7) + 2 ´ 3(5 - 2) + 2(5 - 7) = -10 V 3W + vy
3W
–
2A 6W
14 V
i1
+ v1 –
2vy
7A
2W
Fig. S1.3.40
41. (D) Let i1 and i2 be two loop current 0.5 vx = 500 i1 + 500( i1 - i2 ), vx = -500 i1 Þ
5 i1 - 2 i2 = 0
K(i)
500( i2 - i1 ) + 900( i2 + 0.3) + 600( i2 - 0.6) = 0 -5 i1 + 20 i2 = 0.9
K(ii)
i1 = 20 mA, vx = -500 ´ 20m = -10 V 42. (C) 30 = 5 ia + 3( ia - 2.5) + 4( ia - 2.5 + 2)
Page 32
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GATE EC BY RK Kanodia
CHAPTER
1.4 NETWORKS THEOREM
1. vTH , RTH = ?
3W
4. A simple equivalent circuit of the 2 terminal network
2W
6W
6V
shown in fig. P1.4.4 is
vTH, RTH
R
i
Fig. P.1.4.1
(A) 2 V, 4 W
(B) 4 V, 4 W
(C) 4 V, 5 W
(D) 2 V, 5 W
2. i N , R N = ?
2W
v
Fig. P.1.4.4
R
2W
R v
4W
15 V
iN, RN
(A)
(B)
Fig. P.1.4.2
10 W 3
(A) 3 A,
(B) 10 A, 4 W
(C) 1,5 A, 6 W
2W
3W
i i
(D) 1.5 A, 4 W
3. vTH , RTH = ?
2A
R R
(C)
(D)
5. i N , R N = ? 1W
2W
vTH, RTH 6A
4W
3W
iN RN
Fig. P.1.4.3
(A) -2 V,
6 W 5
5 (C) 1 V, W 6
(B) 2 V,
5 W 6
6 (D) -1 V, W 5
Fig. P.1.4.5
(A) 4 A, 3 W
(B) 2 A, 6 W
(C) 2 A, 9 W
(D) 4 A, 2 W
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UNIT 1
6. vTH , RTH = ?
Networks
The value of the parameter are 25 W
30 W 20 W
vTH, RTH
5V 5A
vTH
RTH
iN
RN
(A)
4 V
2 W
2 A
2 W
(B)
4 V
2 W
2 A
3 W
(C)
8 V
1.2 W
30 3
A
1.2 W
8 V
5 W
8 5
A
5 W
(D) Fig. P.1.4.6
10. v1 = ? (A) -100 V, 75 W
(B) 155 V, 55 W
(C) 155 V, 37 W
(D) 145 V, 75 W
2W
3W
1W
6W
2W
8V
7. RTH = ?
1W
+ v1
6W
18 V
–
6W
Fig. P.1.4.10
6W
2A
RTH 5V
Fig. P.1.4.7
(A) 6 V
(B) 7 V
(C) 8 V
(D) 10 V
11. i1 = ?
(A) 3 W
(B) 12 W
(C) 6 W
(D) ¥
4 kW
8. The Thevenin impedance across the terminals ab of
12 V
i1
20 V
4 kW
6 kW
24 V
3 kW
4 kW
the network shown in fig. P.1.4.8 is a 3W
6W
2A
Fig. P.1.4.11
2V
8W
(A) 3 A
(B) 0.75 mA
(C) 2 mA
(D) 1.75 mA
8W
Statement for Q.12–13:
b
Fig. P.1.4.8
(A) 2 W
(B) 6 W
(C) 6.16 W
4 (D) W 3
A circuit is given in fig. P.1.4.12–13. Find the Thevenin equivalent as given in question.. 10 W
16 W
x
y
40 W
5V
8W
9. For In the the circuit shown in fig. P.1.4.9 a network and its Thevenin and Norton equivalent are given 2W
x’
3W
y’
Fig. P.1.4.12–13 RTH
4V
iN
2A vTH
RN
12. As viewed from terminal x and x¢ is (A) 8 V, 6 W
(B) 5 V, 6 W
(C) 5 V, 32 W
(D) 8 V, 32 W
Fig. P.1.4.9
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1A
GATE EC BY RK Kanodia
Network Theorems
13. As viewed from terminal y and y¢ is (A) 8 V, 32 W
(B) 4 V, 32 W
(C) 5 V, 6 W
(D) 7 V, 6 W
Chap 1.4
19. vTH , RTH = ? 6W
3i1
14. A practical DC current source provide 20 kW to a
i1
iN,
4W
RN
50 W load and 20 kW to a 200 W load. The maximum power, that can drawn from it, is
Fig. P1.4.19
(A) 22.5 kW
(B) 45 kW
(C) 30.3 kW
(D) 40 kW
Statement for Q.15–16: In the circuit of fig. P.1.4.15–16 when R = 0 W , the
(A) 0 W
(B) 1.2 W
(C) 2.4 W
(D) 3.6 W
20. vTH , RTH = ?
current iR equals 10 A. 2W
4W
4V
2W
+
E
4W
2W
R
5W
0.1v1
4A
vTH RTH
v1
iR
–
Fig. P.1.4.20
Fig. P.1.4.15–16.
15. The value of R, for which it absorbs maximum
(A) 8 V, 5 W
(B) 8 V, 10 W
power, is
(C) 4 V, 5 W
(D) 4 V, 10 W
(A) 4 W
(B) 3 W
(C) 2 W
(D) None of the above
21. RTH = ? 3W
2W
+
16. The maximum power will be (A) 50 W
(B) 100 W
(C) 200 W
(D) value of E is required
vx 4
4V
vx
RTH
–
17. Consider a 24 V battery of internal resistance
Fig. P.1.4.21
r = 4 W connected to a variable resistance RL . The rate of heat dissipated in the resistor is maximum when the current drawn from the battery is i . The current drawn form the battery will be i 2 when RL is equal to
(A) 3 W
(B) 1.2 W
(C) 5 W
(D) 10 W
(A) 2 W
(B) 4 W
22. In the circuit shown in fig. P.1.4.22 the effective
(C) 8 W
(D) 12 W
resistance faced by the voltage source is
18. i N , R N = ?
4W 5W
10 W i1
20i1
iN,
30 W
vs
RN
i 4
i
Fig. P.1.4.22 Fig. P.1.4.18
(A) 2 A, 20 W
(B) 2 A, -20 W
(A) 4 W
(B) 3 W
(C) 0 A, 20 W
(D) 0 A, -20 W
(C) 2 W
(D) 1 W
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Page 35
GATE EC BY RK Kanodia
UNIT 1
23. In the circuit of fig. P1.4.23 the value of RTH at
Networks
26. The value of RL will be
terminal ab is
ix
0.75va 16 V
–
a
Fig. P.1.4.26–27
va +
4W
b
Fig. P.1.4.23
(A) -3 W
9 W 8
(B)
8 W 3
(C) -
RL
2W
8W
9V
3W
0.9 A
(A) 2 W
(B) 3 W
(C) 1 W
(D) None of the above
27. The maximum power is
(D) None of the above
(A) 0.75 W
(B) 1.5 W
(C) 2.25 W
(D) 1.125 W
28. RTH = ?
24. RTH = ?
-2ix
200 W – va 100
va +
100 W
50 W
RTH
100 W
300 W
Fig. P.1.4.24
(A) ¥ (C)
(D)
125 W 3
maximum power if RL is equal to
(B) 136.4 W
(C) 200 W
(D) 272.8 W
29. Consider the circuits shown in fig. P.1.4.29 ia
100 W
200 W
2W 6W
6W 2W
2W
3i
–
(A) 100 W
i 6V
RTH
Fig. P.1.4.28
25. In the circuit of fig. P.1.4.25, the RL will absorb 40 W
+ vx
800 W
ix
(B) 0
3 W 125
100 W
0.01vx
RL
12 V 12 V
8V
Fig. P.1.4.25
6W
(A)
400 W 3
(B)
2 kW 9
(C)
800 W 3
(D)
4 kW 9
ib
2W 6W
6W
2W
Statement for Q.26–27:
18 V
2W
6W
3A
In the circuit shown in fig. P1.4.26–27 the maximum power transfer condition is met for the load RL . Page 36
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Fig. P.1.4.29a & b
12 V
GATE EC BY RK Kanodia
Network Theorems
Chap 1.4
33. If vs1 = 6 V and vs 2 = - 6 V then the value of va is
The relation between ia and ib is (A) ib = ia + 6
(B) ib = ia + 2
(A) 4 V
(B) -4 V
(C) ib = 15 . ia
(D) ib = ia
(C) 6 V
(D) -6 V
34. A network N feeds a resistance R as shown in fig.
30. Req = ? 12 W
P1.4.34. Let the power consumed by R be P. If an
4W
identical network is added as shown in figure, the Req
6W
power consumed by R will be
2W
18 W
6W 9W
R
N
N
N
R
Fig. P.1.4.30 Fig. P.1.4.34
72 (B) W 13
(A) 18 W 36 (C) W 13
(D) 9 W
31. In the lattice network the value of RL for the maximum power transfer to it is
(A) equal to P
(B) less than P
(C) between P and 4P
(D) more than 4P
35. A certain network consists of a large number of ideal linear resistors, one of which is R and two constant ideal source. The power consumed by R is P1
7W
when only the first source is active, and P2 when only
6 W
the second source is active. If both sources are active RL
5
W
simultaneously, then the power consumed by R is
9W
(A) P1 ± P2
(B)
(C) ( P1 ± P2 ) 2
(D) ( P1 ± P2 ) 2
P1 ± P2
Fig. P.1.4.31
(A) 6.67 W
(B) 9 W
36. A battery has a short-circuit current of 30 A and an
(C) 6.52 W
(D) 8 W
open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2 W, the power dissipated by the bulb is
Statement for Q.32–33: A circuit is shown in fig. P.1.4.32–33. 12 W 1W
3W
3W
1W
(A) 80 W
(B) 1800 W
(C) 112.5 W
(D) 228 W
37.
+
The
following
results
were
obtained
from
measurements taken between the two terminal of a vs1
1W
va
vs2
resistive network
–
Terminal voltage
12 V
0V
Terminal current
0A
1.5 A
Fig. P.1.4.32–33
32. If vs1 = vs 2 = 6 V then the value of va is (A) 3 V
(B) 4 V
(C) 6 V
(D) 5 V
The Thevenin resistance of the network is (A) 16 W
(B) 8 W
(C) 0
(D) ¥
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Page 37
GATE EC BY RK Kanodia
UNIT 1
SOLUTIONS
38. A DC voltmeter with a sensitivity of 20 kW/V is used to find the Thevenin equivalent of a linear network. Reading on two scales are as follows
1. (B) vTH =
(a) 0 - 10 V scale : 4 V
The Thevenin voltage and the Thevenin resistance
(C) 18 V,
32 V, 3
1 MW 15
(D) 36 V,
200 kW 3
(B)
2 MW 15
( 6)( 6) = 4 V, 3+ 6
RTH = ( 3||6) + 2 = 4 W
(b) 0 -15 V scale : 5 V
of the network is 16 200 (A) V, kW 3 3
Networks
2. (A)
2W
isc
Fig. S.1.4.2
R N = 2 ||4 + 2 = + RL
4W
15 V
39. Consider the network shown in fig. P.1.4.39.
Linear Network
2W
v1
10 W, 3
15 2 v1 = =6 V 1 1 1 + + 2 2 4 v isc = i N = 1 = 3 A 2
vab –
Fig. P.1.4.39
The power absorbed by load resistance RL is shown in table :
(2)( 3)(1) = 1 V, 3+ 3 5 = 1||5 = W 6
3. (C) vTH =
RL
10 kW
30 kW
P
3.6 MW
4.8 MW
RTH
4. (B) After killing all source equivalent resistance is R The value of RL , that would absorb maximum power, is (A) 60 kW
(B) 100 W
(C) 300 W
(D) 30 kW
Open circuit voltage = v1 5. (D) The short circuit current across the terminal is 2W isc
40. Measurement made on terminal ab of a circuit of
6A
4W
3W
fig.P.1.4.40 yield the current-voltage characteristics shown in fig. P.1.4.40. The Thevenin resistance is Fig. S1.4.5
i(mA) +
30 Resistive Network
20 10 -4
-3 -2
-1
0
vab –
1
2
v
Fig. P.1.4.40
a
b
isc =
6´ 4 = 4 A = iN , 4+2
R N = 6 ||3 = 2 W 6. (B) For the calculation of RTH if we kill the sources then 20 W resistance is inactive because 5 A source will
(A) 300 W
(B) -300 W
be open circuit
(C) 100 W
(D) -100 W
RTH = 30 + 25 = 55 W, vTH = 5 + 5 ´ 30 = 155 V
***********
Page 38
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GATE EC BY RK Kanodia
Network Theorems
Chap 1.4
12. (B) We Thevenized the left side of xx¢ and source
7. (C) After killing the source, RTH = 6 W
transformed right side of yy¢
6W
8W
16 W
x
y
8W
6W RTH 4V
8V
x’
Fig. S.1.4.7
y’
Fig. S1.4.12
8. (B) After killing all source, RTH = 3||6 + 8 ||8 = 6 W
4 8 + = 8 24 = 5 V, 1 1 + 8 24
vxx ¢ = vTH
9. (D) voc = 2 ´ 2 + 4 = 8 V = vTH RTH = 2 + 3 = 5 W = R N ,
iN =
RTH = 8 ||(16 + 8) = 6 W
vTH 8 = A RTH 5
13. (D) Thevenin equivalent seen from terminal yy¢ is 4 8 + 24 8 = 7 V, = 1 1 + 24 8
10. (A) If we solve this circuit direct, we have to deal with three variable. But by simple manipulation
v yy¢ = vTH
variable can be reduced to one. By changing the LHS and RHS in Thevenin equivalent 1W
1W
1W
RTH = ( 8 + 16)||8 = 6 W
2W
14. (A)
+ 6W
4V
12 V
v1 –
i
RL
r
Fig. S1.4.10 Fig. S1.4.14
4 12 + v1 = 1 + 1 1 + 2 = 6 V 1 1 1 + + 1+1 6 1+2
2
æ ir ö ç ÷ 50 = 20 k, è r + 50 ø
( r + 200) 2 = 4( r + 50) 2
11. (B) If we solve this circuit direct, we have to deal with three variable. But by simple manipulation
Þ
r = 100 W
i = 30 A,
Pmax =
variable can be reduced to one. By changing the LHS and RHS in Thevenin equivalent 2 kW
i1
4 kW
2
æ ir ö ç ÷ 200 = 20 k è r + 200 ø
20 V
6V
( 30) 2 ´ 100 = 22.5 kW 4
15. (C) Thevenized the circuit across R, RTH = 2 W 4W
2 kW
2W
2W
4W
8V
2W
Fig. S1.4.15 Fig. S1.4.11
i1 =
20 - 6 - 8 = 0.75 mA 2k + 4k + 2k
16. (A) isc = 10 A, RTH = 2 W, 2
æ 10 ö Pmax = ç ÷ ´ 2 = 50 W è 2 ø www.gatehelp.com
Page 39
GATE EC BY RK Kanodia
UNIT 1
Now in this circuit all straight-through connection
Networks
i 2 R = ( P1 ± P2 ) 2
have been cut as shown in fig. S1.4.32b 6W
36. (C) r =
1W 3W 2W
+
P=
voc = 1. 2 W isc
24 2 ´ 2 = 112.5 W (1. 2 + 2) 2
6V
va
37. (B) RTH =
–
voc 12 = =8W isc 15 .
Fig. S.1.4.32b
va =
6 ´ (2 + 3) =5 V 2 + 3+1
38. (A) Let
33. (B) Since both source have opposite polarity, hence short circuit the all straight-through connection as shown in fig. S.1.4.33 6W
2W
For 0 -10 V scale Rm = 10 ´ 20 k = 200 kW For 0 -50 V scale Rm = 50 ´ 20 k = 1 MW 4 For 4 V reading i = ´ 50 = 20 mA 10 vTH = 20mRTH + 20m ´ 200 k = 4 + 20mRTH 5 For 5 V reading i = ´ 50m = 5 mA 50
1W 3W
1 1 = = 50 mA sensitivity 20 k
+
vTH = 5m ´ RTH + 5m ´ 1M = 5 + 5mRTH 6V
va
Solving (i) and (ii) 16 200 V, RTH = vTH = kW 3 3
–
Fig. S1.4.33
39. (D) v10 k = 10 k ´ 3.6m = 6
6 ´ ( 6 ||3) va = = -4 V 2 +1
v30 k = 30 k ´ 4.8m = 12 V
34. (C) Let Thevenin equivalent of both network RTH
RTH
vTH
R
vTH
RTH
R
vTH
6 =
10 vTH 10 + RTH
12 =
30 vTH 30 + RTH
Þ
Þ
10 vTH = 6 RTH + 60
5 vTH = 2 RTH + 60
RTH = 30 kW 40. (D) At v = 0 , isc = 30 mA
Fig. S1.4.34
At i = 0, voc = - 3 V v -3 RTH = oc = = - 100 W isc 30m
2
æ VTH ö ÷÷ R P = çç è RTH + R ø æ ç V TH P¢ = ç çç R + RTH 2 è
2
ö ÷ æ VTH ÷ R = 4çç ÷÷ è 2 R + RTH ø
2
ö ÷÷ R ø
************
Thus P < P ¢ < 4 P 35. (C) i1 =
P1 P2 and i2 = R R
using superposition i = i1 + i2 =
Page 42
P1 ± R
P2 R www.gatehelp.com
...(i)
...(ii)
GATE EC BY RK Kanodia
CHAPTER
1.6 THE RLC CIRCUITS
1. The natural response of an RLC circuit is described
(A) iL¢¢ ( t) + 1100 i¢¢L ( t) + 11 ´ 108 iL ( t) = 108 is ( t)
by the differential equation
(B) i¢¢L ( t) + 1100 i¢¢L ( t) + 11 ´ 108 iL ( t) = 108 is ( t)
d 2v dv dv(0) +2 + v = 0, v(0) = 10, = 0. dt 2 dt dt The v( t) is (A) 10(1 + t) e - t V
(B) 10(1 - t) e - t V
(C) 10e - t V
(D) 10te - t V
vs
2W
1 mH
. i¢¢L ( t) iL¢¢ ( t) 11 . iL ( t) = is ( t) + + 11 108 10 4
(D)
i¢¢L ( t) 11i¢L ( t) + + 11iL ( t) = is ( t) 108 10 4
4. In the circuit of fig. P.1.6.4 vs = 0 for t > 0. The initial
2. The differential equation for the circuit shown in fig. P1.6.2. is
(C)
condition are v(0) = 6 V and dv(0) dt = -3000 V s. The v( t) for t > 0 is
v
1H
10 mF
100 W
vs
25 mF
80 W
+ vC –
Fig. P1.6.2
(A) v¢¢( t) + 3000 v¢( t) + 102 . ´ 108 v( t) = 108 vs ( t)
Fig. P1.6.4
(B) v¢¢( t) + 1000 v¢ ( t) + 102 . ´ 10 v( t) = 10 vs ( t) 8
8
(A) -2 e -100 t + 8 e -400 t V
(B) 6 e -100 t + 8 e -400 t V (D) None of the above
(C)
2 v¢ ( t) v¢¢( t) . v( t) = vs ( t) + + 102 108 10 5
(C) 6 e -100 t - 8 e -400 t V
(D)
2 v¢ ( t) v¢¢( t) . v( t) = vs ( t) + + 198 8 10 10 5
5. The circuit shown in fig. P1.6.5 has been open for a long time before closing at t = 0. The initial condition is
3. The differential equation for the circuit shown in fig.
v(0) = 2 V. The v( t) for t > is
P1.6.3 is
t=0 1H
10 W is
3 4W
1 3F
+ vC –
10 mF
100 W
Fig. P.1.6.5 iL
Fig. P.1.6.3
Page 54
-t
(A) 5 e - 7 e
-3t
V
(C) - e - t + 3e -3t V www.gatehelp.com
(B) 7 e - t - 5 e -3t V (D) 3e - t - e -3t V
GATE EC BY RK Kanodia
The RLC Circuits
Statement for Q.6–7:
Chap 1.6
10. The switch of the circuit shown in fig. P1.6.10 is
Circuit is shown in fig. P.1.6. Initial conditions are
opened at t = 0 after long time. The v( t) , for t > 0 is
i1 (0) = i2 (0) = 11 A
t=0
3W
i1
i2
2H
1W
1 2H
1W
6V
+ vC –
1 4F
2W
3H
Fig. P1.6.10 Fig. P1.6.6–7
6. i1 (1 s) = ?
(A) 4 e -2 t sin 2 t V
(B) -4 e -2 t sin 2 t V
(C) 4 e -2 t cos 2 t V
(D) -4 e -2 t cos 2 t V
(A) 0.78 A
(B) 1.46 A
11. In the circuit of fig. P1.6.23 the switch is opened at
(C) 2.56 A
(D) 3.62 A
t = 0 after long time. The current iL ( t) for t > 0 is 4H
iL
7. i2 (1 s) = ? (A) 0.78 A
(B) 1.46 A
(C) 2.56 A
(D) 3.62 A
t=0 2W
1 4F
8W
8. vC ( t) = ? for t > 0
4W
7A
25 mH
Fig. P1.6.11
10 mF
100 W
30u(-t) mA
+ vC –
(A) e -2 t (2 cos t + 4 sin t) A
(B) e -2 t ( 3 sin t - 4 cos t) A
(C) e -2 t ( -4 sin t + 2 cos t) A
(D)e -2 t (2 sin t - 4 cos t) A
Statement for Q.12–14: Fig. P1.6.8
In the circuit shown in fig. P1.6.12–14 all initial
(A) 4 e -1000 t - e -2000 t V
(B) ( 3 + 6000 t) e -2000 t V
(C) 2 e -1000 t + e -2000 t V
(D) ( 3 - 6000 t) e -2000 t V
condition are zero. iL
9. The circuit shown in fig. P1.6.9 is in steady state with switch open. At t = 0
10 mH
1 mF
+ vL –
the switch is closed. The
output voltage vC ( t) for t > 0 is
Fig. P1.5.12-14
12. If is ( t) = 1 A, then the inductor current iL ( t) is
0.8 H
250 W
100 W 65
isu(t) A
500 W t=0 9V
5 mF
+ vC –
(A) 1 A
(B) t A
(C) t + 1 A
(D) 0 A
13. If is ( t) = 0.5 t A,
Fig. P1.6.9
A
(B) 2 t - 3250 A
-3
A
(D) 2 t + 3250 A
(A) 0.5 t + 3. 25 ´ 10 (C) 0.5 t - 0. 25 ´ 10
then iL ( t) is
-3
(B) e
-400 t
[ 3 cos 300 t + 4 sin 300 t ]
14. If is ( t) = 2 e -250 t A then iL ( t) is 4000 -250 t 4000 -250 t (A) A (B) A te e 3 3
(C) e
-300 t
[ 3 cos 400 t + 4 sin 300 t ]
(C)
(A) -9 e
-400 t
+ 12 e
-300 t
(D) e -300 t [ 3 cos 400 t + 2. 25 sin 300 t ]
200 -250 t A e 7
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(D)
200 -250 t A te 7
Page 55
GATE EC BY RK Kanodia
UNIT 1
15. The forced response for the capacitor voltage v f ( t) is
Networks
19. In the circuit shown in fig. P 1.5.19 v( t) for t > 0 is
100 W
2u(-t) A iL
– vx +
avx
50 W
1H
20 mH
0.04 F + vC – 2W
4W
Fig. P1.6.15
(A) 0. 2 t + 117 . ´ 10 -3 V
(B) 0. 2 t - 117 . ´ 10 -3 V
(C) 117 . ´ 10 -3 t - 0. 2 V
(D) 117 . ´ 10 -3 t + 0. 2 V
50u(t) V
Fig. P1.6.19
16. For a RLC series circuit R = 20 W , L = 0.6 H, the value of C will be
(B) 50 + ( 46.5 sin 3t + 62 cos 3t) e -4 t V
[CD =critically damped, OD =over damped,
(C) 50 + ( 62 cos 4 t + 46.5 sin 4 t) e -3t V
UD =under damped]. CD
OD
UD
(A) C = 6 mF
C >6 mF
C 6 mF
(C) C >6 mF
C = 6 mF
C < 6 mF
(D) C < 6 mF
C =6 mF
C > 6 mF
(A) 50 - ( 46.5 sin 3t + 62 cos 3t) e -4 t V
(D) 50 - ( 62 cos 4 t + 46.5 sin 4 t) e -3t V 20. In the circuit of fig. P1.6.20 the switch is closed at t = 0 after long time. The current i( t) for t > 0 is 1 16 F + vC –
17. The circuit shown in fig. P1.6.17 is critically damped. The value of R is
20 V
t=0
1 4H
Fig. P1.6.20
120 W
R
5W
iL
10 mF 4H
(A) -10 sin 8 t A
(B) 10 sin 8 t A
(C) -10 cos 8 t A
(D) 10 cos 8 t A
21. In the circuit of fig. P1.6.21 switch is moved from 8 Fig. P1.6.17
(A) 40 W
(B) 60 W
(C) 120 W
(D) 180 W
V to 12 V at t = 0. The voltage v( t) for t > 0 is 2W t=0
18. The step response of an RLC series circuit is given
8V
12 V
by
1H
d 2 i( t) 2 di( t) di(0 + ) = 4. + + 5 i( t) = 10, i(0 + ) = 2, dt dt dt
Fig. P1.6.21
(A) 12 - ( 4 cos 2 t + 2 sin 2 t) e - t V
The i( t) is (A) 1 + e - t cos 4 t A
(B) 4 - 2 e - t cos 4 t A
(B) 12 - ( 4 cos 2 t + 8 sin 2 t) e - t V
(C) 2 + e - t sin 4 t A
(D) 10 + e - t sin 4 t A
(C) 12 + ( 4 cos 2 t + 8 sin 2 t) e - t V (D) 12 + ( 4 cos 2 t + 2 sin 2 t) e - t V
Page 56
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1 F 6
+ vC –
GATE EC BY RK Kanodia
The RLC Circuits
22. In the circuit of fig. P1.5.22 the voltage v( t) is
5W
3u(t) A
25. In the circuit shown in fig. P1.6.25 a steady state has been established before switch closed. The i( t) for
1W
5H
t > 0 is 5W
+ vC –
0.2 F
20 V
Chap 1.6
20 W
1H
2W
i
t=0
1 F 25
5W
100 V
Fig. P1.6.22
(A) 40 - (20 cos 0.6 t + 15 sin 0.6 t) e -0 .8 t V (B) 35 + (15 cos 0.6 t + 20 sin 0.6 t) e
-0 .8 t
(C) 35 - (15 cos 0.6 t + 20 sin 0.6 t) e (D) 35 - 15 cos 0.6 t e
-0 .8 t
-0 .8 t
Fig. P1.6.25
V
(A) 0.73e -2 t sin 4.58 t A
V
(B) 0.89 e -2 t sin 6.38 t A
V
(C) 0.73e -4 t sin 4.58 t A
23. In the circuit of fig. P1.6.23 the switch is opened at t = 0 after long time. The current i( t) for t > 0 is
(D) 0.89 e -4 t sin 6.38 t A 26. The switch is closed after long time in the circuit of fig. P1.6.26. The v( t) for t > 0 is
2A
2A 3 4H 10 W
1 3F
t=0
1W
1H 6W
5W
10 W
4V
t=0
1 25 F
+ vC –
Fig. P1.6.23
(A) e-2 .306 t + e-0 . 869t A
Fig. P1.6.26
(B) -e -2 .306 t + 2 e -0 . 869t A
(A) -8 + 6 e -3t sin 4 t V
(C) e -4 .431 t + e -0 .903t A
(B) -12 + 4 e -3t cos 4 t V
(D) 2 e -4 .431 t - e -0 .903t A
(C) -12 + ( 4 cos 4 t + 3 sin 4 t) e -3t V
24. In the circuit of fig. P1.6.24 switch is moved from position a to b at t = 0. The iL ( t) for t > 0 is 0.02 F
(D) -12 + ( 4 cos 4 t + 6 sin 4 t) e -3t V 27. i( t) = ? 2 kW
14 W
b
i
2 H t=0 2W
12 V
5 mF
12u(t) V
8 mH
a iL
6W
Fig. P1.6.27
(A) 6 - ( 6 cos 500 t + 6 sin 5000 t) e -50 t mA (B) 8 - ( 8 cos 500 t + 0.06 sin 5000 t) e -50 t mA
4A
Fig. P1.6.24
(A) ( 4 - 6 t) e 4 t A
(B) ( 3 - 6 t) e -4 t A
(C) ( 3 - 9 t) e -5t A
(D) ( 3 - 8 t) e -5t A
(C) 6 - ( 6 cos 5000 t + 0.06 sin 5000 t) e -50 t mA (D) 6 e -50 t sin 5000 t mA www.gatehelp.com
Page 57
GATE EC BY RK Kanodia
UNIT 1
Networks
SOLUTIONS
28. In the circuit of fig. P1.6.28 i(0) = 1 A and v(0) = 0. The current i( t) for t > 0 is i 4u(t) A
1. (A) s 2 + 2 s + 1 = 0 2W
1H
s = -1, - 1,
v( t) = ( A1 + A2 t) e dv(0) v(0) = 10 V, = 0 = - 1 A1 + A2 dt
0.5 F
A1 = A2 = 10
Fig. P1.6.28
(A) 4 + 6.38 e -0 .5t A
Þ
-t
(B) 4 - 6.38 e -0 .5t A
(C) 4 + ( 3 cos 1.32 t + 113 . sin 1.32 t) e -0 .5t A
2. (A) iL =
(D) 4 - ( 3 cos 1.32 t + 113 . sin 1.32 t) e -0 .5t A
v dv + 10 ´ 10 -6 100 dt iL
29. In the circuit of fig. P1.6.29 a steady state has been
2W
1 mH
v
established before switch closed. The vo( t) for t > 0 is 10 W
vs
t=0 5W
3A
+
10 mF
1H
Fig. S1.6.2
vo –
vs = 2 iL + 10 -3
Fig. P1.6.29
(A) 100 te -10 t V
(B) 200 te -10 t V
(C) 400 te -50 t V
(D) 800 te -50 t V
108 vs ( t) = v¢¢( t) + 3000 v¢( t) + 102 . v( t) t > 0 is
established before switch closed. The i( t) for i 1H 1 4F
2W t=0
(B) -e
sin 2 t A
(C) -2(1 - t) e
-2 t
-2 t
sin 2 t A
(D) 2(1 - t) e -2 t A
A
4. (A)
31. In the circuit of fig. P1.6.31 a steady state has been established. The i( t) for t > 0 is i 3A
10 W
6u(t) A
diL di d 2 iL + iL + 10 -4 L + 10 -8 dt dt dt 2 . i¢¢L ( t) 11 . iL ( t) = is ( t) + iL¢ ( t) + 11 108 10 4
Þ
Fig. P1.6.30
(A) 2 e
vC dvC + iL + 10m 100 dt di vC = 10 iL + 10 -3 L dt di d di is = 0.1iL + 10 -5 L + iL + 10 -5 (10 iL + 10 -3 L ) dt dt dt 3. (C) is =
= 0.1iL + 10 -5
6V
-2 t
diL +v dt
æ 1 dv dv ö d2 v ö æ v ÷+ v ÷ + 10 -3 çç = 2ç + 10 -6 ´ 10 - t + 10 ´ 10 -6 dt ø dt 2 ÷ø è 100 è 100 dt
30. In the circuit of fig. P1.6.30 a steady state has been
1W
10 mF
100 W
40 W
10 mF
4H
v dv + 25m + ( v - vs ) dt = 0 80 dt ò
d 2v dv + 500 + 40000 = 0 2 dt dt
Þ
s 2 + 500 s + 40000 = 0 Þ
s = -100, - 400,
v( t) = Ae -100 t + Be -400 t A + B = 6, -100 A - 400 B = -3000
Fig. P1.6.31
(A) 9 + 2 e
-10 t
- 8e
-2 .5t
(B) 9 - 8 e
10 t
A
(C) 9 + (2 cos 10 t + sin 10 t) e
-2 .5t
A
(D) 9 + (cos 10 t + 2 sin 10 t) e -2 .5t A ***************
Page 58
+ 2e
-2 .5t
Þ
B = 8, A = -2
A 5. (C) The characteristic equation is s 2 + After putting the values, v( t) = Ae - t + Be -3t , www.gatehelp.com
s2 + 4 s + 3 = 0
1 s + =0 RC LC
GATE EC BY RK Kanodia
The RLC Circuits
v(0 + ) = 2 V
Þ
iL (0 + ) = 0Þ
iR (0) =
-C
9. (B) vC (0 + ) = 3 V , iL (0 + ) = -12 mA
A + B =2
dv(0 + ) 8 = dt 3
vC dvC + iL + 5 ´ 10 -6 =0 250 dt
2 8 = , 34 3
dvC (0 + ) 3 dvC (0 + ) - 12m + 5 ´ 10 -6 =0 Þ =0 dt 250 dt 1 s + =0 s2 + -6 250 ´ 5 ´ 10 0.8 ´ 5 ´ 10 -6
dv(0 + ) = - 8, dt
Þ
- A - 3B = -8, B = 3, A = -1
Þ
di1 di - 3 2 = 0, dt dt 3di2 di 2 i2 + - 3 1 =0 dt dt 6. (D) i1 + 5
Þ
i1 = A e
+ Be
-2 t
10. (B) v(0 + ) = 0, s2 + 4 s + 8 = 0 A1 = 0,
, i(0) = A + B = 11
-
1 6
i2 = - e
2W
1 4F
-
t 6
+ De
+ 12 e
di2 (0) -143 C = = - - 2D dt 6 6
A, i2 (1 s) = e
+ vC –
8W
diL (0 + ) diL (0 + ) = 8 - ( -4) ´ 8 Þ = 10 dt dt 1 s vC + vC + iL = 0, vC = 4 siL + 8 iL 4 2
4
-2 t
s 2 iL + 4 siL + 5 = 0, s = -2 ± j
and D = 12 -2 t
4H
Fig. S1.6.11
-
1 6
+ 12 e
-2
= 0.78 A
iL ( t) = e-2 t ( A1 cos t + A2 sin t) A1 = -4,
8. (B) vC (0 + ) = 30m ´ 100 = 3 V dvC (0 - ) dvC (0 + ) = iL (0 - ) = 0 = iL (0 + ) = C dt dt 100 1 s2 + s+ 25 ´ 10 -3 25 ´ 10 -3 ´ 10 ´ 10 -6 C
Þ
vC (0 + ) = 8 V iL
+ 8 e -2 = 3.62 A
i2 (0) = 11 = C + D,
t 6
11. (D) iL (0 + ) = -4,
+ 8e ,
7. (A) i2 = Ce
-
dvC (0 + ) = -8 = -2 (0 + 0) + (0 + 2 A2 ), A2 = -4 dt
-2 t
i1 (1 s) = 3e
C = -1
s = - 2 ± j2
vC ( t) = e ( A1 cos 2 t + A2 sin 2 t)
di1 (0 + ) 33 di2 (0 + ) 143 ==, dt 2 dt 6 A 33 - 2 B = - , Þ A = 3, B = 8, 6 2 i1 = 3e
Þ
1 dvC (0 + ) = -2 4 xdt
iL (0 + ) = 2 A,
-2 t
In differential equation putting t = 0 and solving
t 6
s = -400 ± j 300
vC ( t) = e-400 t ( A1 cos 300 t + A2 sin 300 t) dvC (0) A1 = 3, = -400 A1 + 300 A2 , A2 = 4 dt
6 s 2 + 13s + 2 = 0 1 s = - , -2 6 1 - t 6
s 2 + 800 s + 25 ´ 10 4 = 0 Þ
(1 + 5 s) i1 - 3si2 = 0, -3si1 + (2 + 3s) i2 = 0 ( 3s)( 3s) i1 (1 + 5 s) i1 =0 2 + 3s Þ
Chap 1.6
s = -2000, -2000
diL (0 + ) = 10 = -2 ( A1 + 0) + A2 , A2 = 2 dt
12. (A) is = is =
v dv di + 10 -3 + iL , v = 10 ´ 10 -3 L 100 65 dt dt
di d 2 iL 65 (10 ´ 10 -3) L + 10 -3(10 ´ 10 -3) + iL = 0 dt dt 100
d 2 iL di + 650 L + 10 5 iL = 10 5 is dt dt
vC ( t) = ( A1 + A2 t) e -2000 t dvC ( t) = A2 e -2000 t + ( A1 + A2 t) e -2000 t ( -2000) dt dvC (0) = A2 - 2000 ´ 3 = 0 vC (0 + ) = A1 = 3, dt
0 + 0 + 10 5 B = 10 5,
Þ
0 + 650 A + ( At + B)10 5 = 10 5(0.5 t), A = 0.5
A2 = 6000
Trying iL ( t) = B B = 1,
iL = 1 A
13. (A) Trying iL ( t) = At + B, www.gatehelp.com
Page 59
GATE EC BY RK Kanodia
The RLC Circuits
Chap 1.6
di(0 + ) -16 = = -4.431 A - 0.903B dt 3
28.(D) i(0 + ) = 1 A, v(0 + ) =
A = 1, B = 1
a=
Ldi(0 + ) dt
1 = 0.5, 2 ´ 2 ´ 0.5
4´ 6 =3 6 +2 dvC (0) dv (0) = 150 0.02 C = iL (0) = 3 Þ dt dt 6 + 14 1 a= = 5, wo = =5 2´2 2 ´ 0.02
1 = 4 + A, Þ A = -3 di(0) = 0 = 0.5 A + 1.32 B, dt
a = wo critically damped
29. (B) Vo(0 + ) = 0 ,iL (0 + ) = 1 A
24. (C) vC (0) = 0,
v( t) = 12 + ( A + Bt) e 0 = 12 + A,
v( t) = 12 + (90 t - 12) e
A = -12,
iL ( t) = 0.02( -5) e (90 t - 12) + 0.02(90) e
-5t
= ( 3 - 9 t) e
-5t
diL (0 + ) = v1 (0) = 0 dt 1 a= = 10, 2 ´ 5 ´ 0.01
B = -113 .
Wo =
1 1 ´ 0.01
= 10
a = Wo, so critically damped response
100 ´ 5 50 , 25. (A) v(0 ) = = 5 + 5 + 20 3 +
+
iL (0 ) = 0
s = -10, - 10 i( t) = 3( A + Bt) e -10 t ,
if = 0 A
i(0) = 1 = 3 + A
+
di(0 ) = -10 A + B dt
diL (0 + ) 50 10 = 20 = dt 3 3 4 1 a= = 2, wo = =5 2´1 1 1´ 25
iL ( t) = 3 - (2 + 20 t) e -10 t ,
i( t) = ( A cos 4.58 t + B sin 4.58 t) e -2 t 26.(A) iL (0 + ) = 0, vL (0 + ) = 4 - 12 = -8 +
LdiL ( t) = 200 te -10 t dt
di(0 + ) -6 vc (0 + ) = 2 ´ 1 = 2 = = -2 A, dt 1+2 1 1 1 a= = = 2, Wo = =2 2 RC 2 ´ 1 ´ 0.25 LC a = Wo, critically damped response s = -2 , -2
1 dvL (0 ) = iL (0 + ) = 0 25 dt 6 1 a = = 3, Wo = =5 2 1 ´ 1 / 25
A = -2 i( t) = ( A + Bt) e -2 t , di( t) = ( -2 + Bt) e 2 t ( -2) + (0 + B) e -2 t dt
b = -3 ± 9 - 25 = -3 ± j 4 v1 ( t) = -12 + ( A cos 4 t + B sin 4 t) e
vo =
30. (C) i(0 + ) =
s = -2 ± 4 - 25 = -2 ± j 4.58
At t = 0, Þ B = -2 -3t
vL (0) = -8 = 12 + A, Þ A =4 dvL (0) = 0 = -3 A + 4 B, Þ B=3 dt 1 1 = = 50 2 RC 2 ´ 2 k ´ 54 1 1 = = 5000 LC 8 m ´ 5m
27. (C) a = Wo =
B = 90
-5t
-5t
= 2
i( t) = 4 + ( A cos 1.32 t + B sin 1.32 t) e -0 .5t
-5t
Þ
1 - 0.5
s = -0.5 ± 0.5 2 - 2 = 0.5 ± j1.323
iL (0) =
150 = -5 A + B
1
Wo =
31. (A) i(0 + ) = 3 A, vC (0 + ) = 0 V = is = 9 A, R = 10||40 = 8 W 1 1 a= = = 6.25 2 RC 2 ´ 8 ´ 0.01 1 1 Wo = = =5 LC 4 ´ 0.01 a > Wo, so overdamped response
a < Wo, underdamped response.
s = -6.25 ± 6.25 2 - 25 = -10, -2.5
s = -50 ± 50 2 - 5000 2 = -50 ± j5000
i( t) = 9 + Ae -10 t + Be -2 .5t
i( t) = 6 + ( A cos 5000 t + B sin 5000 t) e -50 t mA
3 = 9 + A + B,
i(0) = 6 = 6 + A, Þ A = -6 di(0) = -50 A + 5000 B = 0, B = -0.06 dt
4 di(0 + ) dt
0 = -10 A - 2.5 B
On solving, A = 2, B = -8 ************ www.gatehelp.com
Page 61
GATE EC BY RK Kanodia
CHAPTER
1.7 SINUSOIDAL STEADY STATE ANALYSIS
1. i( t) = ?
20cos 300t V
(A)
3W
i
(C)
~
25 mH
1 2 1 2
cos (2 t - 45 ° ) V
(B)
sin (2 t - 45 ° ) V
(D)
1 2 1 2
cos (2 t + 45 ° ) V sin (2 t + 45 ° ) V
4. vC ( t) = ? Fig. P1.7.1
3H
(A) 20 cos ( 300 t + 68. 2 ° ) A (B) 20 cos( 300 t - 68. 2 ° ) A
8cos 5t V
(C) 2.48 cos( 300 t + 68. 2 ° ) A
~
+ vC –
50 mF 9W
(D) 2.48 cos( 300 t - 68. 2 ° ) A
Fig. P1.7.4
2. vC ( t) = ? (A) 2. 25 cos (5 t + 150 ° ) V 3
cos 10 t A
~
2W
+ vC –
1 mF
(B) 2. 25 cos (5 t - 150 ° ) V (C) 2. 25 cos (5 t + 140.71° ) V (D) 2. 25 cos (5 t - 140.71° ) V
Fig. P1.7.2
(A) 0.89 cos (10 3 t - 63.43° ) V
5. i( t) = ?
(B) 0.89 cos (10 3 t + 63.43° ) V
1W
4W
(C) 0.45 cos (10 t + 26.57 ° ) V 3
i
(D) 0.45 cos (10 t - 26.57 ° ) V 3
10cos 2t V
~
0.25 F
4H
3. vC ( t) = ? 5W
cos 2t V
~
0.1 F
Fig. P1.7.5 + vC –
(A) 2 sin (2 t + 5.77 ° ) A
(B) cos (2 t - 84. 23° ) A
(C) 2 sin (2 t - 5.77 ° ) A
(D) cos (2 t + 84. 23° ) A
Fig. P1.7.3
Page 62
www.gatehelp.com
GATE EC BY RK Kanodia
UNIT 1
13. In the bridge shown in fig. P1.7.13, Z1 = 300 W, Z 2 = ( 300 - j 600) W, Z 3 = (200 + j 100)W. The Z 4
Networks
Statement for Q.17-18:
at
The circuit is as shown in fig. P1.7.17-18
balance is
1W
1W
1H
Z
Z
3
1
1H
1W
~
i1
o
10cos (4t-30 ) V
1F
Z
Z
~
i2
4
2
5cos 4t V
Fig. P1.7.17–18
~
17. i1 ( t) = ? (A) 2.36 cos ( 4 t - 4107 . °) A
Fig. P1.7.13
(A) 400 + j 300 W
(B) 400 - j 300 W
(C) j100 W
(D) - j900 W
(B) 2.36 cos ( 4 t + 4107 . °) A (C) 1.37 cos ( 4 t - 4107 . °) A (D) 2.36 cos ( 4 t + 4107 . °) A
14. In a two element series circuit, the applied voltage
and
the
resulting
current
are
v( t) = 60 + 66 sin (10 3 t) V, i( t) = 2.3 sin (10 3 t + 68.3° ) A. The nature of the elements would be (A) R - C
(B) L - C
(C) R - L
(D) R - R
(A) 2.04 sin ( 4 t + 92.13° ) A (B) -2.04 sin ( 4 t + 2.13° ) A (C) 2.04 cos ( 4 t + 2.13° ) A (D) -2.04 cos ( 4 t + 92.13° ) A 19. I x = ?
15. Vo = ? 40 W
j20
120Ð-15o V
18. i2 ( t) = ?
~
~
50 W
-j30
Ix 6Ð30 A
10Ð30 V
o
o
~
Fig. P1.7.15
-j2 W
(B) 223Ð56 ° V
(C) 124 Ð - 154 ° V
(D) 124 Ð154 ° V
16. vo( t) = ?
(A) 394 . Ð46. 28 ° A
(B) 4.62 Ð97.38 ° A
(C) 7.42 Ð92.49 ° A
(D) 6.78 Ð49. 27 ° A
20. Vx = ?
3W
j10 W
20 W 10sin (t+30o) V
~
1F
+ vo –
j3 W
Fig. P1.7.19
(A) 223Ð - 56 ° V
1H
0.5Ix
4W
Vo
~
o
20cos (t-45 ) V
4Vx
3Ð0 A o
~
20 W
+ Vx –
Fig. P1.7.16 Fig. P1.7.20
(A) 315 . cos ( t + 112 ° ) V (B) 43. 2 cos ( t + 23° ) V
(A) 29.11Ð166 ° V
(B) 29.11Ð - 166 ° V
(C) 315 . cos ( t - 112 ° ) V
(C) 43. 24 Ð124 ° V
(D) 43. 24 Ð - 124 ° V
(D) 43. 2 cos ( t - 23° ) V Page 64
www.gatehelp.com
GATE EC BY RK Kanodia
UNIT 1
Statement for Q.27–32:
Networks
35. In the circuit shown in fig. P1.7.35 power factor is
Determine the complex power for hte given values
-j2
4W
in question. 27. P = 269 W, Q = 150 VAR (capacitive) (A) 150 - j269 VA
(B) 150 + j269 VA
(C) 269 - j150 VA
(D) 269 + j150 VA
-j2
j5
Fig. P1.7.35
28. Q = 2000 VAR, pf = 0.9 (leading) (A) 4129.8 + j2000 VA
(B) 2000 + j 4129.8 VA
(C) 2000 - j 4129 . .8 VA
(D) 4129.8 - j2000 VA
(A) 56.31 (leading)
(B) 56.31 (lagging)
(C) 0.555 (lagging)
(D) 0.555 (leading)
36. The power factor seen by the voltage source is 29. S = 60 VA, Q = 45 VAR (inductive) (A) 39.69 + j 45 VA
(B) 39.69 - j 45 VA
(C) 45 + j 39.69 VA
(D) 45 - j 39.69 VA
4W
1W
+ v1 – 10cos 2t V
30. Vrms = 220 V, P = 1 kW, |Z |= 40 W (inductive) (A) 1000 - j 68125 . VA
(B) 1000 + j 68125 . VA
(C) 68125 . + j1000 VA
(D) 68125 . - j1000 VA
31. Vrms = 21Ð20 ° V, Vrms = 21Ð20 ° V, I rms = 8.5 Ð - 50 ° A (A) 154.6 + j 89.3 VA
(B) 154.6 - j 89.3 VA
(C) 61 + j167.7 VA
(D) 61 - j167.7 VA
~
1 3F
3v 4 1
Fig. P1.7.36
(A) 0.8 (leading)
(B) 0.8 (lagging)
(C) 36.9 (leading)
(D) 39.6 (lagging)
37. The average power supplied by the dependent source is
32. Vrms = 120 Ð30 ° V, Z = 40 + j 80 W
Ix
(A) 72 + j144 VA
(B) 72 - j144 VA
(C) 144 + j72 VA
(D) 144 - j72 VA
2Ð90 A o
~
j1.92
4.8 W
8W
1.6Ix
33. Vo = ? Fig. P1.7.37
+
6Ð0 A o
~
16 kW 0.9 pf lagging
VO
20 kW 0.8 pf lagging
(A) 96 W
(B) -96 W
(C) 92 W
(D) -192 W
–
38. In the circuit of fig. P1.7.38 the maximum power
Fig. P1.7.33
absorbed by Z L is (A) 7.1Ð32. 29 ° kV
(B) 42.59 Ð32.29 ° kV
(C) 38.49 Ð24.39 ° kV
(D) 38.49 Ð32. 29 ° kV
10 W
120Ð0o V
34. A relay coil is connected to a 210 V, 50 Hz supply. If
~
j15
-j10
it has resistance of 30 W and an inductance of 0.5 H, the Fig. P1.7.38
apparent power is
Page 66
(A) 30 VA
(B) 275.6 VA
(C) 157 VA
(D) 187 VA
(A) 180 W
(B) 90 W
(C) 140 W
(D) 700 W
www.gatehelp.com
ZL
GATE EC BY RK Kanodia
Sinusoidal Steady State Analysis
Chap 1.7
39. The value of the load impedance, that would
The line impedance connecting the source to load is
absorbs the maximum average power is
0.3 + j0.2 W. If the current in a phase of load 1 is
j100
3Ð20o A
~
I = 10 Ð20 ° A rms , the current in source in ab branch is
-j40
80 W
ZL
(A) 15 Ð - 122 ° A rms
(B) 8.67 Ð - 122 ° A rms
(C) 15 Ð27.9 ° A rms
(D) 8.67 Ð - 57.9 ° A rms
45. Fig. P1.7.39
(A) 12.8 - j 49.6 W
(B) 12.8 + j 49.6 W
(C) 339 . - j 86.3 W
(D) 339 . + j 86.3 W
An
abc
phase
sequence
3-phase
balanced
Y-connected source supplies power to a balanced D –connected load. The impedance per phase in the load is 10 + j 8 W. If the line current in a phase is I aA = 28.10 Ð - 28.66 ° A rms and the line impedance is zero, the load voltage V AB is
Statement for Q.40–41: In a balanced Y–connected three phase generator Vab = 400 Vrms 40. If phase sequence is abc then phase voltage Va , Vb , and Vc are respectively
(A) 207.8 Ð - 140 ° Vrms
(B) 148.3Ð40 ° Vrms
(C) 148.3Ð - 40 ° Vrms
(D) 207.8 Ð40 ° Vrms
46. The magnitude of the complex power supplied by a 3-phase balanced Y-Y system is 3600 VA. The line voltage is 208 Vrms . If the line impedance is negligible
(A) 231Ð0 ° , 231Ð120 ° , 231Ð240 °
and the power factor angle of the load is 25°, the load
(B) 231Ð - 30 ° , 231Ð - 150 ° , 231Ð90 °
impedance is
(C) 231Ð30 ° , 231Ð150 ° , 231Ð - 90 °
(A) 5.07 + j10.88 W
(B) 10.88 + j5.07 W
(D) 231Ð60 ° , 231Ð180 ° , 231Ð - 60 °
(C) 432 . + j14.6 W
(D) 14.6 + j 432 . W
41. If phase sequence is acb then phase voltage are (A) 231Ð0 ° , 231Ð120 ° , 231Ð240 ° (B) 231Ð - 30 ° , 231Ð - 150 ° , 231Ð90 °
***********
(C) 231Ð30 ° , 231Ð150 ° , 231Ð - 90 ° (D) 231Ð60 ° , 231Ð180 ° , 231Ð - 60 ° 42. A balanced three-phase Y-connected load has one phase voltage Vc = 277 Ð45 ° V. The phase sequence is abc. The line to line voltage V AB is (A) 480 Ð45 ° V
(B) 480 Ð - 45 ° V
(C) 339 Ð45 ° V
(D) 339 Ð - 45 ° V
43. A three-phase circuit has two parallel balanced D loads, one of the 6 W resistor and one of 12 W resistors. The magnitude of the total line current, when the line-to-line voltage is 480 Vrms , is (A) 120 A rms
(B) 360 A rms
(C) 208 A rms
(D) 470 A rms
44. In a balanced three-phase system, the source has an abc phase sequence and is connected in delta. There are two parallel Y-connected load. The phase impedance of load 1 and load 2 is 4 + j 4 W and 10 + j 4 W respectively. www.gatehelp.com
Page 67
GATE EC BY RK Kanodia
UNIT 1
SOLUTIONS
æ -j ö 7. (C) Z = çç ÷÷||( 6 + j(27m) w) è w(22m ) ø
1. (D) Z = 3 + j(25m)( 300) = 3 + j7.5 W = 8.08 Ð68. 2 ° I=
20 Ð0 = 2.48 Ð - 68. 2 ° A 8.08 Ð68. 2 °
i( t) = 2.48 cos ( 300 t - 68. 2 ° ) A 1 + j(1m)(10 3) = 0.5 + j = 112 . Ð63.43° 2 (1Ð0) VC = = 0.89 Ð - 63.43° V 112 . Ð63.43° 2. (A) Y =
vC ( t) = 0.89 cos (10 3 t - 63.43° ) V
(1Ð0) (5 Ð - 90 ° )
vC ( t) =
5 2 Ð - 45 ° 1 2
1
=
2
27 ´ 10 3 j 6 ´ 106 - j106 ( 6 + j27 ´ 10 -3 w) 22 22 w = 22 w = 6 10 æ 106 ö 6 + j(27mw ) ÷ 6 + jwçç 27m 22 w 22 w2 ÷ø è j27 ´ 10 3 - j 36 ´ 106 22 w22
æ 106 wçç 27m 22 w2 è
Þ w = 1278 1278 w Hz = = 203 Hz f = 2p 2p
V1 = Vs - V2 = 7.68 Ð47° - 7.51Ð35° = 159 . Ð125° 9. (B) vin = 32 + (14 - 10) 2 = 5
Ð - 45 ° V
10. (C) I1 = 744 Ð - 118 ° mA,
cos (2 t - 45 ° ) V
I 2 = 540 Ð100 ° mA
4. (D) Z = 9 + j( 3)(5) +
I = I1 + I 2 = 744 Ð - 118 ° + 540.5 Ð100 °
-j = 9 + j11 (50m) (5)
= 460 Ð - 164 ° i( t) = 460 cos ( 3t - 164 ° ) mA
Þ
Z = 14.21Ð50.71° W ( 8 Ð0)( 4 Ð - 90 ° ) VC = = 2. 25 Ð140.71° V 14. 21Ð50.71°
11. (A)
2 Ð45° =
VC V - 20 Ð0 + C - j4 j5 + 10
vC ( t) = 2. 25 cos (5 t - 140.71° ) V
j50
10 Ð0 10 Ð0 1 5. (B) Va = V = 1 1 1 105 . + j0.4 + + 1 - j2 4 + j 8 I=
ö ÷÷ = 0 ø
8. (C) Vs = 7.68 Ð47° V, V2 = 7.51Ð35°
-j 3. (A) Z = 5 + = 5 - j5 = 5 5 Ð - 45 ° (0.1)(2) VC =
Networks
Ö2Ð45 A o
~
+ -j4
Va 10 Ð0 = = 1Ð - 84. 23 A 4 + j 8 1 + j10 1W
Va
10Ð0 V
~
–
~
20Ð0 V o
Fig. S1.7.11 4W I
o
VC
10 W
-j2
j8
(1 + j)( - j 4)(10 + j5) = VC (10 + j5 - j 4) + j 8 Þ
60 - j100 = VC (10 + j) Þ
VC = 11.6 Ð - 64.7°
12. (D) X = X L + X C = 0 Fig. S1.7.5
i( t) = cos (2 t - 84.23° ) A 6. (D) w = 2 p ´ 10 ´ 10 3 = 2 p ´ 10 4 -j 1 Y = j(1m )(2 p ´ 10 4 ) + + 4 (160m )(2 p ´ 10 ) 36 = 0.0278 - j0.0366 S 1 Z= = 1316 . + j17.33 W Y
Page 68
So reactive power drawn from the source is zero. 13. (B) Z1 Z 4 = Z 3Z 2 300 Z 4 = ( 300 - j 600)(200 + j100) Þ Z 4 = 400 - j 300 14. (A) R - C causes a positive phase shift in voltage Z =|Z |Ðq , -90 ° < q < 0 , I=
V V = Ð-q Z |Z |
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Sinusoidal Steady State Analysis
120 Ð15 ° - 6 Ð30 ° 40 + j20 15. (C) Vo = = 124 Ð - 154 ° 1 1 1 + + 40 + j20 - j 30 50
20. (B) Let Vo be the voltage across current source Vo - 4 Vx Vo - Vx + =3 20 j10 Vo(20 + j10) - (20 + j 40) Vx = j 600 Vo(20) V Vx = Þ Vo = x (2 + j) 20 + j10 2
16. (C) 10 sin ( t + 30 ° ) = 10 cos ( t - 60 ° ) 10 Ð - 60 ° 20 Ð - 45 ° + j 3 Vo = 1 1 1 + + j -j 3
æ (2 + j)(20 + j10) ö Vx = ç - 20(1 + j2) ÷ = j 600 2 è ø j 600 Vx = = 29. 22 Ð - 166 ° -5 - j20
= 30 Ð - 150 °+20 Ð - 45 ° . Ð - 112 ° V Vo = 315
10Ð-60 V
æ j ö V - V2 21. (A) I1 = V3ç ÷ + 3 = j0.1V2 + j0.4 V3 j10 è2 ø
3W
j1
o
+
~
-j1 W
~
Vo –
= (0.1Ð90 ° )(0.757 Ð66.7 ° ) + (0.4 Ð90 ° )(0.606 Ð - 69.8 ° ) Þ
20Ð-45 V o
I1 = 0.196 Ð35.6 °
22. (A) Fig. S.1.7.16
1W
5Ð0 V o
~
1W
10Ð-30 V o
12Ð0 V o
-j0.25 W
j4
+
~
2W
I1
Vo
I2
–
2W
Þ
( 8 + j15) I1 - ( 4 - j) I 2 = 20 Ð0 j j -10 Ð - 30 ° = I 2 (1 + j 4 + 1 - ) - I1 (1 - ) 4 4 ( 4 - j) I1 - ( 8 + j15) I 2 = 40 Ð - 30 °
~
...(i)
2Ð0o A
Fig. P1.7.23
...(ii) 12 Ð0 ° = I1 ( - j 3 + 2 + 2) + 8 Ð90 °-4 Ð0 °
= (20 Ð0)( 8 + j15) - ( 40 Ð - 30 ° )( 4 - j)
Þ
I1 ( -176 + j248) = 41.43 + j 414.64
. + j0.64 + j 4) = 11.65 Ð52.82 ° V Vo = 2( 352
Þ
2
. - j0.9 = 1.37 Ð - 4107 . I1 = 103
. + j0.64 I1 = 352
24. (D) I 2 = 3Ð0 ° A , I 4 - I 3 = 6 Ð0 ° A
( 8 + j15)(103 . - j0.9) - 20 Ð0 ° 18. (B) I 2 = 4-j = -0.076 + j2.04
Þ
o
2W
I1 [( 8 + j15) - ( 4 - j) ] 2
4Ð90 V
~
Fig. S.1.7.17
Þ
Vo = 5.65 Ð - 75 °
-j3
~
I2
Þ
23. (D) I 2 = 4 Ð90 ° , I 3 = 2 Ð0 °
j4
1W I1
Vo Vo - 3Vo + = 4 Ð - 30 ° 2 j4
Vo(0.5 + j0.5) = 3.46 - j2
jö jö æ æ 17. (C) 5 Ð0 ° = I1 ç j 4 + 1 + 1 - ÷ - I 2 ç 1 - ÷ 4ø 4ø è è j4
Chap 1.7
Io
I 2 = 2.04 Ð92.13°
15Ð90o V
~
2W
I2
-j4
j2
~
3Ð0o A
19. (B) 10 Ð30 ° = 4 I1 - 0.5 I x + ( - j2) I x ( - j2) I x = ( I1 - I x ) j 3, I1 = æ4 ö 10 Ð30 ° = ç - 0.5 - j2 ÷I x è3 ø
Ix 3
1W
Þ
Ix =
10 Ð30 ° 2.17 Ð - 67.38 °
I3
~
6Ð0 A I4 o
1W
Fig. S.1.7.24
I 3(1) + ( I 3 - I o)( - j 4) + ( I 4 + I 2 )( j2) + I 4 = 0 www.gatehelp.com
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GATE EC BY RK Kanodia
UNIT 1
I 3 + ( I 3 - I o)( - j 4) + ( I 3 + 6 Ð0 °+3Ð0 ° )( j2) + I 3 + 60 ° = 0 I 3(2 - j2) + I o( j 4) = -18 j - 6 -I ( j2) - 3 - 9 j I3 = o (1 - j)
S2 = 20 + j
15 Ð90 ° = ( I o + 3Ð0 ° )(2) + ( I o - I 3)( - j 4) Þ
VTH =
|V |2 (120) 2 = = 72 + j144 VA Z* 40 - j 80
20 sin (cos -1 (0.8)) = 20 + j15 0.8
S = VoI * = 6 Vo
( j10)( 8 - j5) = = 9 + j 4.4 8 + j10 - j5
Þ
S=
|V |2 (210) 2 = * Z 30 - j157
...(i)
Apparent power =|S |=
...(ii)
35. (D) Z = 4 +
300 I 2 = 3V1 , V1 = ( - j 300)( I1 - I 2 ) I 2 = - j 3( I1 - I 2 )
Þ
3I1 = ( 3 + j) I 2
Solving (i) and (ii) I 2 = 12.36 Ð - 16 ° mA
36. (A)
q = 25.84 °
Þ
3V 4 1
-j1.5
Fig. S.1.7.36
I1 = 1Ð36.9 ° (1Ð36.9 ° )(10 Ð0 ° ) S= = 5 Ð - 36.9 ° 2
S = 4129.8 - j2000 29. (A) Q = S sin q
1W
~
Q 2000 = = 4588.6 VA sin q sin 25.84 °
P = S cos q = 4129.8,
V1 = 4 Ð36.9 °,
+ V1 –
I1 o
S=
Þ
4W
10Ð0 V
Þ
= 275.6 VA
( - j2)( j5 - j2) - j2 + j5 - j2
10 - V1 V1 3 + V1 = 4 4 1 - j15 .
27. (C) S = P - jQ = 269 - j150 VA
Q = S sin q
30 2 + 152 2
pf = cos 56.31° = 0.555 leading
-2 V1 - V1 = 0 Þ V1 = 0 9 Ð0 ° Þ I sc = = 15 Ð0 ° mA 600 311 . Ð - 16 ° V Z TH = oc = = 247 Ð - 16 ° W I sc 15 Ð0 °´10 -3
Þ
(210) 2
= 4 - j 6 = 7.21Ð - 56.31°,
. Ð - 16 ° Voc = 300 I 2 = 371
28. (D) pf = cos q = 0.9
Vo = 7.1Ð32.29 °
34. (B) Z = 30 + j(0.5)(2 p)(50) = 30 + j157,
( 32 Ð0 ° )( j10) = 339 . Ð58 ° V 8 + j10 - j5
26 (D) ( 600 - j 300) I1 + j 300 I 2 = 9 Þ
16 sin (cos -1 (0.9)) = 16 + j7.75 0.9
S = S1 + S2 = 36 + j2.75 = 42.59 Ð32.29 °
j15 = 2 I o + 6 + ( j 4)( 3 - j 6)
25. (A) Z TH
32. (A) S =
33. (A) S1 = 16 + j
I3 = Io + 3 - j6
Þ
Networks
Þ
sin q =
Q 45 or = S 60
q = 48.59 °,
pf = cos 36.9 ° = 0.8 leading 37. (A) (2 Ð - 90 ° ) 4.8 = - I x ( 4.8 + j192 . ) + 0.6 I x ( 8)
P = S cos q = 39.69,
Ix
j1.92
Va
S = 39.69 + j 45 VA 4.8 W
|V |2 (220 2 ) 30. (B) S = rms = = 1210 |Z | 40
1.6Ix (2Ð90o)4.8 V
P 1000 cos q = = = 0.8264 or q = 34.26 °, S 1210
Fig. S.1.7.37
Q = S sin q = 68125, . S = 1000 + j 68125 . VA * 31. (C) S = Vrms I rms = (21Ð20 ° )( 8.5 Ð50 ° )
I x = 5 Ð0 °, Va = 0.6 ´ 5 ´ 8 = 24 Ð0 °, 1 Pave = ´ 24 ´ 1.6 ´ 5 = 96 2
= 61 + j167.7 VA Page 70
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GATE EC BY RK Kanodia
Sinusoidal Steady State Analysis
38. (A) Z TH = VTH =
( - j10)(10 + j15) = 8 - j14 W 10 + j15 - j10
120( - j10) = 107.3Ð - 116.6 ° V 10 + j5
40. (B) Va =
3
I aA 3
Ð( q + 30 ° ) = 16.22 Ð1.34 ° A rms
= 207.8 Ð40 ° Vrms 46. (B) |S |= 3VL I L ZY =
( - j 40)( 80 + j100) = 12.8 - j 49.6 W 80 + j 60
400
45. (D) I AB =
V AB = I AB × Z D = (16. 22 Ð1.340 ° )(10 + j 8)
107.3Ð - 116.6 ° IL = = 6.7 Ð - 116.6 ° 16 1 PLmax = ( 6.7) 2 ´ 8 = 180 W 2 39. (B) Z TH =
Chap 1.7
208 10 3
Þ
IL =
3600 208 3
= 10 A rms
Ð25 ° = 12 Ð25 ° = 10.88 + j5.07 W
********
Ð - 30 ° = 231Ð - 30 ° V
Vb = 231Ð - 150 ° V, Vc = 231Ð - 270 ° V 41. (C) For the acb sequence Vab = Va - Vb = Vp Ð0 ° - Vp Ð120 ° æ 1 3 400 = Vp çç 1 + - j 2 2 è 400 Þ Vp = Ð30 ° 3
ö ÷ = Vp 3Ð - 30 ° ÷ ø
Va = Vp Ð0 ° = 231Ð30 ° V, Vb = Vp Ð120 ° = 231Ð150 ° V Vc = Vp Ð240 ° = 231Ð - 90 ° V 42. (B) V A = 277 Ð( 45 °-120 ° ) = 277 Ð - 75 ° V VB = 277 Ð( 45 ° + 120 ° ) = 277 Ð165 ° V V AB = V A - VB = 480 Ð - 45 ° V 43. (C) Z A = 6 ||12 = 4, IP =
480 = 120 A rms 4
I L = 3I P = 208 A rms 44 (B) I =
I aA (10 + j 4) = 10 Ð20 ° (10 + j 4) + ( 4 + j 4) IaA
a
Iac
c
Iab Ibc
IbB b
icC
Fig. S.1.7.44
I aA = 15 Ð - 27.9 ° A rms |I | I ab = - aA Ð( q + 30 ° ) = 8.67 Ð - 122.1° A rms 3 www.gatehelp.com
Page 71
GATE EC BY RK Kanodia
CHAPTER
1.8 CIRCUIT ANALYSIS IN THE S-DOMAIN
s2 + 1 s2 + 2 s + 1 2 s2 + 1 (C) 2 s + 2s + 2
1. Z ( s) = ?
2( s 2 + 1) ( s + 1) 2 s2 + 1 (D) 3s + 2
(A) 1F
2H
Z(s)
1W
1W
(B)
4. Z ( s) = ?
1W
Z(s)
1H
1W
2
s 2 + 3s + 1 s( s + 1) 2 s 2 + 3s + 1 (D) 2 s( s + 1) (B)
W
Fig. P1.8.1
s 2 + 15 . s+1 (A) s( s + 1) 2 s 2 + 3s + 2 (C) s( s + 1)
0.5 F
Fig. P1.8.4
2. Z ( s) = ?
3s 2 + 8 s + 7 s(5 s + 6) 3s 2 + 7 s + 6 (C) s(5 s + 6)
s(5 s + 6) 3s 2 + 8 s + 7 s(5 s + 6) (D) 3s 2 + 7 s + 6 (B)
(A) 1F Z(s)
1W
1H
1W
5. The s-domain equivalent of the circuit of Fig.P1.8.5. is t=0
Fig. P1.8.2
s + s+1 s( s + 1) s( s + 1) (C) 2 s2 + s + 1
2s + s + 1 s( s + 1) s( s + 1) (D) 2 s + s+1
2
3W
2
(A)
(B)
6V
3F
+ vC –
Fig. P1.8.5
3. Z ( s) = ?
3W
3W + 1 3s
1H Z(s)
2W 1F
Fig. P1.8.3
6 V s
(A) (C) Both A and B
Page 72
1 3s
VC(s)
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+ VC(s) -
-
(B) (D) None of these
2A
GATE EC BY RK Kanodia
Circuit Analysis in the s-Domain
Chap 1.8
6. The s–domain equivalent of the circuit shown in Fig.
9. For the network shown in fig. P1.8.9 voltage ratio
P1.8.6 is
transfer function G12 is t=0
+ 12 W
2A
2H
1H
1H
1F
1F
vL v1
–
1F
1F
1F
+ v2 -
Fig. P1.8.6 Fig. P1.8.9
(A)
(B) +
+
(A)
( s 2 + 2) 5 s + 5 s2 + 1
(B)
s2 + 1 5 s + 5 s2 + 1
(C)
( s 2 + 2) 2 5 s4 + 5 s2 + 1
(D)
( s 2 + 1) 2 5 s4 + 5 s2 + 1
2s 12 W
12 W
VL
2A s
VL
2s
4
4
4V -
-
(C) Both A and B
10. For the network shown in fig. P1.8.10, the admittance transfer function is
(D) None of these
Y12 = Statement for Q.7-8:
K ( s + 1) ( s + 2)( s + 4)
3 W 2
The circuit is as shown in fig. P1.8.7–8. Solve the
i1
problem and choose correct option.
i2
1W
+ +
is
1W
1H
io
v1
+ vs
1F
1F
1W
2F
2F
3
-
Fig. P1.8.10
The value of K is (A) -3
V ( s) 7. H1 ( s) = o =? Vs ( s)
(C)
(A) s( s3 + 2 s2 + 3s + 1) -1
(B) 3
1 3
(D) -
1 3
11. In the circuit of fig. P1.8.11 the switch is in position
(B) ( s 3 + 3s 2 + 2 s + 1) -1
1 for a long time and thrown to position 2 at t = 0. The
(C) ( s 3 + 2 s 2 + 3s + 2) -1
equation for the loop currents I1 ( s) and I 2 ( s) are
(D) s( s 3 + 3s 2 + 2 s2 + 2) -1
1F
1
I o( s) =? Vs ( s)
2
t=0
12 V
(A)
-s ( s + 3s + 2 s + 1)
(B) -( s 3 + 3s 2 + 2 s + 1) -1
(C)
-s 3 2 ( s + 2 s + 3s + 1)
(D) ( s 3 + 2 s 2 + 3s + 2) -1
2
v2
-
Fig. P1.8.7–8
3
6
vo –
8. H 2 ( s) =
1W
i1
3H
2W i2 1F
Fig. P1.8.11
1 é ê2 + 3s + s (A) ê ê -3s ë
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ù - 3s ú 1 ú 2+ ú s û
é12 ù é I1 ( s) ù ê s ú êI ( s) ú = ê ú ë 2 û ê0 ú ë û
Page 73
GATE EC BY RK Kanodia
UNIT 1
1 é ê2 + 3s + s (B) ê ê -3s ë
ù - 3s ú 1 ú 2+ ú s û
é 12 ù é I1 ( s) ù ê- s ú ú êI ( s) ú = ê ë 2 û ê 0 ú ë û
1 é -3s ê2 + 3s + s (C) ê 2 + 3s + ê -3s ë
é 12 ù ù ú é I1 ( s) ù ê- s ú = ú 1 ú êëI 2 ( s) úû ê ê 0 ú ú ë û sû
1 é -3s ê2 + 3s + s (D) ê 2 + 3s + ê -3s ë
ù é12 ù ú é I1 ( s) ù ê s ú = 1 ú êëI 2 ( s) úû ê ú ú ê0 ú sû ë û
Networks
3F
S1 -
4F
Va +
5V
+
10 V
2F
1V
S2
-
+ 6V
5V
-
Fig. P1.8.14
(A)
9 t
(B) 9e - t V
V
(C) 9 V
(D) 0 V
15. A unit step current of 1 A is applied to a network whose driving point impedance is
12. In the circuit of fig. P1.8.12 at terminal ab
1
The steady state and initial values of the voltage
s
a
4 A
+ Vo(s) -
2W
(s+1)
developed across the source would be respectively
2Vo(s)
b
(A)
3 4
(C)
3 4
V, I V
1 4
V,
3 4
V
(D) 1 V,
3 4
V
(B)
V, 0 V
16. In the circuit of Fig. P1.8.16 i(0) = 1 A, vC (0) = 8
Fig. P1.8.12
4
V and v1 = 2 e -2 ´10 t u( t). The i( t) is
(A) VTH ( s) =
-8( s + 2) -(2 s + 1) , Z TH ( s) = 3s( s + 1) 3s
(B) VTH ( s) =
8( s + 2) (2 s + 1) , Z TH ( s) = 3s( s + 1) 3s
(C) VTH ( s) =
4( s + 3) (2 s + 1) , Z TH ( s) = 3s( s + 1) 6s
50 W
i
1m H
2.5 mF
v1
+ vC –
Fig. P1.8.16 4
-4( s + 3) -(2 s + 1) (D) VTH ( s) = , Z TH ( s) = 3s( s + 1) 6s
(A)
1 15
[10 e-10
(B)
1 15
[ -10 e -10
13. In the circuit of fig. P1.8.13 just before the closing of
(C) 13 [10 e-10
switch at t = 0, the initial conditions are known to be
(D) 13 [ -10 e -10
-
V ( s) ( s + 3) = I ( s) ( s + 2) 2
Z ( s) =
Thevenin equivalent is
-
vC1 (0 ) = 1 V, vC2 (0 ) = 0. The voltage vC1 ( t) is
4
- 3e-2 ´10
t 4
t
4
t
4
- 22 e-4 ´10 t ]u( t) A
t
+ 3e -2 ´10
+ 3e-2 ´10
t
4
4
4
4
+ 22 e -4 ´10 t ]u( t) A 4
+ 22 e-4 ´10 t ]u( t) A
t
+ 3e -2 ´10
t
4
t
4
- 22 e -4 ´10 t ]u( t) A
17. In the circuit shown in Fig. P1.8.18 v(0 - ) = 8 V and + vC1 -
iin ( t) = 4d( t). The vC ( t) for t ³ 0 is
t=0 1F
1F
+ vC2 -
iin
50 W
20 mF
Fig. P1.8.13
(A) u( t) V (C) 0.5 e
-t
(B) 0.5 u( t) V V
Fig. P1.8.17
(D) e - t V -
14. The initial condition at t = 0 of a switched capacitor
(A) 164e- t V
(B) 208e- t V
(C) 208(1 - e -3t ) V
(D) 164 e -3t V
circuit are shown in Fig. P1.8.14. Switch S1 and S2 are closed at t = 0. The voltage va ( t) for t > 0 is Page 74
+ vC –
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GATE EC BY RK Kanodia
Circuit Analysis in the s-Domain
18. The driving point impedance Z ( s) of a network has
Chap 1.8
The steady state voltage across capacitor is
the pole zero location as shown in Fig. P1.8.18. If
(A) 6 V
(B) 0 V
Z(0) = 3, the Z ( s) is
(C) ¥
(D) 2 V
jw
23.
1
-3
transformed
VC ( s) = -1
4( s + 3) s2 + s + 1
(B)
2( s + 3) (C) 2 s + 2s + 2
2( s + 3) 2 s + 2s + 2
4( s + 3) (D) 2 s + s+2
20 s + 6 (10 s + 3)( s + 4) (B) - 0.12 mA
(C) 0.48 mA
(D) - 0.48 mA
24. The current through an 4 H inductor is given by I L ( s) =
The circuit is as shown in the fig. P1.8.19–21. All initial conditions are zero. io
1H
60 mF
the
(A) 0.12 mA
Statement for Q.19-21:
iin
across
The initial current through capacitor is
Fig. P1.8.18
(A)
voltage
capacitor is given by
s
-1
The
1W
1F
10 s( s + 2)
The initial voltage across inductor is (A) 40 V
(B) 20 V
(C) 10 V
(D) 5 V
25. The amplifier network shown in fig. P1.8.25
1W
is stable if 4W
1F
1H
Fig. P1.8.19–21
19.
I o( s) =? I in ( s)
(A)
( s + 1) 2s
+ + v1 Amplifier v 2 gain=K -
2W
(B) 2 s( s + 1) -1
(C) ( s + 1) s -1
Fig.P1.8.25
(D) s( s + 1) -1 (A) K £ 3
20. If iin ( t) = 4d( t) then io( t) will be
(C) K £
(A) 4d( t) - e- t u( t) A (B) 4 d( t) - 4 e- t u( t) A
(B) K ³ 3
1 3
(D) K ³
1 3
26. The network shown in fig. P1.8.26 is stable if
(C) 4 e - t u( t) - 4 d( t) A
2F
1W
(D) e - t u( t) - d( t) A
+
21. If iin ( t) = tu( t) then io( t) will be (A) e - t u( t) A
(B) (1 - e - t ) u( t) A
(C) u( t) A
(D) (2 - e - t ) u( t) A
Kv2
1W
1F
v2 -
Fig.P1.8.26
22. The voltage across 200 mF capacitor is given by VC ( s) =
2s + 6 s( s + 3)
5 2 2 (C) K ³ 5 (A) K ³
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5 2 2 (D) K £ 5 (B) K £
Page 75
GATE EC BY RK Kanodia
UNIT 1
27. A circuit has a transfer function with a pole s = - 4 and a zero which may be adjusted in position as s = - a The response of this system to a step input has a term of form Ke -4 t . The K will be aö æ (A) H ç 1 - ÷ 4ø è
(H= scale factor) aö æ (B) H ç 1 + ÷ 4ø è
aö æ (C) H ç 4 - ÷ 4ø è
aö æ (D) H ç 4 + ÷ 4ø è
io( t) = 2 sin 2 t u( t) A. The circuit had no internal stored energy at t = 0. The admittance transfer function is 2 s (A) (B) s 2 (D)
31. The current ratio transfer function
Io is Is
(A)
s( s + 4) s + 3s + 4
(B)
s( s + 4) ( s + 1)( s + 3)
(C)
s 2 + 3s + 4 s( s + 4)
(D)
( s + 1)( s + 3) s( s + 4)
2
32. The response is
28. A circuit has input vin ( t) = cos 2 t u( t) V and output
(C) s
Networks
1 s
(A) Over damped
(B) Under damped
(C) Critically damped
(D) can’t be determined
33. If input is is 2u( t) A, the output current io is (A) (2 e - t - 3te -3t ) u( t) A
(B) ( 3te - t - e -3t ) u( t) A
(C) ( 3e - t - e -3t ) u( t) A
(D) ( e -3t - 3e - t ) u( t) A
34. In the network of Fig. P1.8.34, all initial condition are zero. The damping exhibited by the network is
29. A two terminal network consists of a coil having an inductance L and resistance R shunted by a capacitor
1F 4
C. The poles of the driving point impedance function Z of this network are at - 12 ± j
3 2
and zero at -1. If
+
Z(0) = 1 the value of R, L, C are 1 (A) 3 W, 3 H, F 3
2H
vs
2W
vo -
1 (B) 2 W, 2 H, F 2 Fig. P1.8.34
1 (C) 1 W, 2 H, F 2
(D) 1 W, 1 H, 1 F
(A) Over damped (B) Under damped
30. The current response of a network to a unit step input is
(C) Critically damped (D) value of voltage is requires
10( s + 2) Io = 2 s ( s + 11s + 30)
35. The voltage response of a network to a unit step input is
The response is (A) Under damped
(B) Over damped
(C) Critically damped
(D) None of the above
Vo( s) =
10 s( s 2 + 8 s + 16)
The response is Statement for Q.31-33:
(A) under damped
(B) over damped
(C) critically damped
(D) can’t be determined
The circuit is shown in fig. P1.8.31-33. 36. The response of an initially relaxed circuit to a 1H
io
(
signal vs is e -2 t u( t). If the signal is changed to vs + , the response would be
1F
is
3
4W
(A) 5 e -2 t u( t)
(B) -3e -2 t u( t)
(C) 4 e -2 t u( t)
(D) -4 e -2 t u( t)
Fig. P1.8.31-33 Page 76
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2 dvs dt
)
GATE EC BY RK Kanodia
Circuit Analysis in the s-Domain
37. Consider the following statements in the circuit shown in fig. P1.8.37 i
4W
2H
1W
10 V
1F
2
+ vC –
42. The network function
(B) RL admittance
(C) LC impedance function
(D) LC admittance
43. A valid immittance function is ( s + 4)( s + 8) s( s + 1) (A) (B) ( s + 2)( s - 5) ( s + 2)( s + 5) s( s + 2)( s + 3) ( s + 1)( s + 4)
(C) 1. It is a first order circuit with steady state value of 10 5 , i= A vC = 3 3 2. It is a second order circuit with steady state of vC = 2 V , i = 2 A V ( s) has one pole. I ( s)
4. The network function
V ( s) has two poles. I ( s)
(D)
44. The network function
(B) 1 and 4
(C) 2 and 3
(D) 2 and 4
38. The network function
(B) RC admittance
(C) LC admittance
(D) Above all
Z ( s) =
3W
s 2 + 10 s + 24 represent a s 2 + 8 s + 15
1F
(D) None of the above
(C) LC impedance
(D) None of these
3W
8F
1F
(A)
(B)
1F
1W
3
1H
1F
8
s( 3s + 8) 40. The network function represents an ( s + 1)( s + 3)
1W
3W
3
(C)
(A) RL admittance
(B) RC impedance
(C) RC admittance
(D) None of the above
41. The network function
1W
3
8
s( s + 4) represents ( s + 1)( s + 2)( s + 3) (B) RL impedance
1W
3
1F
(C) LC impedance
(A) RC impedance
3( s + 2)( s + 4) s( s + 3)
The network for this function is
(B) RL impedance
an
s 2 + 8 s + 15 is a s2 + 6 s + 8
45. A impedance function is given as
(A) RC admittance
39. The network function
s( s + 2)( s + 6) ( s + 1)( s + 4)
(A) RLadmittance
The true statements are (A) 1 and 3
s2 + 7 s + 6 is a s+2
(A) RL impedance function
Fig. P1.8.37
3. The network function
Chap 1.8
( s + 1)( s + 4) is a s( s + 2)( s + 5)
1H
8
3W
(D)
************
(A) RL impedance function (B) RC impedance function (C) LC impedance function (D) Above all
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Page 77
GATE EC BY RK Kanodia
UNIT 1
18. (B) Z ( s) =
K ( s + 3) K ( s + 3) = 2 ( s - ( -1 + j))( s - (1 - j)) s + 2 s + 2
3K Z (0) = =3 2
Þ
I ( s) 19. (D) o = I in ( s)
26. (B) Let v1 be the node voltage of middle node V1 ( s) =
K =2 s s+1
s 1 + s+1 s+1
=
4s 4 =4s+1 s+1
Þ
s s+1
Þ (2 s + 1) V2 ( s) = 2 sV1 ( s) Þ ( 3s + 1)(2 s + 1) = 2 s(2 s + K ) 2 s 2 + (5 - 2 K ) s + 1 = 0, 5 5 - 2 K > 0, K < 2
io( t) = 4 d( t) - 4 e - t u( t)
H ( s + a) s+4
27. (A) H ( s) =
1 , s2 1 1 1 I o( s) = = s( s + 1) s s + 1 21. (B) I in ( s) =
aö æ Hç 1 - ÷ H ( s + a) Ha 4ø R( s) = = + è s( s + 4) 4s s+4
io( t) = u( t) - e - t u( t) = (1 - e - t ) u( t) 22. (D) vC ( ¥) = lim sVC ( s) = lim s ®0
KV2 ( s) + 2 sV2 ( s) 1 + 2s + s
Þ ( 3s + 1) V1 ( s) = (2 s + K ) V2 ( s) 2 sV1 ( s) Þ V2 ( s) = 2s + 1
20. (B) I in ( s) = 4 I o( s) =
Networks
s ®0
r ( t) =
2s + 6 =2 V s+3
Ha aö æ u( t) + H ç 1 - ÷e - 4 t 4 4ø è
28. (A) Vin ( s) = 23. (D) vC (0 + ) = lim sVC ( s)= s ®¥
CdvC iC = dt
Þ
s(20 s + 6) =2 V (10 s + 3)( s + 4)
I C ( s) = C[ sVC ( s) - vC (0 )]
æ s(20 s + 6) ö -480 ´ 10 -6 (10 s + 3) = 60 ´ 10 -6 çç - 2 ÷÷ = 10 s 2 + 43s + 12 è (10 s + 3)( s + 4) ø s ®¥
Rö 1æ 1 çs + ÷ C Lø è sC = 29. (D) Z ( s) = R 1 1 s2 + sL + R + + L LC sC K ( s + 1) K ( s + 1) Z ( s) = = 2 ( s + s + 1) æ 1 3 öæ 1 3ö çs + + j ÷ç s + - j ÷ ç ÷ ç ÷ 2 2 øè 2 2 ø è ( sL + R)
+
iC (0 + ) = lim sI C ( s) = - 480 ´ 10 -6 = - 0.48 mA
I o( s) 2 s 2 , I o( s) = 2 , = s2 + 1 s + 1 Vin ( s) s
d iL Þ VL ( s) = L [ sI L ( s) - iL (0 + )] dt 10 iL (0 + ) = lim sI L ( s) = =0 s ®¥ s+2
24. (A) vL = L
VL ( s) =
s ®¥
s 40 = 40 s+2
25. (A) V2 ( s) = KV1 ( s) V1 ( s) V1 ( s) - KV1 ( s) + =0 1 2 4+s+ s 1 4 + s + + 2 - 2K = 0 s Þ
s2 + ( 6 - 2 K ) s + 1 = 0
(6 - 2K ) > 0
Page 80
Þ
1 Cs R
40 s 40 = s( s + 2) s + 2
vL (0 + ) = lim sVL ( s) =
Þ
sL Z(s)
Fig. S1.8.29
Since Z (0) = 1, thus K = 1 1 R 1 = 1, = 1, =1 C L LC Þ
C = 1, L = 1, R = 1
30. (B) The characteristic equation is s 2 ( s 2 + 11s + 30) = 0
Þ
s 2 ( s + 6) ( s + 5) =0
s = -6, - 5, Being real and unequal, it is overdamped.
K 16.67
1 , R and C should be as small as possible. RC (1.8) R = ( 3.3) = 1165 . kW 3.3.+1.8 ( 30) C = (10) = 7.5 pF (10 + 30)
24. (B) wo =
w=
1 = 114.5 ´ 106 rad s . 1165 ´ 7.5 ´ 10 -9
25. (D) R¢ = K m R = 800 ´ 12 ´ 10 3 = 9.6 MW L¢ =
800 Km 40 ´ 10 -6 = 32 mF = K f L 1000
C¢ =
C 10 -9 K F = 30 ´ ´ 1000 = 0.375 pF Km 80
26. (A) L¢C¢ =
LC K 2f
Þ
K 2f =
4 ´ 20 ´ 10 -3 ´ 10 -6 1´ 6
Þ K f = 2 ´ 10 -4 (1)(20 ´ 10 -6 ) L¢ L 2 Þ K m2 = = Km (2) ( 4 ´ 10 -3) C¢ C Þ
K m = 0.05 1 RC 1 = 15.9 W R= 2 p ´ 20 ´ 10 3 ´ 0.5 ´ 10 -6
27. (D) wc = 2 pfc = Þ
28. (A) RTH across the capacitor is RTH = (1k + 4 k)||5 k = 2.5 kW 1 . kHz fc = = 106 2 p ´ 2.5 ´ 10 3 ´ 40 ´ 10 -9 29. (B) wc = 2 pfc = Page 106
1 RC
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GATE EC BY RK Kanodia
CHAPTER
2.1 SEMICONDUCTOR PHYSICS
In the problems assume the parameter given in
3. Two semiconductor material have exactly the same
following table. Use the temperature T = 300 K unless
properties except that material A has a bandgap of 1.0
otherwise stated.
eV and material B has a bandgap energy of 1.2 eV. The ratio of intrinsic concentration of material A to that of
Property
Si
Ga As
Ge
material B is
Bandgap Energy
1.12
1.42
0.66
(A) 2016
(B) 47.5
Dielectric Constant
11.7
13.1
16.0
(C) 58.23
(D) 1048
Effective density of 2.8 ´ 1019 states in conduction band N c (cm -3)
4.7 ´ 1017
104 . ´ 1019
Effective density of 104 . ´ 1019 states in valence band N v(cm -3)
7.0 ´ 1018
Intrinsic carrier concertration ni (cm -3) Mobility Electron Hole
15 . ´ 1010
4. In silicon at T = 300 K the thermal-equilibrium concentration of electron is n0 = 5 ´ 10 4 cm -3. The hole concentration is
1.8 ´ 106
6.0 ´ 1018
2.4 ´ 1013
(A) 4.5 ´ 1015 cm -3
(B) 4.5 ´ 1015 m -3
(C) 0.3 ´ 10 -6 cm -3
(D) 0.3 ´ 10 -6 m -3
5. In silicon at T = 300 K if the Fermi energy is 0.22 eV above the valence band energy, the value of p0 is
1350 480
8500 400
3900 1900
(A) 2 ´ 1015 cm -3
(B) 1015 cm -3
(C) 3 ´ 1015 cm -3
(D) 4 ´ 1015 cm -3
6. The thermal-equilibrium concentration of hole p0 in 1. In germanium semiconductor material at T = 400 K
silicon at T = 300 K is 1015 cm -3. The value of n0 is
the intrinsic concentration is
(A) 3.8 ´ 108 cm -3
(B) 4.4 ´ 10 4 cm -3
(C) 2.6 ´ 10 4 cm -3
(D) 4.3 ´ 108 cm -3
(A) 26.8 ´ 1014 cm -3
(B) 18.4 ´ 1014 cm -3
(C) 8.5 ´ 1014 cm -3
(D) 3.6 ´ 1014 cm -3
7. In germanium semiconductor at T = 300 K, the
2. The intrinsic carrier concentration in silicon is to be
acceptor concentrations is N a = 1013 cm -3 and donor
no greater than ni = 1 ´ 1012 cm -3. The maximum
concentration is
temperature allowed for the silicon is ( E g = 112 . eV)
concentration p0 is
(A) 300 K
(B) 360 K
(A) 2.97 ´ 10 9 cm -3
(B) 2.68 ´ 1012 cm -3
(C) 382 K
(D) 364 K
(C) 2.95 ´ 1013 cm -3
(D) 2.4 cm -3
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N d = 0. The thermal equilibrium
Page 109
GATE EC BY RK Kanodia
UNIT 2
Statement for Q.8-9:
Electronics Devices
15. For a particular semiconductor at T = 300 KE g = 15 .
In germanium semiconductor at T = 300 K, the
eV, mp* = 10 mn* and ni = 1 ´ 1015 cm -3. The position of Fermi level with respect to the center of the bandgap is
impurity concentration are N d = 5 ´ 1015 cm -3 and N a = 0
(A) +0.045 eV
(B) - 0.046 eV
(C) +0.039 eV
(D) - 0.039 eV
8. The thermal equilibrium electron concentration n0 is (A) 5 ´ 1015 cm -3 (C) 115 . ´ 10 cm 9
(B) 115 . ´ 1011 cm -3 -3
(D) 5 ´ 10 cm 6
16. A silicon sample contains acceptor atoms at a
-3
concentration of N a = 5 ´ 1015 cm -3. Donor atoms are
9. The thermal equilibrium hole concentration p0 is (A) 396 . ´ 1013
(B) 195 . ´ 1013 cm -3
(C) 4.36 ´ 1012 cm -3
(D) 396 . ´ 1013 cm -3
conduction band edge. The concentration of donors atoms added are
10. A sample of silicon at T = 300 K is doped with boron at a concentration of 2.5 ´ 1013 cm -3 and with arsenic at a concentration of 1 ´ 1013 cm -3. The material is
(B) 4.6 ´ 1016 cm -3
(C) 39 . ´ 1012 cm -3
(D) 2.4 ´ 1012 cm -3
doped with phosphorus atoms at a concentrations of 1015
(B) p - type with p0 = 15 . ´ 10 7 cm -3 (C) n - type with n0 = 15 . ´ 10 cm
(A) 12 . ´ 1016 cm -3
17. A silicon semiconductor sample at T = 300 K is
(A) p - type with p0 = 15 . ´ 1013 cm -3 13
added forming and n - type compensated semiconductor such that the Fermi level is 0.215 eV below the
cm -3. The position of the Fermi level with respect to the
-3
intrinsic Fermi level is
(D) n - type with n0 = 15 . ´ 10 7 cm -3 11. In a sample of gallium arsenide at T = 200 K, n0 = 5 p0 and N a = 0. The value of n0 is
(A) 0.3 eV
(B) 0.2 eV
(C) 0.1 eV
(D) 0.4 eV
(A) 9.86 ´ 10 9 cm -3
(B) 7 cm -3
18. A silicon crystal having a cross-sectional area of
(C) 4.86 ´ 10 3 cm -3
(D) 3 cm -3
0.001 cm 2 and a length of 20 mm is connected to its ends to a 20 V battery. At T = 300 K, we want a current of
12. Germanium at T = 300 K is uniformly doped with an acceptor concentration of N a = 10 cm 15
-3
and a donor
100 mA in crystal. The concentration of donor atoms to be added is
concentration of N d = 0. The position of fermi energy
(A) 2.4 ´ 1013 cm -3
(B) 4.6 ´ 1013 cm -3
with respect to intrinsic Fermi level is
(C) 7.8 ´ 1014 cm -3
(D) 8.4 ´ 1014 cm -3
(A) 0.02 eV
(B) 0.04 eV
(C) 0.06 eV
(D)0.08 eV
19. The cross sectional area of silicon bar is 100 mm 2 . The length of bar is 1 mm. The bar is doped with
13. In germanium at T = 300 K, the donor concentration are N d = 1014 cm -3 and N a = 0. The Fermi energy level with respect to intrinsic Fermi level is
Page 110
(A) 0.04 eV
(B) 0.08 eV
(C) 0.42 eV
(D) 0.86 eV
arsenic atoms. The resistance of bar is (A) 2.58 mW
(B) 11.36 kW
(C) 1.36 mW
(D) 24.8 kW
20. A thin film resistor is to be made from a GaAs film doped n - type. The resistor is to have a value of 2 kW.
14. A GaAs device is doped with a donor concentration of 3 ´ 1015 cm -3. For the device to operate properly, the intrinsic carrier concentration must remain less than 5% of the total concentration. The maximum temperature on that the device may operate is
10 -6 cm 2 . The doping efficiency is known to be 90%. The
(A) 763 K
(B) 942 K
(A) 8.7 ´ 1015 cm -3
(B) 8.7 ´ 10 21 cm -3
(C) 486 K
(D) 243 K
(C) 4.6 ´ 1015 cm -3
(D) 4.6 ´ 10 21 cm -3
The resistor length is to be 200 mm and area is to be mobility of electrons is 8000
cm 2 V - s. The doping
needed is
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Semiconductor Physics
21. A silicon sample doped n - type at 1018 cm -3 have a -6
Chap 2.1
27. For a sample of GaAs scattering time is t sc = 10 -13 s
2
and electron’s effective mass is me* = 0.067 mo . If an
and a length of 10 mm . The doping efficiency of the
electric field of 1 kV cm is applied, the drift velocity
sample is (m n = 800 cm V - s)
produced is
resistance of 10 W . The sample has an area of 10
cm
2
(A) 43.2%
(B) 78.1%
(A) 2.6 ´ 106 cm s
(B) 263 cm s
(C) 96.3%
(D) 54.3%
(C) 14.8 ´ 106 cm s
(D) 482
22.
Six
volts
is
applied
across
a
2
cm
long
semiconductor bar. The average drift velocity is 10 4 cm s. The electron mobility is (A) 4396 cm 2 V - s
(B) 3 ´ 10 4 cm 2 V - s
(C) 6 ´ 10 4 cm 2 V - s
(D) 3333 cm 2 V - s
28. A gallium arsenide semiconductor at T = 300 K is doped with impurity concentration N d = 1016 cm -3. The mobility m n is 7500 cm 2 V - s. For an applied field of 10 V cm the drift current density is
23. For a particular semiconductor material following
(A) 120 A cm 2
(B) 120 A cm 2
(C) 12 ´ 10 4 A cm 2
(D) 12 ´ 10 4 A cm 2
parameters are observed: 29. In a particular semiconductor the donor impurity
m n = 1000 cm 2 V - s ,
concentration is N d = 1014 cm -3. Assume the following
m p = 600 cm V - s , 2
N c = N v = 10 cm 19
parameters,
-3
m n = 1000 cm 2 V - s,
These parameters are independent of temperature.
æ T ö N c = 2 ´ 1019ç ÷ è 300 ø
The measured conductivity of the intrinsic material is s = 10 -6 (W - cm) -1 at T = 300 K. The conductivity at
æ T ö N v = 1 ´ 1019ç ÷ è 300 ø
T = 500 K is (A) 2 ´ 10
-4
(C) 2 ´ 10
-5
(W - cm) (W - cm)
-1
-1
(B) 4 ´ 10
-5
-1
(W - cm)
(D) 6 ´ 10
-3
(W - cm) -1
24. An n - type silicon sample has a resistivity of 5 W - cm at T = 300 K. The mobility is m n = 1350
32
cm -3,
32
cm -3,
. eV. E g = 11 An electric field of E = 10 V cm is applied. The electric current density at 300 K is
cm 2 V - s. The donor impurity concentration is
(A) 2.3 A cm 2
(B) 1.6 A cm 2
(A) 2.86 ´ 10 -14 cm -3
(B) 9.25 ´ 1014 cm -3
(C) 9.6 A cm 2
(D) 3.4 A cm 2
(C) 11.46 ´ 1015 cm -3
(D) 11 . ´ 10 -15 cm -3
25.
Statement for Q.30-31:
In a silicon sample the electron concentration 18
drops linearly from 10
cm
-3
16
to 10
cm
-3
A semiconductor has following parameter
over a length
m n = 7500 cm 2 V - s,
of 2.0 mm. The current density due to the electron diffusion current is ( Dn = 35 cm 2 s).
m p = 300 cm 2 V - s,
(A) 9.3 ´ 10 4 A cm 2
(B) 2.8 ´ 10 4 A cm 2
ni = 3.6 ´ 1012 cm -3
(C) 9.3 ´ 10 9A cm 2
(D) 2.8 ´ 10 9 A cm 2
30.
26. In a GaAs sample the electrons are moving under an
electric
field
of
5
kV cm
and
the
carrier
concentration is uniform at 1016 cm -3. The electron velocity is the saturated velocity of 10
7
(A) 1.6 ´ 10 A cm (C) 1.6 ´ 108 A cm 2
conductivity
is
minimum,
the
hole
(A) 7.2 ´ 1011 cm -3
(B) 1.8 ´ 1013 cm -3
(C) 1.44 ´ 1011 cm -3
(D) 9 ´ 1013 cm -3
cm s. The drift 31. The minimum conductivity is
current density is 4
When
concentration is
2
2
(A) 0.6 ´ 10 -3 (W - cm) -1
(B) 17 . ´ 10 -3 (W - cm) -1
(D) 2.4 ´ 108 A cm 2
(C) 2.4 ´ 10 -3 (W - cm) -1
(D) 6.8 ´ 10 -3 (W - cm) -1
(B) 2.4 ´ 10 A cm 4
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GATE EC BY RK Kanodia
UNIT 2
32.
A
particular
intrinsic
semiconductor
has
Electronics Devices
æ -x ö ÷ ç ÷ ç 15 è L p ø
a
Hole concentration p0 = 10 e
resistivity of 50 (W - cm) at T = 300 K and 5 (W - cm) at T = 330 K. If change in mobility with temperature is
cm -3,x ³ 0 ö æ ç -x ÷ ÷ ç
neglected, the bandgap energy of the semiconductor is
Electron concentration n0 = 5 ´ 1014 e è L n ø cm -3,x £ 0
(A) 1.9 eV
(B) 1.3 eV
Hole diffusion coefficient Dp = 10 cm 2 s
(C) 2.6 eV
(D) 0.64 eV
Electron diffusion coefficients Dn = 25 cm 2 s a
Hole diffusion length Lp = 5 ´ 10 -4 cm,
semiconductor. If only the first mechanism were
Electron diffusion length Ln = 10 -3 cm
33.
Three
scattering
mechanism
exist
in
present, the mobility would be 500 cm V - s. If only the 2
second mechanism were present, the mobility would be
The total current density at x = 0 is
750 cm 2 V - s. If only third mechanism were present,
(A) 1.2 A cm 2
(B) 5.2 A cm 2
the mobility would be 1500 cm 2 V - s. The net mobility
(C) 3.8 A cm 2
(D) 2 A cm 2
is (A) 2750 cm 2 V - s
(B) 1114 cm 2 V - s
37. A germanium Hall device is doped with 5 ´ 1015
(C) 818 cm 2 V - s
(D) 250 cm 2 V - s
donor atoms per cm 3 at T = 300 K. The device has the geometry d = 5 ´ 10 -3 cm, W = 2 ´ 10 -2 cm and L = 0.1 cm.
34. In a sample of silicon at T = 300 K, the electron
The current is I x = 250 mA, the applied voltage is
concentration varies linearly with distance, as shown in
Vx = 100
fig. P2.1.34. The diffusion current density is found to be
tesla. The Hall voltage is
J n = 0.19 A cm 2 . If the electron diffusion coefficient is
(A) -0.31mV
(B) 0.31 mV
Dn = 25 cm 2 s, The electron concentration at is
(C) 3.26 mV
(D) -3.26 mV
5´1014 n(cm-3)
Statement for Q.38-39: A silicon Hall device at T = 300 K has the geometry d = 10 -3 cm , W = 10 -2 cm, L = 10 -1 cm. The following parameters are measured: I x = 0.75 mA,
n(0) 0
mV, and the magnetic flux is Bz = 5 ´ 10 -2
0.010
Vx = 15 V, V H = +5.8 mV, tesla
x(cm)
Fig. P2.1.34
38. The majority carrier concentration is (A) 4.86 ´ 10 cm 8
-3
(C) 9.8 ´ 10 26 cm -3
-3
(A) 8 ´ 1015 cm -3, n - type
(D) 5.4 ´ 1015 cm -3
(B) 8 ´ 1015 cm -3, p - type
(B) 2.5 ´ 10 cm 13
35. The hole concentration in p - type GaAs is given by xö æ p = 10 ç 1 - ÷ cm -3 for 0 £ x £ L L è ø
(C) 4 ´ 1015 cm -3, n - type (D) 4 ´ 1015 cm -3, p - type
16
39. The majority carrier mobility is
where L = 10 mm. The hole diffusion coefficient is 10 cm s. The hole diffusion current density at x = 5 mm 2
(A) 430 cm 2 V - s
(B) 215 cm 2 V - s
(C) 390 cm 2 V - s
(D) 195 cm 2 V - s
is (A) 20 A cm 2
(B) 16 A cm 2
40. In a semiconductor n0 = 1015 cm -3 and ni = 1010 cm -3.
(C) 24 A cm 2
(D) 30 A cm 2
The excess-carrier life time is 10 -6 s. The excess hole
36. For a particular semiconductor sample consider following parameters:
Page 112
concentration is dp = 4 ´ 1013 cm -3. The electron-hole recombination rate is (A) 4 ´ 1019 cm -3s -1
(B) 4 ´ 1014 cm -3s -1
(C) 4 ´ 10 24 cm -3s -1
(D) 4 ´ 1011 cm -3s -1
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Semiconductor Physics
41. A semiconductor in thermal equilibrium, has a hole concentration
cm -3
p0 = 1016
of
and
an
SOLUTIONS
intrinsic
-3
concentration of ni = 10 cm . The minority carrier life 10
time
4 ´ 10 -7s.
is
The
thermal
equilibrium
recombination rate of electrons is (A) 2.5 ´ 10 22 cm -3 s -1
(B) 5 ´ 1010 cm -3 s -1
(C) 2.5 ´ 1010 cm -3 s -1
(D) 5 ´ 10 22 cm -3 s -1
n-type
1. (C) ni2 = N c N v e
For Ge at 300 K, . ´ 1019, N v = 6.0 ´ 1018 , E g = 0.66 eV N c = 104
silicon
sample
contains
a
donor
concentration of N d = 106 cm -3. The minority carrier hole lifetime is t p 0 = 10 m s.
-3
Þ
(A) 5 ´ 10 cm s
-1
-3
-1
3
-3
-1
(B) 10 cm s
-3
(C) 2. 25 ´ 10 cm s 9
4
-1
(D) 10 cm s
ni = 8.5 ´ 1014 cm -3
2. (C) n = N c N v e 2 i
electron is
T e
(A) 1125 . ´ 10 cm s (C) 8.9 ´ 10 44.
-10
-3
cm s
A n -type
-1
-3
(B) 2.25 ´ 10 cm s 9
-1
sample
concentration of N d = 10
16
contains
a
-13´10 3 T
= 9.28 ´ 10 -8 , By trial T = 382 K
-1
(D) 4 ´ 10 9 cm -3 s -1
silicon
æ 1 .12 e ö ÷÷ kT ø
æ T ö - ççè (1012 ) 2 = 2.8 ´ 1019 ´ 104 . ´ 1019ç ÷ e è 300 ø 3
-3
æ -Eg ö ÷ - çç ÷ è kT ø 3
43. The thermal equilibrium generation rate for 9
æ 0 .66 ö
3
÷÷ - çç æ 400 ö è 0 .0345 ø ni2 = 104 . ´ 1019 ´ 6.0 ´ 1018 ´ ç ÷ ´e è 300 ø
42. The thermal equilibrium generation rate of hole is 8
æ -Eg ö ÷ - çç ÷ è kT ø
æ 400 ö Vt = 0.0259ç ÷ = 0.0345 è 300 ø
Statement for Q.42-43: A
Chap 2.1
-
donor
E gA
æ E - E gB ö ç gA ÷ ÷ kT ø
-ç n2 e kT 3. (B) iA = E gB = e è 2 niB e kT
= 2257.5
Þ
niA = 47.5 niB
-3
cm . The minority carrier
hole lifetime is t p 0 = 20 m s. The lifetime of the majority carrier is ( ni = 15 . ´ 1010 cm -3) (A) 8.9 ´ 106 s
(B) 8.9 ´ 10 -6 s
(C) 4.5 ´ 10 -17 s
(D) 113 . ´ 10 -7 s
concentration are N a = 10
cm
-3
and N a = 0. The
equilibrium recombination rate is Rp 0 = 1011 cm -3s -1 . A uniform generation rate produces an excess- carrier -3
concentration of dn = dp = 10 cm . The factor, by which 14
the total recombination rate increase is
ni2 (15 . ´ 1010 ) 2 = = 4.5 ´ 1015 cm -3 n0 5 ´ 10 4
5. (A) p0 = N v e
45. In a silicon semiconductor material the doping 16
4. (A) p0 =
6. (B) p0 = N v e
-
-
( EF - Ev ) kT
( EF - Ev ) kT
-0 .22
= 104 . ´ 1019 e 0 .0259 = 2 ´ 1015 cm -3 Þ
At 300 K, N v = 10 . ´ 1019 cm -3 æ 104 . ´ 1019 ö ÷÷ = 0. 239 eV EF - Ev = 0.0259 lnçç 1015 ø è ( Ec - EF )
(A) 2.3 ´ 1013
(B) 4.4 ´ 1013
n0 = N c e
(C) 2.3 ´ 10 9
(D) 4.4 ´ 10 9
At 300 K, N c = 2.8 ´ 1019 cm -3
***********
æN ö EF - Ev = kT ln çç v ÷÷ è p0 ø
kT
Ec - EF = 112 . - 0. 239 = 0.881 eV n0 = 4.4 ´ 10 4 cm -3 2
7. (C) p0 =
Na - Nd N ö æ + ç N a - d ÷ + ni2 2 2 ø è
For Ge ni = 2.4 ´ 10 3 2
æ 1013 ö 1013 ÷÷ + (2.4 ´ 1013) 2 = 2.95 ´ 1013 cm -3 p0 = + çç 2 è 2 ø www.gatehelp.com
Page 113
GATE EC BY RK Kanodia
Semiconductor Physics
=
(1.6 ´ 10
Þ n0 =
26. (A) J = evn = (1.6 ´ 10 -19)(10 7)(1016 ) = 1.6 ´ 10 4 A cm 2
0.1 = 11.36 kW )(1100)(5 ´ 1016 )(100 ´ 10 -8 )
-19
L , sA
20. (A) R =
s » e m n n0 ,
R=
27. (A) vd =
L em n n0 A
et sc E (1.6 ´ 10 -19)(10 -13)(10 5) = (0.067)(9.1 ´ 10 -31 ) me*
= 26.2 ´ 10 3 m s = 2.6 ´ 106 cm s
L em n AR
28. (A) N d >> ni
n0 = 0.9 N d =
Chap 2.1
-4
20 ´ 10 = 8.7 ´ 1015 cm -3 (0.9)(1.6 ´ 10 -19)( 8000)(10 -6 )(2 ´ 10 3)
21. (B) s » e m n n0 , R =
L L , n0 = sA em n AR
10 ´ 10 -4 = 7.81 ´ 1017 cm -3 = (1.6 ´ 10 -19)( 800)(10 -6 )(10)
Þ
J = em n n0 E = (1.6 ´ 10 29. (D) ni2 = N c N v e
)(7500)(1016 )(10) = 120 A cm 2
æ -Eg ö ÷ - çç ÷ è kT ø
= (2 ´ 1019)(1 ´ 1019) e Þ
n0 = N d -19
æ 1 .1 ö ÷÷ - çç è 0 .0259 ø
= 7.18 ´ 1019
ni = 8.47 ´ 10 9 cm -3
N d >> ni
Þ
N d = n0
Efficiency =
n0 7.8 ´ 1017 ´ 100 = ´ 100 = 78.1 % Nd 1018
J = sE = em n n0 E
22. (D) E =
V 6 = = 3 V/cm, vd = m n E, L 2
30. (A) s = em n n0 + em p p0 and n0 =
vd 10 4 = 3333 cm 2 V - s = E 3
mn =
= (1.6 ´ 10 -19)(1000)(1014 )(100) = 1.6 A cm 2
Þ
10 -6 = (1.6 ´ 10 -19)(1000 + 600)ni At T = 300 K, ni = 391 . ´ 10 cm 9
ni2 = N c N v e
ni2 + em p p0 , p0
( -1) em n ni2 ds =0 = + em p dp0 p02
23. (D) s1 = eni (m n + m p )
æ Eg ö ÷ - çç ÷ è kT ø
s = em n
ni2 p0
1 -3
æN N ö Þ E g = kT ln çç c 2 v ÷÷ è ni ø
Þ
æm p0 = ni ç n çm è p
1
ö2 2 ÷ = 3.6 ´ 1012 æç 7500 ö÷ ÷ 300 è ø ø
= 7. 2 ´ 1011 cm -3
ö æ 1019 eV ÷ = 1122 Þ E g = 2(0.0259) ln çç . 9 ÷ 391 . ´ 10 ø è
31. (B) smin =
æ 500 ö At T = 500 K , kT = 0.0259ç ÷ = 0.0432 eV, è 300 ø
= 2 ´ 1.6 ´ 10 -19( 3.6 ´ 1012 ) (7500)( 300)
ni2 = (1019) 2 e Þ
æ 1 .122 ö ÷÷ - çç è 0 .0432 ø
32. (B) s =
= (1.6 ´ 10 -19)(2.29 ´ 1013)(1000 + 600) = 5.86 ´ 10 -3(W - cm) -1 24. (B) r = Nd =
1 1 = s em n N d
1 1 = 9.25 ´ 1014 cm -3 = 5(1.6 ´ 10 -19)(1350) rem n
25. (B) J n = eDn
dn dx
æ 1018 - 1016 = (1.6 ´ 10 -19)( 35)çç -4 è 2 ´ 10
mp + mn
= 2 en i m pm n
= 17 . ´ 10 -3(W - cm) -1
cm -3,
ni = 2.29 ´ 1013 cm -3
2 s i m pm n
1 = emni , r
Eg 1 ni1 e 2 kT1 r1 = = Eg 1 ni 2 2 kT2 e r2
0.1 = e
-
Eg æ 1 1 ö÷ ç T2 ÷ø 2 k çè T1
E g æ 330 - 300 ö ç ÷ = ln 10 2 k çè 330 ´ 300 ÷ø E g = 22( k300) ln 10 = 1.31 eV
ö ÷÷ = 2.8 ´ 10 4 A cm 2 ø
33. (D) www.gatehelp.com
1 1 1 1 = + + m m1 m 2 m 3 Page 115
GATE EC BY RK Kanodia
CHAPTER
2.2 THE PN JUNCTION
In this chapter, N d and N a denotes the net donor
5. The Fermi level on p - side is
p-region.
(A) 0.2 eV (C) 0.4 eV
1. An abrupt silicon in thermal equilibrium at T = 300
Statement for Q.6–8:
and acceptor concentration in the individual n and
K is doped such that Ec - EF = 0.21 eV in the n - region
(B) 0.1 eV (D) 0.3 eV
A silicon pn junction at T = 300 K with zero
and EF - Ev = 0.18 eV in the p - region. The built-in
applied bias has doping concentrations of N d = 5 ´ 1016
potential barrier Vbi is
cm -3 and N a = 5 ´ 1015 cm -3.
(A) 0.69 V (C) 0.61 V
(B) 0.83 V (D) 0.88 V
6. The width of depletion region extending into the
2. A silicon pn junction at T = 300 K has N d = 1014
n-region is
cm and N a = 10 cm . The built-in voltage is
(A) 4 ´ 10 -6 cm (C) 4 ´ 10 -5 cm
(A) 0.63 V (C) 0.026 V
7. The space charge width is
-3
17
-3
(B) 0.93 V (D) 0.038 V
3. In a uniformly doped GaAs junction at T = 300 K, at zero bias, only 20% of the total space charge region is to be in the p-region. The built in potential barrier is Vbi = 1.20 V. The majority carrier concentration in n-region is (A) 1 ´ 1016 cm -3 (C) 1 ´ 10
22
cm
-3
(B) 4 ´ 10 16 cm -3 (D) 4 ´ 10
22
cm
(B) 3 ´ 10 -6 cm (D)3 ´ 10 -5 cm
(A) 32 . ´ 10 -5 cm (C) 4.5 ´ 10 -4 cm
(B) 4.5 ´ 10 -5 cm (D) 32 . ´ 10 -4 cm
8. In depletion region maximum electric field | Emax | is (A) 1 ´ 10 4 V cm (C) 3 ´ 10 4 V cm
(B) 2 ´ 10 4 V cm (D) 4 ´ 10 4 V cm
9. An n – n isotype doping profile is shown in fig. P2.2.9.
-3
The built-in potential barrier is
(ni = 15 . ´ 1010 cm -3 )
Nd(cm-3)
Statement for Q.4–5:
10
An abrupt silicon pn junction at zero bias and
10
T = 300 K has dopant concentration of N a = 1017 cm -3
16
15
and N d = 5 ´ 1015 cm -3. 0
4. The Fermi level on n - side is (A) 0.1 eV
(B) 0.2 eV
(C) 0.3 eV
(D) 0.4 eV
Fig. P2.2.9
(A) 0.66 V (C) 0.03 V www.gatehelp.com
(B) 0.06 V (D) 0.33 V Page 117
GATE EC BY RK Kanodia
UNIT 2
Electronics Devices
concentration are N d = 4 ´ 1016 cm -3 and N a = 4 ´ 1017
Statement for Q.10–11: A silicon abrupt junction has dopant concentration
cm -3. The magnitude of the reverse bias voltage is
N a = 2 ´ 1016 cm -3 and N d = 2 ´ 10 15 cm -3. The applied
(A) 3.6 V
(B) 9.8 V
reverse bias voltage is VR = 8 V.
(C) 7.2 V
(D) 12.3 V
10. The maximum electric field | Emax | in depletion region is
17. An abrupt silicon pn junction has an applied
(A) 15 ´ 10 V cm
(B) 7 ´ 10 V cm
reverse bias voltage of VR = 10 V. it has dopant
(C) 35 . ´ 10 4 V cm
(D) 5 ´ 10 4 V cm
concentration of N a = 1018 cm -3 and N d = 1015 cm -3. The
4
4
pn junction area is 6 ´ 10 -4 cm 2 . An inductance of 2.2
11. The space charge region is
mH is placed in parallel with the pn junction. The
(A) 2.5 mm
(B) 25 mm
resonant frequency is
(C) 50 mm
(D) 100 mm
(A) 1.7 MHz
(B) 2.6 MHz
(C) 3.6 MHz
(D) 4.3 MHz
12.
A uniformly
doped
silicon
pn junction
has
N a = 5 ´ 1017 cm -3 and N d = 10 17 cm -3. The junction has
18. A uniformly doped silicon p+ n junction is to be
a cross-sectional area of 10 -4 cm -3 and has an applied
designed such that at a reverse bais voltage of V R = 10
reverse-bias voltage of VR = 5 V. The total junction
V the maximum electric field is limited to Emax = 106
capacitance is
V cm. The maximum doping concentration in the
(A) 10 pF
(B) 5 pF
n-region is
(C) 7 pF
(D) 3.5 pF
(A) 32 . ´ 1019 cm -3
(B) 32 . ´ 1017 cm -3
(C) 6.4 ´ 1017 cm -3
(D) 6.4 ´ 1019 cm -3
Statement for Q.13–14: An ideal one-sided silicon n+ p junction has
19. A diode has reverse saturation current I s = 10 -10 A
uniform doping on both sides of the abrupt junction.
and non ideality factor h = 2. If diode voltage is 0.9 V,
The doping relation is N d = 50 N a . The built-in potential
then diode current is
barrier is Vbi = 0.75 V. The applied reverse bias voltage
(A) 11 mA
(B) 35 mA
is V R = 10.
(C) 83 mA
(D) 143 mA
13. The space charge width is
20. A diode has reverse saturation current I s = 10 -18 A
(A) 1.8 mm
(B) 1.8 mm
and nonideality factor h = 105 . . If diode has current of 70
(C) 1.8 cm
(D) 1.8 m
mA, then diode voltage is
14. The junction capacitance is (A) 3.8 ´ 10 -9 F cm 2
(B) 9.8 ´ 10 -9 F cm 2
(C) 2.4 ´ 10 -9 F cm 2
(D) 5.7 ´ 10 -9 F cm 2
(A) 0.63 V
(B) 0.87 V
(C) 0.54 V
(D) 0.93 V
21. An ideal pn junction diode is operating in the forward bais region. The change in diode voltage, that
+
15. Two p n silicon junction is reverse biased at VR = 5
will cause a factor of 9 increase in current, is
V. The impurity doping concentration in junction A are
(A) 83 mV
(B) 59 mV
(C) 43 mV
(D) 31 mV
N a = 10 cm 18
-3
and N d = 10
-15
-3
cm , and those in junction
B are N a = 1018 cm -3 and N d = 1016 cm -3. The ratio of the 22. An pn junction diode is operating in reverse bias
space charge width is (A) 4.36
(B) 9.8
region. The applied reverse voltage, at which the ideal
(C) 19
(D) 3.13
reverse current reaches 90% of its reverse saturation current, is
16. The maximum electric field in reverse-biased silicon pn junction Page 118
is
| Emax | = 3 ´ 10
5
V cm.
The
doping
(A) 59.6 mV
(B) 2.7 mV
(C) 4.8 mV
(D) 42.3 mV
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The pn Junction
p+ n junction diode the doping
23. For a silicon
cross-sectional area is 10 -3 cm 2 . The minority carrier
-3
cm .
lifetimes are t n 0 = 1 ms and t p 0 = 0.1 m s. The minority
The minority carrier hole diffusion coefficient is Dp = 12
carrier diffusion coefficients are Dn = 35 cm 2 s and
cm 2 s and the minority carrier hole life time is t p 0 = 10 -7
Dp = 10 cm 2 s. The total number of excess electron in
s. The cross sectional area is A = 10 -4 cm 2 . The reverse
the p - region, if applied forward bias is Va = 0.5 V, is
saturation current is
(A) 4 ´ 10 7 cm -3
(B) 6 ´ 1010 cm -3
(C) 4 ´ 1010 cm -3
(D) 6 ´ 10 7 cm -3
concentrations are N a = 10
18
(A) 4 ´ 10
-12
cm
-3
Chap 2.2
and N d = 10
16
(B) 4 ´ 10
A
(C) 4 ´ 10 -11 A
-15
A
(D) 4 ´ 10 -7 A
28. Two ideal pn junction have exactly the same
24. For an ideal silicon pn junction diode
electrical and physical parameters except for the band gap of the semiconductor materials. The first has a
t no = t po = 10 -7 s ,
bandgap energy of 0.525 eV and a forward-bias current
Dn = 25 cm 2 s ,
of 10 mA with Va = 0.255 V. The second pn junction
Dp = 10 cm 2 s
diode is to be designed such that the diode current
The ratio of N a N d , so that 95% of the current in
I = 10 mA at a forward-bias voltage of Va = 0.32 V. The
the depletion region is carried by electrons, is
bandgap energy of second diode would be
(A) 0.34
(B) 0.034
(A) 0.77 eV
(B) 0.67 eV
(C) 0.83
(D) 0.083
(C) 0.57 eV
(D) 0.47 eV
Statement for Q.25–26:
29. A pn junction biased at Va = 0.72 V has DC bias
A ideal long silicon pn junction diode is shown in
current I DQ = 2 mA. The minority carrier lifetime is 1 ms
fig. P2.2.25–26. The n - region is doped with 10
is both the n and p regions. The diffusion capacitance is
organic atoms per cm 3 and the p - region is doped with
in
5 ´ 10
cm . The minority carrier
(A) 49.3 nF
(B) 38.7 nF
lifetimes are D n = 23 cm s and Dp = 8 cm s. The
(C) 77.4 nF
(D) 98.6 nF
16
16
3
boron atoms per 2
2
forward-bias voltage is Va = 0.61 V.
30. A p+ n silicon diode is forward biased at a current of 1 mA. The hole life time in the n - region is 0.1 ms.
W
Neglecting p
Va
n
x=0
x
the
diode
(A) 38.7 + j12.1 W
(B) 235 . + j7.5 W
(C) 38.7 - j12.1 mW
(D) 235 . - j7.5 W
forward-bias current of a p+ n diode is 2.5 ´ 10 -6 F A.
(A) 6.8 ´ 1012 e -246 x cm -3, x ³ 0
The hole lifetime is
(B) 6.8 ´ 1012 e -246 x cm -3, x ³ 0 (C) 3.8 ´ 1014 e -3534 x cm -3, x ³ 0 (D) 3.8 ´ 10 e
capacitance
31. The slope of the diffusion capacitance verses
25. The excess hole concentration is
+3534 x
depletion
impedance at 1 MHz is
Fig. P.2.2.25-26
14
the
(A) 1.3 ´ 10 -7 s
(B) 1.3 ´ 10 -4 s
(C) 6.5 ´ 10 -8 s
(D) 6.5 ´ 10 -4 s
-3
cm , x ³ 0
32. A silicon pn junction with doping profile of N a = 1016
26. The hole diffusion current density at x = 3 mm is (A) 0.6 A cm 2
(B) 0.6 ´ 10 -3 A cm 2
(C) 0.4 A cm 2
(D) 0.4 ´ 10 -3 A cm 2
cm -3 and N d = 10 15 cm -3 has a cross sectional area of 10 -2 cm 2 . The length of the p - region is 2 mm and length of the n - region is 1 mm. The approximately series resistance of the diode is
27. The doping concentrations of a silicon pn junction are
N d = 10
16
cm
-3
and
N a = 8 ´ 10
15
-3
cm .
The
(A) 62 W
(B) 43 W
(C) 72 W
(D) 81 W
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GATE EC BY RK Kanodia
UNIT 2
Electronics Devices
33. A gallium arsenide pn junction is operating in
39. A GaAs laser has a threshold density of 500 A cm 2 .
reverse-bias voltage VR = 5 V. The doping profile are
The laser has dimensions of 10 mm ´ 200 mm. The active
N a = N d = 10
16
-3
cm . The minority carrier life- time are
t p 0 = t n 0 = t 0 = 10
-8
s. The reverse-biased generation (e r = 131 . , ni = 1.8 ´ 10 ) 6
current density is (A) 19 . ´ 10
-8
(C) 1.4 ´ 10
-8
A cm
(B) 19 . ´ 10
2
A cm
-9
(D) 1.4 ´ 10
2
-9
A cm
region is dLas = 100 A °. The electron-hole recombination time at threshold is 1.5 ns. The current density of 5J th is injected into the laser. The optical power emitted, if
2
emitted photons have an energy of 1.43 eV, is
2
(A) 143 mW
(B) 71.5 mW
(C) 62.3 mW
(D) 124.6 mW
A cm
34. For silicon the critical electric field at breakdown is approximately Ecrit = 4 ´ 10 5 V cm. For the breakdown voltage
of 25
V,
maximum n - type
the
doping
+
concentration in an abrupt p n-junction is (A) 2 ´ 1016 cm -3
(B) 4 ´ 10 16 cm -3
(C) 2 ´ 1018 cm -3
(D) 4 ´ 10 18 cm -3
35. A uniformly doped silicon pn junction has dopant profile of N a = N d = 5 ´ 1016 cm -3. If the peak electric field in the junction at breakdown is E = 4 ´ 10 5 V cm, the breakdown voltage of this junction is (A) 35 V
(B) 30 V
(C) 25 V
(D) 20 V
36. An abrupt silicon p+ n junction has an n - region doping
concentration n - region
minimum
of
N d = 5 ´ 10 15
width,
such
that
cm -3.
The
avalanche
breakdown occurs before the depletion region reaches an ohmic contact, is (VB » 100 V) (A) 5.1 mm
(B) 3.6 mm
(C) 7.3 mm
(D) 6.4 mm
37. A silicon pn junction diode has doping profile N a = N d = 5 ´ 1019 cm -3. The space charge width at a forward bias voltage of Va = 0.4 V is (A) 102 A°
(B) 44 A°
(C) 153 A°
(D) 62 A°
38. A
GaAs
pn+
junction LED has following
parameters Dn = 25 cm 2 s, Dp = 12 cm 2 s N d = 5 ´ 1017 cm -3, N a = 1016 cm -3 t n 0 = 10 ns , t p 0 = 10 ns The injection efficiency of the LED is
Page 120
(A) 0.83
(B) 0.99
(C) 0.64
(D) 0.46
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***************
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UNIT 2
= 301 . ´ 10 -5 cm, CT =
-14
. ´ 8.85 ´ 10 ´ 10 eA 117 = 301 . ´ 10 -5 W
-4
= 35 . ´ 10 -12 F
ö ÷÷ ø
=
2 eEmax 2e
æ 1 1 ö ÷÷ çç + è Na Nd ø
(117 . ´ 8.85 ´ 10 -14 )( 3 ´ 10 5) 2 æ 1 1 ö V + ç 17 ÷ 16 ´ ´ 2 ´ 1.6 ´ 10 -19 4 10 4 10 è ø
= 8.008 V
1
ì 2 e ( V + VR ) é 1 1 ùü 2 W = í s bi + ê úý , e ë Na Na û þ î
VR = 8.008 - 0.826 = 7.18 V æN N ö 17. (B) Vbi = Vt ln çç a 2 d ÷÷ è ni ø
1
Þ
1
Vbi + VR =
N a = 4. 2 ´ 1015 cm -3, N d = 2.1 ´ 1017 cm -3
Nd > > Na
æ 4 ´ 1016 ´ 4 ´ 10 17 ö ÷÷ = 0.826 V = 0.0259 ln çç 2. 25 ´ 10 20 ø è é2 e( Vbi + VR ) Na Nd ù 2 | Emax | = ê e ( N a + N d ) úû ë
æN N ö 13. (A) Vbi = Vt ln çç a 2 d ÷÷ è ni ø æ 50 N a2 0.751 = 0.0259 ln çç 20 è 2.25 ´ 10
Electronics Devices
é2 e( Vbi + VR ) æ 1 öù 2 ÷÷ú çç W » ê e è N a øû ë 1 2
é2 ´ (117 . ´ 8.85 ´ 10 -4 )(10.752) ù -4 =ê ú = 1.8 ´ 10 cm 1.6 ´ 10 -19 ´ 4.2 ´ 1015 ë û = 1.8 mm
æ 1018 ´ 1015 = 0.0259 ln çç 20 è 2.25 ´ 10
ö ÷÷ = 0.754 V ø 1
é ù2 eeN a N d C¢ = ê ú ë2 ( Vi + VR )( N a + N d ) û 1
é ù ee N a N d 14. (D) C¢ = ê ú ë2( Vbi + VR )( N a + N d ) û é ee N a ù For N d >> N a , C¢ = ê ú ë2( Vbi + VR ) û é1.6 ´ 10 =ê ë
-19
. ´ 8.85 ´ 10 ´ 117 2 (10 + 0.754)
-4
é ù2 eeN d For N a >> N d , C¢ = ê ú ë2 ( Vbi + VR ) û
1 2
1
é1.6 ´ 10 -19 ´ 117 . ´ 8.85 ´ 10 -4 ´ 1015 ù 2 =ê ú 2 (10 + 0.754) ë û
1 2
´ 4.2 ´ 10 ù ú û 15
= 5.7 ´ 10 -9 F cm 2
1 2
= 2.77 ´ 10 -9 F cm 2 C = AC¢ = 6 ´ 10 -4 ´ 2.77 ´ 10 -9 = 1.66 ´ 10 -12 F 1 1 fo = = = 2.6 MHz. 2p LC 2 p 2. 2 ´ 10 -3 ´ 1.66 ´ 10 -12
1
ì 2 e ( V + VR ) é 1 1 ùü 2 15. (D) W = í s bi + êN úý e ë a Na û þ î 1
W A é( Vbia + V R ) ( N aA + N dA ) N aB N dB ù 2 = WB êë ( Vbib + R ) ( N aB + N dB ) N aA N dB úû æN N ö Vbi = Vt ln çç a 2 d ÷÷ è ni ø æ 1018 ´ 1015 = 0.0259 ln çç 20 è 2.25 ´ 10
ö ÷÷ = 0.754 V ø
æ 1018 ´ 1016 VbiB = 0.0259 ln çç 20 è 2.25 ´ 10
ö ÷÷ = 0.814 V ø
W A éæ 5.754 öæ 1018 + 1015 = êç ÷ç WB ëè 5.814 øçè 10 18 + 1016
öæ 1016 öù 2 . ÷÷çç 15 ÷÷ú = 313 øè 10 øû
VbiA
1
æN N ö 16. (C) Vbi = Vt ln çç a 2 d ÷÷ è ni ø Page 122
18. (B) | Emax | =
eN a xn e
é2e( Vbi + VR ) ù For a p+ n junction, xn » ê ú eN d ë û 1
é2 eN d ù2 So that | Emax | = ê ( Vbi + VR ) ú ë es û Assuming Vbi > I n 0
¶ ( dpn ) 26. (A) J p = - eDp = eDp ( 3.8 ´ 1014 )( 3534) e -3534 x ¶x
= 0.6 A cm 2
æ E g2 - 0 .59 ö ç ÷ ç 0 .0259 ÷ è ø
30. (D) g d =
ù ú ú û
ni2 (15 . ´ 1010 ) 2 = = 2. 25 ´ 10 4 cm -3 Nd 1016
-3534 x
( 0 .0259)
æ I p 0 t p 0 + I n 0 tn 0 29. (B) Cd = çç 2 Vt è
Cd =
Na = 0.083 Nd
14
( 0 .255- 0 .32 - 0 .525+ E g2 )
E g 2 = 0.59 + 0.0259 ln 10 3 = 0.769 EV
Dp
æ é æçç eVa ö÷÷ ù é çç - Lx = pn 0 êe è kt ø - 1ú êe è p êë úû ê ë
1
ø t ( Va1 - Va2 - E g1 + E g2 ) I1 e è = æ V - E ö = e Vt a2 g2 ÷ ç I2 ç ÷ Vt ø eè
10 = e 1 J p = en Nd
= 0.95,
æ V -Eg ö ç a ÷ ç V ÷ ø t
3
2 i
25. (C) dpn = pn - pn 0
pn 0 =
12 = 394 . ´ 10 -15 A 10 -7
æ Va ö ç ÷ çV ÷ è tø
I µ eè
æ V a - E g1 ö ç ÷ ç ÷ V
D t po
1 23. (B) I s = Aen Nd 2 i
e
rp L A
=
L A( em p N a )
0.2 = 26 W (10 -2 )(1.6 ´ 10 -19)( 480)(10 16 )
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GATE EC BY RK Kanodia
UNIT 2
Rn = =
rn L L = A Ae(m n N d )
æ 5 ´ 1019 = 2(0.0259) ln çç . ´ 1010 è 15
0.10 = 46.3 W (10 )(1.6 ´ 10 -19)(1350)(1015)
1
R = Rp + Rn = 72.3 W
1
é2 ´ (117 . ´ 8.85 ´ 10 -14 )(114 . - 0.4) æ 2 öù 2 =ê ç -14 19 ÷ú 1.6 ´ 10 è 5 ´ 10 øû ë
æN N ö 33. (B) Vbi = Vt ln çç a 2 d ÷÷ è ni ø æ 10 = 2(0.0259) ln çç 6 è 1.8 ´ 10
= 6.19 ´ 10 -7 cm = 62 A°
ö ÷÷ =1.16 V ø
38. (B) Ln = Dn t n 0 , Lp = Dp t p 0
1
ì 2 e ( V + VR ) W = í s bi e î
Dn np 0
é 1 1 ùü 2 ê N + N úý ë a a ûþ
é2 ´ (131 . ´ 8.85 ´ 10 -14 )( 6.16) æ 2 öù -4 =ê ç 16 ÷ú = 1.34 ´ 10 cm -19 1 . 6 10 10 ´ è øû ë
=
eniW 2t o
1.6 ´ 10 -19 ´ 1.8 ´ 106 ´ 1.34 ´ 10 -4 = 193 . ´ 10 -9 A cm 2 2 ´ 10 -8
34. (A) VB = 25 =
Dn np 0 Ln
35. (D) Emax =
eN d xn e
= np 0
Lp
Dp Dn + pn 0 tn 0 tp 0
ni2 (1.8 ´ 106 ) 2 = = 324 . ´ 10 -4 cm -3 Na 1016
pn 0 =
ni2 (1.8 ´ 106 ) 2 = = 6.48 ´ 106 cm -3 Nd 5 ´ 1017
Dn = tn 0
25 = 5 ´ 10 4 , 10 ´ 10 -9
Dp
12 . ´ 10 4 = 35 10 ´ 10 -9
tp 0
=
(5 ´ 10 4 )( 3. 24 ´ 10 -4 ) (5 ´ 10 )( 3. 24 ´ 10 -4 ) + ( 35 . ´ 10 4 )( 6.48 ´ 10 -6 )
hinj =
N B = N d = 2 ´ 1016 cm -3
+
Dp pn 0
Dn tn 0
np 0 =
2 eEcrit 2 eN B
(117 . ´ 8.85 ´ 10 -4 )( 4 ´ 10 5) 2 2 ´ 1.6 ´ 10 -19 ´ N B
np 0
Ln
hinj = 1 2
J gen =
ö ÷÷ = 114 . V ø
ì 2 e ( V + VR ) é 1 1 ùü 2 W = í s bi + êN úý e ë a Na û þ î
-3
16
Electronics Devices
4
= 0.986 Þ
xn =
eEmax eN d
39. (B) The areal density at threshold is n2 D =
(117 . ´ 8.85 ´ 10 -14 )( 4 ´ 10 5) = = 5.18 ´ 10 -5 cm (1.6 ´ 10 -19)(5 ´ 1016 ) æN N ö æ 5 ´ 106 Vbi = Vt ln çç a 2 d ÷÷ = 2(0.0259) ln çç . ´ 1010 è 15 è ni ø
ö ÷÷ = 0.778 V ø
1
. ´ 10 -9) J th t r (500)(15 = = 4.69 ´ 1012 cm -3 1.6 ´ 10 -19 e
The carrier density is nth =
n2 D 4.69 ´ 1012 = = 4.69 ´ 1018 cm -3 dLas 10 -6
ì 2 e V æ N öæ öü 2 1 ÷ý xn = í s bi çç a ÷÷çç ÷ î e è N d øè N a + N d øþ
Once the threshold is reached, the carrier density does
é2(117 . ´ 8.85 ´ 10 -4 )( Vbi + VR ) ù (5.18 ´ 10 ) = ê -19 6 ú ë (1.6 ´ 10 )(2 ´ 5 ´ 10 ) û
15 . ´ 10 -9 J th = 3 ´ 10 -10 s t r ( J th ) = 5 J JA The optical power produced is p = hw e
not
Vbi + VR = 20.7,
VR = 19.9 V,
xn
J > J th
the
VR = VB
1
é2 eVB ù 2 é2(117 . ´ 8.85 ´ 10 -14 )(100) ù 2 » ê = -19 15 ú ê ú = 5.1 mm ë eN d û ë (1.6 ´ 10 )(5 ´ 10 ) û
=
(5 ´ 500)(2 ´ 10 -5)(1.43 ´ 1.6 ´ 10 -19) = 715 . MW 1.6 ´ 10 -19
****************
æN N ö 37. (D) Vbi = Vt ln çç a 2 d ÷÷ è ni ø Page 124
electron
tr ( J ) =
36. (A) For a p+ n diode, Neglecting Vi compared to VB , 1
When
recombination is
-5 2
Þ
change.
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hole
GATE EC BY RK Kanodia
CHAPTER
2.3 THE BIPOLAR JUNCTION TRANSISTOR
Statement for Q.1-2:
5. A uniformly doped npn bipolar transistor is biased in
The parameters in the base region of an npn bipolar transistor are as follows Dn = 20 cm 2 s, nB0 = 10 4 cm -3, xB = 1 mm, ABE = 10 -4 cm 2 .
the
forward-active
region.
The
transistor
doping
concentration are N E = 5 ´ 1017 cm -3, N B = 10 16 cm -3 and N C = 1015 cm -3. The minority carrier concentration pE0 , nB0 and pC0 are
1. If VBE = 0.5 V, then collector current I C is
(A) 4.5 ´ 10 2 , 2. 25 ´ 10 4 , 2. 25 ´ 10 5 cm -3
(A) 7.75 mA
(B) 1.6 mA
(B) 2. 25 ´ 10 4 , 2. 25 ´ 10 5, 4.5 ´ 10 2 cm -3
(C) 0.16 mA
(D) 77.5 mA
(C) 2. 25 ´ 10 4 , 2. 25 ´ 10 5, 4.5 ´ 10 4 cm -3 (D) 4.5 ´ 10 4 , 2.25 ´ 10 4 , 2. 25 ´ 10 5 cm -3
2. If VBE = 0.7 V, then collector current I C is (A) 418 mA
(B) 210 mA
(C) 17.5 mA
(D) 98 mA
6. A uniformly doped silicon pnp transistor is biased in the forward-active mode. The doping profile is N E = 10 18 cm -3, N B = 5 ´ 1016 cm -3 and N C = 1015 cm -3. For VEB = 0.6 V, the pB at x = 0 is
(See fig. P2.3.7-8)
3. In bipolar transistor biased in the forward-active
(A) 5.2 ´ 10 cm
-3
(B) 5.2 ´ 10 13 cm -3
region the base current is I B = 50 mA and the collector
(C) 5.2 ´ 1016 cm -3
(D) 5.2 ´ 10 11 cm -3
currents is I C = 2.7 mA. The a is (A) 0.949
(B) 54
(C) 0.982
(D) 0.018
19
Statement for Q.7-8: An npn bipolar transistor having uniform doping of N E = 10 18 cm -3 N B = 1016 cm -3 and N C = 6 ´ 10 15 cm -3
4. A uniformly doped silicon npn bipolar transistor is to be biased in the forward active mode with the B-C junction reverse biased by 3 V. The transistor doping
is operating in the inverse-active mode with VBE = - 2 V and VBC = 0.6 V. The geometry of transistor is shown in fig P2.3.7-8. Emitter -n-
are N E = 1017 cm -3, N B = 1016 cm -3 and N C = 10 15 cm -3.
Base -p-
Collector -n-
The BE voltage, at which the minority carrier electron concentration at x = 0 is 10% of the majority carrier hole
xE
xB
xC
concentration, is (A) 0.94 V (C) 0.48 V
(B) 0.64 V
x' = xE
(D) 0.24 V
x'=0 x'
x=0 x
x = xB x'' = 0
x'' = xC x''
Fig. P2.3.7-8
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GATE EC BY RK Kanodia
UNIT 2
Electronics Devices
7. The minority carrier concentration at x = x B is
N E = 1018 cm -3, N B = 5 ´ 1016 cm -3,
(A) 4.5 ´ 1014 cm -3
(B) 2.6 ´ 10 12 cm -3
N C = 2 ´ 1019 cm -3,
(C) 2.6 ´ 1014 cm -3
(D) 39 . ´ 10 14 cm -3
DE = 8 cm 2 s , DB = 15 cm 2 s , DC = 14 cm 2 s xE = 0.8 mm, xB = 0.7 mm
8. The minority carrier concentration at x¢¢ = 0 is (A) 39 . ´ 1014 cm -3 (C) 2.7 ´ 10
14
cm
(B) 2.7 ´ 10 12 cm -3
-3
(D) 4.5 ´ 10
14
cm
The emitter injection efficiency g is
-3
(A) 0.999
(B) 0.977
(C) 0.982
(D) 0.934
9. An pnp bipolar transistor has uniform doping of N E = 6 ´ 1017 cm -3, N B = 2 ´ 10 16 cm -3 and N C = 5 ´ 1014 -3
15. A uniformly doped silicon epitaxial npn bipolar
cm . The transistor is operating is inverse-active mode.
transistor
The maximum V CB voltage, so that the low injection
N B = 3 ´ 10
condition applies, is
with N C = 5 ´ 10
fabricated
cm
-3
(A) 0.86 V
(B) 0.48 V
xB = 0.7 mm
(C) 0.32 V
(D) 0.60 V
punch-through is
Statement for Q.10-12: The
following
currents
are
measured
in
with
a
base
doping
of
and a heavily doped collector region cm -3. The neutral base width is
17
when
VBE = VBC = 0.
The
(A) 26.3 V
(B) 18.3 V
(C) 12.2 V
(D) 6.3 V
VBC
at
a 16. A silicon npn transistor has a doping concentration
uniformly doped npn bipolar transistor:
of
I nE = 120 . mA, I pE = 0.10 mA, I nC = 118 . mA
N B = 1017
cm -3
and
N C = 7 ´ 10 15
cm -3.
The
metallurgical base width is 0.5 mm. Let VBE = 0.6 V.
I R = 0.20 mA, I G = 1 mA, I pC0 = 1 mA
Neglecting the B–E junction depletion width the VCE at punch-through is
10. The a is (A) 0.667
(B) 0.733
(C) 0.787
(D) 0.8
(A) 146 V
(B) 70 V
(C) 295 V
(D) 204 V
17. A uniformly doped silicon pnp transistor is to
11. The b is (A) 3.69
(B) 0.44
(C) 2.27
(D) 8.39
designed with N E = 1019 cm -3 and N C = 10 16 cm -3. The metallurgical base width is to be 0.75 mm. The minimum
base
doping,
so
that
the
minimum
punch-through voltage is Vpt = 25 V, is
12. The g is (A) 0.816
(B) 0.923
(C) 1.083
(D) 0.440
(A) 4.46 ´ 1015 cm -3
(B) 4.46 ´ 10 16 cm -3
(C) 195 . ´ 1015 cm -3
(D) 195 . ´ 1016 cm -3
13. A silicon npn bipolar transistor has doping
18. For a silicon npn transistor assume the following
concentration of N E = 2 ´ 1018 cm -3, N B = 10 17 cm -3 and
parameters:
. ´ 1016 cm -3. The area is 10 -3 cm 2 and neutral N C = 15
I E = 0.5 mA, b = 48
base width is 1 mm. The transistor is biased in the active
xB = 0.7 mA, xdc = 2 mm
region at VBE = 0.5 V. The collector current is
Cs = Cm = 0.08 pF, C je = 0.8 pF
(DB = 20 cm 2 s)
Page 126
is 16
(A) 9 mA
(B) 17 mA
(C) 22 mA
(D) 11 mA
Dn = 25 cm 2 s, rc = 30 W The carrier cross the space charge region at a speed of 10 7 cm s. The total delay time t ec is
14. A uniformly doped npn bipolar transistor has
(A) 164.2 ps
(B) 234.4 ps
following parameters:
(C) 144.2 ps
(D) 298.4 ps
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The Bipolar Junction Transistor
19. In a bipolar transistor, the base transit time is 25%
Statement for Q.24-26:
of the total delay time. The base width is 0.5 mm and base diffusion coefficient is DB = 20 cm 2 s. The cut-off
Chap 2.3
For the transistor in circuit of fig. P2.3.24-26. The parameters are bR = 1 , bF = 100 , and I s = 1 fA .
frequency is (A) 637 MHz
(B) 436 MHz
(C) 12.8 GHz
(D) 46.3 GHz
5V
20. The base transit time of a bipolar transistor is 100 ps and carriers cross the 1.2 mm B–C space charge at a speed of 10 7 cm s . The emitter-base junction charging time is 25 ps and the collector capacitance and resistance are 0.10 pF and 10 W, respectively. The cutoff frequency is (A) 43.8 GHz
(B) 32.6 GHz
(C) 3.26 GHz
(D) 1.15 GHz
Fig. P2.3.24-26
24. The current I C is (A) 1 fA (C) 1.384 fA
(B) 2 fA (D) 0 A
25. The current I E is
Statement for Q.21-22: Consider the circuit shown in fig. P2.3.21-22. If voltage Vs = 0.63 V, the currents are I C = 275 mA and I B = 5 mA .
(A) 1 fA
(B) -1 fA
(C) 2 fA
(D) -2 fA
26. The current I B is (A) 2 fA
(B) -2 fA
(C) 1 fA
(D) -1 fA
27. For the transistor in fig. P2.3.27, I S = 10 -15 A, bF = 100 , bR = 1. The current I CBO is Vs 5V
Fig.P2.3.21-22
21. The forward common-emitter gain bF is (A) 56
(B) 55
(C) 0.9821
(D) 0.9818
Fig.P2.3.27
(A) 101 . ´ 10 -14 A (C) 101 . ´ 10 -15 A
22. The forward current gain a F is (A) 0.9821
(B) 0.9818
(C) 55
(D) 56
(B) 2 ´ 10 -14 A (D) 2 ´ 10 -15 A
Statement for Q.28-31: Determine
23. Consider the circuit shown in fig P2.3.23. If Vs = 0.63 V, I1 = 275 mA and I 2 = 125 mA, then the value of I 3 is
the
region
of
operation
for
the
transistor shown in circuit in question. 28.
I1
6V I2
I3
Vs
Fig. P2.3.23
(A) - 400 mA
(B) 400 mA
(C) - 600 mA
(D) 600 mA
Fig.P2.3.28
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(B) Reverse-Active (D) Cutoff Page 127
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UNIT 2
29.
Electronics Devices
33. The current I1 is (A) -12.75 mA
(B) 12.75 mA
(C) 12.5 mA
(D) - 12.5 mA
Statement for Q.34–35:
6V
The leakage current of a transistor are I CBO = 5 m A and I CEO = 0.4 mA, and I B = 30 mA.
Fig. P2.3.29
(A) Forward-Active
(B) Reverse-Active
(C) Saturation
(D) Cutoff
34. The value of b is
30.
(A) 79
(B) 81
(C) 80
(D) None of the above
35. The value of I C is (A) 2.4 mA
(B) 2.77 mA
(C) 2.34 mA
(D) 1.97 mA
6V
Statement for Q.36–37: Fig.P2.3.30
(A) Forward-Active
(B) Reverse-Active
(C) Saturation
(D) Cutoff
For a BJT, I C = 5 mA, I B = 50 mA and I CBO = 0.5 mA. 36. The value of b is
31. 3V
(A) 103
(B) 91
(C) 83
(D) 51
37. The value of I E is
6V
(A) 5.25 mA
(B) 5.4 mA
(C) 5.65 mA
(D) 5.1 mA
Fig.P2.3.31
(A) Forward-Active
(B) Reverse-Active
(C) Saturation
(D) Cutoff
********
Statement for Q.32-33: For the circuit shown in fig. P2.3.32-33, let the value of b R = 0.5 and bF = 50. The saturation current is 10 -16 A.
+3 V
250 mA
I1
Fig. P2.3.32-33
32. The base-emitter voltage is
Page 128
(A) 0.53 V
(B) 0.7 V
(C) 0.84 V
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GATE EC BY RK Kanodia
The Bipolar Junction Transistor
SOLUTIONS 1. (A) I C = I s e
æ çç
ö ÷÷
0 .5
2. (C) I C = 32 . ´ 10
-14
e
æ 0 .7 ö çç ÷÷ è 0 .0259 ø
= 17.5 mA
pC (0) = 0.10 N C = 5 ´ 1013 cm -3, ni2 (15 . ´ 1010 ) 2 = = 4.5 ´ 10 5 NC 5 ´ 1014 ö æ ç VCB ÷ ç V ÷ t ø
pC (0) = pC 0 e è
10. (C) a =
ni2 (15 . ´ 1010 ) 2 = = 2. 25 ´ 10 4 cm -3 NB 1016 æ VBE ö ÷ ç ç V ÷ t ø
At x = 0, np (0) = np 0 e è
VBE
æ np (0) ö ÷ VBE = VT ln ç ç n ÷ è p0 ø
=
æ 1015 = 0.0259 ln çç 4 è 2.25 ´ 10
ö ÷÷ = 0.635 V ø
ni2 (15 . ´ 10 10 ) 2 = = 450 cm -3 NE 5 ´ 1017
nB 0 =
ni2 (15 . ´ 10 10 ) 2 = = 2. 25 ´ 10 4 cm -3 NB 5 ´ 1016
pC 0 =
ni2 (15 . ´ 10 10 ) 2 = = 2. 25 ´ 10 5 cm -3 NE 5 ´ 1015
6. (B) pB 0 = pB (0) = pB 0 e 7. (C) nB 0 =
ni2 (15 . ´ 1010 ) 2 = = 4.5 ´ 10 3 cm -3 NB 5 ´ 1016
æ VEB ç ç V è t
ö ÷ ÷ ø
æ 0 .6 ö ÷÷ çç 3 è 0 .0259 ø
= 4.5 ´ 10 e
= 5. 2 ´ 1013 cm -3
ni2 (15 . ´ 10 10 ) 2 = = 2. 25 ´ 10 4 cm -3 NB 1016
nB ( x = xB ) = nB 0 e
Þ
æ p (0) ö VCB = Vt ln çç C ÷÷ è pC 0 ø
ö æ ç VBC ÷ ç V ÷ è t ø
0 .6
= 2. 25 ´ 10 4 e 0 .0259 = 2.6 ´ 10 14 cm -3
J nE
J nC I nC = + J R + J pE I nE + I R + I pE
118 . = 0.787 12 . + 0.2 + 0.1
11. (A) b =
a 0.787 == = 3.69 1-a 1 - 0.787
12. (B) g =
J nE I nE 1. 2 = = = 0.923 J nE + J pE I nE + I pE 1. 2 + 0.1
10 1016 ´ NB = = 10 15 100 10
5. (A) pE 0 =
ö ÷÷
æ 5 ´ 1013 ö ÷ = 0.48 V = 0.0259 ln çç 5÷ è 4.5 ´ 10 ø
IC 2.7m = = 0.982 I C + I B 2.7m + 50m
np (0) =
0 .6
9. (B) Low injection limit is reached when
pC 0 =
I bF 3. (C) bF = C , a F = IB 1 + bF
Þ
æ VBC ö ÷ ç ç V ÷ t ø
æ çç
I C = 3. 2 ´ 10 -14 e è 0 .0259 ø = 7.75 mA
4. (B) np 0 =
ni2 (15 . ´ 1010 ) 2 = = 375 . ´ 10 4 cm -3 NC 6 ´ 1015
= 375 . ´ 10 4 e è 0 .0259 ø = 4.31 ´ 1014 cm -3
(1.6 ´ 10 -19)(20)(10 -4 )(10 4 ) = 32 . ´ 10 -14 A 10 -4
aF =
8. (D) pC 0 =
pC ( x¢¢ = 0) = pC 0 e è
æ VBE ö ÷ ç ç V ÷ è b ø
eDn ABE nB 0 Is = xB =
Chap 2.3
13. (B) nB 0 =
ni2 (15 . ´ 10 10 ) 2 = = 2.25 ´ 10 3 cm -3 NB 1017
æ VBE ö ÷ ç ç V ÷ t ø
nB (0) = nB 0 e è IC = =
0 .5
ö ÷÷
eDB AnB (0) xB
(1.6 ´ 10 -19)(20)(10 -3)(5.45 ´ 10 11 ) = 17.4 mA 10 -4
14. (B) g =
=
æ çç
= 2.25 ´ 10 3 e è 0 .0259 ø = 5.45 ´ 1011 cm -3
1 N B D E xB 1+ × × N E D B xE
1 = 0.977 5 ´ 1016 8 0.7 1+ 1018 15 0.8
æN N ö 15. (B) Vbi = Vt ln çç B 2 C ÷÷ è ni ø æ 3 ´ 1016 ´ 5 ´ 1017 = 0.0259 ln çç . ´ 1010 ) 2 è (15
ö ÷÷ = 0.824 V ø
At punch-through
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The Bipolar Junction Transistor
ö æ æ VBE ö ç VBC ÷ ö æ çæç VBE ÷ö÷ ö ÷ ç I æ çç ÷÷ 27. (C) I E = I S ç e è Vt ø - e è Vt ø ÷ + S ç e è Vt ø - 1 ÷ = 0 ç ÷ bF ç ÷ è ø è ø
Þ
ö æ ç VBE ÷ ç V ÷ t ø
eè
=
33. (A) I E = (bF + 1) I B = 12.75 mA I1 = - I E = - 12.75 mA.
æ VBC ö ÷ Vt ÷ø
ç ç 1 bR + eè 1 + bF 1 + BF
34. (A) I CEO = (b + 1) I CBO b+1=
ö æ æ VBC ö ç VBC ÷ ù é æçç VBE ö÷÷ ù ç V ÷ I é çç ÷÷ I C = I S êe è Vt ø - e è t ø ú - S êe è Vt ø - 1ú ê ú bR ê úû ë ë û
I CBO =
IS é ê1 - e 1 + bF ê ë
æ VBC ö ÷ ç ç V ÷ è t ø
ù I é ú - S êe úû bR êë
æ VBC ç ç V è t
ö ÷ ÷ ø
0.4m = 80 5m
Þ
b = 79
35. (B) I C = bI B + I CEO = 79( 30m ) + 0.4m =2.77 mA
ù - 1ú úû
36. (A) I C = bI B + I CEO = b I B + (b + 1) I CBO b=
VBC = - 5 V, Vt = 0.0259 V I CBO =
Chap 2.3
Is I (1 - 0) - S (0 - 1) = 101 . I S = 101 . ´ 10 -15 A 101 1
I C - I CBO 5.2m - 0.5m = » 10396 . I B + I CBO 50m + 0.5m
37. (A) a =
28. (D)
IE =
b = 0.9904 b+1
I C - I CBO 5.2m - 0.5m = = 5.25 MA a 0.9904
B-C Junction VBC
B-E junction VBE
Reverse Bias
Forward bias
Forward bias
Forward-Active
Saturation
Reverse Bias
Cut-off
Reverse-Active
*******
VBE = 0 , VBC < 0, Thus both junction are in reverse bias. Hence cutoff region. 29.(A) VBE > 0 , VBC = 0, Base-Emitter junction forward bais,
Base-collector
junction
reverse
bias,
Hence
forward-active region. 30. (B) VBE = 0 , VBC > 0, Base-Emitter junction reverse bais , Base-collector junction forward bias, Hence reverse-active region. 31. (C) VBE = 6 V, VBC = 3 V, Both junction are forward biase, Hence saturation region. 32. (C) The current source will forward bias the base-emitter junction and the collector base junction will then be reverse biased. Therefore the transistor is in the forward active region IC = ISe
æ VBE ö ÷ ç ç V ÷ è t ø
I C = bF I B = 50 ´ 250 ´ 10 -6 = 12.5 ´ 10 -3 A æ 12.5 ´ 10 -3 ö æI ö VBE = Vt ln çç C ÷÷ = 0.0259 ln çç -16 ÷÷ = 0.84 V ø è 10 è IS ø
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UNIT 2
electronics Devices
11. In n-well CMOS fabrication substrate is
18. Monolithic integrated circuit system offer greater
(A) lightly doped n - type
reliability than discrete-component systems because
(B) lightly doped p - type
(A) there are fewer interconnections
(C) heavily doped n - type
(B) high-temperature metalizing is used
(D) heavily doped p - type
(C) electric voltage are low (D) electric elements are closely matched
12. The chemical reaction involved in epitaxial growth in IC chips takes place at a temperature of about (A) 500° C
(B) 800° C
(C) 1200° C
(D) 2000° C
19. Silicon dioxide is used in integrated circuits (A) because of its high heat conduction (B) because it facilitates the penetration of diffusants
13. A single monolithic IC chip occupies area of about
(C) to control the location of diffusion and to protect and insulate the silicon surface.
(A) 20 mm 2
(B) 200 mm 2
(D) to control the concentration of diffusants.
(C) 2000 mm 2
(D) 20,000 mm 2 s
20. Increasing the yield of an IC
14. Silicon dioxide layer is used in IC chips for
(A) reduces individual circuit cost
(A) providing mechanical strength to the chip
(B) increases the cost of each good circuit (C) results in a lower number of good chips per wafer
(B) diffusing elements
(D) means that more transistor can be fabricated on the same size wafer.
(C) providing contacts (D) providing mask against diffusion
21. The main purpose of the metalization process is 15. The p-type substrate in a monolithic circuit should be connected to
(A) to act as a heat sink (B) to interconnect the various circuit elements
(A) any dc ground point
(C) to protect the chip from oxidation
(B) the most negative voltage available in the circuit
(D) to supply a bonding surface for mounting the chip
(C) the most positive voltage available in the circuit 22. In a monolithic-type IC
(D) no where, i.e. be floating 16. The collector-substrate junction in the epitaxial
(A) each transistor is diffused into a separate isolation region (B) all components are fabricated into a single crystal of silicon
collector structure is, approximately (A) a step-graded junction
(C) resistors and capacitors of any value may be made
(B) a linearly graded junction (C) an exponential junction
(D) all isolation problems are eliminated
(D) None of the above 23. Isolation in ICs is required 17. The sheet resistance of a semiconductor is
(A) to make it simpler to test circuits
(A) an important characteristic of a diffused region especially when used to form diffused resistors
(B) to protect the transistor from possible ``thermal run away’’
(B) an undesirable parasitic element
(C) to protect the components mechanical damage
(C) a characteristic whose value determines the required area for a given value of integrated capacitance
(D) to minimize electrical interaction between circuit components
(D) a parameter whose value is important in a thin-film resistance Page 140
24. Almost all resistor are made in a monolithic IC (A) during the base diffusion
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GATE EC BY RK Kanodia
Integrated Circuits
Chapc2.5
(B) during the collector diffusion
30. For the circuit shown in fig. P2.5.30, the minimum
(C) during the emitter diffusion
number and the maximum number of isolation regions
(D) while growing the epitaxial layer
are respectively
25. The equation governing the diffusion of neutral
(A)
¶N ¶2 N =D ¶t ¶x 2
(C)
¶ N ¶N =D 2 ¶t ¶x
R1
R2
atom is (B)
¶N ¶2 N =D ¶x ¶t 2
(D)
¶ N ¶N =D 2 ¶x ¶t
2
Vo2
Q1 Q2
2
Fig. P2.5.32
26. The true statement is (A) thick film components are vacuum deposited (B) thin film component are made by screen-and- fire process
(A) 2, 6
(B) 3, 6
(C) 2, 4
(D) 3, 4
31. For the circuit shown in fig. P2.5.31, the minimum number of isolation regions are
(C) thin film resistor have greater precision and are more stable (D) thin film resistor are cheaper than the thin film resistor 27. The False statement is (A) Capacitor of thin film capacitor made with proper dielectric is not voltage dependent (B) Thin film resistors and capacitor need to be biased for isolation purpose (C) Thin film resistors and capacitor have smaller stray capacitances and leakage currents.
Fig. P2.5.31
(A) 2
(B) 3
(C) 4
(D) 7
(D) None of the above
*******
28. Consider the following two statements S1 : The dielectric isolation method is superior to junction isolation method. S2 : The beam lead isolation method is inferior to junction isolation method. The true statements is (are) (A) S1 , S2
(B) only
(C) only
(D) Neither nor S2
29. If P is passivation, Q is n-well implant, R is metallization and S is source/drain diffusion, then the order in which they are carried out in a standard n-well CMOS fabrication process is (A) S - R - Q - P
(B) R - P - S - Q
(C) Q - S - R - P
(D) P - Q - R - S
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Page 141
GATE EC BY RK Kanodia
UNIT 2
SOLUTIONS 1. (D)
2. (D)
3. (B)
4. (B)
5. (C)
6. (B)
7. (C)
8. (B)
9. (C)
10. (D)
11. (B)
12. (C)
13. (C)
14. (D)
15. (B)
16. (A)
17. (A)
18. (A)
19. (C)
20. (A)
21. (B)
22. (B)
23. (D)
24. (A)
25. (A)
26. (C)
27. (B)
28. (B)
29. (C)
30. (D) The minimum number of isolation region is 3 one containing Q1 , one containing and one containing both and . The maximum number of isolation region is 4, or one per component. 31. (A) The minimum number of isolation region is two. One for transistor and one for resistor.
*******
Page 142
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electronics Devices
GATE EC BY RK Kanodia
CHAPTER
3.1 DIODE CIRCUITS
vo
Statement for Q.1–4:
vo 22
18
In the question a circuit and a waveform for the t
input voltage is given. The diode in circuit has cutin (A)
voltage Vg = 0. Choose the option for the waveform of
(B)
vo
output voltage vo .
vo t
-3
(C)
vi
+ 20
2.2 kW
3.
vo 5V
(D)
2 kW
t
_
t -7
1.
vi
t
vi
+
16
-5
vo
vi
Fig.3.1.1
2 T
4V 20
_
15
t
t
-5
-10
(A)
(B)
Fig. P3.1.3
vo
vo
16
4
4
t
T
T 2
vo
20
(A)
20
(A) 2.
(D)
10 kW
T
T 2
-4
(D)
R
vi + D1
t -5
t
4.
vi
vo _
T
4
t
(C) 20
vi
T 2
t
16
12
2V +
T
vo
5
t
T 2
(B)
vo t
t
-16
12
vo
T
6V vo
vi 8V
Fig.3.1.2
10
D2
_
t -10
Fig.3.1.4
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GATE EC BY RK Kanodia
UNIT 3 vo
vo
Analog Electronics
vo
vo
10
8
9.42
9.42 5.7 t
4.3
t
5.7
-6
15
vi
(C)
-10
(A)
vi
(D)
7. For the circuit shown in fig. P3.1.7, each diode has
(B)
vo
15
4.3
Vg = 0.7 V. The vo for -10 £ vs +10 V is
vo
+10 V
8
6
t
10 kW
t
-6
D1
-8
(C)
D2
vs
(D)
vo
5. For the circuit in fig. P.3.1.5, let cutin voltage Vg = 0.7 V. The plot of vo verses vi for -10 £ vi £ 10 V is
10 kW
D4
D3
10 kW
+ 10 kW
20 kW
-10 V
vo
vi
Fig. P3.1.7.
10 V
10 V
vo
_
vo
8.43
7.48
Fig. P3.1.5 vo
vo
9.3
-10
10
9.3
3.33 3.33
-10
10
4.03
10
(B)
vo
vi
vo
5.68
vo 10
10
4.65
-10 -6.8
6.8 10
vs
-10
-5.68
4.65 -4.65
(C) 3.33
10
vi
4.33
-10
(C)
10
vi
(D)
clipping circuit shown in fig. P3.1.8. If diode has rf = 0
(D)
and rr = 2 MW and Vg = 0, the output waveform is
is Vg = 0.7 V. The plot of vo versus vi is
+ 1 MW
+
vo
vi
1 kW
2.5 V
vo
1 kW
_
vi
15 V
_
Fig. P3.1.8 vo
Fig. P3.1.6 vo
vo
10
vo 19.6
19.6
t
-5
5
5.7
(A) Page 146
15
(B) vo
vo
4.3 vi
15
4.3
5
vi
(B)
t
-5
(C) www.gatehelp.com
t
-10
(A) 5.7
vs
varies between +10 V and -10 V is impressed upon the
6. For the circuit in fig. P3.1.6 the cutin voltage of diode
2 kW
10
4.65
8. A symmetrical 5 kHz square wave whose output
3.33
3.33 -10
vs
-7.48
(B)
vo
10
(A)
-10
(A)
-10
-8.43
3.33 vi
vs
t
(D)
GATE EC BY RK Kanodia
Diode Circuits
Chap 3.1
C
9. In the circuit of fig. P3.1.9, the three signals of fig are
vi
+
10
impressed on the input terminals. If diode are ideal then the voltage vo is
vi
D1
v D2
+
D3
+ v1
v2
+
vo
v3
10 kW
-
-
t
vo
R 5V
_
v3 v2
-20
Fig. P.3.1.11
v1
vo
t
35
vo
25
-
Fig. P.3.1.9
vo
5 t
vo
t
-5
(A)
(B)
vo t
t
(A)
15
(B)
vo
(D) None of the above
t
vo -15
(C) t
t
(D)
12. In the circuit of fig. P3.1.12, D1 and D2 are ideal diodes. The current i1 and i2 are
(C)
(D) D1
10. For the circuit shown in fig. P3.1.10 the input
i1
voltage vi is as shown in fig. Assume the RC time
D2
i2
500 W
constant large and cutin voltage of diode Vg = 0. The
3V
5V
output voltage vo is
5V
C vi
+
Fig. P3.1.12
10 vi
t
vo
R
-10 _
Fig. P.3.1.10
vo
vo
(A) 0, 4 mA
(B) 4 mA, 0
(C) 0, 8 mA
(D) 8 mA, 0
13. In the circuit of Fig. P3.1.13 diodes has cutin
20
voltage of 0.6 V. The diode in ON state are
10 t
t
(A)
D1
(B)
vo
12 W
6W
D2
vo t
5.4 V
t
18 W
5V
-10 -20
(C)
Fig. P3.1.13
(D)
11. For the circuit shown in fig. P.3.1.11, the input
(A) only D1
(B) only D2
voltage vi is as shown in fig. Assume the RC time
(C) both D1 and D2
(D) None of the above
constant large and cutin voltage Vg = 0. The output voltage vo is www.gatehelp.com
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GATE EC BY RK Kanodia
Diode Circuits
Chap 3.1
22. The diodes in the circuit in fig. P3.1.22 has
26. If v2 = 0, then output voltage vo is
parameters Vg = 0.6 V and rf = 0. The current iD2 is
(A) 6.43 V
(B) 9.43 V
(C) 7.69 V
(D) 8.93 V
+10 V 9.5 kW
27. If v2 = 5 V, then vo is
D2
0.5 kW 0V
iD2
vo
+5 V D1
0.5 kW
D3
Fig. P3.1.22
(B) 10 mA
(C) 7.6 mA
(D) 0 mA
(B) 12.63 V
(C) 18.24 V
(D) 10.56 V
28. If v2 = 10 V, then vo is
+5 V
(A) 8.4 mA
(A) 8.93 V
(A) 10 V
(B) 9.16 V
(C) 8.43 V
(D) 12.13 V
Statement for Q.29–30: The diode in the circuit of fig. P3.1.29–30 has the
Statement for Q.23–25:
non linear terminal characteristic as shown in fig. Let
The diodes in the circuit in fig. P3.1.23-25 have linear parameter of Vg = 0.6 V and rf = 0.
the voltage be vs = cos wt V. 100 W vi
9.5 kW 0.5 kW
D2
iD 4
+ vD
100 W
-
2V
vo
v2
~
iD(mA)
a
+10 V
b
v1
vD(V)
Fig. P3.1.29–30
D1
0.5 kW
0.5 0.7
Fig. P3.1.23–25
29. The current iD is
23. If v1 = 10 V and v2 = 0 V, then vo is (A) 8.93 V
(B) 7.82 V
(C) 1.07 V
(D) 2.18 V
(A) 2.5(1 + cos wt) mA
(B) 5(0.5 + cos wt) mA
(C) 5(1 + cos wt) mA
(D) 5(1 + 0.5 cos wt) mA
30. The voltage vD is
24. If v1 = 10 V and v2 = 5 V, then vo is (A) 9.13 V
(B) 0.842 V
(A) 0.25( 3 + cos wt) V
(C) 5.82 V
(D) 1.07 V
(C) 0.5( 3 + 1 cos wt) V
(B) 0.25(1 + 3 cos wt) V (D) 0.5(2 + 3 cos wt) V
25. If v1 = v2 = 0, then output voltage vo is
31. The circuit inside the box in fig. P3.1.31. contains
(A) 0.964 V
(B) 1.07 V
only resistor and diodes. The terminal voltage vo is
(C) 10 V
(D) 0.842 V
connected to some point in the circuit inside the box. The largest and smallest possible value of vo most
Statement for Q.26–28:
nearly to is respectively +15 V
The diodes in the circuit of fig. P3.1.26–28 have linear parameters of Vg = 0.6 V and rf = 0. 500 W
D2
-9 V
v2
Circuit Containing Diode and Resistor
vo
vo
+10 V 500 W
D1
Fig. 3.1.31
9.5 kW
(A) 15 V, 6 V
(B) 24 V, 0 V
(C) 24 V, 6 V
(D) 15 V, -9 V
Fig. P3.1.26–28
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Page 149
GATE EC BY RK Kanodia
UNIT 3
32. In the voltage regulator circuit in fig. P3.1.32 the
Statement for Q.36–38:
maximum load current iL that can be drawn is
In the voltage regulator circuit in fig. P3.1.36–38
15 kW
the Zener diode current is to be limited to the range iL
30 V
Analog Electronics
Vz = 9 V Rz = 0
5 £ iz £ 100 mA. 12 W
RL
iL
iz
Fig. 3.1.32
(A) 1.4 mA
(B) 2.3 mA
(C) 1.8 mA
(D) 2.5 mA
power dissipation in the Zener diode is 150 W
Vz = 15 V Rz = 0
Fig. P3.1.33
(B) 1.5 W
(C) 2 W
(D) 0.5 W
36. The range of possible load current is (A) 5 £ iL £ 130 mA
(B) 25 £ iL £ 120 mA
(C) 10 £ iL £ 110 mA
(D) None of the above
37. The range of possible load resistance is
75 W
(A) 1 W
RL
Fig. P3.1.36–38
33. In the voltage regulator shown in fig. P3.1.33 the
50 V
Vz = 4.8 V Rz = 0
6.3 V
(A) 60 £ RL £ 372 W
(B) 60 £ RL £ 200 W
(C) 40 £ RL £ 192 W
(D) 40 £ RL £ 360 W
38. The power rating required for the load resistor is
34. The Q-point for the Zener diode in fig. P3.1.34 is
(A) 576 mW
(B) 360 mW
(C) 480 mW
(D) 75 mW
11 kW
39. The secondary transformer voltage of the rectifier 20 V
Vz = 4 V Rz = 0
circuit shown in fig. P3.1.39 is vs = 60 sin 2 p60 t V. Each
3.6 kW
diode has a cut in voltage of Vg = 0.6 V. The ripple voltage is to be no more than Vrip = 2 V. The value of filter capacitor will be
Fig. P3.1.34
(A) (0.34 mA, 4 V)
(B) (0.34 mA, 4.93 V)
(C) (0.94 mA, 4 V)
(D) (0.94 mA, 4.93 V)
+
vi
35. In the voltage regulator circuit in fig. P3.1.35 the
+ vo 10 kW
+ vs -
power rating of Zener diode is 400 mW. The value of RL
C -
that will establish maximum power in Zener diode is
Fig. P3.1.39
222 W
20 V
Vz = 10 V Rz = 0
RL
(A) 48.8 mF
(B) 24.4 mF
(C) 32.2 mF
(D) 16.1 mF
40. The input to full-wave rectifier in fig. P3.1.40 is Fig. P3.1.35
Page 150
(A) 5 kW
(B) 2 kW
(C) 10 kW
(D) 8 kW
vi = 120 sin 2 p60 t V. The diode cutin voltage is 0.7 V. If the output voltage cannot drop below 100 V, the required value of the capacitor is www.gatehelp.com
GATE EC BY RK Kanodia
Diode Circuits
+
1 : 1
vi
SOLUTIONS
+ + vs + vs -
2.5 kW
C
vo -
1. (D) Diode is off for vi < 5 V. Hence vo = 5 V. For vi > 5 V, vo = vi , Therefore (D) is correct option.
-
2. (C) Diode will be off if vi + 2 > 0.Thus vo = 0
Fig. P3.1.40
(A) 61.2 mF
(B) 41.2 mF
(C) 20.6 mF
(D) 30.6 mF
For vi + 2 < 0 V,
voltage is Vin = 0. The ripple voltage is to be no more than vrip = 4 V. The minimum load resistance, that can be connected to the output is +
~
50 mF
vo
vi < - 2, vo = vi + 2 = -3 V
Thus (C) is correct option.
41. For the circuit shown in fig. P3.1.41 diode cutin
75sin 2p60t V
Chap 3.1
3. (D) For vi < 4 V the diode is ON and output vo = 4 V. For vi > 4 V diode is off and output vo = vi . Thus (D) is correct option. 4. (C) During positive cycle when vs < 8 V, both diode are OFF vo = vs . For vs > 8 V , vo = 8 V, D1 is ON. During
RL
negative cycle when |vs | < 6 V, both diode are OFF,
-
vo = vs . For vs > 6 V, D2 is on vo = -6 V. Therefore (C) is
Fig. P3.1.41
correct.
(A) 6.25 kW
(B) 12.50 kW
(C) 30 kW
(D) None of the above
10 10 5. (B) For D off , vo = 20 20 = 3.33 V. 1 1 + 20 10 For vi £ 3.33 + 0.7 = 4.03 V, vo = 3.33 V
****************
For vi > 4.03 V , vo = vi - 0.7 For vi = 10 V, vo = 9.3 V 6. (C) Let v1 be the voltage at n-terminal of diode, v1 =
15 ´ 1 =5 V 2 +1
For vi £ 5.7 V, vo = vi v1 - 15 v1 v - vi + + o =0 2k 1k 1k
Þ
3v1 + 2 vo - 2 vi = 15
Þ
vo = 0.4 vi + 3.42
vo = v1 + 0.7 5 vo - 2 v1 = 15 + 2.1 = 17.9
7. (D) For vs > 0, when D1 is OFF, Current through D2 is i=
10 - 0.7 = 0.465 mA, vo = 10 ki = 4.65 V 10 + 10
vo = vs for 0 < vs < 4.65 V. For negative values of vs , the output is negative of positive part. Thus (D) is correct option. 8. (B) The diode conducts (zero resistance) when vi < 2.5 V and vo = vi . Diode is open (2 MW resistance) when v - 2.5 vi > 2.5 V and vo = 2.5 + i = 5 V. 3
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GATE EC BY RK Kanodia
UNIT 3
= 50 ´ 5 (1 + cos wt) ´ 10 -3 + 0.5
C=
= 0.75 + 0. 25 cos wt = 0. 25( 3 + cos wt) V 31. (D) The output voltage cannot exceed the positive power supply voltage and cannot be lower than the negative power supply voltage. 32. (A) At regulated power supply is =
vmax 2 fkVrip
RL = 30 - 9 = 1.4 mA iL 15 k
75(50) 50 V = 75 + 150 3
50 > VZ , RTH = 150 ||75 = 50 W 3 1 æ 50 ö iZ = - 15 ÷ = 33 mA, P = 15 iZ = 0.5 W ç 50 è 3 ø 3.6(20) = 4.93 V > VZ , 11 + 3.6 4.93 - 4 = 11 || 3.6 = 2.71 kW, iZ = = 0.34 mA 2.71k
34. (A) vTH = RTH
35. (B) iZ ( max ) = iL + iZ =
400m = 40 mA 10
20 - 10 = 45 mA 222
iL ( min ) = 45 - 40 = 5 mA, RL =
10 = 2 kW 5m
36. (B) Current through 12 W resistor is 6.3 - 4.8 i= = 125 mA 12 iL = i - iZ = 125 - iZ
Þ
25 £ iL £ 120 mA
37. (C) 25 £ iL £ 120 mA, iL RL = 4.8 V 25 £
4.8 £ 120 mA RL
Þ
40 £ RL £ 192 W
38. (A) PL = iL VZ = (120m)( 4.8) = 576 mV 39. (B) vs = 60 sin 2 p60 t V vmax = 60 - 1.4 = 58.6 V 58.6 vmax C= = = 24.4 mF 2 fRVrip 2( 60)10 ´ 10 3 ´ 2 40. (C) Full wave rectifier vs = vi = 120 sin 2 p60 t V vmax = 120 - 0.7 = 119.3 V Vrip = 119.3 - 100 = 19.3 V Page 154
=
41. (A) Vrip =
will remain less than 1.4 mA. 33. (D) vTH =
Analog Electronics
www.gatehelp.com
119.3 = 20.6 mF 2( 60)2.5 ´ 10 3 ´ 14.4
vmax fRL C
75 vmax = 6.25 kW = fCVrip 60 ´ 50 ´ 10 -5 ´ 4 ***************
GATE EC BY RK Kanodia
CHAPTER
3.2 BASIC BJT CIRCUITS
VBE ( ON ) = 0.7
VCE ( Sat ) = 0.2
(A) 8.4 V
(B) 6.2 V
transistor if not given in problem.
(C) 4.1 V
(D) None of the above
Statement for Q.1-4:
3. I C , RC = ?
Use
V,
V
for
npn
+5 V
The common-emitter current gain of the transistor is b =75. The voltage VBE in ON state is 0.7 V. 1. I E , RC = ?
RC VC = 2 V
50 kW +12 V
10 kW
IQ = 1 mA
+ VEC = 6 V -
-5 V
Fig. P3.3.3
RC
(A) 0.987 mA, 3.04 kW (B) 1.013 mA, 2.96 kW
-12 V
(D) 0.946 mA, 4.18 kW
Fig. P3.3.1
(A) 1.46 mA, 6.74 kW
(B) 0.987 mA, 3.04 kW
(C) 1.13 mA,, 5.98 kW
(D) None of the above
(D) 1.057 mA, 3.96 kW 4. VC = ?
2. VEC = ?
+5 V
+8 V 10 kW 20 kW
10 kW
VC
10 kW -2 V 2 kW
3 kW
-8 V
Fig. P3.3.2
Fig. P3.3.4
(A) 1.49 V
(B) 2.9 V
(C) 1.78 V
(D) 2.3 V
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GATE EC BY RK Kanodia
UNIT 3
Statement for Q.5-6:
Analog Electronics
9. VB = 1 V
In the circuit of fig.P3.3.5-6 VB = - 1 V
(A) 4 V (C) 1 V
+3 V
(B) 3 V (D) 1.9 V
10. VB = 2 V 500 kW
4.8 kW
(B) 1.5 V
(C) 2.6 V
(D) None of the above
Statement for Q.11-12:
-3 V
The transistor in circuit shown in fig. P3.3.11-12 has b = 200. Determine the value of voltage Vo for given value of VBB .
Fig. P3.3.5-6
5. b = ? (A) 103.4 (C) 134.5
(A) -7 V
(B) 135.5 (D) 102.4
+5 V
5 kW
6. VCE = ? (A) 6.4 V (C) 1.3 V
Vo
50 kW
(B) 4.7 V (D) 4.2 V
10 kW VBB
7. In the circuit shown in fig. P3.3.7 voltage VE = 4 V. The value of a and b are respectively +5 V
Fig. P3.3.11-12
11. VBB = 0
2 kW
(A) 2.46 V (C) 3.33 V
100 kW VE
(B) 1.83 V (D) 4.04 V
12. VBB = 1 V 8 kW
(A) 4.11 V (C) 2.46 V
(B) 1.83 V (D) 3.44 V
-5 V
13. VBB = 2 V
Fig. P3.3.7
(A) 0.943, 17.54 (C) 0.914, 11.63
(B) 0.914, 17.54 (D) 0.914, 11.63
(A) 3.18 V (C) 0.2 V
(B) 1.46 V (D) None of the above
Statement for Q.14-16:
Statement for Q.8-10: For the transistor in circuit shown in fig. P3.3.8-10, b = 200. Determine the value of I E and I C for given value of VB in question.
The transistor shown in the circuit of fig. P3.3.14-16 has b = 150. Determine Vo for given value of I Q in question. +5 V
+6 V
5 kW 10 kW
Vo
VC VB
IQ 1 kW
-5 V
Fig. P3.3.14-16
Fig. P3.3.8-10
14. I Q = 0.1 mA
8. VB = 0 V (A) 6.43 mA, 2.4 V (C) 0 A, 6 V Page 156
(B) 2.18 mA, 3.4 V (D) None of the above
(A) 1.4 V
(B) 4.5 V
(C) 3.2 V
(D) None of the above
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GATE EC BY RK Kanodia
Basic BJT Circuits
15. I Q = 0.5 mA (A) 3.16 V
(B) 2.52 V
(C) 2.14 V
(D) 394 . V
Chap 3.2
(A) 0.991
(B) 0.939
(C) 0.968
(D) 0.914
20. For the transistor in fig. P3.3.20 , b = 50. The value
16. I Q = 2 mA
of voltage VEC is
(A) 4.9 V
(B) -4.9 V
(C) 0.5 V
(D) -0.5 V
+9 V
1 mA
17. For the circuit in fig. P3.3.17 VB = VC and b = 50. The value of VB is
+6 V 50 kW
10 kW
4.7 kW
VC
-9 V
VB
Fig. P3.3.20 1 kW
Fig. P3.317
(A) 0.9 V
(B) 1.19 V
(C) 2.14 V
(D) 1.84 V
(A) 3.13 V
(B) 4.24 V
(C) 5.18 V
(D) 6.07 V
21. In the circuit shown in fig. P3.3.21 if b = 50, the power dissipated in the transistor is +9 V
18. For the circuit shown in fig. P3.3.18, VCB = 0.5 V and b = 100. The value of I Q is
0.5 mA
+5 V
5 kW Vo
50 kW
4.7 kW
-9 V
IQ
Fig. P3.3.21 -5 V
Fig. P3.3.18
(A) 1.68 mA
(B) 0.909 mA
(C) 0.134 mA
(D) None of the above
19. For the circuit shown in fig. P3.3.19 the emitter voltage is VE = 2 V. The value of a is
(A) 3.87 mW
(B) 10.46 mW
(C) 7.49 mW
(D) 18.74 mW
22. For the circuit shown in fig. P3.3.22 the Q-point is VCEQ = 12 V and I CQ = 2 A when b = 60. The value of resistor RC and RB are
+24 V
+10 V RC
RB
10 kW VE
50 kW
10 kW
-10 V
Fig. P3.3.19
Fig. P3.3.22
(A) 10 kW, 241 kW
(B) 10 kW, 699 kW
(C) 6 kW, 699 kW
(D) 6 kW, 241 kW
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GATE EC BY RK Kanodia
Basic BJT Circuits
Chap 3.2
29. For the transistor in the circuit of fig. P3.3.29,
32. The current gain of the transistor shown in the
b = 100. The voltage VB is
circuit of fig.P3.3.32 is b = 100. The values of Q-point ( I CQ , VCEQ) is
+10 V
+5 V
20 kW
1 kW 5 kW
12 kW
15 kW 0.5 kW
2 kW
Fig. P3.3.29 -5 V
(A) 3.6 V
(B) 4.29 V
(C) 3.9 V
(D) 4.69 V
Fig. P3.3.32
30. The current gain of the transistor shown in the
(A) (1.8 mA, 2.1 V)
(B) (1.4 mA, 2.3 V)
(C) (1.4 mA , 1.8 V)
(D) (1.8 mA, 1.4 V)
circuit of fig. P3.3.30 is b = 125. The Q-point values ( I CQ , VCEQ) are
33. For the circuit in fig. P3.3.33, let b = 60. The value of +24 V
VECQ is +5 V
+10 V
58 kW
42 kW
10 kW
2 kW
20 kW
2.2 kW
10 kW
Fig. P3.3.30
-5 V
(A) (0.418 mA, 20.4 V)
(B) (0.915 mA, 14.8 V)
(C) (0 .915 mA, 16.23 V)
(D) (0.418 mA, 18.43 V)
31. For the circuit shown in fig. P3.3.31, let b = 75. The
-10 V
Fig.P3.3.33
(A) 2.68 V
(B) 4.94 V
(C) 3.73 V
(D) 5.69 V
34. In the circuit of fig. P3.3.34 Zener voltage is VZ = 5
Q-point (I CQ , VCEQ) is +24 V
V and b = 100. The value of I CQ and VCEQ are +12 V
25 kW
3 kW 500 W
8 kW
1 kW
Fig. P3.3.34 Fig. P3.3.31
(A) ( 4.68 mA, 16.46 V)
(B) ( 312 . mA , 1.86 V)
(C) ( 312 . mA, 8.46 V)
(D) ( 4.68 mA , 5.22 V)
(A) 12.47 mA, 4.3 V
(B) 12.47 mA, 5.7 V
(C) 10.43 A, 5.7 V
(D) 10.43 A , 4.3 V
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UNIT 3
35. The two transistor in fig. P3.2.35 are identical. If b = 25, the current I C2 is
Analog Electronics
(A) 36.63 mA (C) 49.32 mA
(B) 36.17 mA (D) 49.78 mA
+5 V IC2
39. In the bipolar current source of fig. P3.2.39 the diode voltage and transistor BE voltage are equal. If
25 mA
base current is neglected then collector current is
Fig. P3.2.35
(A) 28 mA
(B) 23.2 mA
(C) 26 mA
(D) 24 mA
10 kW
36. In the shunt regulator of fig. P3.2.26, the VZ = 8.2 V
4.7 kW
and VBE = 0.7 V. The regulated output voltage Vo is
10 kW
120 W
Vo
+22 V
-20 V
Fig. P3.2.39
(A) 6.43 mA (C) 1.48 mA
100 W
(B) 2.13 mA (D) 9.19 mA
40. In the current mirror circuit of fig. P3.2.40. the transistor parameters are VBE = 0.7 V, b = 50 and the Early voltage is infinite.
Fig. P3.2.36
(A) 11.8 V
(B) 7.5 V
(C) 12.5 V
(D) 8.9 V
Assume transistor are
matched. The output current I o is +5 V
37. In the series voltage regulator circuit of fig. P3.2.37
1 mA
VBE = 0.7 V, b = 50, VZ = 8.3 V. The output voltage Vo is +25 V
Io
Vo 220 W 20 kW
50 kW
Fig. P3.2.40 50 kW
(A) 1.04 mA (C) 962 mA
30 kW
(B) 1.68 mA (D) 432 mA
41. All transistor in the N output mirror in fig. P3.2.41 are matched with a finite gain b and early voltage
Fig. P3.2.37
(A) 25 V
(B) 25.7 V
(C) 15 V
(D) 15.7 V
V A = ¥. The expression for each load current is V
38. In the regulator circuit of fig. P3.2.38 VZ = 12 V, b = 50, VBE = 0.7 V. The Zener current is +20 V
+
Io1 Iref
Io2
Io3
R1
Vo
QS
220 W
QR
1 kW
Q1
V
Q2
-
Fig. P32.41 Fig. P3.2.38
Page 160
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QN
GATE EC BY RK Kanodia
Basic BJT Circuits
(A)
(C)
I ref
(B)
æ (1 + N ) ö çç 1 + ÷ b (b + 1) ÷ø è b I ref
(D)
æ (1 + N ) ö çç 1 + ÷ (b + 1) ÷ø è
I ref
Chap 3.2
SOLUTIONS
æ N ö çç 1 + ÷ ( b + 1) ÷ø è
1. (C) I E =
b I ref æ N ö çç 1 + ÷ + 1 ÷ø b è
12 - 0.7 10 k
Þ
. mA I E = 113
æ 75 ö I C = çç . ) = 112 . mA ÷÷(113 è 75 + 1 ø
42. Consider the basic three transistor current source in
VCE = 12 - 1.13 ´ 10 - 1.12 RC - ( -12) = 6 V
fig. P3.2.42. Assume all transistor are matched with
RC = 5.98 kW
finite gain and early voltage V A = ¥. The expression for I o is V
2. (C) 8 = 10 ´ (75 + 1) I B + 0.7 + 10 I B - 2
+
IB = Io
I C = bI B = 0.906 mA, I E = (b + 1) I B = 0.918 mA
R1
Iref
9.3 = 12.08 mA, 10 + 760
8 = 10(0.918) + VEC + 3(0.906) - 8 Þ
VEC = 4.1 V
75 æ 75 ö 3. (A) I C = ç (1m) = 0.987 mA ÷ IE = 76 è 75 + 1 ø V
-
RC =
Fig. P3.2.42
(A)
(C)
I ref
(B)
æ 2 ö çç 1 + ÷ (1 + b) ÷ø è I ref
(D)
æ ö 2 çç 1 + ÷÷ b ( 1 + b ) è ø
I ref æ 1 ö çç 1 + ÷ (2 + b) ÷ø è
4. (A) 5 = (1 + b)10 kI B + 20 kI B + 0.7 + b2 kI B 5 = (760 k + 20 k + 150 k) I B + 0.7
I ref
Þ
æ ö 1 çç 1 + ÷÷ b ( 2 + b ) è ø
VC = 5 - (b + 1) I B RC = 5 - 760 ´ 4.62 ´ 10 -3 = 1.49 V
Both of transistor are identical and b >> 1 and VBE1 = 0.7 V. The value of resistance R1 and RE to produce I ref = 1 mA and I o = 12 mA is ( Vt = 0.026)
5. (C) VB = - I B RB Þ
IB =
-VB 1 = = 2.0 mA RB 500k
VE = -1 - 0.7 = - 17 . V V - ( -3) -17 . +3 IE = E = = 0.271 mA 4.8 k 4.8 k IE 0.271m = (b + 1) = IB 2m
+5 V Iref Io Q1
I B = 4.62 mA,
I C = bI B = 0.347 mA
43. Consider the wilder current source of fig. P3.2.43.
R1
5.2 = 304 . kW 0.987m
Þ
Q2
b = 134.5
6. (B) VCE = 3 - VE = 3 - ( -17 . ) = 4.7 V
RE
-5 V
7. (C) I E =
Fig. P3.2.43
5.4 = 0.5 mA 2k
(A) 9.3 kW, 18.23 kW
(B) 9.3 kW , 9.58 kW
4 = 0.7 + I B RB + I C RC - 5, I C » I E ,
(C) 15.4 kW , 16.2 kW
(D) 15.4 kW , 32.4 kW
8.3 = 100 I B + 0.5 ´ 8
*****************
Þ
I B = 43 mA , IE 0.5m =1 + b = = 11.63 IB 43m
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UNIT 3
b = 10.63, a =
b ® a = 0.914 1+b
Vo = VCE ( sat ) - VBE = 0.2 - 0.7 = - 0.5 V 17. (B) I E =
8. (C) VB = 0 Transistor is in cut-off region.
VB - 0.7 1k
æ b ö æ 50 ö I C = çç ÷÷I E = ç ÷ ( VB - 0.7) mA è 51 ø èb + 1ø
I E = 0, VC = 6 V 9. (B) VB = 1 V , I E =
Analog Electronics
1 - 0.7 = 0.3 mA 1k
IC =
I C » I E = 0.3 mA
6 - VC mA, VC = VB 10
50 6 - VB ( VB - 0.7) = 51 10
VC = 6 - I C RC = 6 - (0.3)(10) = 3 V
10.8 VB = 12.86 , VB = 119 . V
2 - 0.7 10. (B) VB = 2 V, I E = = 1.3 mA, 1
18. (B) VCB = 0.5 V , VC = 0.5 V
I C » I E = 1.3 mA VC = 6 - (1.3)(10) = - 7 V Transistor is in saturation. The saturation voltage
IC =
5 - 0.5 æ 101 ö =0.9 mA, I Q = ç ÷0.9 = 0.909 mA 5k è 100 ø
VCE = 0.2 V 19. (C) I E =
11. (C) VBB = 0, Transistor is in cutoff region
VB = VE - 0.7 = 1.3 V
RL 10(5) Vo = VCC = + 5 = 3.33 V RC + RL 10
IB =
12. (B) I B =
1 - 0.7 = 6 mA 50 k
I C = bI B = 75 ´ 6m = 0.45 mA 5 - Vo V = IC + o 5k 10 k Vo Vo Vo = 1.83 V (1 - 0.45) = + , Þ 5 10 13. (C) I B =
2 - 0.7 = 26 mA 50 k
I C = b I B = 75 ´ 26 mA =1.95 mA . = -4.75 V VC = 5 - I C RC = 5 - 5 ´ 195 Transistor is in saturation, VCE = 0.2 V = VC = Vo 14. (B) I E = 0.1 mA IC =
b 150 IE = (0.1) = 0.099 mA (b + 1) 151
VB 1.3 = = 26 mA RB 50 k
b+1= a=
I E 0.8m = = 30.77 I B 26m
Þ
b = 29.77
b 29.77 = = 0.968 b + 1 30.77
æ b ö 50 20. (D) I C = çç mA = 0.98 mA I E ÷÷ = b + 1 è ø 51 VC = I C RC - 9 = (0.98)( 4.7) - 9 = - 4.394 V IB =
IE 1 = mA = 19.6 mA (b + 1) 51
VE = I B RB + VEB = 50(0.0196) + 0.7 = 1.68 V VEC = 1.68 - ( -4.394) = 6.074 V æ b ö 50 21. (A) I C = çç (0.5) = 0.49 mA ÷÷I E = b + 1 51 è ø
Vo = 5 - RC I C = 5 - 5(0.099) = 4.50 V
IB =
15. (B) I E = I Q = 0.5 mA
VE = I B RB + VEB = (0.0098)(50) + 0.7 = 1.19 V
æ 150 ö IC = ç ÷ (0.5m) = 0.497 mA è 150 + 1 ø
VC = I C RC - 9 = (0.49)( 4.7) - 9 = - 6.7 V VEC = 119 . - ( -6.7) = 7.89 V
Vo = 5 - RC I C = 2.517 V
PQ = I C VEC + I B VEB
16. (D) Transistor is in saturation Page 162
10 - VE = 0.8 mA 10k
. V VE = (1.3)(1) = 1.3 V , VC = VCE + VE = 15
0.5 = 9.8 mA 51
= (0.49)(7.89) + (0.0098)(0.7) mW = 3.87 mW www.gatehelp.com
GATE EC BY RK Kanodia
UNIT 3
æ R2 VTH = çç è R1 + R2
ö æ 15 ö ÷÷VCC = ç ÷(10) = 4.29 V è 20 + 15 ø ø
10 = I E (1k) + VEB + 10 = I E + 0.7 + Þ IB =
IE ( 8.57 k) + 4.29 b+1
Analog Electronics
5.82 - 0.7 = ( 6.06 k + 76 k) I BQ Þ
I BQ = 62.4 mA
I EQ = (b + 1) I BQ = 4.74 mA I CQ = bI BQ = 4.68 mA VCEQ = 24 - I CQ RC - I EQ RE
IE ( 8.57 k) + 4.29 101
= 24 - ( 4.68)( 3) - ( 4.74)(1) = 5.22 V
I E = 4.62 MA
32. (B) R1 = 12 kW, R2 = 2 kW
IE = 0.046 mA b+1
RTH = R1||R2 = 12 ||2 = 171 . kW æ 2 ö VTH = ç . V ÷(10) - 5 = - 357 è 12 + 2 ø
VB = ( 8.57)(0.046) + 4.29 = 4.69 V 30. (B) R1 = 58 kW, R2 = 42 kW
+5 V
+24 V 5 kW 1.71 kW -3.57 V
24.36 kW +10.1 V 0.5 kW
10 kW -5 V
Fig. S3.3.32 Fig. S3.3.30
-357 . = I BQ(171 . k) + VBE + (b + 1) I BQ(0.5 k) - 5
RTH = 58 ||42 = 24.36 kW VTH
5 - 357 . - 0.7 = (171 . + 50.5)I BQ
æ 42 ö =ç ÷(24) = 10.1 V è 42 + 58 ø
Þ
10.1 = I BQ(24.36 k) + VBE + (b + 1) I BQ(10 k) 10.1 - 0.7 = I BQ(24.36 k + 1260 k)
I BQ = 14 mA
I EQ = (100 + 1) I BQ = 1.412 mA I CQ = 100 I BQ = 1.4 mA VCEQ = 5 - RC I CQ - RE I EQ + 5
I BQ = 7.32 mA
= 5 - (5)(1.4) - (0.5)(1.412) + 5 = 2.3 V
I CQ = bI BQ = 0.915 mA I EQ = (b + 1) I BQ = 0.922 mA
33. (B) RTH = 20 ||10 = 6.67 kW
VCEQ = 24 - (0.922)(10) = 14.8 V
æ 20 ö VTH = ç ÷10 - 5 = 1.67 V è 10 + 20 ø
31. (D) R1 = 25 kW, R2 = 8 kW
+10 V +24 V 2 kW 3 kW 6.67 kW
6.06 kW
+1.67 V
+5.82 V 2.2 kW 1 kW -10 V
Fig. S3.3.33
Fig. S3.3.31
RTH = 25 ||8 = 6.06 kW, VTH
æ 8 ö =ç ÷(24) = 5.82 V è 25 + 8 ø
5.82 = ( 6.06 k)( I BQ) + VBE + (b + 1) I B (1k) Page 164
10 = (1 + b) I BQ(2) + VEB + I BQ( 6.67) + 1.67 10 - 1.67 - 0.7 = I BQ( 6.67 + 122) www.gatehelp.com
GATE EC BY RK Kanodia
UNIT 3
41. (A) I ref = I CR + I BS = I CR +
æ I ref I o RE = Vt ln çç è Io
I ES 1+b
I ES = I BR + I B1 + I B 2 + ....... I BN I BR = I Bi , I CR = I Ci = I oi (1 + N ) I CR I ES = (1 + N ) I BR = b Then I ref = I CR +
æ 1 ´ 10 -3 0.026 ln çç -6 -6 12 ´ 10 è 12 ´ 10
R1 =
V + - VBE1 - V - 5 - 7 - ( -5) = = 9.3 kW I ref 1m
(1 + N ) I CR I ES = I CR + b (b + 1) b+1
+
V
Iref
Io = IC2 Q3
IC1
IE3 Q2
Q1
IB1 IB2
V
-
Fig. S3.2.42
I E 3 = (1 + b) I B 3 2 IB2 IE3 I ref = I C1 + = I C1 + (1 + b) (1 + b) I C1 = I C 2 = bI B 2 æ ö 2 IC2 2 = I C2 çç 1 + ÷÷ b(1 + b) b ( 1 + b ) è ø I ref
æ ö 2 çç 1 + ÷÷ 1 b ( + b ) è ø
43. (B) If b >> 1 and transistor are identical VBE1
VBE2
I ref » I C1 = I S e
Vt
, Io = IC2 = IS e
Vt
æ I ref VBE1 = Vt lnçç è IS
ö æI ÷÷ , VBE 2 = Vt lnçç o è IS ø
ö ÷÷ ø
æ I ref VBE1 - VBE 2 = Vt ln çç è Io
ö ÷÷ ø
From the circuit, VBE1 - VBE 2 = I E 2 RE » I o RE
Page 166
ö ÷÷ = 9.58 kW ø
***********
42. (C) I ref = I C1 + I B 3 , I B1 = I B 2 , I E 3 = 2 I B 2
IC2 = Io =
ö ÷÷ ø
RE =
æ (1 + N ) ö = I Oi çç 1 + ÷ b(b + 1) ÷ø è I ref I oi = æ (1 + N ) ö çç 1 + ÷ b(b + 1) ÷ø è
I ref = I C 2 +
Analog Electronics
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GATE EC BY RK Kanodia
CHAPTER
3.3 BASIC FET CIRCUITS
Statement for Q.4–6:
Statement for Q.1–3: In the circuit shown in fig. P3.3.1–3 the transistor
parameter are as follows:
parameters are as follows: Threshold voltage
In the circuit shown in fig. P3.3.4–6 the transistor
VTN = 2 V
Conduction parameter K n = 0.5 mA / V
VTN = 2 V, k¢n = 60 mA / V 2 ,
2
W = 60 L
+10 V
+10 V
32 kW
4 kW
14 kW
1.2 kW
18 kW
2 kW
6 kW
0.5 kW
-10 V
Fig. P3.3.1–3
Fig. P3.3.4–6
4. VGS = ?
1. VGS = ? (A) 2.05 V
(B) 6.43 V
(A) -3.62 V
(B) 3.62 V
(C) 4.86 V
(D) 3.91 V
(C) -0.74 V
(D) 0.74 V
5. I D = ?
2. I D = ? (A) 1.863 mA
(B) 1.485 mA
(C) 0.775 mA
(D) None of the above
(A) 13.5 mA
(B) 10 mA
(C) 19.24 mA
(D) 4.76 mA
6. VDS = ?
3. VDS =? (A) 4.59 V
(B) 3.43 V
(A) 2.95 V
(B) 11.9 V
(C) 5.35 V
(D) 6.48 V
(C) 3 V
(D) 12.7 V
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Basic FET Circuits
Chap 3.3
16. The parameter of the transistor in fig. P3.3.16 are
19. In the circuit of fig. P.3.3.19 the PMOS transistor
VTN = 1.2 V, K n = 0.5 mA / V 2 and l = 0. The voltage VDS
has parameter VTP = -1.5 V, kp¢ = 25 mA / V 2 , L = 4 mm
is
and l = 0. If I D = 0.1 mA and VSD = 2.5 V, then value of W will be
+5 V
+9 V
50 mA
R
Fig. P3.3.16
(A) 1.69 V
(B) 1.52 V
(C) 1.84 V
(D) 0
Fig. P3.3.19
17. The parameter of the transistor in fig. P3.3.17 are VTN = 0.6 V and K n = 0. 2 mA / V 2 . The voltage VS is
(A) 15 mm
(B) 1.6 mm
(C) 32 mm
(D) 3.2 mm
20. The PMOS transistor in fig. P3.3.20 has parameters
+9 V
VTP = -1.2 V,
24 kW
W = 20, and k¢p = 30 mA / V 2 . L +5 V
Rs
0.25 mA
RD
-9 V
Fig. P3.3.17 -5 V
(A) 1.72 V
(B) -1.72 V
(C) 7.28 V
(D) -7.28 V
Fig. P3.3.20
If I D = 0.5
mA and VD = -3 V, then value of RS
18. In the circuit of fig. P3.3.18 the transistor
and RD are
parameters are VTN = 17 . V and K n = 0.4 mA / V 2 .
(A) 4 kW, 5.8 kW
(B) 4 kW, 5 kW
(C) 5.8 kW, 4 kW
(D) 5 kW, 4 kW
+5 V
21. The parameters for the transistor in circuit of fig.
RD
P3.3.21 are VTN = 2 V and K n = 0.2 mA / V 2 . The power dissipated in the transistor is +10 V
RS
50 kW
-5 V
Fig. P3.3.18 10 kW
If I D = 0.8 mA and VD = 1 V, then value of resistor Fig. P3.3.21
RS and RD are respectively (A) 2.36 kW, 5 kW
(B) 5 kW, 2.36 kW
(C) 6.43 kW, 8.4 kW
(D) 8.4 kW, 6.43 kW
(A) 5.84 mW
(B) 2.34 mW
(C) 0.26 mW
(D) 58.4 mW
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UNIT 3
Statement for Q.22–23:
Analog Electronics
25. The transistors in the circuit of fig. P3.3.25 have
Consider the circuit shown in fig. P3.2.22–33.
parameter VTN = 0.8 V, kn¢ = 40 mA / V 2 and l = 0. The width-to-length ratio of M 2
+5 V
when Vi = 5 V, then
( )
W L 1
is
( WL )2 = 1.
If Vo = 0.10 V
for M1 is +5 V
M1 Vo
M1
M2
Vo
The both transistor have parameter as follows VTN = 0.8 V,
M2
Vi
Fig. P3.3.22-23
Fig. P3.3.25
k¢n = 30 mA / V 2
22. If the width-to-length ratios of M1 and M 2 are æW ö æW ö ç ÷ = ç ÷ = 40 è L ø1 è L ø2
(A) 47.5
(B) 28.4
(C) 40.5
(D) 20.3
Statement for Q.26–27:
The output Vo is
All transistors in the circuit in fig. P3.3.26–27
(A) -2.5 V
(B) 2.5 V
(C) 5 V
(D) 0 V
have parameter VTN = 1 V and l = 0. +5 V
æW ö æW ö 23. If the ratio is ç ÷ = 40 and ç ÷ = 15, then Vo is è L ø1 è L ø2 (A) 2.91 V
(B) 2.09 V
(C) 3.41 V
(D) 1.59 V
RD M4
M1 ID1
24. In the circuit of fig. P3.324. the transistor
RG
ID4 M3
M2
parameters are VTN = 1 V and k¢n = 36 mA / V 2 . If I D = 0.5 mA, V1 = 5 V and V2 = 2 V then the width to-length
-5 V
Fig. P3.3.26–27
ratio required in each transistor is +5 V
The conduction parameter are as follows: M1
K n1 = 400 mA / V 2 V1
K n 2 = 200 mA / V 2
M2
K n 3 = 100 mA / V 2
V2
K n 4 = 80 mA / V 2
M3
Fig. P3.3.24
Page 170
26. I D1 = ? (A) 0.23 mA
(B) 0.62 mA
(C) 0.46 mA
(D) 0.31 mA
æW ö ç ÷ è L ø1
æW ö ç ÷ è L ø2
æW ö ç ÷ è L ø3
(A)
1.75
6.94
27.8
27. I D4 = ?
(B)
4.93
10.56
50.43
(A) 0.62 mA
(B) 0.31 mA
(C)
35.5
22.4
8.53
(C) 0.46 mA
(D) 0.92 mA
(D)
56.4
38.21
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GATE EC BY RK Kanodia
Basic FET Circuits
Chap 3.3
28. For the circuit in fig. P3.3.28 the transistor
(A) 7.43 V
(B) 8.6 V
parameter are VTN = 0.8 V and kn¢ = 30 mA / V 2 . If output
(C) -1.17 V
(D) 1.17 V
voltage is Vo = 0.1 V, when input voltage is Vi = 4.2 V, the required transistor width-to length ratio is
32. A p-channel JFET biased in the saturation region with VSD = 5 V has a drain current of I D = 2.8 mA, and
+5 V
I D = 0.3 mA at VGS = 3 V. The value of I DSS is 10 kW Vo
(A) 10 mA
(B) 5 mA
(C) 7 mA
(D) 2 mA
Vi
Statement for Q.33–34: For the p-channel transistor in the circuit of fig.
Fig. P3.3.28
(A) 1.568
(B) 0.986
(C) 0.731
(D) 1.843
P3.3.33–34 the parameters are I DSS = 6 mA, VP = 4 V and l = 0. 1 kW
29. For the transistor in fig. P3.3.29 parameters are VTN = 1 V and K n = 12.5 mA / V 2 . The Q-point ( I D , VDS ) is +10 V
0.4 kW
20 kW -5 V
Fig. P3.3.33–34
10 kW
33. The value of I DQ is Fig. P3.3.29.
(A) 8.86 mA
(B) 6.39 mA
(C) 4.32 mA
(D) 1.81 mA
(A) (1 mA, 8 V)
(B) (0.2 mA, 4 V)
34. The value of VSD is
(C) (1.17 mA, 8 V)
(D) (0.23 mA, 3.1V)
(A) -4.28 V
(B) 2.47 V
30. For an n-channel JFET, the parameters are I DSS = 6
(C) 4.28 V
(D) 2.19 V
mA and VP = -3 V. If VDS > VDS ( sat ) and VGS = -2 V, then
35. The transistor in the circuit of fig. P3.3.35 has
I D is
parameters I DSS = 8 mA and VP = -4 V. The value of VDSQ
(A) 16.67 mA
(B) 0.67 mA
(C) 5.55 mA
(D) 1.67 mA
is
31. For the circuit in fig. P3.3.32 the transistor
+20 V
140 kW
2.7 kW
parameters are Vp = - 35 . V, I DSS = 18 mA, and l = 0. The value of VDS is
+15 V 60 kW
2 kW
0.8 kW
Fig. P3.3.35 IQ = 8 mA
(A) 2.7 V
(B) 2.85 V
(C) -1.30 V
(D) 1.30 V
-15 V
Fig. P3.3.32
******************
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GATE EC BY RK Kanodia
UNIT 3
10 - ( 4.67 + VSG) = (0.5)(1)( VGS )
SOLUTIONS 1. (A) R1 = 32 kW, æ R2 VG = çç è R1 + R2
R2 = 18 kW,
Þ
VDD = 10 V
ö æ 18 ö ÷÷VDD = ç ÷10 = 3.6 V è 18 + 32 ø ø
=
Þ
VGS = 2.05 V
10. (C) Assume transistor in saturation. I D = 0.4 mA, VG = 0,
Þ
VSG = 2 + 0.8 = 2.21 V
VSG = VS - VG = VS
11. (A) VD = I D RD - 5 = (0.4)(5) - 5 = -3 V
= 10 - 0.775( 4 + 2) = 5.35 V
VSD = VS - VD = 2. 21 - ( -3) = 5. 21 V
VDS ( sat ) = VGS - VTN = (2.05 - 0.8) = 1. 25 V VDS > VDS ( sat ) as assumed.
12. (C) R1 = 14.5 kW, R2 = 5.5 kW,
4. (B) R1 = 14 kW, R2 = 6 kW, RS = 0.5 kW, RD = 1.2 kW
RS = 0.6 kW, RD = 0.8 kW, æ RL VG = çç è R1 + R2
ö æ 6 ö ÷÷(20) - 10 = ç ÷(20) - 10 = -4 V è 14 + 6 ø ø
ö 5.5 æ ö ÷÷(10) - 5 = ç ÷(10) - 5 = -2.25 V è 14.5 + 5.5 ø ø
Assume transistor in saturation. V - ( -5) ID = S = K n ( VGS - VTN ) 2 RS
Assume transistor in saturation V - ( -10) VG - VGS + 10 ID = S = = K n ( VGS - VTN ) 2 RS RS
VS = VG - VGS
kn¢ W ( 60)( 60 ´ 10 -6 ) = = 1.8 mA / V 2 2 L 2
-2.25 - VGS + 5 = (0.6)(0.5) ( VGS - ( -1)) 2
Þ
- 4 - VGS + 10 = (0.5)(1.8)( VGS - 2) 2
Þ
Þ
VGS = 3.62, - 0.74 V, VGS will be positive.
VGS is positive. Thus (D) is correct option.
5. (D) I D =
VG - VGS + 10 -4 - 3.62 + 10 = = 4.76 mA RS 0.5 k
VGS = 124 . , - 6.58 V
13. (D) I D =
VS + 5 VG - VGS + 5 -2.25 - 124 . +5 = = RS RS 0.6 k
= 2.52 mA, Therefore (D) is correct option.
6. (B) 10 = I D( RS + RD) + VDS - 10 VDS = 20 - 4.76(12 . + 0.5) = 119 . V
14. (B) 5 = I D( RS + RD) + VDS - 5
VDS ( sat ) = VGS - VTN = 3.62 - 2 = 1.62 V
VDS = 10 - I D( RS + RD) = 10 - 2.52(0.8 + 0.6) = 6.47 V
. V > VDS ( sat ) , Assumption is correct. VDS = 119
VDS ( sat ) = VGS - VTH = 124 . - ( -1) =2.24
7. (B) R1 = 8 kW, RL = 22 kW, RS = 0.5 kW, RD = 2 kW æ R2 VG = çç è R1 + R2
ö æ 22 ö ÷÷(20) - 10 = ç ÷(20) - 10 = 4.67 V è 8 + 22 ø ø
Assume transistor in saturation 10 - VS ID = = K P ( VSG + VTP ) 2 RS VS = VG + VSG
VDS > VDS ( sat ) ,Assumption is correct. 15. (B) I S = 50 mA = I D ,I D = K n ( VGS - VTN ) 2 Þ
50 ´ 10 -6 = 0.5 ´ 10 -3( VGS - 12 . )2
V VG = 0, VS = VG - VGS = -1516 . VDS = VD - VS = 5 - ( -1516 . ) = 6.516 V 16. (B) I D = 50m = K n ( VGS - VTN ) 2 Þ
Page 172
0.4 = K P ( VGS + VTP ) 2
0.4 = (0.2)( VSG - 0.8) 2
3. (C) VDS = VDD - I D( RD + RS )
Kn =
10 - ( 4.67 + 377 . ) = 312 . mA 0.5
VSD = 20 - I D ( RS + RD)= 20 - 2.12(2 + 0.5) = 12.2 V
V - VGS 3.6 - 2.05 2. (C) I D = G = = 0.775 mA RS 2k
æ RL VG = çç è R1 + R2
10 - VS 10 - ( VG + VGS ) = RS RS
9. (C) 10 = I D( RS + RD) + VSD - 10
K n = 0.5 mA / V 2
3.6 - VGS = (2)(0.5)( VGS - 0.8) 2
VSG = 377 . V, - 177 . V, VSG is positive voltage.
8. (A) I D =
Assume that transistor in saturation region V V - VGS ID = S = G = K n ( VGS - VTN ) 2 RS RS RS = 2 kW,
Analog Electronics
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50 ´ 10 -6 = 0.5 ´ 10 -3( VGS - 12 . )2
Þ
V, VGS = 1516 .
GATE EC BY RK Kanodia
Basic FET Circuits
VGS = 152 . V,
æW ö æW ö 23. (A) ç ÷ > ç ÷ thus VGS1 < VGS 2 è L ø1 è L ø2
VGS = VDS
17. (B) I D = K n ( VGS - VTN ) 2 Þ
0.25 = 0.2( VGS - 0.6)
VGS = VG - VS , 18. (A) I D =
Þ
2
VG = 0,
40( VGS1 - 0.8) 2 = 15( VGS 2 - 0.8) 2
V, VGS = 172 .
VGS 2 = 5 - VGS1
V VS = -172 .
5 - VD = 0.8 mA, RD
RD =
1.63( VGS1 - 0.8) = (5 - VGS1 - 0.8) VGS1 = 2.09, VGS 2 = 2.91 V, Vo = VGS 2 = 2.91 V
6 -1 = 5 kW 0.8m
24. (A) Each transistor is biased in saturation because
I D = K n ( VGS - VTN ) 2 Þ 0.8 = (0.4)( VGS - 17 . )2
Þ
VDS = VGS and VDS > VGS - VTN
VGS = 311 . V
For M 3 , V2 = 2 V = VGS 3
VGS = VG - VS , VG = 0, VS = -311 . V -311 . - ( -5) I D = 0.8 mA = Þ RS = 2.36 kW RS
æ 25 ö æ W 10 -4 = ç ÷ç è 2 øè 4
ö . )2 ÷ (2.5 - 15 ø
æ 30 ´ 10 -6 20. (D) K p = çç 2 è I D = K p ( VSG + VTP ) 2 Þ
æ 36 ´ 10 -3 öæ W ö ÷÷ç ÷ (2 - 1) 2 I D = 0.5 = çç 2 øè L ø3 è
Þ
æ 36 ´ 10 -3 öæ W ö ÷÷ç ÷ ( 3 - 1) 2 I D = 0.5 = çç 2 øè L ø2 è
25. (D) M 2 is in saturation because
0.5 = 0.3( VSG - 12 . )2
VGS 2 = VDS 2 > VGS 2 - VTN
VSG = 2.49 V, VG = 0
VD - ( -5) RD
Þ RD =
M1 is in non saturation because VGS1 = Vi = 5 V, VDS1 = VD = 0 V VDS1 < VGS1 - VTN , I D1 = I D2 æW ö æW ö 2 2 ç ÷ [2( VGS1 - VTN1 ) VDS1 - VDS 2 ]= ç ÷ ( VGS 2 - VTN 2 ) L L è ø1 è ø2
-3 + 5 = 4 kW 0.5m
æW ö Þ ç ÷ [2(5 - 0.8)(0.1) - (0.1) 2 ] = (1)(5 - 0.1 - 0.8) 2 è L ø1
21. (B) Assume transistor in saturation 10 - VGS ID = = K n ( VGS - VTN ) 2 10 k
æW ö ç ÷ (0.83) = 16.81 è L ø1
10 - VGS = (10)(0.2)( VGS - 2) 2 Þ
. V VGS = VDS = 377 10 - 377 . ID = = 0.623 mA 10 k
VGS1 = 5 - VGS 2 Þ
VDS > VGS - VTN assumption is correct.
transistor
I D1 = I D2
Þ
are
K n 1 ( VGS1 - VTN1 ) = K n 2 ( VGS 2 - VTN 2 )
K n 1 = K n 2 , VTN1 = VTN 2 5 VGS1 = VGS 2 = V 2 Vo = VGS 2 = 2.5 V
-6
(5 - VGS 2 - 1) 2 = 200 ( VGS 2 - 1) 2 VGS1 = 2.24 V
(2.24 - 1) 2 =0.62 mA
I D4 = K n 4 ( VGS 4 - VTN ) 2 = K n 3 ( VGS 3 - VTN ) 2 in
saturation. 2
æW ö ç ÷ =20.3 è L ø1
27. (B) VGS 2 = VGS 3 = 2.76 V
22. (B) For both transistor VDS = VGS , both
Þ
VGS 2 = 2.76 V,
I D1 = 400 ´ 10
Power = I DVDS = 2.35 mW
Therefore
Þ
26. (B) I D1 = K n 1 ( VGS1 - VTN ) 2 = K n 2 ( VGS 2 - VTN ) 2
. V, - 0.27 V, VGS will be 3.77 V VGS = 377
VDS > VGS - VTN
æW ö ç ÷ = 6.94 è L ø2
æ 36 öæ W ö æW ö I D = 0.5 = ç ´ 10 -5 ÷ç ÷ (5 - 1) 2 Þ ç ÷ = 174 . è 2 øè L ø1 è L ø1
VS = VSG = 2.49 V 5 - VS 5 - 2.49 ID = Þ RS = = 5.02 kW RS 0.5m ID =
Þ
For M1 , VGS1 = 10 - V1 = 10 - 5 = 5 V
W = 32 mm
ö ÷÷(20) = 0.3 mA / V 2 ø
Þ
æW ö ç ÷ = 27.8 è L ø3
Þ
For M 2 , VGS 2 = V1 - V2 = 5 - 2 = 3 V
k¢p W ( VGS + VTP ) 2 2 L
19. (C) VSD = VSG, I D =
Chap 3.3
= 100 ´ 10 -6 (2.76 - 1) 2 = 0.31 mA 28. (C) VGS = 4.2 V,
2
VDS = 0.1 V
VDS < VGS - VTN , Thus transistor is in non saturation. 5 - 0.1 ID = = 0.49 mA 10 k k¢ W ID = n {2 ( VGS - VTN ) VDS - VDS2 } 2 L
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GATE EC BY RK Kanodia
CHAPTER
3.4 AMPLIFIERS
1. If the transistor parameter are b = 180 and Early
3. The small signal votlage gain Av = vo vs is
voltage V A = 140 V and it is biased at I CQ = 2 mA, the
(A) -4.38
(B) 4.38
values of hybrid-p parameter g m , rp and ro are
(C) -1.88
(D) 1.88
respectively 4. The nominal quiescent collector current of a
(A) 14 A V, 2.33 kW, 90 kW
transistor is 1.2 mA. If the range of b for this transistor
(B) 14 A V, 90 kW , 2.33 kW
is 80 £ b £ 120 and if the quiescent collector current
(C) 77 mA V, 2.33 kW , 70 kW
changes by ±10 percent, the range in value for rp is
(D) 77.2 A V, 70 kW, 2.33 kW
(A) 1.73 kW < rp < 2.59 kW (B) 1.93 kW < rp < 2.59 kW
Statement for Q.2–3.
(C) 1.73 kW < rp < 2.59 kW
Consider the circuit of fig. P3.4.2–3. The transistor
(D) 1.56 kW < rp < 2.88 kW
parameters are b = 120 and V A = ¥. Statement for Q.5–6: +5 V
Consider the circuit in fig. P3.4.5.6. The transistor parameter are b = 100 and V A = ¥.
4 kW 250 kW vs
+10 V
vo
~
RC 50 kW
2V
vs
Fig. P3.4.2–3
~
vBB
2. The hybrid-p parameter values of g m , rp and ro are (A) 24 mA V, ¥, 5 kW
Fig. P3.4.5–6
(B) 24 mA V, 5 kW , ¥
5. If Q-point is in the center of the load line and I CQ = 0.5
(C) 48 mA V, 10 kW , 18.4 kW
mA, the values of VBB and RC are
(D) 48 mA V, 18.4 kW, 10 kW
(A) 10 kW , 0.95 V
(B) 10 kW , 1.45 V
(C) 48 kW , 0.95 V
(D) 48 kW , 1.45 V
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GATE EC BY RK Kanodia
Amplifiers
Statement for Q.14–15:
Chap 3.4
19. For an n-channel MOSFET biased in the saturation
Consider the common Base amplifier shown in fig. P3.4.14–15. The parameters are
g m = 2 mS
and
region, the parameters are VTN = 1 V, 12 m n Cox = 18 mA V 2 and l = 0.015 V -1 and I DQ = 2 mA. If transconductance is
ro = 250 kW. Find the Thevenin equivalent faced by load
g m = 3.4 mA V, the width-to-length ratio is
resistance RL .
(A) 80.6
(B) 43.2
(C) 190
(D) 110
Thevenin equivalent
20. In the circuit of fig. P3.4.20, the parameters are
270 W vi
~
g m = 1mA / V, ro = 50 kW. The gain Av = vo vs is
RL
VDD
Fig. P3.4.14–15
60 kW
14. The Thevenin voltage vTH is (A) 263vi
(B) 132vi
(C) 346vi
(D) 498vi
2 kW
vs
~
300 kW
15. The Thevenin equivalent resistance RTH is (A) 384 kW
(B) 697 kW
(C) 408 kW
(D) 915 kW
Fig. P3.4.20
Statement for Q.16–17:
(A) -8.01
(B) 8.01
(C) 14.16
(D) -14.16
Statement for Q.21–23:
The common-base amplifier is drawn as a two-port in fig. P3.4.16–17. The parameters are b = 100, g m = 3 mS, and ro = 800 kW. +
For the circuit shown in fig. P3.4.21–23 transistor parameters are VTN = 2 V, K n = 0.5 mA / V 2 and l = 0. The transistor is in saturation.
i2
i1
v1
+ 3.9 kW
10 kW
18 kW
_
+10 V
v2
10 kW vo
_
Fig. P3.4.16–17
vi
16. The h-parameter h21 is
~
VGG
(A) 2.46
(B) 0.9
(C) 0.5
(D) 0.67
Fig. P3.4.21–23
21. If I DQ is to be 0.4 mA, the value of VGSQ is
17. The h-parameter h12 is (A) 3.8 ´10 -4
(B) 4.83 ´10 -3
(C) 3.8 ´10 4
(D) 4.83 ´10 3
18. For an n-channel MOSFET biased in the saturation region, the parameters are K n = 0.5 mA V 2 , VTN = 0.8 V and l = 0.01 V -1 , and I DQ = 0.75 mA. The value of g m and ro are
(A) 5.14 V
(B) 4.36 V
(C) 2.89 V
(D) 1.83 V
22. The values of g m and ro are (A) 0.89 mS, ¥
(B) 0.89 mS, 0
(C) 1.48 mS, 0
(D) 1.48 mS, ¥
23. The small signal voltage gain Av is
(A) 0.68 mS, 603 kW
(B) 1.22 mS, 603 kW
(A) 14.3
(B) -14.3
(C) 1.22 mS, 133 kW
(D) 0.68 mS, 133 kW
(C) -8.9
(D) 8.9
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Page 177
GATE EC BY RK Kanodia
Amplifiers
(A) 4.44
(B) -4.44
(C) 2.22
(D) -2.22
SOLUTIONS
the
source-follower
circuit
in
fig.
P3.4.33-34. The values of parameter are g m = 2 mS and ro = 100 kW. +5 V
Ro
vs
~
I CQ
1. (C) g m =
Statement for Q.33–34: Consider
Chap 3.4
IQ
500 kW
bVt b 180 = = = 2.33 kW I CQ g m 77.2m
ro =
V A 140 = = 70 kW I CQ 2m
gm =
Fig. P3.4.33-34
rp =
33. The voltage gain Av is (B) -0.89
(C) 2.79
(D) -2.79
2 - 0.7 = 5.2 m A 250 k
I CQ
=
Vt
0.624 = 24 mA V 0.0259
bVt b 120 = = = 5 kW, ro = ¥ I CQ g m 24m
ö æ rp bRC 3. (C) Av = - g m RC çç ÷÷ = r + R r B ø p + RB è p
34. The output resistance Ro is
æ ö 5k = - (24m)( 4 k) çç ÷÷ = -1.88 5 k + 250 k è ø
(A) 100 kW
(B) 0.498 kW
(C) 1.33 kW
(D) None of the above 4. (D) rp =
rp(min) =
bVT , I CQ
(120)(0.0259) = 2.88 kW, 108 . m
rp( max ) = *******************
2m = 77.2 mA V 0.0259
I CQ = bI B = (120)(5.2m ) = 0.642 mA
4 kW
-5 V
(A) 0.89
=
rp =
2. (B) I BQ =
vo
Vt
( 80)(0.0259) = 156 . kW 1.32m
5. (A) VECQ =
1 VCC = 5 V 2
VECQ = 10 - I CQ RC = 5 Þ
10 - (0.5m) RC = 5
RC = 10 kW, I BQ =
I CQ b
=
0.5 =5 mA 100
VEB ( ON ) + I BQ RB = VBB Þ
0.7 + (5m ) (50 k) = 0.95 V
6. (D) g m = rp =
I CQ Vt
=
0.5 = 19.3 mA V 0.0259
bVt (100)(0.0259) = = 5.18 kW , ro = ¥ 0.5m I CQ
æ 100 ö 7. (B) I CQ = ç ÷(0.35) = 0.347 mA è 1001 ø
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GATE EC BY RK Kanodia
UNIT 3
The small-signal equivalent circuit is as shown in fig. S3.4.7 500 W
vo +
~
vs
C
B
10 kW
rp
Vp _
gmVp
ro
7 kW
Analog Electronics
9. (A) DC Analysis: VTH =
RTH = 11||50 = 8.33 kW 12 - 0.7 - 10 I BQ = = 119 . mA 8.33k + (101)1k . mA, I EQ = 12 . mA I CQ = bI BQ = 119 . )2 = 8.42 V VECQ = 12 - (1. 20)1 - (119
E
AC Analysis: Ib
Fig. S3.4.7
rp vs
ö æ rp ||10 k vo ÷÷ ( ro ||7 k) = - g m çç vs 500 r | | 10 k + ø p è I CQ 0.347m gm = = = 1313 . mA V Vt 0.0259
rp =
288 ´ 7 7.6 ´ 10 = 6.83 kW, rp ||10k = = 4.32 kW 288 + 7 7.6 + 10
vo
10||50 kW
bIb E 2 kW (b+1)Ib
bVt (100)(0.0259 = = 2.18 kW 119 . m I CQ
vo = bI b (2k) , vs = - (b + 1) I b (1k) + I b ( rp) v -b(2 k) -(100)(2 k) Av = o = = = -196 . vs rp + (b + 1)1k 2.18 k + (100)(1k) 10. (B) VECQ = 8.42 V,
4.32 k æ ö Av = -1313 . mç ÷( 6.83k) = - 80 è 500 + 4.32 k ø
Þ Output voltage swing = 5.16 V peak to peak.
8. (C) DC Analysis: I CQ = I EQ
11. (B) Since the B–C junction is not reverse biased, the transistor continues to operate in the forward-active Þ
I CQ = 3.57 mA
-mode
AC Analysis: Ib
C
B
vo
+ rp vs
~
R1 || R2
Ie
+
+ vce _
vce
rp
gmvce
_
Vp bIb _ E
Fig. S 3.4.11 1.2 kW
bVt (0.0259) = (150) = 109 . kW , ro = ¥ 357 . m I CQ
æ 1 ö vce 1 , So rp || çç ÷÷ || ro = g m Vce g m è gm ø (100)(0.0259) rp = = 2.33 kW 2m I CQ 2m gm = = = 77.2 mA V Vt 0.0259
vo -(b I b ) RC , vs = I b rp + (b + 1) RE I b = vs vs
1 V 150 = 12.95 , ro = A = = 75 kW gm I CQ 2m
0.2 kW
Fig. S3.4.8
rp =
DvEC = 11 - 8.42 = 2.58 V
For 1 £ vEC £ 11 V,
VCEQ = 5 = 10 - I CQ ( RC + RE ) Þ 5 = 10 - I CQ(1. 2 k + 0. 2 k) 3.57 I BQ = = 23.8 m A 150
Av = Av =
Page 180
C
Fig. S3.4.9
VA 100 = = 288 kW I CQ 0.347m
ro ||7k =
~
Vp +
1 kW
b bV 100 = t = = 7.6 kW rp = . m g m I CQ 1313 ro =
Ic
B _
rp ||10 k ( vs ), vo = - g m Vp( ro ||7k) 500 + rp ||10 k
Vp =
50 (12) = 10 V 10 + 50
-bRC -(150)(12 . )k = = -5.75 rp + (1 + b) RE 109 . k + (151)(0.2 k)
r=
re = (2.33k)||(12.95)||(75 k) =12.87 W
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ro
GATE EC BY RK Kanodia
Amplifiers
VA I CQ
12. (C) ro =
Þ
I CQ =
Vp Vp v + Vpi + + g m Vp + i =0 rp ro 270
VA 75 = = 0.375 mA ro 200 k
vi + Vpi Vp Vp + + 2mVp + =0 50 k 250 k 270
bV 75(0.0259) 13. (B) rp = t = = 194 . kW I CQ 1m E
bIb
Ie Ib
C
2.7 kW
B
Þ
16. (B) The equivalent small-signal circuit is shown in
+ vo _
fig. S3.4.16
1.5 kW
ro
Fig. S 3.4.13
i1
vi = I b ( rp + 15k . ), I in = I e = (b + 1) I b Vi ( rp + 15 . k) 194 . + 15 . k Rin = = = = 45 W Ie (b + 1) 76
E
~
vs
3.9 kW rp
~
rp = gmVp
Vp +
vi rp Removing the RL , -Vp = 270 + rp
rp =
vi rp(1 + g m ro) 270 + rp
i2 i1
, i2 = v2 = 0
Vp V V - p - p - g m Vp , 39 . k rp ro
h21 =
i2 = i1
. k - g m rp 39 - gm = = 0.91 1 1 39 . k . k r + + g p m rp 39 + + gm 39 . k rp
i1 = 0
E
3.9 kW rp
fig. S3.4.15
C
Vp +
gmVp 18 kW
i2
~
v2
B
ro
Fig. S 3.4.17
270 W _ rp
gmVp
Vp +
Fig. S 3.4.15
I sc = g m Vp +
v1 v v - v2 + 1 + 1 = g m Vp 39 . k rp ro
_
15. (A) The equivalent small-signal circuit is shown in
~
Vp can be neglected ro
ro
vi 50 k(1 + (2m)(250 k)) = 498 vi 270 + 50 k
vi
Vp + g m Vp r0
i1 = -
17. (A) v1 = -Vp ,
b 100 = = 50 kW g m 2m
vTH =
v2 = 0
b 100 = = 33.3 kW gm 3m
h21 =
Fig. S 3.4.14
vTH = -ro g m Vp - Vp =
18 kW
Vp +
Fig. S 3.4.16
270 W _
gmVp
B
ro
rp
C
_
14. (D) The equivalent circuit is shown below
vi
Vp = -0.647 vi ,
. mvi I sc = 1297 vTH 498 vi RTH = = = 384 kW I SC 1297 . mvi
rp
~
vi
Chap 3.4
Vp Vp = 2.004 mVp = 2mVp + ro 250 k
Isc
æ 1 1 1ö v + + ÷÷ - 2 = - g m v1 v1 çç 39 . k r ro ø ro è p 1 1 v1 ro 800 k = = 1 1 1 1 1 1 v2 + + + 3m + + + gm 39 . k 33.3k 800 k 39 . k rp ro = 3.8 ´ 10 -4 www.gatehelp.com
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GATE EC BY RK Kanodia
Amplifiers
RS = 4 kW, v gs = 0.84 vi ,
v gs = vi ,
vo = - g m v gs ( ro ||RD)
Chap 3.4
vo = Av = - g m (7k) vi
g m = 2 K n ( VGS - VTN )
= -(1.41m)(0.84 vi )(100 k || 5 k) vo Þ = Av = - 5.6 vi
= 2 (1m)(151 . - 0.8) = 1.42 mS
28. (A) Ro = RD ||ro || 100k =4.76 kW
32. (A) The small-signal equivalent circuit is shown in
Av = -(1.42m) (7 k) = - 9.9
fig. S.3.4.34 29. (A) As shown in fig. S3.4.27, Ri = R1 || R2 = 20.6 kW
. V, I DQ = 0.5 mA VGSQ = 15
vi
g m = 2 K n ( VGS - VTN ) = 2(1m) (15 . - 0.8) = 1.4 mA V ro = [ lI DQ ]
vo
vgs
~
10 kW
RS
5 kW +
RD
RL
4 kW
G
=¥ Fig. S3.4.32
The resulting small-signal equivalent circuit is shown in fig. S5.4.30 G
D
+
~
vi
D
_
30. (C) From the DC analysis:
-1
gmvgs
S
vgs gmvgs
RTH
RD
vo = - g m v gs ( RD || RL ), vi = -v gs v Av = o = g m ( RD || RL ) = (2m)(5 k ||4 k) = 4.44 vi
vo 7 kW
_
33. (A) The small-signal equivalent circuit is shown in
S
fig. S3.4.33 RS
0.5 kW G
D
+
Fig. S 3.4.30
vi
vo = - g m v gs RD, vi = v gs + g m v gs RS vo - g m RD (7 k) Þ = = - (1.4m) = -5.76 vi 1 + g m RS 1 + (1.4m) (0.5 k)
~
ro
vgs gmvgs
500 kW
_ vo
S
4 kW
31. (B) Since the DC gate current is zero, VS = -VGSQ Fig. S3.4.33
I DQ = I Q = K n ( VGSQ - VTN ) 2 Þ
0.5 = 1( VGSQ - 0.8) 2
vo = g m v gs ( RL || ro)
VGSQ = 151 . V = -VS . ) = 301 . V VDSQ = 5 - (0.5m)(7 k) - ( -151 The transistor is therefore biased in the saturation region. The small-signal equivalent circuit is shown in fig.S3.4.31. D
G + vs
~
vgs _
gmvgs
vo
7 kW
vi = v gs + vo = v gs + g m v gs ( RL || ro) 1 v gs = 1 + g m ( RL || ro) vo g m ( RL || ro) = Av = vi 1 + g m ( RL || ro) RL ||ro = 4 k ||100 k = Av =
(2m)( 3.85 k) = 0.89 1 + (2m)( 3.85 k)
S
34. (B) Ro = Fig. S3.4.31
vo = - g m v gs (7k)
100 k = 3.86 kW 26
1 ||ro gm
1 = æç ö÷ ||(100) » 0.498 kW è2ø ********************
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GATE EC BY RK Kanodia
CHAPTER
3.5 OPERATIONAL AMPLIFIERS
1. Av =
vo =? vi
400 kW vi
(A) -2 sin wt m A
(B) -7 sin wt m A
(C) -5 sin wt m A
(D) 0
40 kW
4. In circuit shown in fig. P3.5.4, the input voltage vi is
vo
0.2 V. The output voltage vo is 50 kW
R
150 kW
10 kW vi
25 kW
Fig. P3.5.1
(A) -10
(B) 10
(C) -11
(D) 11
2. Av =
vo
Fig. P3.5.4
vo =? vi
400 kW 40 kW
vi
vo
(A) 6 V
(B) -6 V
(C) 8 V
(D) -8 V
5. For the circuit shown in fig. P3.5.5 gain is
60 kW
Av = vo vi = -10. The value of R is R
Fig. P3.5.2
(A) -10
(B) 10
(C) 13.46
(D) -13.46
100 kW vi
3.
The
input
to
the
circuit
in
fig.
100 kW
P3.5.3
100 kW vo
is
vi = 2 sin wt mV. The current io is 10 kW vi
Fig. P3.5.5
1 kW io vo
(A) 600 kW
(B) 450 kW
(C) 4.5 MW
(D) 6 MW
4 kW
Fig. P3.5.3
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Operational Amplifiers
6. For the op-amp circuit shown in fig. P3.5.6 the
Chap 3.5
10. In the circuit of fig. P3.5.10 the output voltage vo is
voltage gain Av = vo vi is
20 kW
R
R
20 kW
+0.5 V
R
40 kW -1 V
R
R
vi
vo
60 kW +2 V
R vo
Fig. P3.5.10
Fig. P3.5.6
(A) -8
(B) 8
(C) -10
(D) 10
(B) -2.67 V
(C) -6.67 V
(D) 6.67 V
11. In the circuit of fig. P3.5.11 the voltage vi1 is
7. For the op-amp shown in fig. P3.5.7 open loop differential gain is Aod = 10 3. The output voltage vo for vi = 2 V is
(A) 2.67 V
(1 + 2 sin wt) mV and vi2 = -10 mV. The output voltage vo is 20 kW 20 kW
100 kW
vi1 vi
2 kW 1 kW
100 kW vo
vo
vi2 1 kW
Fig. P3.5.11
Fig. P3.5.7
(A) -1.996
(B) -1.998
(A) -0.4(1 + sin wt) mV
(B) 0.4(1 + sin wt) mV
(C) -2.004
(D) -2.006
(C) 0.4(1 + 2 sin wt) mV
(D) -0.4(1 + 2 sin wt) mV
8. The op-amp of fig. P3.5.8 has a very poor open-loop
12. For the circuit in fig. P3.5.12 the output voltage is
voltage gain of 45 but is otherwise ideal. The closed-loop
vo = 2.5 V in response to input voltage vi = 5 V. The finite
gain of amplifier is
open-loop differential gain of the op-amp is
100 kW
vi
2 kW
500 kW
vo
vo
1 kW
vi
Fig. P3.5.8
(A) 20
(B) 4.5
(C) 4
(D) 5
Fig. P3.5.12
(A) 5 ´ 10 4
(B) 250.5
(C) 2 ´ 10
(D) 501
4
9. For the circuit shown in fig. P3.5.9 the input voltage 13. vo = ?
vi is 1.5 V. The current io is 10 kW vi
100 kW
6 kW
100 kW
8 kW io
vo
vo
20 kW +18 V
5 kW
40 kW +15 V
Fig. P3.5.13 Fig. P3.5.9
(A) -1.5 mA
(B) 1.5 mA
(A) 34 V
(B) -17 V
(C) -0.75 mA
(D) 0.75 mA
(C) 32 V
(D) -32 V
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GATE EC BY RK Kanodia
UNIT 3
28. For the circuit shown in fig. P3.5.28 the input
Analog Electronics
31. io = ?
resistance is
6 kW
io
2 kW vo
6A 2 kW
4 kW is
Fig. S3.5.31 2 kW 10 kW
(A) -18 A
(B) 18 A
(C) -36 A
(D) 36 A
Fig. P3.5.28
(A) 38 kW
(B) 17 kW
(C) 25 kW
(D) 47 kW
Statement for Q.32–33: Consider the circuit shown below
29. In the circuit of fig. P3.5.29 the op-amp slew rate is SR = 0.5 V ms. If the amplitude of input signal is 0.02 V, then the maximum frequency that may be used is vi
3 kW
D1
6 kW
D2
2 kW vo
240 kW vi
10 kW vo
Fig. P3.5.32–33
32. If vi = 2 V, then output vo is
Fig. P3.5.29
(A) 0.55 ´ 106 rad/s
(B) 0.55 rad/s
(C) 1.1 ´ 106 rad/s
(D) 1.1 rad/s
(B) -4 V
(C) 3 V
(D) -3 V
33. If vi = -2 V, then output vo is
30. In the circuit of fig. P3.5.30 the input offset voltage and input offset current are Vio = 4 mV and I io = 150 nA. The total output offset voltage is
(A) 4 V
(A) -6 V
(B) 6 V
(C) -3 V
(D) 3 V
34. vo( t) = ?
500 kW vi
8 mF
5 kW vo 5u(t) mA
250 W
vo 50 W
1 kW
5 kW
Fig. P3.5.34 Fig. P3.5.30
(A) e (A) 479 mV (C) 168 mV
Page 188
(B) 234 mV (D) 116 mV
(C) e
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-
t 10
-
t 1 .6
-
u( t) V
(B) -e
u( t) V
(D) -e
t 10
-
t 1 .6
u( t) V u( t) V
GATE EC BY RK Kanodia
Operational Amplifiers
Chap 3.5
35. The circuit shown in fig. P3.5.35 is at steady state
(A) vs vss
(B) -vs vss
before the switch opens at t = 0. The vC ( t) for t > 0 is t=0
v (C) - s vss
(D)
20 kW
39. If the input to the ideal comparator shown in fig.
vs vss
P3.5.39 is a sinusoidal signal of 8 V (peak to peak)
20 kW 20 kW
without any DC component, then the output of the +
4 mF
5V
comparator has a duty cycle of
vC -
Input Output
Fig. P3.5.35
(A) 10 - 5 e -12 .5t V (C) 5 + 5 e
-
t 12 .5
Vref = 2 V
(B) 5 + 5 e -12 .5t V (D) 10 - 5 e
V
-
t 12 .5
Fig. P3.5.39
V
36. The LED in the circuit of fig. P3.5.36 will be on if vi is 10 kW +10 V
470W 10 kW
vi
(A)
1 2
(B)
1 3
(C)
1 6
(D)
1 12
40. In the op-amp circuit given in fig. P3.5.40 the load current iL is
Fig. P3.5.36
(A) > 10 V
(B) < 10 V
(C) > 5 V
(D) < 5 V
R1 R1 vs
37. In the circuit of fig. P3.5.37 the CMRR of the op-amp is 60 dB. The magnitude of the vo is
R2
2V
IL
R2 RL
100 kW R
R
1 kW
Fig. P3.5.40 R
R
vo
1 kW
(A) -
vs R2
(B)
vs R2
(C) -
vs RL
(D)
vs RL
100 kW
Fig. P3.5.37
(A) 1 mV
(B) 100 mV
(C) 200 mV
(D) 2 mV
41. In the circuit of fig. P3.5.41 output voltage is |vo| = 1
V for a certain set of w, R, an C. The |vo| will be 2 V if R1
38. The analog multiplier X of fig. P.3.5.38 has the R1
characteristics vp = v1 v2 . The output of this circuit is vss
vi = sin wt V
vo
10 kW X vs
R
R
C vo
Fig. P3.5.41 Fig. P3.5.38
(A) w is doubled
(B) w is halved
(C) R is doubled
(D) None of the above
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GATE EC BY RK Kanodia
UNIT 3
Analog Electronics
42. In the circuit of fig. P3.5.42. the 3 dB cutoff
+15 V
frequency is 50 nF
47 kW
6 kW
3 kW
vo vi
vo
Fig. P3.5.42
(A) 10 kHz
(B) 1.59 kHz
(C) 354 Hz
(D) 689 Hz
Vz = 5 V
100 W
Fig. P3.5.45
43. The phase shift oscillator of fig. P3.5.43 operate at
46. In the circuit in fig. P3.5.46 both transistor Q1 and
f = 80 kHz. The value of resistance RF is
Q2 are identical. The output voltage at T = 300 K is
RF 100 pF
100 pF
100 pF
R
R
R1
R2
v1
R
v2
vo
333 kW 20 kW
Fig. P3.5.43
(A) 148 kW
(B) 236 kW
(C) 438 kW
(D) 814 kW
vo
20 kW
333 kW
44. The value of C required for sinusoidal oscillation of Fig. P3.5.46
frequency 1 kHz in the circuit of fig. P3.5.44 is 2.1 kW
1 kW
C 1 kW
æv R ö (A) 2 log10 çç 2 1 ÷÷ è v1 R2 ø
æv R ö (B) log10 çç 2 1 ÷÷ è v1 R2 ø
æv R ö (C) 2.303 log10 çç 2 1 ÷÷ è v1 R2 ø
æv R ö (D) 4.605 log10 çç 2 1 ÷÷ è v1 R2 ø
47. In the op-amp series regulator circuit of fig. P8.3.47 Vz = 6.2 V, VBE = 0.7 V and b = 60. The output voltage vo is
C
1 kW
vo
+36 V 1 kW
Fig. P3.5.44
(A) (C)
1 mF 2p 1 2p 6
30 kW
(B) 2p mF mF
(D) 2 p 6 mF 10 kW
45. In the circuit shown in fig. P3.5.45 the op-amp is ideal. If bF = 60, then the total current supplied by the 15 V source is (A) 123.1 mA
(B) 98.3 mA
(C) 49.4 mA
(D) 168 mA
Fig. P3.5.47
(A) 35.8 V
(B) 24.8 V
(C) 29.8 V
(D) None of the above
*******
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Operational Amplifiers
26. (C) v2 + = v2 - = 0 V, current through 6 V source i=
6 = 2 mA, vo = -2m( 3k + 2 k) = -10 V 3k
33. (D) If vi < 0, then vo > 0, D2 blocks and D1 conduct Av = -
vo(1) vo v (2) vo(1) = , v- = i + 1+ 3 4 2+1 2 +1 2v v v v v+ = v- , o = o + i , o = -8 4 3 3 vi
Chap 3.5
3k . )=3 V = -15 . , vo = ( -2)( -15 2k
34. (A) Voltage follower vo = v- = v+
27. (D) v+ =
v+ (0 + ) = 5m(250 ||1000) = 1 V, v+ ( ¥) = 0 t = 8m(1000 + 250) = 10 s 35. (A) vc (0 - ) = 5 V = vc (0 + ) = 5 V
28. (B) Since op-amp is ideal
For t > 0 the equivalent circuit is shown in fig. S3.5.35 20 kW
i1 is
+ vC –
is 2 kW
i2
Fig. S3.5.35 10 kW
t = 20 k ´ 4m = 0.08 s vc = 10 + (5 - 10) e
Fig. S3.5.28
v- = v+ , 2 kis = 4 ki1
Þ
is = 2 i1
vs = 2 kis + 10 ki2
36. (C) v- =
i2 = is + i1 , vs = 2 kis + 10 k( is + i1 ), i1 = i ö æ vs = 2 kis + 10 kç is + s ÷ 2ø è
Þ
is 2
-
t 0 .08
= 10 - 5 e -12 .5t V for t > 0
(10)(10 k) =5 V 10 k + 10 k
When v+ > 5 V, output will be positive and LED will be on. Hence (C) is correct.
vs = 17k = Rin is
R R = 1 V, v- = (2) = 1 V, vd = 0 2R 2R v + vR VCM VCM = + = 1, vo = F 2 1 CMRR 100 1 CMRR = 60 dB = 10 3 , vo = = 100 mV 1 10 3
37. (B) v+ = (2)
½R ½ 240 k 29. (C) Closed loop gain A =½ F½ = = 24 ½ R1½ 10 k The maximum output voltage vom = 24 ´ 0.02 = 0.48 V w £
4 mF
10 V
4 kW
0.5 / m SR . ´ 106 rad/s = = 11 0.48 vom
38. (C) v+ = 0 = v- ,
æ R ö 30. (A) The offset due to Vio is vo = çç 1 + 1 ÷÷Vio R1 ø è
Let output of analog multiplier be vp . vp vs =Þ vs = -vp , vp = vss vo R R v vs = -vss vo , vo = - s vss
500 ö æ = ç1 + ÷ 4m = 404 mV 5 ø è Due to I io, vo = RF I io = (500 k)(150n) = 75 mV Total offset voltage vo = 404 + 75 = 479 mV
39. (B) When vi > 2 V, output is positive. When vi < 2 V,
-vo v , io = - 6 + o 6k 3k -6( 6 k) io = - 6 + = -18 A. 3k
output is negative.
31. (A) 6 =
V 4V 2V
32. (B) If vi > 0, then vo < 0, D1 blocks and D2 conducts Av = -
6k = -2 3k
p 6
Þ
2p 5p 6
t
vo = ( -2)(2) = -4 V Fig. S3.5.39
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UNIT 3
5p p TON 1 Duty cycle = = 6 6 = 2p T 3 vs - v- v- - vo = R1 R1
40.(A)
v+ v v - vo + + + + =0 R2 RL R2
Þ
Analog Electronics
44. (A) This is Wien-bridge oscillator. The ratio R2 2.1k = = 2.1 is greater than 2. So there will be R1 1k oscillation
2 v1 = vs + vo
R2 R1
æ R ö vo = çç 2 + 2 ÷÷ v+ R è L ø
Þ
æ R ö 2 v- = vs + çç 2 + 2 ÷÷ v+ , v- = v+ R è L ø R Þ 0 = vs + 2 v+ RL v+ = -
RL vs , R2
iL =
v+ , RL
iL = -
R R
vs R2
Fig. S3.5.44
Frequency =
41. (D) This is a all pass circuit vo 1 - jwRC , = H ( jw) = vi 1 + jwRC
| H( jw)| =
1 + ( wR 2 C) 2 1 + ( wRC) 2
C
C
=1
C=
1 2pRC
Þ
1 ´ 10 3 =
1 mF 2p
Thus when w and R is changed, the transfer function is
45. (C) v+ = 5 V = v- = vE ,
unchanged.
The input current to the op-amp is zero. i+15V = iZ + iC = iZ + a F iE
42. (B) Let R1 = 3 kW , R2 = 6 kW , C = 50 nF vi v - vo + i =0 1 R2 æ ö R1 ||ç ÷ è sC ø
Þ
vi v v + i = o ö R2 R2 æ R1 ÷÷ çç 1 + sR C ø 1 è
éR ù vi ê 2 (1 + sR1 C) + 1ú = vo R ë 1 û vi [R2 + R1 + sR1 R2 C ] = vo R1 vo R + R1 = 2 vi R1 Þ f3dB =
é sR1 R2 C ù ê1 + R + R ú 1 2 û ë
vo æ R ö = ç 1 + 2 ÷÷(1 + s( R1 || R2 ) C) vi çè R1 ø 1 = 2 p( R1 || R2 ) C
1 1 = = 159 . kHz 2 p( 3k ||6 k)50n 2 p(2 k)50n
Þ R=
1 ( 80 k)(2 p 6 )(100 p)
RF = 29 R
Þ
=
15 - 5 60 æ 5 ö + ç ÷ = 49.4 mA 47 k 61 è 100 ø
46. (B) vo =
333 ( vo1 - vo2 ) 20
æi vo1 = -vBE1 - Vt ln çç c1 è is
43. (B) The oscillation frequency is 1 1 f = Þ 80 k = 2 p 6 RC 2 p 6 R(100 p) = 8.12 kW
RF = ( 8.12 k)(29) = 236 kW
æi vo1 - vo2 = -Vt ln çç c1 è ic 2 v1 , R1
ic1 =
ic 2 =
ö æi ÷÷, vo2 = -vBE 2 - Vt ln çç c 2 ø è is
æi ö ÷÷ = Vt ln çç c 2 è ic1 ø
ö ÷÷ ø
ö ÷÷ ø
v2 R2
æv R ö vo1 - vo2 = Vt ln çç 2 1 ÷÷, Vt = 0.0259 V è R2 v1 ø vo =
æv R ö 333 333 ( vo1 - vo2 ) = (0.0259) ln çç 2 1 ÷÷ 20 20 è v1 R2 ø
æv R ö æv R ö = 0.4329 lnçç 2 1 ÷÷ = 0.4329(2.3026) log10 çç 2 1 ÷÷ è v1 R2 ø è v1 R2 ø æv R ö = log10 çç 2 1 ÷÷ è v1 R2 ø 47. (B) v+ = v- , vZ =
10 vo v = o 10 + 30 4
vo = 4 vz = 6.2 ´ 4 = 24.8 V
************
Page 194
1 2 p(1k) C
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GATE EC BY RK Kanodia
CHAPTER
4.1 NUMBER SYSTEMS & BOOLEAN ALGEBRA
1. The 100110 2 is numerically equivalent to 1. 2616
2. 3610
3. 468
5. A computer has the following negative numbers stored in binary form as shown. The wrongly stored
4. 212 4
number is
The correct answer are (A) 1, 2, and 3
(B) 2, 3, and 4
(C) 1, 2, and 4
(D) 1, 3, and 4
(B) 5
(C) 7
(D) 9
(B) -89 as 1010 0111
(C) -48 as 1110 1000
(D) -32 as 1110 0000
6. Consider the signed binary number A = 01010110
2. If (211) x = (152)8 , then the value of base x is (A) 6
(A) -37 as 1101 1011
and B = 1110 1100 where B is the 1’s complement and MSB is the sign bit. In list-I operation is given, and in list-II resultant binary number is given.
3. 11001, 1001 and 111001 correspond to the 2’s List–I
complement representation of the following set of
List-II
numbers (A) 25, 9 and 57 respectively
P. A + B
(B) -6, -6 and -6 respectively
Q. B - A
(C) -7, -7 and -7 respectively
R. A - B
(D) -25, -9 and -57 respectively
1. 0 1 0 0 2. 0 1 1 0 3. 0 1 0 0 4. 1 0 0 1 5. 1 0 1 1 6. 1 0 0 1 7. 1 0 1 1 8. 0 1 1 0
S. - A - B
4. A signed integer has been stored in a byte using 2’s
0011 1001 0010 0101 1100 0110 1101 1010
complement format. We wish to store the same integer The correct match is
in 16-bit word. We should copy the original byte to the less significant byte of the word and fill the more
P
Q
R
S
significant byte with (A) 0
(A)
3
4
2
5
(B) 1
(B)
3
6
8
7
(C) equal to the MSB of the original byte
(C)
1
4
8
7
(D) complement of the MSB of the original byte.
(D)
1
6
2
5
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GATE EC BY RK Kanodia
UNIT 4
28.
The
simplified
form
of
a
logic
Digital Electronics
A
function
A
B
Y = A( B + C( AB + AC)) is
Z
C
(A) A B
(B) AB
(C) AB
(D) AB
D
(B) A + B
(C) AB + AB
(D) A B + AB
B
(C) 0
(D) 1
Z
D
B
Z
C D
(C)
(D)
35. In fig. P.4.1.35 the input condition, needed to
30. If X Y + XY = Z then XZ + XZ is equal to (B) Y
A
C
(A) A + B
(B)
A
Y = ( AB ) × ( AB) is
Z
D
(A)
29. The reduced form of the Boolean expression of
(A) Y
B C
produce X = 1, is A B
31. If XY = 0 then X Å Y is equal to (A) X + Y
(B) X + Y
(C) XY
(D) X Y
Fig. P4.1.34
32. From a four-input OR gate the number of input condition, that will produce HIGH output are (A) 1
(B) 3
(C) 15
(D) 0
(A) A = 1, B = 1, C = 0
(B) A = 1, B = 1, C = 1
(C) A = 0, B = 1, C = 1
(D) A = 1, B = 0, C = 0
36. Consider the statements below:
33. A logic circuit control the passage of a signal according to the following requirements : 1. Output X will equal A when control input B and
1. If the output waveform from an OR gate is the same as the waveform at one of its inputs, the other input is being held permanently LOW. 2. If the output waveform from an OR gate is always HIGH, one of its input is being held permanently HIGH.
C are the same.
The statement, which is always true, is
2. X will remain HIGH when B and C are different. The logic circuit would be A X
C
X
C
(B)
A
(C) Only 2
(D) None of the above
is (A) 3
(B) 4
(C) 5
(D) 6
A
B
X
C
B
X
number of two-input NAND required to implement
(D)
34. The output of logic circuit is HIGH whenever A and B are both HIGH as long as C and D are either both LOW or both HIGH. The logic circuit is
38. If the X and Y logic inputs are available and their complements X and Y are not available, the minimum
C
(C)
Page 200
(B) Only 1
NAND gates, minimum number of requirement of gate
B
(A)
(A) Both 1 and 2
37. To implement y = ABCD using only two-input
A
B
X
C
X Å Y is (A) 4
(B) 5
(C) 6
(D) 7
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Number Systems & Boolean Algebra
Statement for Q.39–40:
Chap 4.1
Assuming complements of x and y are not
A Boolean function Z = ABC is to be implement
available, a minimum cost solution for realizing f using
using NAND and NOR gate. Each gate has unit cost.
2-input NOR gates and 2-input OR gates (each having
Only A, B and C are available.
unit cost) would have a total cost of
39. If both gate are available then minimum cost is (A) 2 units
(B) 3 units
(C) 4 units
(D) 6 units
(A) 1 units
(B) 2 units
(C) 3 units
(D) 4 units
44. The gates G1 and G2 in Fig. P.4.2.44 have propagation delays of 10 ns and 20 ns respectively.
40. If NAND gate are available then minimum cost is (A) 2 units
(B) 3 units
(C) 4 units
(D) 6 units
G1
1 Vi 0
G2
Vo
Vi
to
Fig. P4.1.44
If the input Vi makes an abrupt change from logic
41. In fig. P4.1.41 the LED emits light when
0 to 1 at t = t0 then the output waveform Vo is VCC = 5 V
1 kW
1 kW
[t1 = t0 + 10 ns, t2 = t1 + 10 ns, t3 = t2 + 10 ns] (B)
(A)
1 kW
t0
1 kW
t2
t1
t3
t1
t2
t3
t0
t1
t2
t3
(D)
(C) t0
Fig. P4.1.41
(A) both switch are closed
t0
t2
t1
t3
45. In the network of fig. P4.1.45 f can be written as
(B) both switch are open
X0
(C) only one switch is closed
1 2
X1
(D) LED does not emit light irrespective of the switch positions
3 X2
42. If the input to the digital circuit shown in fig.
n-1 X3
Xn-1
n
F
Xn
Fig. P4.1.45
P.4.1.42 consisting of a cascade of 20 XOR gates is X,
(A) X 0 X1 X 3 X 5 + X 2 X 4 X 5 .... X n -1 + .... X n -1 X n
then the output Y is equal to
(B) X 0 X1 X 3 X 5 + X 2 X 3 X 4 .... X n + .... X n -1 X n (C) X 0 X1 X 3 X 5 .... X n + X 2 X 3 X 5 K X n + .... + X n -1 X n
1
(D) X 0 X1 X 3 X 5... X n -1 + X 2 X 3 X 5 K X n +..+ X n -1 X n - 2 + X n Y X
Fig. P4.1.42
(A) X
(B) X
(C) 0
(D) 1
*******
43. A Boolean function of two variables x and y is defined as follows : f (0, 0) = f (0, 1) = f (1, 1) = 1; f (1, 0) = 0
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GATE EC BY RK Kanodia
UNIT 4
Digital Electronics
A - B = A + B,
SOLUTIONS
A
010 10110
B
+ 00010011 0110 1001
1. (D) 100110 2 = 2 5 + 2 2 + 21 = 3810
- A - B = A + B,
2616 = 2 ´ 16 + 6 = 3810 468 = 4 ´ 8 + 6 = 3810
A
1010 1001
B
+ 00010011 10111100
212 4 = 2 ´ 4 2 + 41 = 3810 7. (B) Here A , B are 2’s complement
So 3610 is not equivalent. 2. (C) 2 x 2 + x + 1 = 64 + 5 ´ 8 + 2
Þ
A + B,
x =7
A
0100 0110
B
+ 1101 0011 1 0001 1001
3. (C) All are 2’s complement of 7 Þ
11001
Discard the carry 1
00110 +
1
Þ
1001
Þ
B - A,
= 710
= 710
1101 0011
A
+ 1011 1010
Discard the carry 1 - A - B = A + B,
42 in a byte
00101010
42in a word
0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0
-42 in a byte
11010110
-42 in a word
1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 0
1011 1010
B
+ 0010 1101
8. (B) 1110 = 10112 2Fi-1
Bi
Fi
0.3
0.6
0
0.6
1.2
1
0.2
1
0.4
0
0.4
11010000
0.8
0
0.8
1.6
1
0.6
00110000 2 1100 1111 +
6. (D) Here A , B are 1’s complement A 01010110 B + 1110 1100 10100 0010 , + 1 0100 0011 B
1110 1100
A
+ 1010 1001 110010101 1 +
Repeat from the second line 0.310 = 0.01001
2
9. (C)
Received
b4
b3
b2
p3
b1
p2
p1
1
1
0
1
1
0
0
C1* = b4 Å b2 Å b1 Å p1 = 0 C2* = b4 Å b3 Å b1 Å p2 = 1 C3* = b4 Å b3 Å b2 Å p3 = 1
10010110 Page 202
A
1110 0111
Therefore (C) is correct.
B - A = B + A,
+ 0010 1101
1 1000 1101
4. (C) See a example
A + B,
B
B
1
000111
-4810 =
010 0 0110 0111 0011
000110 +
5. (C) 4810 =
A
0110 + 1 0111
111001
A - B = A + B,
= 710
00111
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GATE EC BY RK Kanodia
Number Systems & Boolean Algebra
Chap 4.1
C3* C2* C1* = 110 which indicate position 6 in error
= ( AA + AB)( A + B + C) = A( A + B + C) = A
Transmitted code 1001100.
Therefore No gate is required to implement this function.
10. (D) X = MNQ + M NQ + M NQ
23. (A)
= MQ + M NQ = Q( M + M N ) = Q( M + N ) 11. (A) The logic circuit can be modified as shown in fig. S. 4.1.11 A B
Z
C+D E
Fig. S4.1.11
Now Z = AB + ( C + D) E
A
B
C
( A + BC)
( A + B)( A + C)
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
0
1
1
1
1
1
0
0
1
1
1
0
1
1
1
Fig. S 4.1.23
12. (D) You can see that input to last XNOR gate is same. So output will be HIGH.
24. (B) X = ABC + ABC + ABC = BC + ABC
13. (D) Z = A + ( AB + BC) + C
25. (B) ( A + B)( B + C ) = ( AB)( BC) = ABC
= A + ( A + B + B + C) + C = A + B + C
( A + B)( B + C ) = ( A + B) + ( B + C) = A + B + C
ABC = A + B + C
( A + B)( B + C) = ( A + B) + ( B + C)
AB + BC + AC = A + B + B + C + A + C = A + B + C
= AB + B + C = A + B + C
14. (C) ( X + Y )( X + Y ) = XY + X Y
From truth table Z = A + B + C
( X + Y )( X + Y )( X + Y ) = ( X + Y )( XY + X Y )
Thus (B) is correct.
= XY + XY = XY
26. (D) AC + BC = AC( B + B) + ( A + A) BC = ABC + ABC + ABC + ABC
15. (B) Using duality ( A + B)( A + C)( B + C) = ( A + B)( A + C)
27. (D) F = A + AB + A BC + A B C( D + DE)
Thus (B) is correct option.
= A + AB + A B( C + C( D + E)) = A + A( B + B( C + D + E)) = A + B + C + D + E
16. (B) Z = ( AB)( CD)( EF ) = AB + CD + EF 17. (A) X = ( A B + AB)( A + B) = ( AB + A B)( AB) = AB 18. (B) Y = ( A Å B) × C = ( AB + AC) × C = ( AB + AB) + C = A B + AB + C 19. (C) Z = A( A + A) BC = ABC 20. (A) Z = AB( B + C) = ABC 21. (A) Z = ( A + B ) × BC = ( AB) × BC = ABC
28. (B) A( B + C ( AB + AC)) = AB + AC ( AB × AC) = AB + AC[( A + B)( A + C)] = AB + AC ( A + AC + AB + B C) = AB 29. (C) ( AB ) × ( AB) = AB + AB = AB + AB 30. (B) X Z + XZ = X ( XY + XY ) + X ( X Y + XY ) = X ( XY + X Y ) + XY = XY + XY = Y 31. (A) X Å Y = X Y + XY = ( XY + XY ) = ( XY ) = X + Y
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Page 203
GATE EC BY RK Kanodia
UNIT 4
Digital Electronics
(A) ( w + y)( x + y + z)( w + x + z)
(A) AC + AB
(B) AC + AB + BC
(B) ( w + x)( w + z )( x + y)( y + z )
(C) AB + BC + ABC
(D) Above all
(C) ( x + z)( w + y)
12. In the logic circuit of fig. P4.2.12 the redundant gate
(D) ( x + z )( w + y)
is
7. A function with don’t care condition is as follows
x y z w x z w y z
f ( a, b, c, d) = Sm(0, 2, 3, 5, 7, 8, 9, 10, 11) + Sdc ( 4, 15) The minimized expression for this function is (A) ab + b d + cd + abc
(B) ab + b d + cd + abd
(C) ab + b d + b c + abd
(D) Above all
8. A function with don’t cares is as follows : g( X , Y , Z ) = Sm(5, 6) + Sdc (1, 2, 4)
1
2
4
X
3
Fig. P4.2.12
(A) 1
(B) 2
(C) 3
(D) 4
For above function consider following expression 1. XYZ + XYZ
2. XY + XZ
3. XZ + XZ + YZ
4. YZ + YZ
13. If function W, X, Y, and Z are as follow W = R + PQ + RS
The solution for g are (A) 1, 2, and 3
(B) 1, 2, and 4
(C) 1, and 4
(D) 1, and 3
X = PQRS + P Q RS + PQ R S Y = RS + PR + PQ + P Q Z = R + S + PQ + P Q R + PQ S
9. A logical function of four variable is given as
Then
f ( A, B, C, D) = ( A + B C)( B + CD) The function as a sum of product is (A) A + BC + ACD + BCD (B) A + BC + ACD + BCD
(A) W = Z , X = Z
(B) W = Z , X = Y
(C) W = Y
(D) W = Y = Z
14. In a certain application four inputs A, B, C, D are
(C) AB + BC + ACD + BCD
fed to logic circuit, producing an output which operates
(D) AB + AB + ACD + BCD
a relay. The relay turns on when f(A, B, C, D) =1 for the
10. A combinational circuit has input A, B, and C and its K-map is as shown in fig. P4.2.10. The output of the circuit is given by
00 00 01
0101, 0110, 1101 and 1110. States 1000 and 1001 do not occur, and for the remaining states the relay is off. The minimized Boolean expression f is
CD
A
following states of the inputs (ABCD) : 0000, 0010,
01
11
1
10 1
1
(A) ACD + BCD + BCD
(B) BCD + BCD + ACD
(C) ABD + BCD + BCD
(D) ABD + BCD + BCD
15. There are four Boolean variables x1 , x2 , x3 and x4 .
1
The following function are defined on sets of them f ( x3 , x2 , x1 ) = Sm( 3, 4, 5)
Fig. P4.2.1
Page 206
(A) ( AB + AB ) C
(B) ( AB + AB ) C
(C) ABC
(D) A Å B Å C
g( x4 , x3 , x2 ) = Sm(1, 6, 7) h( x4 , x3 , x2 , x1 ) = fg Then h( x4 , x3 , x2 , x1 ) is
11. The Boolean Expression Y = ( A + B)( A + C) is equal
(A) Sm(3, 12, 13)
(B) Sm(3, 6)
to
(C) Sm(3, 12)
(D) 0
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GATE EC BY RK Kanodia
Combinational Logic Circuits
Statement for Q.16–17:
Chap 4.2
21. For a binary half subtractor having two input A and
A switching function of four variable, f ( w, x y, z) is to equal the product of two other function f1 and f2 , of
B, the correct set of logical expressions for the outputs D = ( A - B) and X (borrow) are
the same variable f = f1 f2 . The function f and f1 are as
(A) D = AB + AB , X = AB
follows :
(B) D = AB + AB , X = AB
f = Sm( 4, 7, 15)
(C) D = AB + AB , X = AB
f1 = Sm(0, 1, 2, 3, 4, 7, 8, 9, 10, 11, 15)
(D) D = AB + AB , X = AB 22. f1 f2 = ?
16. The number of full specified function, that will satisfy the given condition, is (A) 32
(B) 16
x0
I0
(C) 4
(D) 1
x1
I1
17. The simplest function for f2 is
x2
I2
(A) x
(B) x
(C) y
(D) y
D0 3-to-8 Decoder D1 D2 D3 D4 D5 D6 D7
and m9. If the literals in these minterms are complemented, the corresponding minterm numbers are (A) m3 and m0
(B) m9 and m6
(C) m2 and m0
(D) m6 and m9
(A) x0 x1 x2
(B) x0 Å x1 Å x2
(C) 1
(D) 0
23. The logic circuit shown in fig. P4.2.23 implements
A
I0
B
I1
C
I2
19. The minimum function that can detect a “divisible by 3’’ 8421 BCD code digit (representation D8 D4 D2 D1 ) is given by (A) D8 D1 + D4 D2 + D8 D2 D1 (B) D8 D1 + D4 D2 D1 + D4 D2 D1 + D8 D4 D2 D1
D0 3-to-8 D Decoder 1 D2 D3 D4 D5 D6 D7 EN
Fig. P4.2.23
(D) D4 D2 D1 + D4 D2 D1 + D8 D4 D2 D1
(A) D( A
I1
x2
I2
u C + AC )
(B) D( B Å C + AC )
(C) D( B Å C + AB )
20. f ( x2 , x1 , x0 ) = ?
x1
Z
D
(C) D4 D1 + D4 D2 + D8 D1 D2 D1
I0
f2
Fig. P4.2.22
18. A four-variable switching function has minterms m6
x0
f1
(D) D( B
u C + AB )
Statement for Q.24–25:
D0 3-to-8 D Decoder 1 D2 D3 D4 D5 D6 D7
The building block shown in fig. P4.2.24–25 is a f
active high output decoder.
Fig. P4.2.21
(A) P(1, 2, 4, 5, 7)
(B) S(1, 2, 4, 5, 7)
(C) S(0, 3, 6)
(D) None of Above
A
I0
B
I1
C
I2
D0 3-to-8 Decoder D1 D2 D3 D4 D5 D6 D7
X
Y
Fig. P4.2.24-25
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GATE EC BY RK Kanodia
UNIT 4
24. The output X is
Digital Electronics
28. Z 2 = ?
(A) AB + BC + CA
(B) A + B + C
(A) ab + bc + ca
(B) a + b + c
(C) ABC
(D) None of the above
(C) abc
(D) a u b u c
25. The output Y is
29. This circuit act as a
(A) A + B
(B) B + C
(A) Full adder
(B) Half adder
(C) C + A
(D) None of the above
(C) Full subtractor
(D) Half subtractor
26. A logic circuit consist of two 2 ´ 4 decoder as shown
30. The network shown in fig. P4.2.30 implements
in fig. P4.2.26. x
1 MUX
1
0
f
D0
A0
D1 D2 y
A
A1
D0
A1
D3
f
D1
B
1 MUX
0
0
S0
D2 z
A0
D3
S0 C
Fig. P4.2.26
Fig. P4.2.30
The output of decoder are as follow D0 = 1 when A0 = 0,
A1 = 0
D1 = 1 when A0 = 1,
A1 = 0
D2 = 1 when A0 = 0,
A1 = 1
D3 = 1 when A0 = 1,
A1 = 1
(A) NOR gate
(B) NAND gate
(C) XOR gate
(D) XNOR gate
31. The MUX shown in fig. P4.2.31 is 4 ´ 1 multiplexer. The output Z is C
The value of f ( x, y, z) is
I3 I2
(A) 0
(B) z
(C) z
(D) 1
MUX
I0
Statement for Q.27-29:
+5 V
S 1 S0 A
B) A Å B Å C
(A) ABC
c
1
0
MUX
Z1
B
Fig. P4.2.31
A MUX network is shown in fig. P4.2.27-29. c
Z
I1
(C) A u B
uC
(D) A + B + C
32. The output of the 4 ´ 1 multiplexer shown in fig.
S0
P4.2.32 is a
1
a
0
MUX Z0 S0 b
Y
c
b
1
S0 MUX
+5 V
Z2
X
0
Page 208
(A) a + b + c
(B) ab + ac + bc
(C) a u b u c
(D) a Å b Å c
MUX
I1 I0
Z
S 1 S0 Y
Fig. P4.2.32
Fig. P4.2.27-29
27. Z1 = ?
I3 I2
(A) X + Y
(B) X + Y
(C) XY + X
(D) XY
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Combinational Logic Circuits
Chap 4.2
33. The MUX shown in fig. P4.2.33 is a 4 ´ 1
36. The 4–to–1 multiplexer shown in fig. P4.2.36
multiplexer. The output Z is
implements the Boolean expression
I3
C
I3
I2
MUX
I1
C
I0
Z
uC (D) B u C
0 1 0 1 1
The input to I1 and I 3 will be (A) yz , y + z
(B) y + z, y u z
(C) y + z, y Å z
(D) x + y ,
yÅ z
37. The 8-to-1 multiplexer shown in fig. P4.2.37 realize 0 1 1
x
I0 I1 MUX I2 I3 EN S 1 S0 w
x
f ( w, x, y, z) = Sm( 4, 5, 7, 8, 10, 12, 15)
34. f = ?
w
S1 S0 w
(B) A
(C) B Å C
f
Fig. P4.2.36
Fig. P4.2.33
I0 I1 MUX I2 I3 EN S 1 S0
I0
0
B
(A) A Å C
0 1 1 0 1
MUX
I1
S1 S0 A
I2
z
I0 I1 MUX I2 I3 EN S1 S0 y
the following Boolean expression 1 z z
f
0 z 0 1 z
z
I0 I1 I2 I3 I4 I5 I6 I7
MUX
EN S2 S1 S0 0
x
w
x
y
Fig. P4.2.37
Fig. P4.2.34
(A) wxyz + wx yz + xy + yz
(A) wxz + w x z + wyz + xy z
(B) wxyz + wxyz + x y + yz
(B) wxz + wyz + wyz + w x y
(C) wx yz + w x yz + yz + zx
(C) w x z + wy z + w yz + wxz
(D) wxyz + wxyz + gz + zx
(D) MUX is not enable
35. For the logic circuit shown in fig. P4.2.35 the output
Statement for Q.38-40:
Y is
A PLA realization is shown in fig. P4.2.38–40 x0 1 0 I 0 I1 I 2 I 3 I 4 I 5 I 6 I 7 C B A
EN S2 S1 S0
x1 x2
MUX
Y
X
Fig. P4.2.35
(A) A Å B
(B) A Å B
(C) A Å B Å C
(D) A Å B Å C
X X
X X X
X
X
f1 f2 f3
Fig. P.4.2.38-40
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UNIT 4
38. f1 ( x2 , x1 , x0 ) = ?
Digital Electronics
The terms P1 , P2 , P3 , P4 and P5 are
(A) x2 x0 + x1 x0
(B) x2 x0 + x1 x2
(C) x2 Å x0
(D) x2
ux
0
39. f2 ( x2 , x1 , x0 ) = ? (A) Sm(1, 2, 5, 6)
(B) Sm(1, 2, 6, 7)
(C) Sm(2, 3, 4)
(D) None of the above
(A) ab, ac, bc, bc, ab
(B) ab , bc, ac, ab , bc
(C) ac, ab, bc, ab , bc
(D) Above all
43. The circuit shown in fig. P.4.2.43 has 4 boxes each described by input P, Q, R and output Y , Z with Y = P Å Q Å R and Z = RQ + PR + QP . Q
40. f3( x2 , x1 , x0 ) = ? (A) P M(0, 4, 6, 7)
(B) P M(2, 4, 5,7)
(C) P M(1, 2, 3, 5)
(D) P M(2, 3, 4, 7)
P P
Q
P
Q
P
Q
P
Q
Z
R
Z
R
Z
R
Z
R
41. If the input X 3 X 2 X1 X 0 to the ROM in fig. P4.2.41 are 8–4–2–1 BCD numbers, then output Y3Y2 Y1 Y0 are X3
X2
X1
X0
Output
Fig. P4.2.43
The circuit act as a 4 bit
BCD to Decimal Decoder
(A) adder giving P + Q
D0 D1 D2 D3 D4 D5 D6 D7 D8 D9
X X X X Y3
(C) subtractor giving Q - P
X X X X X X Y2
(D) adder giving P + Q + R
X X X
(B) subtractor giving P - Q
X
X X Y1 X
X
44. The circuit shown in fig. P4.2.44 converts
X Y0
MSB
Fig. P4.2.41
(A) 2–4–2–1 BCD number
(B) gray code number
(C) excess 3 code converter
(D) none of the above +
42. It is desired to generate the following three Boolean
+
function f1 = ab c + abc + bc
MSB
Fig. P4.2.44
f2 = ab c + ab + abc, f3 = ab c + abc + ac
(A) BCD to binary code
by using an OR gate array as shown in fig. P4.2.42 where P1 and P5 are the product terms in one or more of the variable a, a , b, b , c and c. P1
X
P2
X
(B) Binary to excess (C) Excess–3 to Gray Code (D) Gray to Binary code
X X
P3
******* X
P4 P5
X F1
F2
F3
Fig. P4.2.42
Page 210
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+
GATE EC BY RK Kanodia
Combinational Logic Circuits
16. (A) f = Sm(4, 7, 15),
RS 00
01
11
00 PQ
Chap 4.2
f1 = Sm(0, 1, 2, 3, 4, 7, 8, 9, 10, 11, 15)
10
f2 = Sm(4, 7, 15) + Sdc(5, 6, 12, 13, 14)
1
01
1
1
1
11
1
1
1
There are 5 don't care condition. So 2 5 = 32 different
1
functions f2 . 17. (A) f2 = Sm(4, 7, 15) + Sdc(5, 6, 12, 13, 14), f2 = x
1
10
yz 00
01
11
10
00
0
0
0
0
01
1
´
1
´
11
´
´
1
´
10
0
0
0
0
Fig. S 4.2.13c
Z = R + S + PQ + P Q R + PQ S= R + S + PQP Q R PQ S
wx
= R + S + ( P + Q )( P + Q + R)( P + Q + S) = R + S + ( PQ + PR + Q P + Q R)( P + Q + S) = R + S + PQ + PQS + PR + PRQ + PRS + Q PS + Q PR + Q RS
Fig. S 4.2.17
RS 00
01
11
10
1
1
1
1
1
1
11
1
1
1
10
1
1
1
00 PQ
01
1
18. (B) m6 = ABCD ,
m9 = ABCD
After complementing literal m6¢ = ABCD = m9 ,
m¢9 = ABCD = m6
19. (B) 0, 3, 6 and 9 are divisible by 3 D2 D1 00 00
Fig. S 4.2.13d
D8 D4 01 11
= R + S + PQ We can see that W = Z , X = Z
01
11
1
10
1 1
´
10
´
´
´
1
´
´
14. (D) ABD + BCD + BCD Fig. S 4.2.19
CD 00 00 AB
01
11
1
10 1
01
1
1
11
1
1
10
´
f = D8 D1 + D4 D2 D1 + D4 D2 D1 + D8 D4 D2 D1 20. (B) f = Sm(0, 3, 6) = Sm(1, 2, 4, 5, 7) 21. (C) D = AB + AB,
´
X = AB
A
B
D
X
0
0
0
0
Fig. S 4.2.14
0
1
1
1
15. (A) f = x3 x2 x1 + x3 x2 x1 + x3 x2 x1 = x3 x2 x1 + x3 x2
1
0
1
0
g = x4 x3 x2 + x4 x3 x2 + x4 x3 x2 = x4 x3 x2 + x4 x3
1
1
0
0
fg = x4 x3 x2 x1 + x4 x3 x2
Fig. S 4.2.20
= x4 x3 x2 x1 + x4 x3 x2 x1 + x4 x3 x2 x1 h = Sm(3, 12, 13) www.gatehelp.com
Page 213
GATE EC BY RK Kanodia
UNIT 4
Digital Electronics
22. (D) f1 = Sm(0, 2, 4, 6),
29. (A) The equation of Z1 is the equation of sum of A
f2 = Sm(1, 3, 5, 7), f1 f2 = 0
and B with carry and equation of 2 is the resultant carry. Thus, it is a full adder.
23. (D) Z = D( ABC + ABC + ABC + ABC + ABC) = D( AB ( C + C) + BC( A + A) + ABC )
30. (B) f1 = CD + CB = CB ,
= D( AB + BC + ABC ) = D( B ( A + AC ) + BC)
f = f 1 + f1 A = CB + CBA = CB + A
= D( BA + BC + BC) = D( B
u C + AB )
24. (A) X = Sm(3, 5, 6, 7),
S = F1
= C + B + A = ABC
X = AB + BC + CA
31. (D) Z = ABC + AB + AB + AB = A ( BC + B) + A( B + B) = A ( B + C) + A = A + B + C
25. (D) Y = Sm(1, 3, 5, 7),
Y =C 32. (A) Z = XY + XY + XY ,
Z=X +Y
BC 00 A
01
11
00
33. (A) Z = ABC + ABC + ABC + ABC
10
= AC + AC = A Å C
1
01
1
1
1
34. (A) The output from the upper first level and from the lower first level
multiplexer is fa multiplexer is fb
Fig. S4.2.24
fa = wx + wx, 26. (D) D0 = A1 A0 , D1 = A1 A0 ,
fb = wx + wx = x
f = fa yz + fb yz + yz = ( wx + wx) yz + xyz + yz
D2 = A1 A0 ,
= wxyz + wx yz + xy + yz BC 00 A
01
11
00
1
1
01
1
1
10
35. (D) Output is 1 when even parity BA 00 00 C
Fig. S4.2.25
01
11
1
10
1 1
01
1
D3 = A1 A0 Fig. S4.2.35
For first decoder A0 = x , A1 = y, D2 = yx , D3 = xy For second decoder A1 = D2 D3 = yxxy = 0,
A0 = z
Therefore Y = A Å B Å C
f = D0 + D1 = A1 A0 + A1 A0 = A1 = 1
36. (B) I1 = y + z ,
27. (D) The output of first MUX is Z o = ab + ab = ( a Å b)
28. (A) Z 2 = bS0 + cS0 = b( ab + ab ) + c( ab + ab ) = ab + ab c + abc
z
yz 00
01
11
10
00
0
0
0
0
I0 = 0
01
1
1
1
0
I1 = y + z
11
1
0
1
0
10
1
0
0
1
This is input to select S0 of both second-level MUX Z1 = CS0 + CS0 = C Å S0 = a Å b Å c
I3 = y
S1 S0 bb wx
= a( b + b c) + abc = ab + ac + abc = ab + ac + bc
Fig. S 4.2.36
Page 214
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I 3 = yz + yz = y u z I2 = z
GATE EC BY RK Kanodia
Combinational Logic Circuits
Chap 4.2
37. (C) Let z = 0, Then
42. (A) f1 = ab c + abc + bc = ac + ab
f = w x y + w xy + wx y + wxy = w x + w y
f2 = ab c + ab + abc = ac + b c
If we put z = 0 in given option then
f3 = ab c + abc + ac = ab + bc
(A) = w x + xy
(B) = wy + w x y
(C) = w x + wy
Since MUX is enable so option (C) is correct.
Thus P1 = ab, P2 = ac,
P3 = bc, P4 = bc, P5 = ab
43. (B) Let P = 1001 and Q = 1010 then
38. (C)f = x0 x2 + x0 x1 x2 + x0 x2 = x0 x2 (1 + x1 ) + x0 x2
Yn = Pn Å Qn Å Rn ,
= x0 x2 + x0 x2
Z n = Rn Qn + Pn Rn + Qn Pn
output is 1111 which is 2’s complement of –1. So it gives
39. (B) f2 = x0 x1 + x1 x2 + x0 x1 x2
P - Q . Let another example
= x0 x1 x2 + x0 x1 x2 + x1 x2 x0 + x1 x2 x0 + x0 x1 x2
then output is 00111. It gives P–Q.
P = 1101 and Q = 0110
So (B) is correct.
= x2 x1 x0 + x2 x1 x0 + x2 x1 x0 + x2 x1 x0 f2 ( x2 , x1 , x0 ) = Sm(1, 2, 6, 7)
Pn
Qn
Rn
Zn
Yn
40. (C) f3 = x0 x1 + x1 x2
n =1
1
0
0
0
1
= x2 x1 x0 + x2 x1 x0 + x2 x1 x0 + x2 x1 x0
n =2
0
1
0
1
1
f3( x2 , x1 , x0 ) = Sm(0, 4, 6, 7)
n=3
0
0
1
1
1
f3( x2 , x1 , x0 ) = PM(1, 2, 3, 5)
n=4
1
1
1
1
1 1
41. (A) Fig. S4.2.43a
Let X 3 X 2 X1 X 0 be 1001 then Y3Y2 Y1 Y0 will be 1111. Let X 3 X 2 X1 X 0 be 1000 then Y3Y2 Y1 Y0 will be 1110 Let X 3 X 2 X1 X 0 be 0110 then Y3Y2 Y1 Y0 will be 1100 bc 00
01
00 a
1
01
Pn
Qn
Rn
Zn
Yn
n =1
1
0
0
0
1
11
10
n =2
0
1
0
1
1
1
1
n=3
1
1
1
1
1
n=4
1
0
1
0
0
1
0 Fig. S4.2.43b
Fig. S4.2.42a
44. (D) Let input be
1010
10
output will be
1101
1
Let input be
0110
1
output will be 0100
bc 00
01
11
00 a
01
1
1
This convert gray to Binary code. Fig. S4.2.42b
bc 00 a
00
01
11
1
1
1
01
*******
10
1 Fig. S4.2.42b
So this converts 2–4–2–1 BCD numbers.
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GATE EC BY RK Kanodia
Sequential Logic Circuits
1
(B)
Q1
D1
D2
13. The counter shown in fig. P4.3.13 counts from
Y
Q2
Chap 4.3
J
C
X Q1
(C)
Q1
D1
D2
Q1
(D)
Y
Q2
1
D2
A CLR K
Fig.P4.3.13
Q2
Q1
D1
B CLR K
A B C
X
1
J
A
Q2 C CLR K
1
J
B
Y
Q2
X
(A) 0 0 0 to 1 1 1
(B) 1 1 1 to 0 0 0
(C) 1 0 0 to 0 0 0
(D) 0 0 0 to 1 0 0
14. The mod-number of the asynchronous counter Q1
Q2
shown in fig. P4.2.13 is J
Q0
J
Q1
J
Q2
Q3
J
J
Q4
11. The circuit shown in fig. P4.3.11 is K CLR
T
T
Q
K CLR
K CLR
K CLR
K CLR
Q All J.K. input are HIGH
CLK
CLK
A
B
Q
Q
Fig.P4.3.14
Fig.P4.3.11
(A) a MOD–2 counter
(A) 24
(B) 48
(C) 25
(D) 36
(B) a MOD–3 counter 15. The frequency of the pulse at z in the network
(C) generate sequence 00, 10, 01, 00.....
shown in fig. P4.3.15. is
(D) generate sequence 00, 10, 00, 10, 00 ......
160 kHZ
12. The counter shown in fig. P4.3.12 is a
10-Bit Ring Counter
w 4-Bit Parallel Counter
x
Mod-25 Ripple Counter
y
4-Bit Johnson Counter
z
Fig.P4.3.15
J
Q
Q
B
C Q
J
Q
K
Q
J
1
Q
K
1
(B) 160 Hz
(C) 40 Hz
(D) 5 Hz
16. The three-stage Johnson counter as shown in fig.
A K
(A) 10 Hz
P4.2.16 is clocked at a constant frequency of fc from the CLK
starting state of Q2Q1Q0 = 101. The frequency of output Q2Q1Q0 will be
Fig.P4.3.12
(A) MOD–8 up counter
J2
Q2
J1
Q1
J0
Q0
K2
Q2
K1
Q1
K0
Q0
(B) MOD–8 down counter (C) MOD–6 up counter
CLK
(D) MOD–6 down counter
Fig.P4.3.16
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GATE EC BY RK Kanodia
UNIT 4
(A)
fc 8
fc 6
21. In the circuit shown in fig. P4.3.21 is PIPO 4-bit
f (D) c 2
input lines are connected to a 4 bit bus. Its output acts
(B)
f (C) c 3
register, which loads at the rising edge of the clock. The as the input to a 16 ´ 4 ROM whose output is floating
17. The counter shown in the fig. P4.3.17 has initially Q2Q1Q0 = 000. The status of Q2Q1Q0 after the first pulse is
J2
Q1
J1
Q2
Digital Electronics
J0
when the enable input E is 0. A partial table of the contents of the ROM is as follows Address
0
2
4
6
8
10
12
Data
0011
1111
0100
1010
1011
1000
0010
The clock to the register is shown below, and the
Q0
data on the bus at time t1 is 0110. K2
Q2
Q1
K1
K0
MSB
Q0
CLK
Fig.P4.3.17
(A) 0 0 1
(B) 0 1 0
(C) 1 0 0
(D) 1 0 1
CLK
A
18. A 4 bit ripple counter and a 4 bit synchronous 1
counter are made by flips flops having a propagation
E
delay of 10 ns each. If the worst case delay in the ripple
ROM
counter and the synchronous counter be R and S respectively, then (A) R = 10 ns, S = 40 ns
(B) R = 40 ns, S = 10 ns
(C) R = 10 ns, S = 30 ns
(D) R = 30 ns, S = 10 ns
19. A 4 bit modulo–6 ripple counter uses JK flip-flop. If the propagation delay of each FF is 50 ns, the maximum clock frequency that can be used is equal to (A) 5 MHz
(B) 10 MHz
(C) 4 MHz
(D) 20 Mhz
t2
Fig. P4.3.21
right-shift, register shown in fig. P4.3.20 is 0 1 1 0. After three clock pulses are applied, the contents of the
CLK
t t1
20. The initial contents of the 4-bit serial-in-parallel-out
shift register will be
CLK
The data on the bus at time t2 is (A) 1 1 1 1
(B) 1 0 1 1
(C) 1 0 0 0
(D) 0 0 1 0
22. A 4-bit right shift register is initialized to value 1000 for (Q3 , Q2 , Q1 , Q0 ). The D input is derived from 0
1
1
0
Q0 , Q2 and Q3 through two XOR gates as shown in fig. P4.2.22. The pattern 1000 will appear at
Fig.P4.3.20
Page 219
(A) 0 0 0 0
(B) 0 1 0 1
(C) 1 1 1 1
(D) 1 0 1 0
D
Q0
Q1
Fig.P4.3.22
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Q2
Q3
GATE EC BY RK Kanodia
Sequential Logic Circuits
Chap 4.3
(A) 3rd pulse
(B) 7th pulse
28. To count from 0 to 1024 the number of required
(C) 6th pulse
(D) 4th pulse
flip-flop is
Statement for Q.23–24: The 8-bit left shift register and D-flip-flop shown in fig. P4.3.22–23 is synchronized with same clock. The
(A) 10
(B) 11
(C) 12
(D) 13
29. Four memory chips of 16 ´ 4 size have their address buses connected together. This system will be of size
b7
b 6 b 5 b4 b 3 b 2 b 1 b0
D
(A) 64 ´ 4
(B) 32 ´ 8
(C) 16 ´ 16
(D) 256 ´ 1
Q
30. The address bus width of a memory of size 1024 ´ 8
CLK
bits is Q
Fig.P4.3.23-24
(A) 10 bits
(B) 13 bits
(C) 8 bits
(D) 18 bits
31. For the circuit of Fig. P4.3.31 consider the
D flip-flop is initially cleared.
statement: 23. The circuit act as
Assertion (A) : The circuit is sequential
(A) Binary to 2’s complement converter
Reason (R) : There is a loop in circuit
(B) Binary to Gray code converter
d
(C) Binary to 1’s complement converter (D) Binary to Excess–3 code converter 24. If initially register contains byte B7, then after 4
c
(B) 72
(C) 7E
(D) 74
b e
z0
clock pulse contents of register will be (A) 73
z1
a b
Fig.P4.3.131
Choose correct option Statement for Q.25–26: A Mealy system produces a 1 output if the input has been 0 for at least two consecutive clocks followed immediately by two or more consecutive 1’s.
(A) Both A and R true and R is the correct explanation of A (B) Both A and R true but R is not a correct explanation on of A (C) A is true but R is false
25. The minimum state for this system is (A) 4
(B) 5
(C) 8
(D) 9
(D) A is false *****************
26. The flip-flop required to implement this system are (A) 2
(B) 3
(C) 4
(D) 5
27. The output of a Mealy system is 1 if there has been a pattern of 11000, otherwise 0. The minimum state for this system is (A) 4
(B) 5
(C) 6
(D) 7
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GATE EC BY RK Kanodia
UNIT 4
Digital Electronics
6. (D) Q + = LM + LMQ
SOLUTIONS
= L( M + MQ )
1. (C) Given FF is a negative edge triggered T flip-flop.
= L M + LQ
So at the negative edge of clock Vi FF will invert the
L
M
Q+
0
0
0
2. (A) At first rising edge of clock, D is HIGH. So Q will
0
1
0
be high till 2nd rising edge of clock. At 2nd rising edge,
1
0
1
D is low so Q will be LOW till 3rd rising edge of clock.
1
1
Q1
output if there is 1 at input.
At 3rd rising edge, D is HIGH, so Q will be HIGH till Fig. S4.3.6
4th rising edge. At 4th rising edge D is HIGH so Q will be HIGH till 5th rising. edge. At 5th rising edge, D is LOW, so Q will be LOW till 6th rising edge.
7. (D) Initially
3. (C)
J
Q
Q
1
0
1
Qn + 1
Qn +1
x
Q
S
R
Q+
Clock 1st
1
1
0
1
1
0
0
0
0
1
0
2nd
0
1
1
0
0
1
0
1
1
0
1
3rd
1
1
0
1
1
0
1
0
1
0
1
4th
0
1
1
0
0
1
1
1
0
1
0
5th
1
1
0
1
1
0
Fig. S4.3.3
Fig. S4.3.7
4. (D) Q + = x Å Q
Therefore sequence is 010101.
Q1+ = x1 Å Q0 = x1 0 + x1 0 = x1
8. (A)
Q2+ = x2 Å x1 ,
Q3+ = x3 Å x2 Å x1
Q4+ = x4 Å x3 Å x2 Å x1
5. (C)
A
B
X
Y
1
1
0
1
1
0
0
1
X and Y are fixed at 0 and 1.
So this generate the even parity and check odd parity.
9. (D) Z = XQ + YQ
A
B
S
R
Q
Q+
0
0
1
0
0
1
0
0
1
0
1
1
0
1
0
1
0
0
0
1
0
1
1
0
1
0
0
0
0
0
1
0
0
0
1
1
1
1
1
1
0
´
Comparing from the truth table of J–K FF
1
1
1
1
1
´
Y = J, X = K
X
Y
Z
0
0
Q
0
1
0
1
0
1
1
1
Q1
Fig. S4.3.9
Fig. S4.3.5
Q + = AB + AQ = AB + BQ Page 221
K
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GATE EC BY RK Kanodia
Sequential Logic Circuits
Chap 4.3
14. (A) It is a 5 bit ripple counter. At 11000 the output
10. (C)
of NAND gate is LOW. This will clear all FF. So it is a t0
t2
t1
Mod–24 counter. Note that when 11000 occur, the
t3
CLR input is activated and all FF are immediately
Q1
cleared. So it is a MOD 24 counter not MOD 25.
0 1
15. (D) 10-bit ring counter is a MOD–10, so it divides
D2=Q1
the 160 kHz input by 10. therefore, w = 16 kHz. The Q2
four-bit parallel counter is a MOD–16. Thus, the t0
frequency at x = 1 kHz. The MOD–25 ripple counter
t1
produces a frequency at y = 40 Hz. (1 kHz/25 = 40 Hz).
Fig.S4.3.10
11. (B)
The four-bit Johnson Counter is a MOD-8. This, the
Present State
FF Input
Next State
Q A QB
TA TB
Q +A QB+
0
0
0
1
0
1
0
1
1
1
1
0
1
0
1
0
0
0
1
1
1
1
0
0
frequency at z = 5 Hz. 16. (D) Q0 Q0
Q2 Q2
Q1 Q1
J2 K2
J1 K 1
J0 K0
Q2+ Q1+ Q0+ 1 0 1
Fig. S4.3.11
From table it is clear that it is a MOD–3 counter. 12. (B) It is a down counter because 0 state of previous
0 1
1 0
0 1
0 1 0
1 0
0 1
1 0
1 0 1
0 1
1 0
0 1
0 1 0
1 0
0 1
1 0
1 0 1
FFs change the state of next FF. You may trace the following sequence, let initial state be 0 0 0
Fig. S4.3.16
We see that 1 0 1 repeat after every two cycles, hence frequency will be fc / 2 .
FF C
FF B
FF A
JK C
JK B
JK A
C+ B+ A+
111
111
111
111
J 2K 2 = 1 0
Þ
Q2 = 1,
000
000
110
110
J1 K 1 = 0 0
Þ
Q1 = 1,
000
110
111
101
J 0K 0 = 0 0
Þ
Q0 = 0
000
001
110
100
18. (B) In ripple counter delay 4Td = 40 ns.
111
111
111
011
The synchronous counter are clocked simultaneously,
001
000
110
010
then its worst delay will be equal to 10 ns.
001
110
111
001
19. (A) 4 bit uses 4 FF
000
001
110
000
Total delay Ntd = 4 ´ 50 ns = 200 ´ 10 -9 1 f = = 5 Mhz 200 ´ 10 -9
Fig. S4.3.12
13. (C) It is a down counter because the inverted FF
17. (C) At first cycle
output drive the clock inputs. The NAND gate will clear
20. (D) At pulse 1 input,
FFs A and B when the count tries to recycle to 111. This
So contents are 1 0 1 1,
will produce as result of 100. Thus the counting
At pules 2 input 1 Å 1 = 0‘
sequence will be 100, 011, 010, 001, 000, 100 etc.
So contents are 0 1 0 1,
1 Å 0 =1
At pules 3 input 0 Å 1 = 1, contents are 1 0 1 0
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CHAPTER
4.4 DIGITAL LOGIC FAMILIES
Statement for Q.1–2:
4. In the circuit shown in fig. P.4.4.4. the output Z is +5 V
Consider the DL circuit of fig. P4.4.1–2.
+5 V
+5 V +5 V Z
A B
+
C
Fig. P4.4.4
+
V1
+ Vo -
V2 -
-
Fig. P4.4.1-2
1. For positive logic the circuit is a
(A) AB + C
(B) ABC
(C) ABC
(D) ABC
Statement for Q.5–7:
(A) AND
(B) OR
(C) NAND
(D) NOR
Consider the AND circuit shown in fig. P4.4.5–7. The binary input levels are V(0) = 0 V and V(1) = 25 V. Assume ideal diodes. If V1 = V (0) and V2 = V (1), then Vo
2. For negative logic the circuit is a
is to be at 5 V. However, if V1 = V2 = V (1), then Vo is to
(A) AND
(B) OR
(C) NAND
(D) NOR
rise above 5 V.
Vss
3. The diode logic circuit of fig. P4.4.3 is a
20 kW 1 kW
D2
D1
V1
V1
Vo
V2
V2
Vo
1 kW
D1
D2
D0 +5 V
Fig. P4.4.5-7 Fig. P4.4.3
5. If Vss = 20 V
and V1 = V2 = V (1), the diode current
I D1 , I D2 , and I D0 are
Page 224
(A) AND
(B) OR
(A) 1 mA, 1 mA, 4 mA
(B) 1 mA, 1 mA, 5 mA
(C) NAND
(D) NOR
(C) 5 mA, 5mA, 1 mA
(D) 0,
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0,
0
GATE EC BY RK Kanodia
Digital Logic Families
Chap 4.4
6. If Vss = 40 V and both input are at HIGH level then,
+VCC
diode current I D1 , I D2 and I D0 are respectively (A) 0.4 mA, 0.4 mA, 0
RC
(B) 0, 0, 1 mA
(C) 0.4 mA, 0.4 mA, 1 mA
Vo RB1
(D) 0, 0, 0
RB2
V1
V2 Q2
Q1
7. The maximum value of Vss which may be used is (A) 30 V
(B) 25 V
(C) 125 V
(D) 20 V
Fig. P4.4.10-11
10. For positive logic the gate is
8. The ideal inverter in fig. P4.4.8 has a reference
(A) AND
(B) OR
voltage of 2.5 V. The forward voltage of the diode is 0.75
(C) NAND
(D) NOR
V. The maximum number of diode logic circuit, that may be cascaded ahead of the inverter without producing logic error, is +5 V
+5 V
11. For negative logic the gate is (A) AND
(B) OR
(C) NAND
(D) NOR
+5 V
Statement for Q.12–13: Consider the RTL circuit of fig. P4.4.12–13.
+5 V A
Z
+VCC
B C
RC
D
n Stages of Diode Logic
RB1
RB2
V1
Fig. P4.4.8
(B) 4
(C) 5
(D) 9
Vo2
RB3
V2 Q2
Q1
(A) 3
Vo1
Q3
Fig. P4.4.12-13
9. Consider the TTL circuit in fig. P4.4.9. The value of
12. If Vo1 is taken as the output, then circuit is a
V H and VL are respectively
(A) AND
(B) OR
(C) NAND
(D) NOR
+5 V
13. If Vo2 is taken as output, then circuit is a 2 kW
4 kW
Vo
(A) AND
(B) OR
(C) NAND
(D) NOR
Vi
Statement for Q.14–15: Consider the TTL circuit of fig. P4.4.14. If either or
Fig. P4.4.9
(A) 5 V, 0 V
(B) 4.8 V, 0 V
(C) 4.8 V, 0.2 V
(D) 5 V, 0.2 V
both V1
and V2
are logic LOW, Q1
saturation.
+VCC
Statement Q.10–11:
R1
R2
Consider the resistor transistor logic gate of fig.
Vo2
R3 V1 V2
P4.4.10-11.
is driven to
Q2
Q1 Vo1
Fig. P4.4.14-15
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Page 225
GATE EC BY RK Kanodia
Digital Logic Families
22. The circuit shown in fig. P4.4.22 is
25. The circuit shown in fig. P4.4.25. implements the function
+VDD
M2
B
M1
+VDD
C
Y A
Chap 4.4
A
B
D Y
Fig. P4.4.22
(A) NAND
(B) NOR
(C) AND
(D) OR
Fig. P4.4.25
+VDD
Y M2
(A) ( A + B) C + D
(B) ( AB + C) D
(C) ( A + B) C + D
(D) ( AB + C) D
26. Consider the CMOS circuit shown in fig. P4.4.26.
M3
M1
D
C
23. The circuit shown in fig. P4.4.23 acts as a
A
B
A
The output Y is
+VDD
B C
A
Fig. P4.4.23 B
(A) NAND
Y
(B) NOR C
(C) AND
(D) OR B
A
24. The circuit shown in fig. P4.4.24 implements the function
Fig. P4.4.26 +VDD A
B
C
A
B
C
+VDD
(A) ( A + C) B
(B) ( A + B) C
(C) AB + C
(D) AB + C
27. The CMOS circuit shown in fig. P4.4.27 implement Y
A
A
B
B
C
C
+5 V Logic Input A to E
PMOS Network Y
A
B
C
D
Fig. P4.4.24 E
(A) ABC + ABC
(B ABC + ( A + B + C)
(C) ABC + ( A + B + C)
(D) None of the above
Fig. P4.4.27
(A) AB + CD + E
(B) ( A + B)( C + D) E
(C) AB + CD + E
(D) ( A + B)( C + D) E
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Page 227
GATE EC BY RK Kanodia
UNIT 4
V1
V2
Digital Electronics
For n attached gate I o = nI B ( sat ) .
Vo
To assure no logic error Vo = VCC - I o RC > V H = 35 . V V - 35 . 5 - 35 . n £ CC = = 15.6 Þ n £ 15 RC I B ( sat ) 640(0.15m)
Actual
Logic
Actual
Logic
Actual
Logic
VH
1
VH
1
VCE ( sat )
0
VL
0
VL
0
VCE
1
VH
1
VL
0
VCE ( sat )
0
19. (A) Let V1 = V2 = 0 V, then M 3 will be ON, M1 and
VL
0
VH
1
VCE ( sat )
0
M 2 OFF and M 4 ON, hence Vo = -VDD . Let V1 = 0 V and V2 = -VDD then M 3 will be ON, M1 OFF M 4 OFF, M 2
13. (B) The Q3 stage is simply an inverter (a NOT gate).
ON, hence Vo = -VDD. Let V1 = -VDD and V2 = 0 V, then
Thus output Vo2
M 3 OFF, M 4 ON, M 2 OFF hence Vo = -VDD. Finally if
is the logic complement of Vo1 .
Therefore this is a OR gate.
V1 = V2 = -VDD, M 3 and M 4 will be OFF and M1 , M 2
14. (A) When Q1
satisfies the function of a negative NAND gate.
will be ON, hence Vo = 0 V. Thus the given CMOS gate
is saturated, Vo1 is logic LOW
otherwise Vo1 is logic HIGH. The following truth table shows AND logic
20. (C) If V A = -VDD then M1
is ON and VY = 0 V. If
VB = VC = -VDD and V A = 0 V then M 3 and M 2 are ON
V1
V2
Vo1
but M1 is OFF hence VY = 0 V. If V A = 0 V and either or
1
1
1
both VB , VC are 0 V then M1 is OFF and either or both
0
1
0
1
0
0
0
0
0
M2
and M 3
will be OFF, which implies no current
flowing through M 4
hence VY = -VDD . Thus given
circuit satisfies the logic equation A + BC . 21. (A) Let V1 = V2 = 0 V = V (0) then M 4 and M 3 will be OFF hence Vo = VDD = V (1). Let
15. (C) The Q2 stage is simply an inverter. Thus output
ON and M 2 , M1
Vo2 is the logic complement of Vo1 .
V1 = 0 V, V2 = VDD then M 4 and M 2 will be ON but M 3 and
16. (C) If V1 = V2 £ VL , Vo1 » VCC . If V1 ( V2 ) > V H , while V2 ( V1 ) £ VL , Q1 (Q2 )
is ON and Q2 (Q1 )
is OFF and
Vo1 » VCC . If V1 = V2 > V H , both Q1 and Q2 are ON and Vo1 » 2 VCE ( sat ) . The truth table shows NAND logic
M1
will
be
OFF
hence
Vo = 0 = V (0).
Let
V1 = VDD , V2 = 0 V , then M 4 and M 3 will be OFF and M1 ON hence Vo = 0 V = V (0). Finally if V1 = V2 = VDD , M1
and M 2
will be ON but M 4 will be OFF hence
Vo = 0 V = V (0). Thus the given CMOS satisfy the function of a positive NOR gate.
V1
V2
Vo
Actual
Logic
Actual
Logic
Actual
Logic
VL
0
VL
0
VCC
1
VH
1
VL
0
VCC
1
VL
0
VH
1
VCC
1
VH
1
VH
1
2VCE ( sat )
0
22. (A) If either one or both the inputs are V(0) = 0 V the corresponding FET will be OFF, the voltage across the load FET will be 0 V, hence the output is VDD . If boths inputs are V (1) = VDD , both M1 and M 2 are ON and the output is V(0) = 0 V. It satisfy NAND gate. 23. (B) If both the inputs are at V(0) = 0 V,
17. (A) The Q3 stage is simple an inverter. Hence AND
transistor M1
and M 2 are OFF, hence the output is
logic.
V (1) = VDD . If either one or both of the inputs are at
18. (C) For each successive gate, that has a transistor in
output will be V (0) = 0 V. Hence it is a NOR gate.
V (1) = VDD , the corresponding FET will be ON and the
saturation, the current required is I B ( sat ) = Page 230
the
I C ( sat ) VCC - VCE ( sat ) 5 - 0. 2 = = = 0.15 mA b bRC 50( 640)
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GATE EC BY RK Kanodia
Digital Logic Families
Chap 4.4
24. (B) If all inputs A, B and C are HIGH, then input to
30. (A) When an output is LOW, it may be as high as
invertor is LOW and output Y is HIGH. If all inputs are
VOL ( max ) = 0.4 V. The maximum voltage that an input will
LOW, then input to inverter is also LOW and output Y
respond to as a LOW is V IL ( max ) = 0.8 V. A positive noise
is HIGH. In all other case the input to inverter is HIGH
spike can drive the actual voltage above the 0.8 V level
and output Y is LOW.
if its amplitude is greater than V NL = V IL ( max ) - VOL ( max ) = 0.8 - 0.4 = 0.4 V
Y = ABC + ABC = ABC + ( A + B + C)
Hence
31. (B) A positive noise spike can drive the voltage
25. (C) The operation of circuit is given below ABCD
PA
PB
´´´ 1
´
´
´ ´0 0
´
´
0010
PC
PD
above 1.0 V level if the amplitude is greater than Y
NA NB NC ND
´ OFF
V NL = V IL ( max ) - VOL ( max ) = 1 - 0.1 = 0.9 V, A negative noise spike can drive the voltage below 3.5 V
´
´
´
´
´
OFF OFF
HIGH
ON ON OFF ON
OFF OFF ON OFF
HIGH
0110
ON OFF OFF ON
OFF ON ON OFF
LOW
32. (B) V IH ( min ) = VOH ( min ) - V NH = - 0.8 - 0.5 = - 1.3 V
1010
OFF ON OFF ON
ON OFF ON OFF
LOW
V IL ( max ) = VOL ( max ) + V NL = 0.5 + ( -2) = -15 . V
1110
OFF OFF OFF ON
ON ON ON OFF
LOW
ON
ON
ON
LOW
if the amplitude is greater than V NH = VOH ( min ) - V IH ( min ) = 4.9 - 35 . = 1.4 V
33. (C) V NH = VOH ( min ) - V IH ( min ) , V NL = V IL ( max ) - VOL ( max ) V NH = 2.7 (for LS) -2.0 (for ALS) = 0.7 V
Y = ( A + B)C + D
26. (B) The operation of this circuit is given below :
V NL = 0.8 (for ALS) -0.5 (for LS) = 0.3 V 34. (B) V NH = 2.5 (for ALS) - 2.0 (for LS) = 0.5 V
A B C
PA
PB
PC
N A NB NC
´
´ 0
´
´
ON
´
0
0 1
´
1 1
1 ´
1
´
Y
V NL = 0.8 (for LS) - 0.4 (for ALS) = 0.4 V
OFF
HIGH
ON ON OFF
OFF OFF ON
HIGH
´
´
OFF OFF
OFF ´
OFF
ON
ON
LOW
´
ON
LOW
ON
35. (D) V NH ( min ) = 0.5 V ,
V NL ( min ) = 0.3 V
36. (B) fanout (LOW) =
I OL ( max ) 8m = = 80 I IL ( max ) 0.1m
fanout (HIGH) =
Y = ( A + B) C
I OH ( max ) 400m = = 20 I IH ( max ) 20m
The fanout is chosen the smaller of the two. 27. (B) If input E is LOW, output will not be LOW. It 37. (B) In HIGH state the loading on the output of gate
must be HIGH. Option (B) satisfy this condition.
1 is equivalent to six 74LS input load. 28. (A) In this circuit parallel combination are OR gate and series combination are AND gate.
Hence load = 6 ´ I IH = 6 ´ 20m = 120 mA 38. (C) The NAND gate represent only a single input
Hence Y = ( A + B)( C + D)( E + F )
load in the LOW state. Hence only five loads in the
29. (A) When an output is HIGH, it may be as low as VOH ( min ) = 2.4 V. The minimum voltage that an input will
LOW state. load = 5I IL = 5 ´ 0.4 = 2 mA
respond to as a HIGH is V IH ( min ) = 2.0 V. A negative noise spike that can drive the actual voltage below 2.0 V if its amplitude is greater than
*******
V NH = VOH ( min ) - V IH ( min ) = 2.4 - 2.0 = 0.4 V
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Page 231
GATE EC BY RK Kanodia
CHAPTER
4.6 MICROPROCESSOR
1. After an arithmetic operation, the flag register of 8085 mP has the following contents D7
D6
D5
D4
D3
D2
D1
D0
1
0
´
1
´
0
´
1
HLT DSPLY : XRA OUT HLT
A PORT1
The output at PORT1 is (A) 00
(B) FEH
(C) 01H
(D) 11H
The contents of accumulator after operation may be (A) 75
(B) 6C
5. Consider the following 8085 assembly program
(C) DB
(D) B6
MVI A, DATA1 MOV B, A SUI 51H JC DLT MOV A, B SUI 82H JC DSPLY DLT : XRA A OUT PORT1 HLT DSPLY : MOV A, B OUT PORT2 HLT
2. In an 8085 microprocessor, the instruction CMP B has been executed while the contents of accumulator is less than that of register B. As a result carry flag and zero flag will be respectively (A) set, reset
(B) reset, set
(C) reset, reset
(D) set, set
3. Consider the following 8085 instruction MVI MVI ADD ORA
This program will display
A, A9H B, 57H B A
(A) the bytes from 51H to 82H at PORT2 (B) 00H AT PORT1
The flag status (S, Z, CY) after the instruction ORA A is executed, is
(C) all byte at PORT1 (D) the bytes from 52H to 81H at PORT 2
(A) (0, 1, 1)
(B) (0, 1, 0)
(C) (1, 0, 0)
(D) (1, 0, 1)
6. It is desired to mask is the high order bits ( D7 - D4 ) of the data bytes in register C. Consider the following set
4. Consider the following set of 8085 instructions MVI ADI JC OUT
A, 8EH 73H DSPLY PORT1
of instruction (a)
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MOV ANI MOV HLT
A, C F0H C, A Page 239
GATE EC BY RK Kanodia
UNIT 4
(b)
(c)
(d)
Digital Electronics
MOV MVI ANA MOV HLT
A, C B, F0H B C, A
(C) 8529H are complemented and stored at location 529H
MOV MVI ANA MOV HLT
A, C B, 0FH B C, A
10. Consider the sequence of 8085 instruction
MOV ANI MOV HLT
A, C 0FH C, A
(D) 5829H are complemented and stored at location 85892H
MVI ADI MOV HLT
A, 5EH A2H C, A
The initial contents of resistor and flag are as follows
The instruction set, which execute the desired A ´´
operation are (A) a and b
(B) c and d
(C) only a
(D) only d
S 0
CY 0
After execution of the instructions the contents of
A
A B, 4AH 4FH B
The contents of register A and B are respectively
C
S
Z
CY
(A) 10H
10H
0
0
1
(B) 10H
10H
1
0
0
(C) 00H
00H
1
1
0
(D) 00H
00H
0
1
1
(A) 05, 4A
(B) 4F, 00
11. It is desired to multiply the number 0AH by 0BH
(C) B1, 4A
(D) None of the above
and store the result in the accumulator. The numbers
8. Consider the following 8085 assembly program : MVI MOV MOV MVI OUT HLT
B, 89H A, B C, A D, 37H PORT1
are available in register B and C respectively. A part of the 8085 program for this purpose is given below : A, 00H MVI LOOP : ----------------------------------------------------------------------HLT END
The output at PORT1 is (A) 89
(B) 37
(C) 00
(D) None of the above
The sequence of instruction to complete the program would be
9. Consider the sequence of 8085 instruction given
(A)
JNZ ADD DCR
LOOP B C
(B)
ADD JNZ DCR
B LOOP C
(C)
DCR JNZ ADD
C LOOP B
(D)
ADD DCR JNZ
B C LOOP
below LXI MOV CMA MOV
H, 9258H A, M M, A
By this sequence of instruction the contents of memory location (A) 9258H are moved to the accumulator (B) 9258H are compared with the contents of the accumulator Page 240
Z 0
register and flags are
7. Consider the following 8085 instruction XRA MVI SUI ANA HLT
C ´´
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GATE EC BY RK Kanodia
Microprocessor
Chap 4.6
12. Consider the following assembly language program:
(A) A7H
(B) 98H
MVI MOV START : JMP MVI XRA OUT HLT NEXT : XRA JP OUT HLT
(C) 47H
(D) None of the above
B, 87H A, B NEXT B, 00H B PORT1
15. The memory requirement for this program is
B START1 PORT2
16. The instruction, that does not clear the accumulator
(A) 20 Byte
(B) 21 Byte
(C) 23 Byte
(D) 18 Byte
of 8085, is
The execution of the above program in an 8085 will result in
(A) XRA A
(B) ANI 00H
(C) MVI A, 00H
(D) None of the above
(A) an output of 87H at PORT1
17. The contents of some memory location of an 8085 mP
(B) an output of 87H at PORT2
based system are shown
(C) infinite looping of the program execution with accumulator data remaining at 00H
Address Hex.
Contents (Hex.)
(D) infinite looping of the program execution with accumulator data alternating between 00H and 87H.
3000
02
3001
30
13. Consider the following 8085 program
3002
00
MVI ORA JM OUT CMA DSPLY : ADI OUT HLT
3003
30
A, DATA1 A, DSPLY PORT1 01H PORT1
Fig. P4.6.17
The program is as follows
If DATA1 = A7H, the output at PORT1 is (A) 47H
(B) 58H
(C) 00
(D) None of the above
LHLD MOV INX MOV LDAX MOV INX LDAX MOV
Statement for Q.14–15: Consider the following program of 8085 assembly language: LXI LDA MOV LDA CMP JZ JC MOV JMP MOV FNSH : HLT
3000H E, M H D, M D L, A D D H, A
The contents if HL pair after the execution of the
H 4A02H 4A00H B, A 4A01H B FNSH GRT M, A FNSH M, B
program will be (A) 0030 H
(B) 3000 H
(C) 3002 H
(D) 0230H
18. Consider the following loop
14. If the contents of memory location 4A00H, 4A01H and 4A02H, are respectively A7H, 98H and 47H, then
XRA LXI LOOP : DCX JNZ
A B, 0007H B LOOP
This loop will be executed
after the execution of program contents of memory
(A) 1 times
(B) 8 times
location 4A02H will be respectively
(C) 7 times
(D) infinite times
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Page 241
GATE EC BY RK Kanodia
UNIT 4
19. Consider the following loop LXI LOOP : DCX MOV ORA JNZ
24. Consider the following program MVI RRC RRC
H, 000AH B A, B C LOOP
(A) 1 time
(B) 10 times
(C) 11 times
(D) infinite times
execution of program will be (A) 08H
(B) 8CH
(C) 12H
(D) None of the above
25. Consider the following program
20. The contents of accumulator after the execution of following instruction will be A, A7H A
(A) CFH
(B) 4FH
(C) 4EH
(D) CEH
RJCT :
21. The contents of accumulator after the execution of following instructions will be MVI ORA RAL
A, BYTE1
If BYTE1 = 32H, the contents of A after the
This loop will be executed
MVI ORA RLC
Digital Electronics
MVI MVI MVI CMP JC CMP JNC OUT HLT SUB A OUT HLT
PORT1
If the following sequence of byte is loaded in accumulator,
A, B7H A
DATA (H)
(A) 6EH
(B) 6FH
(C) EEH
(D) EFH
58
64
73
C8
FA
(A) 00, 00, 73, B4, 00, FA (B) 58, 64, 00, 00, C8, FA (C) 58, 00, 00, 00, C8, FA
of the following program will be
(D) 00, 64, 73, B4, 00, FA
A, C5H A
26. Consider the following instruction to be executed by a 8085 mp. The input port has an address of 01H and has a data 05H to input:
(A) 45H
(B) C5H
(C) C4H
(D) None of the above
IN ANI
23. Consider the following set of instruction MVI RLC MOV RLC RLC ADD
B4
then sequence of output will be
22. The contents of the accumulator after the execution
MVI ORA RAL
01H 80H
After execution of the two instruction the contents
A, BYTE1
of flag register are
B, A
(A)
1
0
´
1
´
1
´
0
B
(B)
0
1
´
0
´
1
´
0
(C)
0
1
´
1
´
1
´
0
(D)
0
1
´
1
´
0
´
0
If BYTE1 = 07H, then content of A, after the execution of program will be
Page 242
A, DATA B, 64H C, C8H B RJCT C RJCT PORT1
(A) 46H
(B) 70H
(C) 38H
(D) 68H www.gatehelp.com
GATE EC BY RK Kanodia
Microprocessor
ORA JM
A DSPLY
OUT DSPLY : CMA ADI OUT HLT
PORT1 01H PORT1
;Set flag ;If negative jump to ;DSPLY ;A ® PORT1 ;Complement A ;A+1 ® A ;A ® PORT1
Chap 4.6
18. (A) The instruction XRA will set the Z flag. LXI and DCX does not alter the flag. Hence this loop will be executed 1 times. 19. (B) LXI LOOP : DCX
This program displays the absolute value of DATA1. If
MOV ORA JNZ
DATA1 is negative, it determine the 2’s complements and display at PORT1.
B, 000AH B A, B C LOOP
;00 ® C, 0AH ® B ; CB – 1 ® B, ;flag not affected ;B ® A ;A OR C ® A, set flag
Hence this loop will be executed 0AH or ten times. 14. (A)
LXI
H, 4A02H
LDA
4A00H
MOV LDA
B, A 4A01H
CMA JZ
B FNSH
JC
GRT
MOV JMP GRT : MOV FNSH : HLT
M, A FNSH M, B
;Store destination address ;in HL pair ;Load A with contents of ;memory location A00H ;A ® B ;Load A with contents of ;memory location 4A01H ;Compare A and B ;Jump to FNSH if two ;number are equal ;If CY = 1, (A 98H
20. (B)
Thus A7H will be stored at 4A02H.
15. (C) Operand R, M or implied : 1–Byte instruction Operand 8–bit
: 2–Byte instruction
Operand 16–bit
: 3–Byte instruction
CY
Before RAL
10110111
0
After RAL
01101110
1
22. (A) RRC instruction rotate the accumulator right and D0 is placed in D7 . MVI ORA RAL
3–Byte instruction are: LXI, LDA, JZ, JC, JMP
A, C5H A
RRC
;C5H ® A ;Reset Carry flag ;Rotate A left through ;carry, A = 8AH ;Rotate A right, A = 45H
P–Byte instruction are : MOV, CMP, HLT Hence memory = 3 ´ 6 + 1 ´ 5 = 23
Byte.
23. (A) This program multiply BYTE1 by 10. Hence content of A will be 46H.
16. (D) All instruction clear the accumulator
17. (C)
XRA ANI MVI
A 00H A
LHLD MOV INX MOV LDAX MOV INX LDAX MOV
3000H E, M H D, M D L, A D D H, A
;A Å A ;A AND 00 ;00 ® A ;(3000A) ® HL = 3002H ;(3002H) ® E = 00 ;HL +1 ® HL = 3003H ;M ® D=(3003H) = 30H ;(DE) ® A=(3000H) = 02H ;A ® L = 02H ;DE +1 ® DE = 3001H ;(DE) ® A = (3001) = 30H ;A ® H = 30H
Hence HL pair contain 3002H.
07H = 0710 ,7 ´ 10 = 70, 7010 = 46H 24. (B) Contents of Accumulator A = 0011 0010 After First RRC
= 0001 1001
After second RRC
= 1000 1100
25. (D) This program will display the number between 64H to C8H including 64H. C8H will not be displayed. Thus (D) is correct option. 26. (C) 05H AND 80H =00
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GATE EC BY RK Kanodia
UNIT 4
After the ANI instruction S, Z and P are modified to reflect the result of operation. CY is reset and AC is set . Thus, S = 0, Z = 1, AC = 1, P =1, CY = 0 27. (B)
ACI
;A + 56H + CY ® A
56H
37H + 56H + 1 =8EH 28. (C) Instruction load the register pairs HL with 01FFH. SHLD instruction store the contents of L in the memory location 2050H and content of H in the memory location 2051H. Contents of HL are not altered. 29. (B) At a time 8085 can drive only a digit. In a second each digit is refreshed 500 times. Thus time given to 1 each digit = = 0.4 ms. (5 ´ 500) 30. (C) The stack pointer register SP point to the upper memory location of stack. When data is pushed on stack, it stores above this memory location. 31. (B) Line 5 push the content of HL register pair on stack. The contents of L will go to 03FFH and contents of H will go to 03FEH. Hence memory location 03FEH contain 22H. 32. (C) Contents of register pair B lie on the top of stack when POP H is executed, HL pair will be loaded with the contents of register pair B. 33. (C) The instruction PUSH B store the contents of BC at stack. The POP PSW instruction copy the contents of BC in to PSW. The contents of register C will be copied into flag register. D0 = 1 = carry flag,
D6 = 0 = zero flag.
Hence zero flag will be reset and carry will be set. 34. (A)
MVI A DATA1 ORA A JP DSPLY
XRA DSPLY OUT HLT
A PORT1
;DATA1 ® A ; Set flag ;If A is positive, then ;jump to DSPLY ; Clear A ; A ® PORT2
If DATA1 is positive, it will be displayed at port1 otherwise 00. ********************
Page 246
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Digital Electronics
GATE EC BY RK Kanodia
UNIT 5
ì e2 t, t < 0 30. y( t) = u( t) * h( t) , where h( t) = í -3t îe , t >0 1 5 1 (A) e -2 t u( - t - 1) + - e -3t u( -t) 2 6 3
Signal & System
(A)
5 3 1 13 -4 t sin t + cos t + e - t e , t ³ 0 34 34 6 61
(B)
5 3 13 -4 t 1 - t sin t + cos t e + e , t ³ 0 34 34 51 6
(B)
1 2t 5 1 e u( -t - 1) + - e-3t u( -t) 2 6 3
(C)
3 5 13 -4 t 1 - t sin t + cos t e + e , t ³ 0 34 34 51 6
(C)
1 2t 1 e + [5 - 3e 2 t - 2 e -3t ]u( t) 2 6
(D)
3 5 1 13 -4 t sin t + cos t + e -4 t e , t ³ 0 34 34 6 51
(D)
1 2t 1 e + [5 - 3e 2 t - 2 e -3t ]u( -t) 2 6
37.
d 2 y ( t) dy ( t) +6 + 8 y( t) = 2 x( t), dt 2 dt
Statement for Q.31-34:
y (0 - ) = -1,
The impulse response of LTI system is given. Determine the step response.
(A)
2 - t 5 -2 t 5 -4 t e - e + e , t ³0 3 2 6
31. h( t) = e - |t |
(B)
2 5 -2 t 5 -4 t + e + e , t ³ 0 3 2 6
(A) 2 + e t - e - t
(B) e t u( -t + 1) + 2 - e - t
(C) e t u( -t + 1) + [2 - e - t ]u( t)
(D) e t + [2 - e - t - e t ]u( t)
(C) 4 + 5( 3e -2 t + e -4 t ) , t ³ 0 (D) 4 - 5( 3e -2 t + e -4 t ), t ³ 0
32. h( t) = d( 2 ) ( t) (A) 1
(B) u( t)
(C) d( 3) ( t)
(D) d( t)
38.
d 2 y( t) 3dx( t) , + y( t) = dt 2 dt y (0 - ) = -1,
33. h( t) = u( t) - u( t - 4) (A) tu( t) + (1 - t) u( t - 4)
(B) tu( t) + (1 - t) u( t - 4)
(C) 1 + t
(D) (1 + t) u( t)
(A) sin t + 4 cos t - 3te -3t + t, t ³ 0 (B) 4 sin t - cos t - 3te - t , t ³ 0 (D) 4 sin t + cos t - 3te - t , t ³ 0
(A) u( t)
(B) t
(C) 1
(D) tu( t)
39. The raised cosine pulse x( t) is defined as ì 1 ï (cos wt + 1) , x ( t) = í 2 ïî 0,
Statement for Q.35-38: The system described by the differential equations has been specified with initial condition. Determine the output of the system and choose correct option. dy( t) + 10 y( t) = 2 x( t), y(0 - ) = 1, x( t) = u( t) dx
(A) 15 (1 + 4 e -10 t ) u( t)
(B) 15 (1 + 4 e -10 t )
(C) - 15 (1 + 4 e -10 t ) u( t)
(D) - 15 (1 + 4 e -10 t )
-
p p £t£ w w otherwise
The total energy of x ( t) is 3p 3p (B) (A) 4w 8w (C)
3p w
(D)
3p 2w
40. The sinusoidal signal x( t) = 4 cos (200 t + p 6) is
d 2 y( t) dy( t) dx( t) , 36. +5 + 4 y( t) = 2 dt dt dt dy( t) = 1, x( t) = sin t u( t) y (0 - ) = 0, dt 0 -
Page 252
dy( t) = 1, x( t) = 2 te- t u( t) dt 0 -
(C) sin t - 4 cos t + 3te -3t + t, t ³ 0
34. h( t) = y( t)
35.
dy( t) = 1, x( t) = e - t u( t) dt 0 -
passed through a square law device defined by the input output relation y ( t) = x 2 ( t). The DC component in the signal is (A) 3.46
(B) 4
(C) 2.83
(D) 8
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Continuous-Time Systems
41. The impulse response of a system is h( t) = d( t - 0.5). If two such systems are cascaded, the impulse response
Chap 5.1
46. The y( t) = x( t) * h( t) is y(t)
y(t)
of the overall system will be (A) 0.58( t - 0.25)
(B) d( t - 0.25)
(C) d( t - 1)
(D) 0.5 d( t - 1)
a
t
1+a
a
t
y(t)
h(t)
a 1
t
5
3
t
1+a 2
1+a
t
a
1-a
(C)
Fig P5.1.42
The output of the system is zero every where except for the
value of a is (A) 1
(B) 2
(C) 1 < t < 5
(D) 1 < t < 8
(C) 3
(D) 0
S1 : h1 ( t) = e - (1 - 2 j ) t u( t)
48. Consider the signal x( t) = d( t + 2) - d( t - 2).The value of E¥ for the signal y( t) =
S2 : h2 ( t) = e cos 2 t u( t) The stable system is (A) S1
(B) S2
(C) Both S1 and S2
(D) None
ò x( t) dt
is
(A) 4
(B) 2
(C) 1
(D) ¥
49. The
response of a system S to a complex input
x( t) = e j 5t is specified as y( t) = te j 5t . The system
44. The non-invertible system is (B) y( t) =
t
ò
(A) is definitely LTI x( t) dt
(B) is definitely not LTI
-¥
dx( t) dt
t
-¥
-t
(A) y( t) = x( t - 4)
t
47. If dy( t) dt contains only three discontinuities, the
(B) 0 < t < 8
43. Consider the impulse response of two LTI system
1
(D)
(A) 0 < t < 5
(C) y( t) =
1
(B)
y(t)
impulse response h( t) of the system.
1-a
(A)
42. Fig. P5.1.40 show the input x( t) to a LTI system and x(t)
1
(D) None of the above
(C) may be LTI (D) information is insufficient
45. A continuous-time linear system with input x( t) and output y( t) yields the following input-output pairs: x( t) = e j 2 t Û y( t) = e j 5t
(B) is definitely not LTI
If x1 ( t) = cos (2 t - 1), the corresponding y1 ( t) is (B) e- j cos (5 t - 1) (D) e j cos (5 t - 1)
Suppose that 0 £ t £1 elsewhere
(C) may be LTI (D) information is insufficient. 51. The auto-correlation of the signal x( t) = e - t u( t) is
Statement for Q.46–47:
ì 1, x( t) = í î 0,
response of a system S to a complex input
x( t) = e j8 t is specified as y( t) = cos 8 t. The system (A) is definitely LTI
x( t) = e - j 2 t Û y( t) = e - j 5t (A) cos (5 t - 1) (C) cos 5( t - 1)
50. The
and
ætö h( t) = xç ÷, where 0 < a £ 1. èaø
(A)
1 t 1 e u( -t) + e - t u( t) 2 2
(B)
et 1 - t + ( e - e t ) u( t) 2 2
(C)
1 -t 1 e u( -t) + e - t u( t) 2 2
(D)
1 t 1 e u( -t) - e - t u( t) 2 2
*********************
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GATE EC BY RK Kanodia
UNIT 5
Signal & System
11. (C)
SOLUTIONS
x(10t)
x(10t-5)
1
1. (A)
2p = 60 p T
2. (C) T1 =
Þ
T=
p 30
1
-0.5 -0.4
0.4 0.5
t
0.1
0.9
1
t
Fig S5.1.11
12. (D) Multiplication by 5 will bring contraction on
2p 2p æ 2p 2p ö s, T2 = s, LCM ç , ÷ = 2p 5 7 7 ø è 5
time scale. It may be checked by x(5 ´ 0.8) = x( 4).
3. (D) Not periodic because of t. 4. (D) Not periodic because least common multiple is
13. (A) Division by 5 will bring expansion on time scale. æ 20 ö It may be checked by y( t) = xç ÷ = x( 4). è 5 ø
infinite. 5. (C) y( t) is not periodic although sin t and 6 cos 2 pt are independently periodic. The fundamental frequency can’t be determined.
¥
ò | x ( t)|dt < ¥
=
-¥
¥
¥
-¥
0
-4 t -4 t ò e u( t) dt = ò e dt =
¥
7. (A) | x( t)| = 1, E¥ =
ò | x( t)| dt 2
1 4
1 T® ¥ 2T
T
5
-5
4
otherwise
=8+ =¥
-¥
ò | x( t)| dt 2
5
4
5
0
0
4
15. (D) E = 2 ò x 2 ( t) dt = 2 ò (1)1 dt + 2 ò (5 - t) 2 dt 2 26 = 3 3
16. (B) Let x1 ( t) = v( t) then y1 ( t) = u{v( t)}
So this is a power signal not a energy. P¥ = lim
-4
for - 5 < t < - 4 for 4 < t 0, y[ n] =
for n - 3 ³ - 3 or
å 1 = n + 1,
k = -3
y[ n] = ( n + 1) u[ n] 31. (A) For n - 1 < 0
12
For
Fig. S5.2.25
26. (B) x[ n] = {1, 2, 1, 1}, h[ n] = {1, -1, 0, 0, 1}
å3
n £5
30. (A) For n - 3 < - 3 or
4
3
k = -¥
Þ
1
3n 6
å 3k =
29. (D) For n - 2 £ 3 or n £ 5 , y[ n] =
Þ
n -1 ³ 0 ì 1, y[ n] = í î 0,
or
or
n -1
(B)
1 , 1+ s
Re ( s) < - 1
(C)
e2 ( s + 1 ) , s+1
Re ( s) < - 1
(D)
1 , 1+ s
Re ( s) > -1
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GATE EC BY RK Kanodia
UNIT 5
35. x( t) = u( -t + 3) 1 - e -3s , (A) s (B)
1-e (C) s (D)
-3s
Re ( s) > 3
(B)
e 3s , s-3
Re ( s) < 3
(C)
e 3( s -1 ) , s-3
Re ( s) > 3
(D)
e 3( s -1 ) , s-3
Re ( s) < 3
Re ( s) < 0
,
-e -3s , s
Re ( s) > 0
(A) e , Re( s) > 0 s
(B) e , Re ( s) < 0
s
(D) None of above
s
(C) e , all s 37. x( t) = sin t u( t) 1 , (A) (1 + s 2 )
Re ( s) > 0
-1 (C) , (1 + s 2 )
Re ( s) < 0
-1 , (1 + s 2 )
Re ( s) > 0
-
t 2
41. x( t) = cos 3t u( -t) * e - t u( t) -s , (A) ( s + 1)( s2 + 9) -s , ( s + 1)( s2 + 9)
-1 < Re ( s) < 0
(C)
s , ( s + 1)( s 2 + 9)
-1 < Re ( s) < 0
(D)
s , ( s + 1)( s 2 + 9)
Re ( s) > 0
42. x( t) = e t sin (2 t + 4) u( t + 2)
-t
38. x( t) = e u( t) + e u( t) + e u( -t) t
(A)
6 s2 + 2 s - 2 , (2 s + 1)( s 2 - 1)
Re ( s) < - 0.5
(B)
6 s2 + 2 s - 2 , (2 s + 1)( s 2 - 1)
-1 > Re ( s) > 1
(C)
1 1 1 , + + s + 0.5 s + 1 s - 1
-1 < Re ( s) < 1
(D)
1 1 1 , + s + 0.5 s + 1 s - 1
-0.5 < Re ( s) < 1
(A)
e 2 ( s -1 ) , ( s - 1) 2 + 4
Re ( s) > 1
(B)
e 2 ( s -1 ) , ( s - 1) 2 + 4
Re ( s) < 1
(C)
e( s - 2 ) , ( s - 1) 2 + 4
Re ( s) > 1
(D)
e( s - 2 ) , ( s - 1) 2 + 4
43. x( t) = e t (A)
1-s , s+1
Re ( s) > -1
(C)
s -1 , s+1
Re ( s) < - 1
s -1 , s+1
Re ( s) > -1
(1 - s) 1 1 , 0.5 < Re ( s) < 1 + + 2 ( s - 1) + 4 s + 1 s - 0.5
(D)
(B)
(1 - s) 1 1 , -1 < Re ( s) < 1 + + ( s - 1) 2 + 4 s + 1 s - 0.5
Statement for Q.44–49:
(C)
( s - 1) 1 1 , 0.5 < Re ( s) < 1 + + ( s - 1) 2 + 4 s + 1 s - 0.5
bilateral Laplace transform.
40. x( t) = e ( 3t + 6 ) u( t + 3) Page 272
Re ( s) < - 1
(A)
( s - 1) 1 1 , -1 < Re ( s) < 1 + + ( s - 1) 2 + 4 s + 1 s - 0.5
Re ( s) < 1
d -2 t [ e u( -t)] dt
1-s , s+1
(B)
t
39. x( t) = e t cos 2 t u( -t) + e - t u( t) + e 2 u( t)
(D)
Re ( s) > 0
(B)
Re ( s) < 0
1 , (1 + s 2 )
(D)
e 3s , s-3
Re ( s) < 0
36. y( t) = d( t + 1)
(B)
(A) Re ( s) > 0
-e -3s , s
Signal & System
Determine the corresponding time signal for given
44. X ( s) =
e 5s with ROC: Re ( s) < -2 s+2
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GATE EC BY RK Kanodia
UNIT 5
(A)
1 -t e sin t u( t) 2
(D)
58.
62. A stable system has input
(B) 2e- t cos t u( t)
(C) 2 e - t cos t u( t) +
is (A) d( t) - ( e -2 t cos t + e -2 t sin t) u( t)
1 -t e cos t u( t - 1) + 2 e - t sin t u( t - 1) 2
(B) d( t) - ( e -2 t cos t + e -2 t sin t) u( t - 2) (C) d( t) - ( e 2 t cos t + e 2 t sin t) u( t)
d 3 y( t) d 2 y( t) dy( t) + 4 +3 = x( t) 3 2 dt dt dt
(D) d( t) - ( e 2 t cos t + e 2 t sin t) u( t + 2)
All initial condition are zero, x( t) = 10 e -2 t
63. The relation ship between the input x( t) and output y( t) of a causal system is described by the differential
5 é5 ù (A) ê + 5 e - t - 5 e -2 t + e -3t ú u( t) 3 ë3 û
equation
5 é5 ù (B) ê - 5 e - t + 5 e -2 t + e -3t ú u( t) 3 ë3 û 5 5 u( t) - 5 u( t - 1) + 5 u( t - 2) + u( t - 3) 3 3
(D)
5 5 u( t) + 5 u( t - 1) - 5 u( t - 2) + u( t - 3) 3 3
dy( t) + 10 y( t) = 10 x( t) dt The impulse response of the system is
59. The transform function H ( s) of a causal system is
(A) -10 e -10 t u( -t + 10)
(B) 10 e -10 t u( t)
(C) 10 e -10 t u( -t + 10)
(D) -10 e -10 t u( t)
64. The relationship between the input x( t) and output y( t) of a causal system is defined as
2 s2 + 2 s - 2 H ( s) = s2 - 1
d 2 y( t) dy( t) dx( t) . - 2 y( t) = -4 x( t) + 5 2 dt dt dt
The impulse response is
The impulse response of system is
(A) 2d( t) - ( e- t + et ) u( -t) -t
(A) 3e - t u( t) + 2 e 2 t u( -t)
(B) 2d( t) - ( e + e ) u( t) t
-t
(B) ( 3e - t + 2 e 2 t ) u( t)
(C) 2d( t) + e u( t) - e u( -t) t
(C) 3e - t u( t) - 2 e 2 t u( -t)
(D) 2d( t) + ( e- t + et ) u( t)
(D) ( 3e - t - 2 e 2 t ) u( -t)
60. The transfer function H ( s) of a stable system is 2s - 1 H ( s) = 2 s + 2s + 1 The impulse response is (A) 2 u( -t + 1) - 3tu( -t + 1) (B) ( 3te- t - 2 e- t ) u( t) (C) 2 u( t + 1) - 3tu( t + 1) (D) (2 e- t - 3te- t ) u( t) 61. The transfer function H ( s) of a stable system is H ( s) =
s2 + 5 s - 9 ( s + 1)( s 2 - 2 s + 10)
The impulse response is (A) -e - t u( t) + ( e t sin 3t + 2 e t cos 3t) u( t) (B) -e - t u( t) - ( e t sin 3t + 2 e t cos 3t) u( -t) (C) -e - t u( t) - ( e t sin 3t + 2 e t cos 3t) u( t) (D) -e - t u( t) + ( e t sin 3t + 2 e t cos 3t) u( -t) Page 274
x( t) and output
y( t) = e -2 t cos t u( t). The impulse response of the system
1 -t e sin t u( t) 2
(C)
Signal & System
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The Laplace Transform
SOLUTIONS ¥
¥
0
2
1. (B) X ( s) = ò x( t) e - st dt = ò e - st dt =
Chap 5.3
r ( t) = e -2 t u( t)
L ¬¾ ®
v( t) = e -2 t u( t - 1)
e -2 s s
L ¬¾ ®
x( t) = q( t) * v( t)
¥
¥
¥
0
0
0
2. (A) X ( s) = ò x( t) e -3t dt = ò u( t + 2) -3t dt = ò e -3t dt =
1 s
L ¬¾ ®
3. (A) X ( s) = ò e -2 t e - st dt = 0
1 s+2
¥
¥
0
0
- 1 1 - e -2 ( 2 - s ) = 2-s s -2
p( t) dt
1 P ( s) p( t) dt + ò s -¥ s
L ¬¾ ®
Þ
X ( s) =
5 ( e j 5t - e - j 5t ) - st e dt = 2 2 25 j s + 0
q( t) =
d p( t) dt
x( t) = tq( t) 6. (B) X ( s) = ò e - st dt =
1 - e -2 s s
7. (B) p( t) = te - t u( t)
L ¬¾ ®
2
0
d p( t) dt
L ¬¾ ®
Þ P ( s) =
1 ( s + 1) 2
s ( s + 1) 2
X ( s) =
8. (A) p( t) = tu( t)
L ¬¾ ®
P ( s) =
1 s2
q( t) = cos 2 pt u( t)
L ¬¾ ®
Q( s) =
s s + 4 p2
x( t) = p( t) * q( t) X ( s) =
L ¬¾ ®
q( t) = - tp( t)
t n u( t)
X ( s) = P ( s)Q( s)
L ¬¾ ®
P ( s) =
Q( s) = X ( s) =
1 s2
d -2 P ( s) = 3 ds s
d 6 Q( s) = 4 ds s
n! sn + 1
L ¬¾ ®
10. (D) p( t) = u( t) q( t) = u( t - 1)
L ¬¾ ®
L ¬¾ ®
Q( s) = X ( s) = -
P ( s) =
s+1 ( s + 1) 2 + 1
s( s + 1) ( s + 1) 2 + 1 d Q( s) ds
-( s 2 + 4 s + 2) ( s 2 + 2 s + 2) 2
13. (B) X ( s) = A=
L ¬¾ ®
s+3 A B = + ( s + 3s + 2) s + 1 s + 2 2
s+3 s+3 = 2, B = = -1 s + 2 s = -1 s + 1 s = -2
x( t) = [2 e - t - e -2 t ]u( t) 14. (A) X ( s) = 2 -
1 1 1 =2 + ( s + 2) ( s + 3) ( s + 2) ( s + 3)
x( t) = 2 d( t) + ( e-3t - e-3t ) u( t) 2s - 1 A B = + s 2 + 2 s + 1 ( s + 1) ( s + 1) 2
B = (2 s - 2) s = -1 = -3, A = 2 x( t) = x( t) = [2 e - t - 3te - t ]u( t) 16. (B) X ( s) =
5s + 4 A B C = + + 2 s + 3s + 2 s s s + 1 s + 2 3
A = sX ( s) s = 0 = 2, B = ( s + 1) X ( s) s = -1 = 1, C = ( s + 2) X ( s) s = -2 = -3 x( t) = [2 + e - t - 3e -2 t ]u( t)
L ¬¾ ®
L ¬¾ ®
L ¬¾ ®
15. (C) X ( s) =
2
9. (C) p( t) = tu( t)
X ( s) =
L ¬¾ ®
2
1 s( s + 4 p2 )
x( t) = - tq( t)
( s + 3) s[( s + 3) 2 + 4 ]
12. (A) p( t) = e - t cos t u( t)
¥
s+3 ( s + 3) 2 + 4
0
ò
5. (C) X ( s) = ò
Þ
X ( s) = Q( s) V ( s)
2 (2 - s )
0
x( t) =
e- ( s + 2 ) s2
e X ( s) = s+2
Þ
t
4. (C) X ( s) = ò x( t) e - st dt = ò e 2 t u( -t + 2) e - st dt e
V ( s) =
-2 ( s + 1 )
-¥
= ò e t ( 2 - s ) dt =
1 s+2
L 11. (B) p( t) = e -3t cos 2 t u( t) ¬¾ ® P ( s) =
¥
2
R( s) =
P ( s) =
Q( s) =
e- s s2
1 s
17. (C) X ( s) = =
s2 - 3 ( s + 2)( s 2 + 2 s + 1)
A B C + + ( s + 2) ( s + 1) ( s + 1) 2
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Page 275
GATE EC BY RK Kanodia
UNIT 5
A = ( s + 2) X ( s) s = -2 = 1, C = ( s + 1) 2 X ( s) A + B =1 x( t) = [ e
-2 t
Þ
B =0
æsö L 23. (C) P ç ÷ ¬¾ ® ap( at) èaø 1 1 -t L ¬¾ ® e sin 2 t u( t) ( s + 1) 2 + 4 2
= -2
s = -1
-t
- te ]u( t) 3s + 2 3( s + 1) 1 = 2 2 ( s + 1) 2 + 32 s + 2 s + 10 ( s + 1) + 3
18. (A) X ( s) =
2
1 é ù x( t) = ê3e - t cos 3t - e - t sin 3t ú u( t) 3 ë û
Q( s) =
Þ
A = ( s + 2) X ( s) s = -2 = 2 Þ
Þ
C = -2
Þ
x( t) = [2 e -2 t + 2 e - t cos 2 t - e - t sin 2 t ]u( t)
A = ( s + 2) X ( s) s = -2 = 1, A + B = 3 x( t) = [ e
-2 t
+ 2e
-3t
21. (D) X ( s) =
Þ
Þ
-3t
sin t ]u( t)
( s + 2)(2 s 2 + 11s + 16) =2 ( s2 + 5 s + 6) s = -2
B=
( s + 3)(2 s + 11s + 16) = -1 ( s2 + 5 s + 6) s = -3
R( s) = sQ( s)
L ¬¾ ®
e-2 t x( t)
X ( s) s
L ¬¾ ®
p( t) =
ò x( t) dt
-¥ t
ò cos 2 t u( t) dt =
P ( s) s
1 sin 3t u( t) 3
sin 2 t 1 - cos 2 t dt = u( t), 2 4 0
ò
L ¬¾ ®
31. (A) p( t) =
q( t) = - p( t)
32. (A) e - t x ( t)
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L ¬¾ ®
d x( t) dt
y( t) = x( t) * p( t)
é2 t ù x( t) = v( t) + r ( t) = ê sin 3t u( t) + t 2 cos 3t u( t) + e -3t ú u( t) ë3 û
p( t) = x( t - 3)
= -t cos 2( t - 3) u( t - 3).
t2 sin 3t u( t) 3
2t sin 3t u( t) + t 2 cos 3t u( t) 3 1 L V ( s) = ¬¾ ® v( t) = e -3t u( t) s+3
L ¬¾ ®
= cos 2( t - 3) u( t - 3) d L Q( s) = P ( s) ¬¾ ® ds
30. (A) x( t - 2)
d q( t) - q(0 - ) dt
sin 2 t 2
t
29. (C) P ( s) = e -3s X ( s)
q( t) = ( -1) 2 t 2 p( t) =
r ( t) =
t
L ¬¾ ®
-¥
=
Page 276
L ¬¾ ®
L ¬¾ ®
x( t) = 2 d( t) + [2 e -2 t - e-3t ]u( t) + [ e -2 ( t - 2 ) - e -3( t - 2 ) ]u( t - 2)
L ¬¾ ®
ax( at)
1 æ2 ö cos ç t ÷ u( t) 3 è3 ø
L ¬¾ ®
28. (B) P ( s) =
2
d2 P ( s) ds 2
L ¬¾ ®
2 s 2 + 11s + 16 + e -2 s ( s2 + 5 s + 6)
A=
Q( s) =
dx( t) + x( t) dt
L ¬¾ ®
C = -6
cos t - 6 e
1 s2 + 9
x( t) = q( t - 2)
y( t) = ( -2 sin 2 t + cos 2 t ) u( t)
27. (D) X ( s + 2)
B =2
A B e -2 s e -2 s =2 + + + ( s + 2) ( s + 3) ( s + 2) ( s + 3)
22. (C) P ( s) =
L ¬¾ ®
x( t) = - ( t - 2) e ( t - 2 ) u( t - 2)
X ( 3s)
A B( s + 3) C = + + ( s + 2) ( s + 3) 2 + 1 ( s + 3) 2 + 1
q( t) = -tp( t) = -t 2 e - t u( t)
L ¬¾ ®
æsö 26. (B) X ç ÷ èaø
3s 2 + 10 s + 10 20. (B) X ( s) = ( s + 2)( s 2 + 6 s + 10)
10 A + 6 B + 2 C = 10
d P ( s) ds
p( t) = te - t u( t)
L ¬¾ ®
25. (A) sX ( s) + X ( s)
B =2
5 A + 2 B + 2 C = 10
1 ( s + 1) 2
X ( s) = e -2 sQ( s)
A B( s + 1) C = + + 2 2 ( s + 2) ( s + 1) + 2 ( s + 1) 2 + 2 2 A + B=4
1 -0 .5t e sin t u( t) 4
L ¬¾ ®
x( t)
24. (C) P ( s) =
4 s 2 + 8 s + 10 ( s + 2)( s 2 + 2 s + 5)
19. (C) X ( s) =
Signal & System
e -2 s X ( s), Y ( s) =
L ¬¾ ®
L ¬¾ ®
L ¬¾ ®
2 se -2 s s2 + 2
P ( s) = sX ( s)
Y ( s) = P ( s) X ( s) = s( X ( s)) 2 X ( s + 1) =
2( s + 1) ( s + 1) 2 + 2
GATE EC BY RK Kanodia
The Laplace Transform
33. (B) 2 tx( t)
L ¬¾ ®
¥
ò x( t) e
34. (A) X ( s) =
-2
- st
d 4 s2 - 8 X ( s) = 2 ds ( s + 2) 2
Chap 5.3
42. (A) x( t) = e -2 e t + 2 sin (2 t + 4) u( t + 2) p( t + 2) X ( s) =
dt
L ¬¾ ®
e 2 s P ( s),
e 2 s e -2 , ( s - 1) 2 + 4
Re ( s) > 1
-¥ ¥
¥
-2
-2
= ò e - t e - st dt = ò e - t ( s + 1 ) dt =
e2 ( s + 1 ) , Re( s) > - 1 s+1
¥
-3s
-e 35. (B) X ( s) = ò u( -t + 3) e - st dt = ò e - st dt = s -¥ -¥ 36. (C) Y ( s) =
3
¥
ò d( t + 1) e
- st
dt = e s ,
43. (A) p( t) = e -2 t u ( -t) q( t) = Re ( s) < 0
( e - e ) - st e dt 2j 0
37. (B) X ( s) = ò ¥
X ( s) = e 5s P ( s)
¥
1 1 1 , e t ( j - s ) dt e - t ( j + s ) dt = = 2 j ò0 2 j ò0 1 + s2 ¥
-
¥
t 2
38. (D) X ( s) = ò e e - st + ò e - t e - st dt + 0
=
0
Þ
Re ( s) > 0
òee
t - st
dt
-¥
òe
-¥
¥
X ( s) = -
0.5 < Re ( s) < 1
q( t) = p( t + 3) X ( s) =
e 3( s -1 ) , s-3
¬¾® L
L ¬¾ ®
Re ( s) > 3
41. (B) p( t) * q( t)
d2 P ( s) ds 2
1 R( s) s
u( -( t + 5))
p( t) = e 3t u( t)
L ¬¾ ®
x( t) = t 2 e 3t u( t)
Þ
L ¬¾ ®
Re ( s) > -1, Re ( s) < 0 -1 < Re ( s) < 0
P ( s)Q( s)
p( t) = u( t)
L ¬¾ ®
r ( t) = -tq( t) = -tu( t - 3) v( t) =
t
ò r( t) dt
-¥
t
3
e 3s s-3
L ¬¾ ®
q( t) = p( t - 3) = u( t - 3)
v( t) = -ò tdt = -
X ( s) =
1 s
L ¬¾ ®
L ¬¾ ®
1 v( s) s
1 2 ( t - 9) 2
L ¬¾ ®
x( t) = -
t
1 2 ò ( t - 9) 2 -¥
9 é -1 ù x( t) = ê ( t 3 - 27) + ( t - 3) ú u( t - 3) 6 2 ë û
48. (B) X ( s) =
-s æ 1 ö X ( s) = 2 ç ÷ s + 9 è s + 1ø Þ
x( t) = p( t + 5)
47. (C) Right-sided, P ( s) =
Þ
1 P ( s) = s-3 Q( s) = e 3s P ( s) =
-
V ( s) =
40. (C) x( t) = e -3e -3( t + 3) u( t + 3) p( t) = e u( t)
-2 ( t + 5)
Q( s) = e-3s P ( s) d R( s) = Q( s) ds
s -1 1 1 + + ( s - 1) 2 + 4 ( s + 1) s - 0.5
3t
L ¬¾ ®
t 2
( e - e ) - st e dt + ò e - t e- st dt + ò e e - st dt 2j 0 0
Re ( s) < 1, Re ( s) > -1, Re ( s) > 0.5 Therefore
p( t) = -e -2 t u( -t)
x( t) = -u( -t) + u ( -t + 1) + d( t + 2) ¥
- jt
jt
t
1-s 1+ s
46. (D) Left-sided
-0.5 < Re ( s) < 1
39. (A) X ( s) =
x( t) = - e
X ( s) =
Re ( s) > -0.5, Re ( s) > -1, Re ( s) < 1
0
X ( s) = Q( s - 1) =
45. (A) Right-sided 1 L P ( s) = ¬¾ ® ( s - 3)
0
1 1 1 + s + 0.5 s + 1 s - 1
Þ
L ¬¾ ®
-1 , Re ( s) < - 2 s+2
Q( s) = sP ( s)
44. (C) Left-sided 1 L P ( s) = ¬¾ ® s+2
- jt
jt
x( t) = e t q( t)
L ¬¾ ®
P ( s) =
Re ( s) < - 1 thus left-sided .
All s
-¥
¥
d p( t) dt
L ¬¾ ®
-s - 4 -3 2 = + s 2 + 3s + 2 ( s + 1) s + 2
Left-sided, x( t) = 3e - t u( -t) - 2 e -2 t u( -t) 49. (A) X ( s) = Left-sided, www.gatehelp.com
5 1 ( s + 1) ( s + 1) 2
x( t) = -5 u( -t) + te - t u( -t) Page 277
GATE EC BY RK Kanodia
UNIT 5
50. (D) x(0 + ) = lim sX ( s) = x ®¥
h( t) = (2 e - t - 3te - t ) u( t).
s =0 s + 5s - 2 2
61. (A) H ( s) =
s2 + 2 s 51. (A) x(0 + ) = lim sX ( s) = 2 =1 s ®¥ s + 2s - 3 52. (D) x(0 + ) = lim sX ( s) = s ®¥
53. (A) x( ¥) = lim sX ( s) = s ®0
e
-2 s
h( t) = - e - t u( t) + (2 e t cos 3t + et sin 3t) u( -t)
(6s + s ) =0 s2 + 2 s - 2 3
s ®0
2
62. (A) X ( s) =
2 s3 + 3s =0 s2 + 5 s + 1
H ( s) = =1 -
1 , s+1
Y ( s) =
( s + 2) ( s + 2) 2 + 1
Y ( s) ( s + 1)( s + 2) = X ( s) ( s + 2) 2 + 1
( s + 2) 1 ( s + 2) 2 + 1 ( s + 2) 2 + 1
h( t) = d( t) - ( e-2 t cos t + e-2 t sin t) u( t)
e -3s (2 s2 + 1) 1 = s2 + 5 s + 4 4
63. (B) sY ( s) + 10 Y ( s) = 10 X ( s) H ( s) =
-
56. (C) sY ( s) - y(0 ) + 10 Y ( s) = 10( s)
Þ
1 y(0 ) = 1, X ( s) = s 10 1 1 Y ( s) = + = s( s + 1) ( s + 1) s -
Þ
-1 2( s - 1) 3 + + ( s + 1) ( s - 1) 2 + 32 ( s - 1) 2 + 32
System is stable
s+2 54. (C) x( ¥) = lim sX ( s) = 2 =2 s ®0 s + 3s + 1 55. (B) x( ¥) = lim sX ( s) =
Signal & System
Y ( s) 10 = X ( s) s + 10
h( t) = 10 e-10 t u( t)
64. (B) Y ( s)( s 2 - s - 2) = X ( s)(5 s - 4) H ( s) =
y( t) = u( t)
Y ( s) 5s - 4 3 2 = = + X ( s) s 2 - s - 2 s + 1 s - 2
h( t) = 3e - t u( t) + 2 e 2 t u( t).
57. (C) s Y ( s) - 2 s + 2 sY ( s) - 2 + 5 Y ( s) = 1 2
( s 2 + 2 s + 5) Y ( s) = 3 + 2 s 2s + 3 2( s + 1) 1 Y ( s) = 2 = + s + 2 s + 5 ( s + 1) 2 + 2 2 ( s + 1) 2 + 2 2 Þ
y( t) = 2 e - t cos t u( t) +
1 -t e sin t u( t) 2
58. (B) s 3Y ( s) + 4 s 2 Y ( s) + 3sY ( s) = Y ( s) =
***********
10 ( s + 2)
10 A B C D = + + + s( s + 1)( s + 2)( s + 3) s ( s + 1) ( s + 2) s + 3
5 A = sY ( s) s = 0 = , B = ( s + 1) Y ( s) s = -1 = -5, 3 C = ( s + 2) Y ( s) s = -2 = 5, D = ( s + 3) Y ( s) s = 0 = Þ
5 é5 ù y( t) = ê - 5 e - t + 5 e -2 t + e -3t ú u( t) 3 3 ë û
59. (D) For a causal system H ( s) = 2 + Þ
h( t) = 0
for t < 0
1 1 + s + 1 s -1
h( t) = 2d( t) + ( e- t + e t ) u( t)
60. (D) H ( s) = Page 278
5 3
2 3 , System is stable s + 1 ( s + 1) 2 www.gatehelp.com
GATE EC BY RK Kanodia
CHAPTER
5.4 THE Z-TRANSFORM
Statement forQ.1-12: Determine the z-transform and choose correct option. 1. x[ n] = d[ n - k] , k > 0 k
(A) z , k
(C) z ,
z >0 z ¹0
(C) (B) z
-k
,
z >0
(D) z
-k
,
z ¹0
(A) z - k , z ¹ 0
(B) z k , z ¹ 0
(C) z - k , all z
(D) z k , all z
(C)
z , |z|< 1 1 - z -1
(B) (D)
1 , |z|< 1 1 - z -1 z , |z|> 1 1 - z -1
(A)
z - 0.25 , z > 0.25 z 4 ( z - 0.25)
(B)
z - 0.25 , z >0 z 4 ( z - 0.25)
(C)
z 5 - 0.25 5 , z < 0.25 z 3( z - 0.25)
(D)
z 5 - 0.25 5 , all z z 4 ( z - 0.25)
5
(B)
-5 z 2 3 , 3 3-z
(C)
1 1 , 1 -1 1 -1 1- z 1- z 2 4
1 1 0 ( -1) (A) k
k -1
Im
( -1) (B) k
d[ n - 1]
( -1) k (C) d[ n - 1] k
k -1
R1 z - plane
d[ n + 1]
R2 R3 2
( -1) k (D) d[ n + 1] k
Re 1 2
35. If z-transform is given by X ( z) = cos ( z -3), |z|> 0,
Fig. P5.4.37
R1
R2
R3
(A)
x1 [ n]
x2 [ n ]
x3[ n]
(B)
x2 [ n ]
x3[ n]
x1 [ n]
(C)
x1 [ n]
x3[ n]
x2 [ n ]
(D)
x3[ n]
x2 [ n ]
x1 [ n]
The value of x[12 ] is (A) -
1 24
(B)
(C) -
1 6
(D)
1 24 1 6
36. X ( z) of a system is specified by a pole zero pattern in fig. P.5.4.36.
38. Given 7 -1 z 6 X ( z) = 1 -1 öæ 1 -1 ö æ ç 1 - z ÷ç 1 + z ÷ 2 3 è øè ø
Im
1+
z - plane
1 3
Re 2
For three different ROC consider there different Fig. P.5.4.36
solution of signal x[ n] :
Consider three different solution of x[ n]
(a) |z|>
n é æ1ö ù x1 [ n] = ê2 n - ç ÷ ú u[ n] è 3 ø úû êë
n é 1 1 æ -1 ö ù , x[ n] = ê n -1 - ç ÷ ú u[ n] 2 è 3 ø úû êë2
(b) |z|<
é -1 æ -1 ö 1 , x[ n] = ê n -1 + ç ÷ 3 è 3 ø êë2
1 u[ n] 3n 1 x3[ n] = - 2 n u[ n - 1] + n u[ -n - 1] 3 Correct solution is x2 [ n] = - 2 n u[ n - 1] -
Page 282
n
ù ú u[ -n + 1] úû n
(c)
1 1 1 æ -1 ö
1 4
The h[ n] is
The difference equation representation for this system is (A) 4 y[ n] - y[ n - 1] = 3 x[ n - 1]
(A) Stable
(B) Causal
(C) Stable and Causal
(D) None of the above
55. The transfer function of a system is given as
(B) 4 y[ n] - y[ n + 1] = 3 x[ n + 1]
1ö æ 2ç z + ÷ 2ø è . H ( z) = 1 öæ 1ö æ ç z - ÷ç z - ÷ 2 øè 3ø è
(C) 4 y[ n] + y[ n - 1] = - 3 x[ n - 1] (D) 4 y[ n] + y[ n + 1] = 3 x[ n + 1]
Consider the two statements
51. The impulse response of a system is given by
Statement(i) : System is causal and stable.
h[ n] = d[ n] - d[ n - 5 ] The difference equation representation for this
Statement(ii) : Inverse system is causal and stable.
system is
The correct option is
(A) y[ n] = x[ n] - x[ n - 5 ]
(B) y[ n] = x[ n] - x[ n + 5 ]
(A) (i) is true
(C) y[ n] = x[ n] + 5 x[ n - 5 ]
(D) y[ n] = x[ n] - 5 x[ n + 5 ]
(B) (ii) is true
52. The transfer function of a system is given by H ( z) =
z( 3z - 2) 1 z2 - z 4
(C) Both (i) and (ii) are true (D) Both are false 56. The impulse response of a system is given by n
(A) Causal and Stable (B) Causal, Stable and minimum phase (C) Minimum phase (D) None of the above
n
æ -1 ö æ -1 ö h[ n] = 10ç ÷ u[ n] - 9ç ÷ u[ n] è 2 ø è 4 ø
The system is
Page 284
3 10 -1 1z + z -2 3
For this system two statement are Statement (i): System is causal and stable Statement (ii): Inverse system is causal and stable.
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GATE EC BY RK Kanodia
The z-Transform
Chap 5.4
62. The impulse response of the system shown in fig.
The correct option is (A) (i) is true
(B) (ii) is true
(C) Both are true
(D) Both are false
P5.4.62 is X(z)
Y(z) z-1
+
z-1
57. The system
z-1
y[ n] = cy[ n - 1] - 0.12 y[ n - 2 ] + x[ n - 1] + x[ n - 2 ]
Fig. P5.4.62
is stable if (A) c < 112 .
(B) c > 112 .
(C) |c|< 112 .
(D) |c|> 112 .
æn ö çç - 2 ÷÷ ø
(A) 2 è 2 (B)
58. Consider the following three systems
y2 [ n] = x[ n] - 0.1 x[ n - 1]
(C) 2 è 2 (D)
y3[ n] = 0.5 y[ n - 1] + 0.4 x[ n] - 0.3 x[ n - 1]
(B) y2 [ n] and y3[ n]
(C) y3[ n] and y1 [ n]
(D) all
1 d[ n] 2
2n 1 [1 + ( -1) n ] u [ n] - d [ n] 2 2
H ( z) =
z z2 + z + 1
is shown in fig. P5.4.63. This system diagram is a X(z)
59. The z-transform of a causal system is given as X ( z) =
(1 + ( -1) n ) u[ n] -
63. The system diagram for the transfer function
The equivalent system are (A) y1 [ n] and y2 [ n]
1 d[ n] 2
2n 1 (1 + ( -1) n ) u[ n] + d[ n] 2 2 æn ö çç - 2 ÷÷ ø
y1 [ n] = 0.2 y[ n - 1] + x[ n] - 0.3 x[ n - 1] + 0.02 x[ n - 2 ]
(1 + ( -1) n ) u[ n] +
. z -1 2 - 15 -1 . z + 0.5 z -2 1 - 15
Y(z)
+
+
+
The x[0 ] is
z-1
z-1
(A) -15 .
(B) 2
(C) 1.5
(D) 0
Fig. P5.4.63
(A) Correct solution 60. The z-transform of a anti causal system is X ( z) =
12 - 21z 3 - 7 z + 12 z 2
The value of x[0 ] is 7 (A) 4
(B) Not correct solution (C) Correct and unique solution (D) Correct but not unique solution
(B) 0
(C) 4
*****************
(D) Does not exist
61. Given the z-transforms X ( z) =
z( 8 z - 7) 4z2 - 7z + 3
The limit of x[ ¥ ] is (A) 1
(B) 2
(C) ¥
(D) 0 www.gatehelp.com
Page 285
GATE EC BY RK Kanodia
The z-Transform
z 2 - 3z 1 - 3z -1 = 3 3 z 2 + z -1 1 + z -1 - z -2 2 2 2 1 1 , ROC : 2 , z1 >
1 gives z1 > 2 2
x2 [ n] is left-sided signal 1 1 z 2 < 2, z 2 < gives z 2 < 2 2 x3[ n] is double sided signal 1 1 z 3 > and z 3 < 2 gives < z3 < 2 2 2 38. (B) X ( z) =
2 -1 , + 1 -1 1 1- z 1 + z -1 2 z n
2 æ -1 ö u[ n] - ç ÷ u[ n] 2n è 3 ø
|z|>
1 (Right-sided) Þ 2
x[ n] =
|z|<
1 (Left-sided) Þ 3
é -2 æ -1 ön ù x[ n] = ê n + ç ÷ ú u[ -n - 1] è 3 ø úû êë 2
www.gatehelp.com
Page 287
GATE EC BY RK Kanodia
UNIT 5
n
1 1 2 æ -1 ö 1, Tn ( s) =
5. (D) T( s) = G2 (1 + G1 ) + 1 = 1 + G1 + G1 G2
TN ( s) =
T( s) =
G2 1 + G1 G2 H
7. (D) Apply the feedback formula to both loop and then
=
öæ G2 ÷÷çç + 1 G2 H 2 øè
ö ÷÷ ø
- H 2 G1 -1 = G1 H 2 H1 H1
G G = 1 + H1 + H 2 + H1 H 2 (1 + H1 )(1 + H 2 )
17. (B) Ga = 1, Gb = 1 + 1 = 2, Gc =
1 1 1 1 + + + =1 4 4 4 4
18. (B) P. P1 = ab, D = 1, L = 0 , T = ab
G1 G2 1 + G1 H1 + G2 H 2 + G1 G2 H1 H 2
8. (C) For positive feedback
Q. P1 = a, P2 = b , D = 1, L = Dk = 0, T = a + b a R. P1 = a, L1 = b, D = 1 - b, D1 = 1, T = a-b a S. P1 = a, L1 = ab, D = 1 - ab, D1 = 1, T = 1 - ab
C 6 = =-6 R 1 - 6 ´31
9. (D) For system (b) closed loop transfer function G G + s+1 G + s+1 s+2 , , Hence G = 1 +1= = s+1 s+1 s+1 s+1
Page 332
- H 2 G1 1 + G1 H 2 H1
There are no loop in any graph. So option (B) is correct.
multiply æ G1 T( s) = çç 1 + G1 H1 è
Tn ( s) =
16. (C) P1 = G , L1 = - H1 , L2 = - H 2 , L1 L2 = H1 H 2 , D1 = 1
6. (A) Open-loop gain = G2 Feed back gain = HG1
15. (B) By putting R ( s) = 0
19. (A) Between e1 and e2 , there are two parallel path. Combining them gives ta + tb . Between e2 and e4 there is a path given by total gain tc td . So remove node e3 and
10. (A) In open loop system change will be 10% in C1
place gain tc td of the branch e2 e4 . Hence option (A) is
also but in closed loop system change will be less
correct.
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GATE EC BY RK Kanodia
Transfer Function
20. (A) Option (A) is correct. Best method is to check the
Chap 6.1
28. (A) SFG:
signal flow graph. In block diagram there is feedback
G3
from 4 to 1 of gain -H1 H 2 . The signal flow graph of
G4 G3
option (A) has feedback from 4 to 1 of gain -H1 H 2 .
R
1
21. (C) Consider the block diagram as SFG. There are
G5
G2 -G1
22. (B) Consider the block diagram as a SFG. Two forward path G1 G2 and G3 and three loops
P1 = G2 G5G6 , P2 = G3G5G6 , P3 = G3G6 , P4 = G4 G6 If any path is deleted, there would not be any loop. Hence D1 = D2 = D3 = D4 = 1 C G4 G6 + G3G6 + G3G5G6 + G2 G5G6 = D R
-G1 G2 H 2 , - G2 H1 , - G3 H 2 . There are no nontouching loop. So (B) is correct. 23. (C) P1 = 5 ´ 3 ´ 2 = 30, D = 1 - ( 3 ´ - 3) = 10
29. (A)
C 30 D1 = 1 , = =3 R 10
R
1
-2 s
1 s2
L3 = -4, L4 = -5,
C
s
-1
L1 L3 = 8, D = 1 - ( -2 - 3 - 4 - 5) + 8 = 23,
Fig. S6.1.29
D1 = 1, D2 = 1 - ( -3) = 4, C 24 + 5 ´ 4 44 = = R 24 23
P1 =
1 50 50 × ×s= s2 ( s + 1) s( s + 1)
P2 =
1 50 -100 × × ( -2) = 2 s2 s + 1 s ( s + 1)
L1 =
50 -2 -100 × = s+1 s s( s + 1)
L2 =
1 50 -50 × × s × ( -1) = s2 s + 1 s( s + 1)
L3 =
1 50 100 × × ( -2) × ( -1) = 2 s2 s + 1 s ( s + 1)
P2 = G3G2
L1 = -G3G2 H1 , L2 = -G1 G2 H1 ,
-2 1
50 (s + 1)
24. (A) P1 = 2 ´ 3 ´ 4 = 24 , P2 = 1 ´ 5 ´ 1 = 5
25. (B) P1 = G1 G2 ,
C
Fig. S6.1.28
forward path G1 G2 G3 . So (D) is correct option.
L2 = -3,
1
-1
two feedback loop -G1 G2 H1 and -G2 G3 H 2 and one
L1 = -2 ,
G6
L3 = G4 , D1 = D2 = 1
There are no nontouching loop. P1 D1 + P2 D2 G1 G2 + G2 G3 T( s) = = 1 - ( L1 + L2 + L3) 1 + G1 G2 H1 + G2 G3 H1 - G4 26. (C) P1 = G1 G2 , L1 = -G1 G2 H1 H 2 , L2 = G2 H 2 C( s) G1 G2 = R( s) 1 + G1 G2 H1 H 2 - G2 H 2 27. (B) There is one forward path G1 G2 .
D =1 +
100 50 100 + s( s + 1) s( s + 1) s 2 ( s + 1)
D1 = D2 = 1 C P1 + P2 50( s - 2) = = 3 R D s + s2 + 150 s - 100
Four loops -G1 G4 , - G1 G2 G8 , - G1 G2 G5G7 30. (D) P1 = G1 G2 G3
and -G1 G2 G3G6 G7 .
L1 = - G1 H1 , L2 = - G2 H 2 , L3 = - G3 H 3 R(s) 1
G2
G1
1 C(s)
D = 1 - ( - G1 H1 - G2 H 2 - G3 H 3 ) + G1 G3 H1 H 3
1 -H1
L1 L3 = G1 G3 H1 H 3
H2
Fig. S6.1.27
There is no nontouching loop. So (B) is correct.
D = 1 + G1 H1 + G2 H 2 + G3 H 3 + G1 G3 H1 H 3 D1 = 1 C G1 G2 G3 = R 1 + G1 H! + G2 H 2 + G3 H 3 + G1 G3 H1 H 3
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CHAPTER
6.2 STABILITY
1. Consider the system shown in fig. P6.2.1. The range of K for the stable system is R(s)
+
E(s)
C(s)
K(s2 - 2s + 2)
The range of K to ensure stability is 6 3 (A) K > (B) K < - 1 or K > 8 4 (C) K < - 1
(D) -1 < K <
3 4
1 s2 + 2s + 1
5. Consider a ufb system with forward-path transfer
Fig. P6.2.1
(A) -1 < K < -
1 2
function
(B) -
(C) -1 < K < 1
1 < K 0
(B) K < 0
(C) K > 1
(D) Always unstable
6. The open-loop transfer function of a ufb
The system is stable for
control
system is
(A) K < - 1
(B) K > -1
(C) K < - 2
(D) K > -2
G( s) =
3. The open-loop transfer function with ufb are given below for different systems. The unstable system is
For K > 6, the stability characteristic of the open-loop and closed-loop configurations of the system
(A)
2 s+2
(B)
2 s 2 ( s + 2)
are respectively
(C)
2 s( s + 2)
(D)
2( s + 1) s( s + 2)
(B) stable and stable
(A) stable and unstable
(C) unstable and stable
4. Consider a ufb system with forward-path transfer function
K ( s + 2) ( s + 1)( s - 7)
(D) unstable and unstable 7. The forward-path transfer function of a ufb system is
G( s) =
K ( s + 3)( s + 5) ( s - 2)( s - 4)
G( s) = www.gatehelp.com
K ( s 2 - 4) s2 + 3 Page 335
GATE EC BY RK Kanodia
UNIT 6
For the system to be stable the range of K is 3 (A) K > -1 (B) K < 4 (C) -1 < K <
3 4
Control & System
13. The open-loop transfer function of a ufb system is R(s)
+
C(s)
1 s(s + 1)(s + 5)
K
(D) marginal stable Fig. P6.2.12
8. A ufb system have the forward-path transfer function G( s) =
K ( s + 6) s( s + 1)( s + 3)
G( s) =
K ( s + 10)( s + 20) s 2 ( s + 2)
The closed loop system will be stable if the value of
The system is stable for (A) K < 6
(B) -6 < K < 0
K is
(C) 0 < K < 6
(D) K > 6
(A) 2
(B) 3
(C) 4
(D) 5
9. The feedback control system shown in the fig. P6.2.8. R(s)
+
K(1 + Ts) s2(1 + s)
Statement for Q.14–15:
C(s)
A feedback system is shown in fig. P6.14-15. 2 s
Fig. P6.2.9 R(s) +
is stable for all positive value of K , if (A) T = 0
(B) T < 0
(C) T > 1
(D) 0 < T < 1
+
C(s)
+ s2
1 s+1
Fig. P6.2.14–15
10. Consider a ufb system with forward-path transfer function
14. The closed loop transfer function for this system is K G( s) = ( s + 15)( s + 27)( s + 38)
The system will oscillate for the value of K equal to (A) 23690
(B) 2369
(C) 144690
(D) 14469
(A)
s 5 + s 4 + 2 s 3 + ( K + 2) s 2 + ( K + 2) s + K s3 + s2 + 2 s + K
(B)
2 s 4 + ( K + 2) s 3 + Ks 2 s3 + s2 + 2 s + K
(C)
s3 + s2 + 2 s + K s 5 + s 4 + 2 s 3 + ( K + 2) s 2 + ( K + 2) s + K
(D)
s3 + s2 + 2 s + K 2 s + ( K + 2) s 3 + Ks 2
11. The forward-path transfer function of a ufb system is G( s) =
K ( s - 2)( s + 4)( s + 5) ( s 2 + 3)
1 3 (B) K < 54 40 (C)
K s2
P6.2.15. The value of K is jw
(D) Unstable s
12. The closed loop system shown in fig. P6.2.12 become marginally stable if the constant K is chosen to be
Page 336
(A) 30
(B) -30
(C) 10
(D) -10
Fig. P6.2.15
(A) 4
(B) -4
(C) 2
(D) -2
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UNIT 6
(A) stable
Control & System
32. The closed loop transfer function of a system is
(B) unstable
(C) marginally stable
T( s) =
(D) More information is required.
( s + 8)( s + 6) s 5 - s 4 + 4 s 3 - 4 s 2 + 3s - 2
The number of poles in RHP and in LHP are 27. The forward path transfer of ufb system is 1 G( s) = 2 2 4 s ( s + 1)
(A) 4, 1
(B) 1, 4
(C) 3, 2
(D) 2, 3
33. The closed loop transfer function of a system is
The system is (A) stable
(B) unstable
T( s) =
(C) marginally stable
s 3 + 3s 2 + 7 s + 24 s - 2 s 4 + 3s 3 - 6 s 2 + 2 s - 4 5
The number of poles in LHP, in RHP, and on jw
(D) More information is required
axis are 28. The forward-path transfer function of a ufb system is G( s) =
G( s) 2 s4 + 5 s3 + s2 + 2 s
(B) 0, 1, 4
(C) 1, 0, 4
(D) 1, 2, 2
34. For the system shown in fig. P6.2.34. the number
The system is (A) stable
(A) 2, 1, 2
of poles on RHP, LHP, and imaginary axis are
(B) unstable
R(s)
(C) marginally stable
+
C(s)
507 s4 + 3s3 + 10s2 + 30s + 169
(D) more information is required. 1 s
29. The open loop transfer function of a system is as
Fig. P6.2.34
K ( s + 0.1) s( s - 0.2)( s 2 + s + 0.6)
G( s) H ( s) =
(A) 2, 3, 0
(B) 3, 2, 0
(C) 2, 1, 2
(D) 1, 2, 2
The range of K for stable system will be (A) K > 0.355
(B) 0.149 < K < 0.355
35. A Routh table is shown in fig. P6.2.36. The location
(C) 0.236 < K < 0.44
(D) K > 0.44
of pole on RHP, LHP and imaginary axis are
30. The open-loop transfer function of a ufb control system is given by G( s) =
K s( sT1 + 1)( sT2 + 1)
s7
1
2
s5
1
2
s5
3
4
s4
1
-1
For the system to be stable the range of K is æ 1 1 (A) 0 < K < çç + è T1 T2
ö ÷÷ ø
æ 1 1 (B) K > çç + è T1 T2
(C) 0 < K < T1 T2
ö ÷÷ ø
(D) K > T1 T2
(A) 1, 2, 4
(B) 1, 6, 0
(C) 1, 0, 6
(D) None of the above
36. For the open loop system of fig. P6.2.35 location of
31. The closed loop transfer function of a system is
poles on RHP, LHP, and an jw-axis are
s + 4 s + 8 s + 16 s + 3s 4 + 5 s 2 + s + 3 3
T( s) =
Fig. P6.2.35
2
R(s)
5
Page 338
(A) 3, 2
(B) 2, 3
(C) 1, 4
(D) 4, 1
C(s)
Fig. P6.2.35
The number of poles in right half-plane and in left half-plane are
-8 s6 + s5 - 6s4 + s2 + s - 6
(A) 3, 3, 0
(B) 1, 3, 2
(C) 1, 1, 4
(D) 3, 1, 2 ************
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UNIT 6
11. (C) T( s) = =
Control & System
K 2 æK 2ö + . Then apply feedback formula with ç 2 + ÷ and 2 s s sø ès 1 , and then multiply with s 2 . ( s + 1)
G( s) 1 + G( s)
K ( s - 2)( s + 4)( s + 5) Ks + (7 K + 1) s 2 + 2 Ks + ( 3 - 40 K ) 3
æK 2ö s2 ç 2 + ÷ 2 s 4 + ( K + 2) s 3 + Ks 2 sø ès T( s) = = 1 æK 2ö s3 + s2 + 2 s + K 1+ ç 2 + ÷ s + 1è s sø
Routh table is as shown in fig. S.6.211 s3
K
2K
s2
7K + 1
3 - 40 K
2
s1
54 K - K 7K + 1
s0
3 - 40 K
15. (C) Denominator = s 3 + s 2 + 2 s + K Routh table is as shown in fig. S.6.2.15
Fig. S.6.2.11
K > 0, 7K + 1 > 0
Þ
54 K 2 - K >0 7K + 1
Þ
3 - 40 K > 0
Þ
12. (A) T( s) =
1ü K >- ï 7 ï 1 ï K > ý 54 ï 3 ï K < 40 ïþ
Þ
1 3 0 ,
K < 35
Auxiliary equation 5 s 2 + 35 = 0,
200 K > 0 ® K > 0, 30 K 2 - 140 K > 0 14 Þ K > , 5 satisfy this condition. 3
Page 340
Þ
18. (C) At K = 35 system will oscillate.
Fig. S.6.2.13
14. (B) First combine the parallel loop
35 - K >0 5
K 2 and giving 2 s s
Þ
19. (B) For inner loop K K , Gi ( s) = = ( s - a)( s + 3a)( s + 4 a) P ( s) For outer loop, Go( s) = Ti ( s) =
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s=± j 7
Ti ( s) =
K , P ( s) + K
K P ( s) + K
GATE EC BY RK Kanodia
Stability
To( s) =
K , P ( s) + 2 K
Therefore if inner loop is stable for X < K < Y , then outer loop will be stable for X < 2 K < Y X Y . Þ 0 2 18 K - 109 >0 2K - 1
Chap 6.2
Þ Þ
1 ü 2 ï ý 109 ï K > 18 þ K >
K ( s + 2) s 4 + 3s 3 - 3s 2 + ( K + 3) s + (2 K - 4)
Routh table is as shown in fig. S.6.2.23
2K - 4
s4
1
20
s3
9
K K
1
-3
s3
3
K +3
s2
180 - K 9
s2
- ( K +312 )
2K - 4
s1
K ( K - 99) K -180
s1
K ( K + 33) K + 12
s0
K
s0
2K - 4
For stability 0 < K < 99
-( K + 12) > 0 Þ K < - 12, 2 K - 4 > 0 3
24. (C) T( s) =
Þ K > 2 and K > -33, These condition can not be met simultaneously. System is unstable for any value of K .
2
2
s4
1
-3
s3
3
K +3
s2
- K + 12 3
2K - 4
4
1
1
3
K
1
s1
K ( K + 33) K + 12
K -1 K
1
s0
2K - 4
s s
K ( s + 2) s + 3s - 3s + ( K + 3) s + (2 K - 4) 4
Routh table is as shown in fig. S.6.2.24
21. (D) Routh table is as shown in fig. S.6.2.21
1
1
K -1 - K 2 K -1
0
2K - 4
Fig. S.6.2.24
For K < - 33, 1 sign change
1
For -33 < K < - 12, 1 sign change
Fig. S.6.2.21
K > 0, K - 1 > 0 Þ
K
Fig. S.6.2.23
Fig. S.6.2.20
s2
109 18
s4 + 9 s3 + 20 s2 + Ks + K = 0
s4
s
K >
23. (B) Characteristic equation
Routh table is as shown in fig. S.6.2.20
s
Þ
K >1 ,
For -12 < K < 0, 1 sign change
K -1 - K2 > 0, K -1
For 0 < K < 2,
3 sign change
But for K > 1 third term is always -ive. Thus the three
For K > 2,
condition cannot be fulfilled simultaneously.
Therefore K > 2 yield two RHP pole.
22. (D) Routh table is as shown in fig. S.6.2.22
25. (B) Routh table is as shown in fig. S.6.2.25
2 sign change
s4
1
8
9
s3
4
20
25
s2
3
15
18 K -109 2 K -1
s1
6
25
s0
15
s4
1
4+K
s3
2
s2
2 K -1 2
s1 s0
25
Fig. S.6.2.22
15
ROZ
Fig. S.6.2.25
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UNIT 6
P ( s) = 3s 2 + 15,
d P ( s) = 6 s, No sign change from s 2 to s 0 ds
on jw-axis 2 roots, RHP 0,
Þ
s s
3
2
s1 s
0
Routh table is as shown in fig. S.6.2.29 2
0.4
s3
0.8
K - 0.12 0.1K
35
10
50
s2
0.55 - 125 . K
30
264
s1
-1 .25K 2 + 0 .63K - 0 .066 0 .55-1 .25K
s0
0.1K
-386
264
s4
1
ROZ
264
K > 0, 0.55 - 125 . K > 0 Þ K < 0.44
Two sign change. RHP-2 poles. System is not stable. 27. (C) Closed loop transfer function G( s) 1 T( s) = = 1 + G( s) 4 s 4 + 4 s 2 + 1
-125 . K 2 + 0.63K - 0.066 > 0 ( K - 0.149)( K - 0.355) < 0, 0.149 < K < 0.355 30. (A) Characteristic equation
Routh table is as shown in fig. S.6.2.27
s( sT1 + 1)( sT2 + 1) + K = 0
s4
4
4
1
s3
16
8
ROZ
s2
2
1
s1
46
s0
1
T1 T2 s 3 + ( T1 + T2 ) s2 + s + K = 0 Routh table is as shown in fig. S.6.2.30 s3
T1 T2
1
s2
T1 + T2
K
Fig. S.6.2.27
s1
( T1 + T2 ) - T1 T2K T1 + T2
dp( s) = 16 s 3 + 3s ds
s0
K
ROZ
Fig. S.6.2.30
There is no sign change. So all pole are on jw–axis. So
æ 1 1 ö ÷÷ 0 < K < çç + T T è 1 2 ø
system is marginally stable.
K > 0, ( T1 + T2 ) - T1 T2 K > 0
28. (B) Closed loop transfer function 1 G( s) T( s) = = 4 3 1 + G( s) 2 s + 5 s + s 2 + 2 s + 1
31. (B) Routh table is as shown in fig. S.6.2.31
Routh table is as shown in fig. S.6.2.28
Þ
s5
1
5
1
s4
3
4
3
s4
2
1
s3
5
2
s3
3.67
0
s2
1 5
1
s2
4
3
s1
-23
s1
-2.75
s0
1
s0
3
Fig. S.6.2.28
Page 342
1
Fig. S.6.2.29
Fig. S.6.2.26
P ( s) = 4 s 4 + 4 s + 1,
s( s - 0.2)( s 2 + s + 0.6) + K ( s + 0.1) = 0
s 4 + 0.8 s 3 + 0.4 s 2 + ( K - 0.12) s + 0.1K = 0
Routh table is as shown in fig. S.6.2.26 s
2 RHP poles so unstable. 29. (B) The characteristic equation is 1 + G( s) H ( s) = 0
LHP 2.
26. (B) Closed-loop transfer function is G( s) 240 T( s) = = 4 3 s + 10 s + 35 s 2 + 50 s + 264 1 + G( s)
4
Control & System
1
In RHP -2 poles. In LHP -3 poles.
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Stability
32. (C) Routh table is as shown in fig. S.6.2.32 e
+
-
+
+
s5 s
- -
4
3
4
-1
-4
-2
3
e
1 -2
s
+
-
s2
1-4e e
+
+
s1
2e2 +1-4e 1-4e
-
-
s0
-2
From s 4 row down to s 0 there is one sign change. So LHP–1 + 1 = 2 pole. RHP–1 pole, jw-axis -2 pole. 35. (A) Notice that in s 5 row there would be zero. In this row coefficient of
Corresponding to this pole there is a pole on LHP. Rest 4 out of 6 poles are on imaginary axis. Rest 1 pole is on LHP. 36. (A) Routh table is as shown in fig. S.6.2.36
1
3
2
-2
-6
-4
-2
-3
ROZ
2
-3
-4
1
-
s
s s
3
s s
e
1 3
-4
0
Fig. S.6.2.33
dP ( s) = - 8 s 3 - 12 s , -2 , -3 P ( s) = - 2 s - 6 s , ds 4
2
4
0
No sign change exist from the s row down to the s row.
+ + + +
+
-
-
-
-
+
-
-
+
-
-
Thus, the even polynomial does not have RHP poles. Therefore 4 pole
RHP
1 pole
LHP
0 pole
s6
1
-6
s5
1
0
s4
-6
0
s3
-24
0
s2
e
s1
- 144 e
s0
-6
-6
ROZ
Fig. S.6.2.36
P ( s) = - 6 s 4 - 6,
because of symmetry all four poles must be on jw-axis. jw-axis
P ( s) = s6 + 2 s 4 - s 2 - 2
Corresponding to this pole there is a pole on LHP.
33. (B) Routh table is as shown in fig. S.6.2.33
4
, where
there is one sign change. So there is one pole on RHP.
2 LHP poles.
s5
dP ( s ) ds
have been entered. From s6 to row down to the s 0 row,
Fig. S.6.2.32
3 RHP,
dP ( s) = 12 s 3 + 60 ds
P ( s) = 3s 4 + 30 s 2 + 507,
1
-
+
Chap 6.2
dP ( s) = -24 s 3 , ds
There is two sign change from the s 4 row down to the s 0 (1 sign change)
row. So two roots are on RHS. Because of symmetry rest two roots must be in LHP. From s6 to s 4 there is 1 sign change so 1 on RHP and 1 on LHP.
34. (D) Closed loop transfer function G( s) T( s) = 1 + G( s) + H ( s) =
Total
LHP 3 root,
RHP 3 root.
507 s s 5 + 3s 4 + 10 s 3 + 30 s 2 + 169 s + 507
Routh table is as shown in fig. S.6.2.34
***********
5
1
10
69
s4
3
30
57
s3
12
60
ROZ
s2
15
507
s1
-345.6
s0
507
s
Fig. S.6.2.33
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Time Response
8. A system is shown in fig. P6.3.8. The rise time and
Chap 6.3
d. T( s) =
s+2 s2 + 9
e. T( s) =
( s + 5) ( s + 10) 2
settling time for this system is R(s)
1 s
10 (s + 10)
C(s)
Consider the following response
Fig. P6.3.8
(A) 0.22s, 0.4s
(B) 0.4s, 0.22s
(C) 0.12s, 0.4s
(D) 0.4s, 0.12s
1. Overdamped
2. Under damped
3. Undamped
4. Critically damped.
The correct match is 9. For a second order system settling time is Ts = 7 s and
1
2
3
4
peak time is Tp = 3 s. The location of poles are
(A)
a
c
d
e
(A) -0.97 ± j0.69
(B) -0.69 ± j0.97
(B)
b
a
d
e
(C) -1.047 ± j0.571
(D) -0.571 ± j1.047
(C)
c
a
e
d
(D)
c
b
e
d
10. For a second order system overshoot = 10% and peak time Tp = 5 s. The location of poles are (A) -0.46 ± j0.63
(B) -0.63 ± j0.46
(C) -0.74 ± j0.92
(D) -0.92 ± j0.74
15. The forward-path transfer of a ufb control system is G( s) =
11. For a second-order system overshoot = 12 % and settling time = 0.6 s. The location of poles are (A) -9.88 ± j6.67
(B) -6.67 ± j9.88
(C) -4.38 ± j6.46
(D) -6.46 ± j4.38
The step, ramp, and parabolic error constants are (A) 0, 1000, 0
(B) 1000, 0, 0
(C) 0, 0, 0
(D) 0, 0, 1000
16. The open-loop transfer function of a ufb control
Statement for Q.12–13: A system has a damping ratio of 1.25, a natural
system is G( s) =
frequency of 200 rad/s and DC gain of 1. 12. The response of the system to a unit step input is (A) 1 + (C) 1 +
1000 (1 + 0.1s)(1 + 10 s)
5 -50 t 2 -150 t e - e 3 3
(B) 1 -
1 -100 t 4 -400 t e - e 3 3
(D) 1 +
4 -100 t 1 -400 t e + e 3 3 2 -50 t 5 -150 t e - e 3 3
K (1 + 2 s)(1 + 4 s) s 2 ( s 2 + 2 s + 8)
The position, velocity and acceleration error constants are respectively K , 0 8
(A) 0, 0, 4K
(B) ¥ ,
(C) 0, 4 K , ¥
(D) ¥, ¥,
K 8
13. The system is (A) overdamped
(B) under damped
17. The open-loop transfer function of a unit feedback
(C) critically damped
(D) None of the above
system is G( s) =
14. Consider the following system a. T( s) =
5 ( s + 3)( s + 6)
10( s + 7) b. T( s) = ( s + 10)( s + 20) 20 c. T( s) = 2 s + 6 s + 144
50 (1 + 0.1s)(1 + 2 s)
The position, velocity and acceleration error constants are respectively (A) 0, 0, 250
(B) 50, 0, 0
(C) 0, 250, ¥
(D) ¥, 50, 0
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GATE EC BY RK Kanodia
UNIT 5
Control Systems
Statement for Q.18–19:
D(s)
The forward-path transfer function of a unity R(s)
feedback system is G( s) =
1 (s + 5)
+
100 s(s + 2)
+ +
C(s)
K s ( s + a) n
Fig. P6.3.22
The system has 10% overshoot and velocity error constant K v = 100.
(A) -
49 11
(B)
49 11
18. The value of K is
(C) -
63 11
(D)
63 11
(A) 237 ´ 10
3
(B) 144
(C) 14.4 ´ 10
(D) 237
3
23. The forward path transfer function of a ufb system
19. The value of a is (A) 237 . ´ 10
is
3
(C) 14.4 ´ 10
(B) 237
3
G( s) =
(D) 144
20. For the system shown in fig. P6.3.20 the steady state error component due to unit step disturbance is 0.000012 and steady state error component due to unit ramp input is 0.003. The values of K 1
and K 2
are
K s( s + 4)( s + 8)( s + 10)
If a unit ramp is applied, the minimum possible steady-state error is (A) 0.16
(B) 6.25
(C) 0.14
(D) 7.25
respectively 24. The forward-path transfer function of a ufb system D(s)
R(s)
K1(s + 2) (s + 3)
+
is K2 s(s + 4)
+ +
C(s)
G( s) =
1000( s 2 + 4 s + 20)( s 2 + 20 s + 15) s 3( s + 2)( s + 10)
The system has r ( t) = t 3 applied to its input. The steady state error is
Fig. P6.3.20
(A) 16.4, 1684
(B) 1250, 2.4
(C) 125 ´ 10 , 0.016
(D) 463, 3981
3
21. The transfer function for a single loop nonunity
(A) 4 ´ 10 -4
(B) 0
(C) ¥
(D) 2 ´ 10 -5
25. The transfer function of a ufb system is
feedback control system is G( s) =
G( s) =
1 1 , H ( s) = s2 + s + 2 ( s + 1)
10 5( s + 3)( s + 10)( s + 20) s( s + 25)( s + a)( s + 30)
The value of a to yield velocity error constant The steady state error due to unit step input is 6 6 (A) (B) 7 5 2 (C) 3
(B) 0
(C) 8
(D) 16
26. A system has position error constant K p = 3. The
error due to a unit step input and a unit step
Page 346
(A) 4
(D) 0
22. For the system of fig. P6.3.22 the total steady state disturbance is
K v = 10 4 is
steady state error for input of 8tu( t) is (A) 2.67
(B) 2
(C) ¥
(D) 0
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Time Response
27. The forward path transfer function of a unity
If
Chap 6.3
the
r ( t) = 1 + t +
feedback system is G( s) =
For input of 60u( t) steady state error is (B) 300
(C) ¥
(D) 10
E(s)
G(s)
subjected
(B) 0.1
(C) 10
(D) ¥
to
an
input
32. The system shown in fig. P6.3.32 has steady-state error 0.1 to unit step input. The value of K is R(s)
function is +
is
(A) 0
28. For ufb system shown in fig. P6.3.28 the transfer
R(s)
system
t 2 , t > 0 the steady state error of the
system will be
1000 ( s + 20)( s 2 + 4 s + 10)
(A) 0
1 2
+
C(s)
K (s + 1)(0.1s + 1)
C(s)
Fig. P6.3.32 Fig. P6.3.28
G( s) =
20( s + 3)( s + 4)( s + 8) s 2 ( s + 2)( s + 15)
(B) 0
(C) ¥
(D) 64
(B) 0.9
(C) 1.0
(D) 9.0
Statement for Q.33–34:
If input is 30 t 2 , then steady state error is (A) 0.9375
(A) 0.1
Block diagram of a position control system is shown in fig.P6.3.33–34.
29. The forward-path transfer function of a ufb control
R(s)
+
Ka
1 s(0.5s + 1)
+
C(s)
system is G( s) =
450( s + 8)( s + 12)( s + 15) s( s + 38)( s 2 + 2 s + 28)
sKt
Fig. P6.3.33–34
The steady state errors for the test input 37tu( t) is (A) 0
(B) 0.061
(C) ¥
(D) 609
unit ramp input is
30. In the system shown in fig. P6.3.30, r ( t) = 1 + 2 t, t > 0. The steady state error e( t) is equal to r(t)
+
10(s + 1) s2(s + 2)
e(t)
1 5
(B) 5
(C) 0
(D) ¥
G( s) =
10(1 + 4 s) s 2 (1 + s)
(B) 0.2
(C) ¥
(D) 0
0.7 without affecting the steady state error, then the
c(t)
value of K a and K t are (A) 86, 12.8
(B) 49, 9.3
(C) 24.5, 3.9
(D) 43, 6.4
35. A system has the following transfer function G( s) =
31. A ufb control system has a forward path transfer function
(A) 5
34. If the damping ratio of the system is increased to
Fig. P6.3.30
(A)
33. If K t = 0 and K a = 5, then the steady state error to
100( s + 15)( s + 50) s 4 ( s + 12)( s 2 + 3s + 10)
The type and order of the system are respectively (A) 7 and 5
(B) 4 and 5
(C) 4 and 7
(D) 7 and 4
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Time Response
SOLUTIONS
Chap 6.3
10 1 1 = s( s + 10) s s + 10
8. (A) C( s) = Þ
1. (D) Characteristic equation is s + 9 s + 18.
c( t) = 1 - e -10 t
2
a = 10,
w = 18, 2 xwn = 9 2 n
Therefore x = 106 . ,
wn = 4.24 rad/s
Settling time Ts =
1 0.6 6 ( s + 0.8) 2 + (0.6) 2
2. (A) T( s) =
wn 1 - x2 = 0.6 , Hence wn = 1,
9. (D) xwn =
xwn = 0.8
Ds = { s - ( -3 + j 4)}{ s - ( -3 - j 4)} = ( s + 3) + 4 . 2
= s + 6 s + 25, w = 25 2 xwn = 6 ,
x=
w2n = 4
wn = 2 ,
5. (D) M p = e xp 1-x
2
T( s) =
=3
xp 1 - x2
Þ
=
x = 0.59
-
xp 1 - x2
Þ
x = 0.56, wn =
4 = 1192 . xTs
Note : 32 . , For 0 < x < 0.69 Ts = xwn
5 = 0.05, 100
Ts =
4.5 , For xwn
12. (B) T( s) =
1 = 1.45 0.69
=
Peak time, Tp =
Þ
Therefore Poles = - xwn ± jwn 1 - x2 = - 6.67 ± j9.88
x = 0.5
2 xwn = 2,
x = 0.69,
wn =
xp 1 - x2
1 - x2 = 0.779,
11. (B) 0.12 = e
1 K = 2 1 + G( s) s + 2 s + K
2 xwn = 2,
p Tp
-
Poles = - xwn ± jwn 1 - x2 = - 0.46 ± j0.63
16 4 = ( 4 s 2 + 8 s + 16) ( s 2 + 2 s + 4)
-
wn =
6 = 0.6 2´5
4. (A) T( s) = Þ
2
wn = 5 rad/s
Þ
4 = 0.4s a
4 p = 0.571, wn 1 - x2 = = 1047 . Ts Tp
10. (A) 0.1 = e
3. (A) Characteristic equation is 2 n
2.2 2.2 = = 0.22s a 10
Poles = - 0.571 ± j1.047
x = 0.8
2
Rise time Tr =
p p p = = = 3 sec wd wn - x2 1.45 (1 - 0.69 2 )
40000 w2n = 2 2 s + 2 xwn s + wn s + 500 s + 40000
40000 ( s + 100)( s + 400)
R( s) =
But the peak time Tp given is 1 sec. Hence these two
x > 0.69
40000 1 4 1 = + s( s + 100)( s + 400) s 3( s + 100) 3( s + 400)
r ( t) = 1 -
4 -100 t 1 -400 t e + e 3 3
specification cannot be met. 13. (A) System has two different poles on negative real
K1 , 6. (C) T( s) = 2 s + ( K 2 + s) + K 1 w2n = K 1 ,
14. (A) 1. Overdamped response (a, b)
2 xwn = 1 + K 2
wd = 0.10, wn = 12.5
axis. So response is over damped.
x = 0.6, Þ
wd = wn 1 - 0.6 2 = 10
2. Underdamped response (c)
K 1 = 156.25,
Poles : Two complex in left half plane
2 wn 3 = K 2 + 1 2 ´ 12.5 ´ 0.6 = K 2 + 1 7. (A) M p = e
-
Þ
K 2 = 14
xp 1 - x2
Poles : Two real and different on negative real axis.
3.Undamped response (d) Poles : Two imaginary. 4.Critically damped (e)
, At x = 0,
M p = 1 = 100%
Poles : Two real and same on negative real axis.
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UNIT 5
15. (B) K p = lim G ( s) = 1000 s ®0
K v = lim sG( s) = 0 s ®0
ess = lim sE( s) = lim s ®0
R( s) =
K a = lim s 2 G( s) = 0 s ®0
16. (D) H ( s) = 1,
K p = lim G( s) × H ( s) = ¥ s ®0
s®
s ®0
17. (B) H ( s) = 1,
K 8
K p = lim G( s) × H ( s) = 50 s ®0
K v = lim sG( s) × H ( s) = 0
ess = lim s ®0
sR( s) , 1 + G ( s)
1 6 = K 1 K 2 ( s + 2) K 1K 2 s+ ( s + 3)( s + 4)
6 = 0 .003 125 ´ 10 3 K 2
Þ
K 2 = 0.016
21. (C) E( s) = R( s) - C( s) H ( s) = R( s) -
s ®0
s ®0
1 K 1 K 2 ( s + 2) , G( s) = s2 s( s + 3)( s + 4)
K v = lim sG( s) × H ( s) = ¥ K a = lim s 2 G ( s) × H ( s) =
Control Systems
K a = lim s 2 G( s) × H ( s) = 0.
R( s) G( s) H ( s) R( s) = 1 + G( s) H ( s) 1 + G( s) H ( s)
s ®0
18. (C) System type = 1, so n = 1 K v = lim sG( s) = s ®0
ess = lim sE( s) = lim s ®0
K = 100 a
s ®0
For 10% overshoot, 0.1 = = e T( s) =
-
22. (A) ess = lim
xp 1 - x2
Þ
s ®0
x = 0.6
G( s) K = 2 1 + G ( s) s + as + K
2 xwn = a ,
w2n = K
K ´ 0.6 K = 100 2
Þ Þ
2 ´ 0.6 K = a
R( s) = D( s) =
K = 14400
K = 100, K = 14400, a
14400 = 100 a
Þ
a = 144
Error in output due to disturbance E( s) = TD( s) D ( s), 1 , s
essD = lim sE( s) = lim s × 3 = 0.000012 2 K1
s ®0
Þ
1 3 × TD( s) = lim TD( s) = s ®0 s 2 K1
1 s+5
and
G2 ( s) =
100 s+2
1 s
23. (A) Using Routh-Hurwitz Criterion, system is stable 2000 = 6. 25 4 ´ 8 ´ 10 1 1 = = 0.16 K v 6.25
K v = lim sG( s) = s ®0
24. (A) R( s) =
6 , s4
E( s) =
R( s) 1 + G ( s)
ess = lim sE( s) s ®0
6s s4 = lim 2 s ®0 1000 ( s + 4 s + 20) ( s 2 + 20 s + 15) 1+ s 3( s + 2) ( s + 10) = lim s ®0
K 1 = 125 ´ 10 3
Error due to ramp input Page 350
sR( s) - sD( s) G2 ( s) 1 + G1 ( s) G2 ( s)
minimum possible error
K 2 ( s + 3) s( s + 3)( s + 4) + K 1 K 2 ( s + 2)
s ®0
2 3
100 -49 2 = = 1 100 11 1+ ´ 5 2
maximum
K2 s( s + 4) TD( s) = K 1 K 2 ( s + 2) 1+ s( s + 4)( s + 3)
If D( s) =
=
for 0 < K < 2000
20. (C) If R ( s) = 0
=
1 1 1+ 2 ( s + s + 2) ( s + 1)
1-
ess 19. (D)
G1 ( s) =
where
s s
=
6 1000 ( s 2 + 4 s + 20) ( s 2 + 20 s + 15) s + ( s + 2) ( s + 10) 3
6 0+
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1000 ´20 ´15 2 ´10
= 4 ´ 10 -4
GATE EC BY RK Kanodia
Time Response
25. (A) K v = lim sG( s) 10 4 ´ 3 ´ 10 ´ 20 25 ´ a ´ 30
10 4 =
Þ
a=4 1 = ¥ K v = 0, ess = Kv
26. (C) System is zero type
34. (C) The equivalent open-loop transfer function Ka Ka s(0.5 s + 1) Ge = = sK t s(0.5 s + 1 + K t ) 1+ s(0.5 s + 1)
27. (D) K p = lim G( s) = 5 s ®0
60 = 10 1 + Kp
T( s) =
28. (A) K a = lim s 2 G( s) = 64 s ®0
ess =
30 ´ 2 = 0.9375 64
=
G( s) Ka = 1 + G( s) 0.5 s 2 + s(1 + K t ) + K a
2K a s 2 + 2 s(1 + K t ) + 2 K a
w2n = 2 K a Þ wn =
29. (B) K v = lim sG( s) = 609.02 s ®0
ess =
1 1 s2 = lim = = 0.2 s ®0 5 5 1+ s(0.5 s + 1 ) s×
ess
For input 60u( t), ess =
5 1 , H ( s) = 1, R( s) = 2 s(0.5 s + 1) s
G( s) =
s ®0
Chap 6.3
2K a
2 x wn = 2(1 + K t ) Kt x =1 + = 0.7 2K a
37 = 0.0607 Kv
30. (C) The system is type 2. Thus to step and ramp
ess = lim s ®0
input error will be zero. E( s) = R( s) - C( s) = R( s) -
G( s) R( s) 1 + G ( s)
R( s) 1 + G( s)
=
... (i)
sR( s) 1 , R ( s) = 2 1 + Ge ( s) s
1 1 + Kt = Ka ö æ Ka ÷÷ s çç 1 + s(0.5 s + 1 + K t ) ø è 1 + Kt = = 0. 2 Ka
ess = lim s ®0
1 2 s+2 + = s s2 s2 s+2 E( s) = 10( s + 1) s2 + ( s + 2) R( s) =
ess
...(ii)
Solving (i) and (ii) K a = 24.5 ,
ess ( t) = lim sE( s) = 0 s ®0
. K t = 39
35. (C) The s has power of 4 and denominator has order
31. (C) System is type 2. Therefore error due to 1 + t
of 7. So Type 4 and Order 7.
t2 1 would be . would be zero and due to 2 Ka
36. (D) For 8u( t),
K a = lim s 2 G( s) = 10,
ess ( t) =
s ®0
1 = 0.1 10
ess =
8 = 2. 1 + Kp
For 8tu( t), ess = ¥, since the system is type 0.
Note that you may calculate error from the formula ess ( t) = lim sE( s) = s ®0
37. (A) For equivalent unit feedback system the forward
sR( s) 1 + G( s)
transfer function is Ge =
32. (D) K p = lim G( s) = K s ®0
1 1 ess ( t) = = = 0.1 1 + Kp 1 + K
Þ
K = 9.
s ®0
When K t = 0 and K a = 5
10 ( s + 10 ) s (s + 2 ) 10 ( s + 10 )( s + 3) s (s + 2 )
10( s + 10) 11s + 132 s + 300 2
The system is of Type 0. Hence step input will produce a
33. (B) ess = lim e( t) = lim sE( s) = lim t ®¥
=
G( s) = 1 + G( s) H ( s) - G( s) 1 +
s ®0
sR( s) 1 + G( s) H ( s)
constant error constant.
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GATE EC BY RK Kanodia
CHAPTER
6.5 FREQUENCY-DOMAIN ANALYSIS
Statement for Q.1–2:
4. The gain-phase plots of open-loop transfer function of
An under damped second order system having a
correct sequence of the increasing order of stability of
transfer function of the form T( s) =
four different system are shown in fig. P6.5.4. The these four system will be
Kw2n 2 s + 2 xwn s + w2n
dB B
A
C
40 dB
has a frequency response plot shown in fig.
D
30 dB
P6.5.1–2.
20 dB 10 dB
|T( jw )|
-270o
2.5
-225o
o
-135
-90o
-45o
-20 dB -30 dB
1.0 w
wn
Fig. P6.5.4
Fig. P6.5.1-2
1. The system gain K is (A) 1 (C)
(B) 2 (D)
2
(A) D, C, B, A
(B) A, B, C, D
(C) B, C, A, D
(D) A, D, B, C
1
5. The open-loop frequency response of a
2
feedback system is shown in following table
2. The damping factor x is approximately
w
|G ( jw)|
ÐG ( jw)
(A) 0.6
(B) 0.2
2
8.5
-119°
(C) 1.8
(D) 2.4
3
6.4
-128°
4
4.8
-142°
5
2.56
-156°
6
1.4
-164°
8
1.00
-172°
10
0.63
-180°
3. For the transfer function G( s) H ( s) =
1 s( s + 1)( s + 0.5)
the phase cross-over frequency is (A) 0.5 rad/sec
(B) 0.707 rad/sec
(C) 1.732 rad/sec
(D) 2 rad/sec
Fig. P6.5.5
Page 362
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unity
GATE EC BY RK Kanodia
Frequency-Domain Analysis
The gain margin and phase margin of the system
Chap 6.5
10. The gain margin of the ufb system
are (A) 2 dB, 8°
(B) 2 dB, -172 °
(C) 4 dB, 8°
(D) 4 dB, -172 °
G( s) =
Statement for Q.6–7: Consider the gain-phase plot shown in fig.
2 is ( s + 1)( s + 2)
(A) 1.76 dB
(B) 3.5 dB
(C) -3.5 dB
(D) -1.76 dB
11. The open-loop transfer function of a system is K s(1 + 2 s)(1 + 3s)
G( s) H ( s) =
P6.5.6–7. dB w=2
2 dB
0
The phase crossover frequency is (B) 2.46 rad/sec
(C) 0.41 rad/sec
(D) 3.23 rad/sec
ÐG( jw)
w = 10
-2 dB
(A) 6 rad/sec
12. The open-loop transfer function of a ufb system is
w = 100 o
-270
-180
o
G( s) = -140
o
-90
o
Fig. P6.5.6-7
1+ s s(1 + 0.5 s)
The corner frequencies are
6. The gain margin and phase margin are
(A) 0 and 2
(B) 0 and 1
(A) -2 dB, 40°
(B) 2 dB, 40°
(C) 0 and -1
(D) 1 and 2
(C) 2 dB, 140°
(D) -2 dB, 140°
13. In the Bode-plot of a unity feedback control system,
7. The gain crossover and phase crossover frequency are
the value of magnitude of G( jw) at the phase crossover
respectively
frequency is
(A) 10 rad/sec, 100 rad/sec
(A) 2
1 2
(B) 100 rad/sec, 10 rad/sec (C)
(C) 10 rad/sec, 2 rad/sec
. The gain margin is 1 (B) 2
1 3
(D) 3
(D) 100 rad/sec, 2 rad/sec 8. The phase margin of a system with the open loop transfer function
14. In the Bode-plot of a ufb control system, the value of phase of G( jw) at the gain crossover frequency is -120 °. The phase margin of the system is
G( s) H ( s) =
(1 - s) is (1 + s)( 3 + s)
(A) 68.3°
(B) 90°
(C) 0
(D) ¥
(A) -120 °
(B) 60°
(C) -60 °
(D) 120°
15. The transfer function of a system is given by G( s) =
9. Consider a ufb system having an open-loop transfer function
K 1 ; K < s( sT + 1) T
The Bode plot of this function is G( s) =
K s(0.2 s + 1)(0.05 s + 1)
dB -20 dB/dec
For K = 1, the gain margin is 28 dB. When gain margin is 20 dB, K will be equal to (A) 2
(B) 4
(C) 5
(D) 2.5
dB -40 dB/dec -40 dB/dec
0 dB
0.1 T
1 T
(A) www.gatehelp.com
w
0 dB
0.1 T
1 T
w
(B) Page 363
GATE EC BY RK Kanodia
UNIT 6
dB
dB -20 dB/dec
(A)
100 s + 10
(B)
(C)
1 s + 10
(D) None of the above
-20 dB/dec 0 dB
w
1 T
0.1 T
0 dB
1 T
0.1 T
w
Control & System
10 s + 10
-40 dB/dec
19. Consider the asymptotic Bode plot of a minimum (C)
(D)
phase linear system given in fig. P6.5.19. The transfer function is dB
16. The asymptotic approximation of the log-magnitude versus frequency plot of a certain system is shown in
32
fig. P6.5.16. Its transfer function is
-20 dB/dec
dB 54 dB
6
-40 dB/dec
-20 dB/dec w1
0.1
-60 dB/dec
w2
w
10 -40 dB/dec
-40 dB/dec
-60 dB/dec
0.1
5
2
w
25
(C)
Fig. P6.5.16
50( s + 5) (A) 2 s ( s + 2)( s + 25) (C)
10 s 2 ( s + 5) ( s + 2)( s + 25)
Fig. P6.5.19
8 s( s + 2) (A) ( s + 5)( s + 10)
(B)
20( s + 5) s 2 ( s + 2)( s + 25)
(D)
20( s + 5) s( s + 2)( s + 25)
4( s + 2) s( s + 5)( s + 10)
(B)
4( s + 5) ( s + 2)( s + 10)
(D)
8 s( s + 5) ( s + 2)( s + 10)
20. The Bode plot shown in fig. P6.5.20 represent dB 100 dB
17. For the Bode plot shown in fig. P6.5.17 the transfer
-60 dB/dec
function is
40 dB/dec dB 4
-4
100 ( s + 4)( s + 10)
100 s 2 (1 + 0.1s) 3
(B)
1000 s 2 (1 + 0.1s) 3
100 s 2 (1 + 0.1s) 5
(D)
1000 s 2 (1 + 0.1s) 5
c
-2 0
de
B/
dB /
0d
(C)
w
Fig. P6.5.20
(A)
(B)
100( s + 4) s( s + 10) 2
(C)
(D)
100 s ( s + 4)( s + 10)
Statement for Q.21–22:
Fig. P6.5.17
100 s (A) ( s + 4)( s + 10) 2
w = 10
w
10
de c
0 dB
2
18. Bode plot of a stable system is shown in the fig. P6.3.18. The open-loop transfer function of the ufb
The Bode plot of the transfer function K (1 + sT) is given in the fig. P6.5.21–22. Phase
dB
system is dB -20 dB/dec 20 dB -20 dB/dec 1 T
w
Page 364
Fig. P6.5.21-22
www.gatehelp.com
-45
10 T
°/d
w
Fig. P6.5.18
0.1 T
ec
w
GATE EC BY RK Kanodia
UNIT 6
Statement for Q.29–30:
Control & System
Im
Im 2
2
Consider the Bode plot of a ufb system shown in fig. P6.5.29–30.
w= 0
w=¥
Re
w= 0
w=¥
Re
dB 32 dB
0 dB
(A)
-20 dB/dec
18 dB
-40 dB/dec w1
0.1
(B)
Im
w
Im 2
2
Fig. P6.5.29-30
Re
w= 0
w=¥
w= 0
w=¥
Re
29. The steady state error corresponding to a ramp input is (A) 0.25
(B) 0.2
(C) 0
(D) ¥
(C)
(D)
35. Consider a ufb system whose open-loop transfer 30. The damping ratio is
function is
(A) 0.063
(B) 0.179
(C) 0.483
(D) 0.639
G( s) =
K s( s 2 + 2 s + 2)
The Nyquist plot for this system is
31. The Nyquist plot of a open-loop transfer function
Im
G( jw) H ( jw) of a system encloses the ( -1, j0) point. The
Im
gain margin of the system is (A) less than zero
(B) greater than zero
(C) zero
(D) infinity
w=¥
Re
Re
w=¥
w= 0
32. Consider a ufb system G( s) =
K s(1 + sT1 )(1 + sT2 )(1 + sT3)
w= 0
(A)
(B)
The angle of asymptote, which the Nyquist plot approaches as w ® 0, is
Im
(A) -90 °
(B) 90°
(C) 180°
(D) -45 °
Im
w=¥
33. If the gain margin of a certain feedback system is
Re
Re
w=¥
w= 0
given as 20 dB, the Nyquist plot will cross the negative real axis at the point
w= 0
(A) s = -0.05
(B) s = -0.2
(C) s = -0.1
(D) s = -0.01
(C)
34. The transfer function of an open-loop system is G( s) H ( s) =
36. The open loop transfer function of a system is
s+2 ( s + 1)( s - 1)
G( s) H ( s) =
K (1 + s) 2 s3
The Nyquist plot for this system is
The Nyquist plot will be of the form
Page 366
(D)
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GATE EC BY RK Kanodia
Frequency-Domain Analysis
Chap 6.5
Im
Im
Im w= 0
w= 0
Re
w=¥
10.64
Re
w=¥
w=¥
Re
w= 0
(A)
(B) Im
Im
Fig. P6.5.38
The no. of poles of closed loop system in RHP are w=¥
w=¥
Re
Re
(C)
(B) 1
(C) 2
D) 4
Statement for Q.39–40:
w= 0
w= 0
(A) 0
The open-loop transfer function of a feedback
(D)
control system is 37. For the certain unity feedback system
G( s) H ( s) =
K G( s) = s( s + 1)(2 s + 1)( 3s + 1)
-1 2 s(1 - 20 s)
39. The Nyquist plot for this system is
The Nyquist plot is
Im Im
Im
Im w= 0
w= 0
w=¥ w=¥
Re
Re
w=¥
Re
Re
w=¥
w= 0
w= 0
(A)
(A)
(B)
Im
(B) Im
Im
Im
w=¥
w=¥
Re
Re
w=¥
Re
Re
w=¥
w= 0 w= 0
w= 0
w= 0
(C) (C)
(D)
(D)
38. The Nyquist plot of a system is shown in fig.
40. Regarding the system consider the statements
P6.5.38. The open-loop transfer function is
1. Open-loop system is stable
G( s) H ( s) =
4s + 1 s ( s + 1)(2 s + 1) 2
2. Closed-loop system is unstable 3. One closed-loop poles is lying on the RHP www.gatehelp.com
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GATE EC BY RK Kanodia
UNIT 6
Control & System
Im
The correct statements are (A) 1 and 2
(B) 1 and 3
(C) only 2
-2 3
Im
w=¥
-3 2
Re
w=¥
Re
(D) All
41. The Nyquist plot shown in the fig. P6.5.41 is for (A) type–0 system
(B) type–1 system
(C) type–2 system
(D) type–3 system
w= 0
w= 0
(C)
Statement for Q.42–43: The open-loop transfer function of a feedback system is
(D)
45. The phase crossover and gain crossover frequencies are
K (1 + s) (1 - s)
(A) 1.414 rad/sec, 0.57 rad/sec
42. The Nyquist plot of this system is
(C) 0.707 rad/sec, 0.57 rad/sec
G( s) H ( s) =
(B) 1.414 rad/sec, 1.38 rad/sec
Im
(D) 0.707 rad/sec, 1.38 rad/sec
Im
46. The gain margin and phase margin are w=¥
w= 0
Re
w=¥
w= 0
(A)
Re
(A) -3.52 dB, -168.5 °
(B) -3.52 dB, 11.6 °
(C) 3.52 dB, -168.5 °
(D) 3.52 dB, 11.6 °
(B) **************** Im
Im
w=¥
w= 0
Re
w=¥
w= 0
(C)
Re
(D)
43. The system is stable for K (A) K > 1
(B) K < 1
(C) any value of K
(D) unstable
Statement for Q.44–46: A unity feedback system has open-loop transfer function G( s) =
1 s(2 s + 1)( s + 1)
44. The Nyquist plot for the system is Im
Im
w=¥
w= 0 2 3
w=¥
(A)
Page 368
Re
3 2
w= 0
Re
(B) www.gatehelp.com
GATE EC BY RK Kanodia
Frequency-Domain Analysis
At w = 10 rad/sec gain is 0 dB. Gain cross over frequency
SOLUTIONS 1. (A) T( s) =
w =10 rad/sec.
Kw2n s + 2 xwn s + w2n
8. (D) |GH( jw)|¹ 1, for any value of w. Thus phase
2
margin is ¥.
Kw2n T( jw) = -w2 + 2 jxwn w + w2n
|T( jw)|
2
=
9. (D) For 28 dB gain Nyquist plot intersect the real axis at a, 1 20 log = 28 a
K w ( w - w ) + 4 x2 w2n w2 2
2 n
4 n
2 2
From the fig. P6.5.1–2, |T( j0)| = 1
|T( j0)|
2
=
K 2 wn4 = K2 =1 wn4
Þ
denominator of function |T( jw)| is minimum i.e. when 2
Þ
w = wn
K w K2 |T( jwn )| = 2 = 2 4x w 4x 2
x=
2
4 n 4 n
Þ
. = 35 . dB = 15
ÐGH ( jw) = -180 ° GH ( jw) =
K jw(1 + 2 jw)(1 + 3 jw)
-90 °- tan -1 2 wp - tan -1 3wp = -180 °
- tan -1 2 w - tan -1 w - 90 ° = -180 °
tan -1 2 wp + tan -1 3wp = 90 ° 2 wp + 3wp = tan 90 ° 1 - (2 wp)( 3wp)
tan -1 2 w + tan -1 w = 90 ° 2w + w = tan 90 ° = ¥ 1 - (2 w)( w) 2
-1
11. (C) For phase crossover frequency
At phase cross over point f = -180 °
w=
1 2
-1
f = -90 °- tan -1 2 w - tan -1 w
1 - (2 w) w = 0 Þ
This is achieved if the system gain is increased by factor 0.1 = 2.5. Thus K = 2.5. 0.04
é KT1 T2 ù é(2)(0.5) ù Gain Margin = ê ú = ê 1 + 0.5 ú + T T ë û 2û ë 1
1 jw( jw + 1)(0.5 + jw)
1
a = 0.04
10. (B) Here K = 2, T1 = 1, T2 =
K |T( jwn )| = = 2.5 2x
K = 0.2 5
3. (B) G( jw) H ( jw) =
Þ
For 20 dB gain Nyquist plot should intersect at b, 1 20 log = 20 Þ b = 0.1. b
K =1
2. (B) The peak value of T( jw) occurs when the
w2n - w2 = 0
Chap 6.5
1 - 6 w2p = 0
= 0.707 rad/sec
Þ
12. (D) G( s) = 4. (B) For a stable system gain at 180° phase must be
wp = 0.41 rad/s
s+1 s+1 = s(1 + 0.5 s) æs ö sç + 1 ÷ è2 ø
negative in dB. More magnitude more stability.
The Bode plot of this function has break at w = 1 and
5. (C) At 180° gain is 0.63. Hence gain margin is 1 = 20 log = 4 dB 0.63
w = 2. These are the corner frequencies. 13. (A) G.M. =
At unity gain phase is -172 °, Phase margin = 180 °-172 ° = 8 °
14. (B) P.M. = 180 ° + ÐGH ( jw1 ) = 180 °-120 ° = 60 °
6. (A) At ÐG( jw) = 180 ° gain is -2 dB. Hence gain margin is 2 dB. At 0 dB gain phase is -140 °. Hence phase margin is 180 °-140 ° = 40 °. 7.
(A) At w = 100 rad/sec phase is 180°. Phase cross-
over frequency wp = 100 rad/sec.
1 1 = =2 GH ( jwp) 1 2
15. (D) Due to pole at origin initial plot has a slope of 1 -20 dB/decade. At s = jw = . Slope increases to -40 T 1 dB/decade. At w = , T
|G( jw)| » KT < 1 ,Gain www.gatehelp.com
in dB < 0. Page 369
GATE EC BY RK Kanodia
Frequency-Domain Analysis
27. (C) Initially slope is -20 dB/decade. Hence there is a
Chap 6.5
ÐGH ( jw) = - tan -1
pole at origin and system type is 1. For type–1 system position error coefficient is ¥. 20 log K = 6
Þ
GH ( jw) =
K = 2,
w (2 - w2 ) 2 + 4 w2
At w = 0, 28. (B) The system is type –0,
At w = ¥
1 1 . 20 log K p = 40, K p = 100, estep ( ¥) = = 1 + K p 101
At w = 1,
-20 dB/dec
0.5
0.1
1.4
K 2 18
Ð - 206.6 °,
correct option. 4
w
36. (B) GH ( jw) =
Fig.S6.5.29
K v = 4, eramp ( ¥) =
GH ( jw) = 0 Ð - 270 °, K GH ( jw) = Ð - 153.43°, 5
Due to s there will be a infinite semicircle. Hence (C) is
-40 dB/dec
18 dB 0 dB
GH ( jw) = ¥ Ð - 90 °,
At w = 2, GH ( jw) =
29. (A) The Bode plot is as shown in fig. S6.5.29 32 dB
2w - 90 ° 2 - w2 K
K (1 + w2 ) w3
|GH( jw)| =
1 1 = = 0.25 Kv 4
K (1 + jw) 2 ( jw) 3
ÐGH ( jw) = -270 °+ 2 tan -1 w For w = 0, GH ( jw) = ¥Ð - 270 °
0.5 w 30. (B) From fig. S6.5.29 x = 2 = = 0.179 2 w3 2(1.4)
For w = 1 , ÐGH ( jw) = -180 ° For w = ¥, GH ( jw) = 0 Ð - 90 °
31. (A) If Nyquist plot encloses the point ( -1, j0), the
As w increases from 0 to ¥, phase goes -270 ° to -90 °. Due to s 3 term there will be 3 infinite semicircle.
system is unstable and gain margin is negative. 37. (A) |GH ( jw)| =
K 32.(A) GH ( jw) = jw(1 + jwT1 )(1 + jwT2 ) (1 + jwT2 )
,
ÐGH ( jw) = -90 °- tan -1 w - tan -1 2 w - tan -1 3w ,
K K lim GH ( jw) = lim = lim Ð - 90 ° w ®0 w ®0 jw w ®0 w
For w = 0, GH ( jw) = ¥Ð - 90 °,
Hence, the asymptote of the Nyquist plot tends to an angle of -90 ° as w ® 0.
K 1 + w2 1 + 4 w2 1 + 9 w2
For w = ¥, GH ( jw) = 0 Ð - 360 °, Hence (A) is correct option. 38. (C) The open-loop poles in RHP are P = 0. Nyquist
33. (C) 20 log
1 = 20 GH ( jw)
path enclosed 2 times the point ( -1 + j0). Taking
1 = 10 GH ( jw)
Þ
N = P - Z,
GH ( jw) = 0.1
Since system is stable, it will cross at s = -0.1.
clockwise encirclements as negative N = -2. -2 = 0 - Z ,
Z = 2 which implies that two
poles of closed-loop system are on RHP. -1 , 2 s(1 - 20 s)
s+2 34. (B) GH ( s) = 2 ( s - 1)
39. (B) G( s) H ( s) =
jw + 2 GH ( jw) = ( -1 - w2 )
GH ( jw) =
At w = 0 ,
GH ( jw) = 2 Ð - 180 °
ÐGH ( jw) = 180 °-90 ° - tan -1
At w = ¥,
GH ( jw) = 0 Ð - 270 °
Hence (B) is correct option. 35. (C) GH ( jw) =
K jw( -w + 2 jw + 2) 2
1 2 w 1 + 400 w2 -20 w , 1
At w = 0 GH ( jw) = ¥Ð90 ° At w = ¥ GH ( jw) = 0 Ð180 ° At w = 0.1 GH ( jw) = 2.24 Ð153.43° At w = 0.01 GH ( jw) = 49 Ð9115 . ° www.gatehelp.com
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GATE EC BY RK Kanodia
UNIT 6
40. (C) One open-loop pole is lying on the RHP. Thus
Control & System
The frequency at which magnitude unity is
open-loop system is unstable and P = 1. There is one
Im
clockwise encirclement. Hence N = -1. Z = P - N = 1 - ( -1) = 2, -2 3
Hence there are 2 closed-loop poles on the RHP and
Re
-1 wp
system is unstable.
w1
41. (B) There is one infinite semicircle. Which represent
Phase Margin
single pole at origin. So system type is1. K 1 + w2
42. (D) |GH ( jw)| =
1 + w2
ÐGH ( jw) = tan -1 w - tan -1 -
=K
Fig.S6.5.44
w (1 + w ) (1 + 4 w ) = 1 2 1
-w 1
At w = 2 GH ( jw) = KÐ127 °,
43. (A) RHP poles of open-loop system P = 1, Z = P - N . For closed loop system to be stable,
w ®0
At unit gain w1 = 0.57 rad/sec,
Phase margin = -168.42 °+180 ° = 11.6 ° Note that system is stable. So gain margin and phase margin are positive value. Hence only possible option is
H ( s) = 1
(D).
***************
1 jw(2 jw + 1)( jw + 1) w ®0
1 = ¥Ð - 90 ° jw
lim GH ( jw) = lim w w ®¥
w ®¥
1 = 0 Ð - 270 ° 2( jw) 3
The intersection with the real axis can be calculated as Im{ GH ( jw)} = 0, The condition gives w (2 w2 - 1) = 0 i.e. w = 0,
1 2
,
æ 1 ö -2 GH ç j ÷= 2ø 3 è
With the above information the plot in option (C) is correct. 45. (C) The Nyquist plot crosses the negative real axis 1 rad/sec. Hence phase crossover frequency is at w = 2 wp = Page 372
1 2
3 = 352 . dB. 2
ÐGH ( jw) = -90 °- tan -1 w - tan -1 2 w ,
1 s(2 s + 1)( s + 1)
lim GH ( jw) = lim
2 3
ÐGH ( jw1 ) = -90 °- tan -1 0.57 - tan -1 2(0.57) = -168.42 °
( -1 + j0). It is possible when K > 1.
GH ( jw) =
|GH( jwp)| =
Phase at this frequency is
N =1
There must be one anticlockwise rotation of point
GH ( s) =
,
|GH( jwp)|
Gain Margin = 20 log
At w = ¥ GH ( jw) = KÐ180 °,
1 , s(2 s + 1)( s + 1)
1
46. (D) G.M. = 20 log
At w = 1 GH ( jw) = KÐ90 °,
44. (C) G( s) =
2 1
w2 = 0.326, w1 = 0.57 rad/sec
At w = 0 GH ( jw) = KÐ0 °,
Z = 0, 0 = 1 - N Þ
2 1
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GATE EC BY RK Kanodia
CHAPTER
6.6 DESIGN OF CONTROL SYSTEMS
1. The term ‘reset control’ refers to
(C) is predictive in nature
(A) Integral control
(B) Derivative control
(D) increases the order of the system
(C) Proportional control
(D) None of the above
6. Consider the List I and List II
2. If stability error for step input and speed of response
List I
List II
be the criteria for design, the suitable controller will be
P. Derivative control
1. Improved overshoot response
(A) P controller
(B) PI controller
Q. Integral control
2. Less steady state errors
(C) PD controller
(D) PID controller
R. Rate feed back control
3. Less stable
S. Proportional control
4. More damping
3. The transfer function
1 + 0.5 s represent a 1+ s
The correct match is
(A) Lag network (B) Lead network (C) Lag–lead network (D) Proportional controller
P
Q
R
S
(A)
1
2
3
4
(B)
4
3
1
2
(C)
2
3
1
4
(D)
1
2
4
3
4. A lag compensation network
7. Consider the List–I (Transfer function) and List–II
(a) increases the gain of the original network without affecting stability.
(Controller) List I
List II
(b) reduces the steady state error.
P.
1. P–controller
(c) reduces the speed of response
Q.
(d) permits the increase of gain of phase margin is acceptable.
2. PI–controller
R.
K1s + K2 K 3s
3. PD–controller
S.
K1 K 2s
4. PID–controller
In the above statements, which are correct (A) a and b
(B) b and c
(C) b, c, and d
(D) all
The correct match is P
Q
R
S
(A)
3
4
2
1
5. Derivative control
(B)
4
3
1
2
(A) has the same effect as output rate control
(C)
3
2
1
4
(D)
4
1
2
3
(B) reduces damping
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GATE EC BY RK Kanodia
UNIT 6
Control Systems
8. The transfer function of a compensating network is of
(B) Two–position controller
form (1 + aTs) (1 + Ts). If this is a phase–Lag network,
(C) Floating controller
the value of a should be
(D) Proportional–position controller
(A) greater than 1 (B) between 0 and 1
14. In case of phase–lag compensation used is system,
(C) exactly equal to 1
gain crossover frequency, band width and undamped
(D) exactly equal to 0
frequency are respectively (A) decreased, decreased, decreased
9.
The
poll–zero
configuration
of
a
phase–lead
compensator is given by jw
(A)
(B) increased, increased, increased (C) increased, increased, decreased
jw
(B)
(D) increased, decreased, decreased s
s
15. A process with open–loop model (C)
jw
(D)
jw
s
G( s) = s
Ke - s TD ts + 1
is controlled by a PID controller. For this purpose 10. While designing controller, the advantage of pole– zero cancellation is
(A) the derivative mode improves transient performance
(A) The system order is increased
(B) the derivative mode improves steady state performance
(B) The system order is reduced
(C) the integral mode improves transient performance
(C) The cost of controller becomes low
(D) the integral mode improves steady state performance.
(D) System’s error reduced to optimum levels
The correct statements are 11. A proportional controller leads to
(A) (a) and (c)
(B) (b) and (c)
(A) infinite error for step input for type 1 system
(C) (a) and (d)
(D) (b) and (d)
(B) finite error for step input for type 1 system (C) zero steady state error for step input for type 1 system (D) zero steady state error for step input for type 0 system 12. The transfer function of a phase compensator is given by (1 + aTs) (1 + Ts) where a > 1 and T > 0. The
16. A lead compensating network (a) improves response time (b) stabilizes the system with low phase margin (c) enables moderate increase in gain without affecting stability. (d) increases resonant frequency In the above statements, correct are
maximum phase shift provided by a such compensator is æ a + 1ö (A) tan -1 çç ÷÷ è a -1ø (C) tan
-1
æ a -1ö çç ÷÷ è a + 1ø
(D) cos
-1
(B) (a) and (c)
(C) (a), (c) and (d)
(D) All
17. A Lag network for compensation normally consists
æ a -1ö çç ÷÷ è a + 1ø
of (A) R, L and C elements
13. For an electrically heated temperature controlled
(B) R and L elements
liquid heater, the best controller is
(C) R and C elements
(A) Single–position controller Page 374
æ a -1ö (B) sin -1 çç ÷÷ è a + 1ø
(A) (a) and (b)
(D) R only
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Design of Control Systems
18. The pole–zero plot given in fig.P6.6.18 is that of a
Chap 6.6
SOLUTIONS
jw s
1. (A)
2. (D)
3. (A)
4. (D)
5. (B)
6. (D)
7. (A)
8. (B)
9. (A)
10. (B)
11. (C)
12. (B)
13. (C)
14. (D)
15. (C)
16. (D)
17. (C)
18. (D)
19. (D)
20. (D)
Fig. P6.6.18
(A) PID controller (B) PD controller (C) Integrator
21. (D)
(D) Lag–lead compensating network 19. The correct sequence of steps needed to improve system stability is (A) reduce gain, use negative feedback, insert derivative action (B) reduce gain, insert derivative action, use negative feedback (C) insert derivative action, use negative feedback, reduce gain (D) use negative feedback, reduce gain, insert derivative action. 20. In a derivative error compensation (A) damping decreases and setting time decreases (B) damping increases and setting time increases (C) damping decreases and setting time increases (D) damping increases and setting time decreases 21. An ON–OFF controller is a (A) P controller (B) PID controller (C) integral controller (D) non linear controller **********
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CHAPTER
6.7 THE STATE-VARIABLE ANALYSIS
1. Consider the SFG shown in fig. P6.7.1 u
1 s
1
1 s
x3
1 s
x2
x1
1
y
-1 -2 -3
Fig. P6.7.1
For this é x& 1 ù é3 (A) ê x& 2 ú = ê0 ê ú ê êë x& 3 úû êë3
system dynamic equation is 1 2 ù é x1 ù é0 ù 1 1ú ê x2 ú + ê0 ú u úê ú ê ú 2 1úû êë x3 úû êë1 úû
1 0ù é-2 ê (A) 0 -2 0 ú ê ú 0 -3úû êë 0
é2 -1 0 ù (B) ê0 2 0 ú ê ú êë0 0 3úû
1 -2 ù é 2 ê (C) 0 -2 0ú ê ú 0 úû êë-3 0
é0 -1 2 ù (D) ê0 2 0 ú ê ú êë3 0 0 úû
3. 5 1 s
1 0 ù é x1 ù é0 ù é x& 1 ù é 0 ê ú ê (B) x& 2 = 0 0 1ú ê x2 ú + ê0 ú u ê ú ê úê ú ê ú êë x& 3 úû êë-3 -2 -1ûú êë x3 úû êë1 ûú
u
-2 1 s
1
1 s
x2
1
x1
5
y
-2
1 x3
5
-3
é x& 1 ù é0 -1 0 ù é x1 ù é0 ù (C) ê x& 2 ú = ê0 0 -1ú ê x2 ú + ê0 ú u ê ú ê úê ú ê ú 1úû êë x3 ûú êë1 ûú êë x& 3 úû êë3 2
Fig. P6.7.3
(D) None of the above Statement for Q.2–4: Represent the given system in state-space equation x& = A ×x + B× u. Choose the correction option for
é 1 0 -2 ù (A) ê 0 -2 0ú ê ú 0 úû êë-3 0
é-1 0 2 ù (B) ê 0 2 0 ú ê ú êë 3 0 0 úû
0 1ù é-2 (C) ê 0 -2 0 ú ê ú 0 -3úû êë 0
é2 0 -1ù (D) ê0 2 0 ú ê ú 3úû êë0 0
matrix A. 4.
2.
1 s
x2
1 s
1
2
x1
5
y
1 u
-2 1 s
u
-2 x3
1 s
x3 1
1 s
x2
Fig. P6.7.4 Fig. P6.7.2
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1
1 s -4
-3
5
-3
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The State-Variable Analysis
Chap 6.7
é 0 1 -4 ù (A) ê 1 0 0ú ê ú 0 úû êë-3 0
é 0 -1 4 ù (B) ê-1 0 0 ú ê ú êë 3 0 0 úû
8. The F( t ) is é cos 2 t sin 2 t ù (A) ê ú ë- sin 2 t cos 2 t û
écos 2 t - sin 2 t ù (B) ê cos 2 t úû ësin 2 t
é- 4 1 0 ù (C) ê 0 0 1ú ê ú êë 0 0 -3úû
é4 - 1 0 ù (D) ê0 0 -1ú ê ú 3úû êë0 0
é sin 2 t cos 2 t ù (C) ê ú ë- cos 2 t sin 2 t û
ésin 2 t - cos 2 t ù (D) ê sin 2 t úû ëcos 2 t
9. The q( t ) is
5. The state equation of a LTI system is represented by 1ù é 0 é0 1ù x& = ê ú x + ê1 0 ú u 2 1 ë û ë û The Eigen values are (A) -1, + 1
(B) -0.5 ± j1.323
(C) -1, - 1
(D) None of the above
é-3 0 ù é0 ù x& = ê x + ê úu ú ë 0 -3û ë1 û The state-transition matrix F( t) is
é-e -3t (C) ê ë 0
0 ù ú e -3t û 0 ù ú -e -3t û
é-e -3t (B) ê ë 0
0 ù ú e -3t û
ée -3t (D) ê ë0
0 ù ú -e -3t û
ésin 2 t ù (B) ê ú ëcos 2 t û
é0.5(1 - cos 2 t) ù (C) ê ú ë 0.5 sin 2 t û
écos 2 t ù (D) ê ú ësin 2 t û
10. From the following matrices, the state-transition
6. The state equation of a LTI system is
ée -3t (A) ê ë0
é0.5(1 - sin 2 t) ù (A) ê ú ë 0.5 cos 2 t û
matrices can be é -e - t 0 ù (A) ê ú 0 1 e- t û ë
é1 - e - t (B) ê ë 0
0ù ú e- t û
é 1 (C) ê -t ë1 - e
é1 - e - t (D) ê ë 0
e- t ù ú e- t û
0ù e - t úû
Statement for Q.11–13: A system is described by the dynamic equations x& ( t) = A ×x ( t) + B ×u( t), y( t) = C × x( t) where 1 0ù é 0 é0 ù ê ú A= 0 0 1 , B = ê0 ú, C = [1 0 0 ] ê ú ê ú êë-1 -2 -3ûú êë1 ûú
7. The state equation of a LTI system is é 0 2ù é0 ù x& = ê x + ê úu ú ë-2 0 û ë1 û
11. The Eigen values of A are (A) 0.325, -1.662 ± j0.562
The state transition matrix is é cos 2 t sin 2 t ù (A) ê ú ë- sin 2 t cos 2 t û
écos 2 t - sin 2 t ù (B) ê cos 2 t úû ësin 2 t
(B) 2.325, 0.338 ± j0.562
é sin 2 t cos 2 t ù (C) ê ú ë- cos 2 t sin 2 t û
ésin 2 t - cos 2 t ù (D) ê sin 2 t úû ëcos 2 t
(D) -0.325, 1.662 ± j0.562
(C) -2.325, -0.338 ± j0.562
12. The transfer-function relation between X ( s) and
Statement for Q.8–9: The state-space representation of a system is given by x& ( t) = A ×x ( t) + B ×u( t), where
U ( s) is (A)
é 1ù 1 ê -s ú s 3 + 3s 2 + 2 s - 1 ê 2 ú êë s úû
(B)
(C)
é 1ù 1 ê -s ú s 3 + 3s 2 + 2 s + 1 ê 2 ú êë s úû
(D) None of the above
é0 ù é 0 2ù , B=ê ú A =ê ú ë1 û ë-2 0 û If x(0) is the initial state vector, and the component of the input vector u( t) are all unit step
é 1ù 1 ê sú s 3 + 3s 2 + 2 s + 1 ê 2 ú êës úû
function, then the state transition equation is given by x& ( t) = F( t )x (0) + q( t ), where F( t ) is a state transition
13. The output transfer function Y ( s) U ( s) is
matrix and q( t ) is a vector matrix.
(A) s( s 3 + 3s 2 + 2 s + 1) -1
(B) s( s 3 + 3s 2 + 2 s - 1) -1
(C) ( s 3 + 3s 2 + 2 s + 1) -1
(D) None of the above
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14. A system is described by the dynamic equation x& ( t) = A ×x ( t) + B ×u( t), y( t) = C × x( t) where é-1 0 ù é1 ù , B = ê ú and C = [1 1] A =ê ú ë 0 -2 û ë0 û
(C)
( s + 1) ( s + 2) 2
(B)
s+1 s+2
( s + 2) ( s + 1)
(D) None of the above
18.
15. The state-space representation of a system is given by
(C) s( s 2 + 3s + 2) -1
(D) ( s + 1) -1
1 0ù é 0 é10 ù x& = ê 0 0 1ú x + ê 0 ú u, y = [1 0 0 ]x ê ú ê ú êë-1 -2 -3úû êë 0 úû
(A)
10(2 s 2 + 3s + 1) s 3 + 3s 2 + 2 s + 1
(B)
10(2 s 2 + 3s + 1) s 3 + 2 s 2 + 3s + 1
(C)
10(2 s 2 + 3s + 2) s 3 + 3s 2 + 2 s + 1
(D)
10(2 s 2 + 3s + 2) s 3 + 2 s 2 + 3s + 1
1 é0 ê0 0 (C) x& = ê ê0 0 ê20 10 ë
0 ù é0 ù ú ê0 ú 0 úx + ê úr 0 1 ú ê0 ú ú ê1 ú 7 100 û ë û 0 1
1 0 0 ù é 0 é0 ù ú ê0 ú ê 0 0 1 0 úx + ê úr (D) x& = ê 0 0 1 ú ê0 ú ê 0 ê-20 -10 -7 -100 ú ê ú ë û ë1 û y = [100 0 0 0 ]x 19. A state-space representation of a system is given by
Statement for Q.17–18: Determine the state-space representation for the transfer function given in question. Choose the state variable as follows 2
3
d c d c dc c = c& , x3 = 2 = &&c , x4 = 2 = &&& dt dt dt
C( s) 24 = 3 R( s) s + 9 s 2 + 26 s + 24
1 0 ù é x1 ù é 0 ù é x& 1 ù é 0 ê ú ê (A) x& 2 = 0 0 1 ú ê x2 ú + ê 0 ú r ê ú ê úê ú ê ú êë x& 3 úû êë-24 -26 -9 ûú êë x3 úû êë24 úû 1 0 ù é x1 ù é 0 ù é x& 1 ù é 0 (B) ê x& 2 ú = ê 0 0 1 ú ê x2 ú + ê 0 ú r ê ú ê úê ú ê ú êë x& 3 úû êë24 26 9 úû êë x3 úû êë24 úû Page 378
0
y = [100 0 0 0 ]x
The transfer function Y ( s) U ( s) is
17.
0ù é 0 ù ú ê 0 ú 0 1 0 úx + ê ú r, 0 0 1ú ê 0 ú ê100 ú 7 10 20 úû ë û 1
y = [1 0 0 0 ]x
16. The state-space representation for a system is
x1 = c = y, x2 =
é 0 ê 0 (A) x& = ê ê 0 ê100 ë
1 0 0 ù é 0 é 0 ù ê 0 ú ê 0 ú 0 1 0 úx + ê úr (B) x& = ê 0 0 1 ú ê 0 ê 0 ú ê-100 -7 -10 -20 ú ê ú ë û ë100 û
The transfer function of this system is (B) ( s + 2) -1
C( s) 100 = 4 R( s) s + 20 s 3 + 10 s2 + 7 s + 100
y = [1 0 0 0 ]x
é-1 0 ù é1ù x& ( t) = ê x( t) + ê ú u( t), y( t) = [1 1]x( t) ú ë 0 -2 û ë1û
(A) ( s 2 + 3s + 2) -1
é x& 1 ù é0 1 0 ù é x1 ù é 0 ù (C) ê x& 2 ú = ê0 0 1 ú ê x2 ú + ê 0 ú r ê ú ê úê ú ê ú êë x& 3 úû êë9 26 24 úû êë x3 úû êë24 úû 1 0 ù é x1 ù é 0 ù é x& 1 ù é 0 (D) ê x& 2 ú = ê 0 0 1 ú ê x2 ú + ê 0 ú r ê ú ê úê ú ê ú êë x& 3 úû êë-9 -26 -24 úû êë x3 úû êë24 úû
The output transfer function Y ( s) U ( s) is (A)
Control & System
é 0 1ù é0 ù x& = ê ú x, y = [1 - 1]x, and x(0) = ê1 ú 2 0 ë û ë û The time response of this system will be 3 (A) sin 2t (B) sin 2 t 2 (C) -
1 2
(D)
sin 2 t
3 sin 2 t
20. For the transfer function Y ( s) s+3 = U ( s) ( s + 1)( s + 2) The state model is given by x& = A ×x + B× u, y = C × x. The A , B, C are www.gatehelp.com
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-4 -2 ù é-5 é1 ù ê ú 29. x& = -3 -10 0 x + ê1 ú r, y = [ -1 2 1]x ê ú ê ú 1 -5 úû êë -1 êë0 ûú estep ( ¥)
Statement for Q.34–36: Consider the system shown in fig. P6.7.34-36. 1 s
eramp ( ¥)
(A) 1.0976
1 u
0
(B) 1.0976
¥
(C) 0
1.0976
(D) ¥
1.0976
0.714
(B) ¥
0.714
(C) 0
4.86
(D) ¥
4.86
x2
1 s
1
-1 1 s
y
-10
x3
34. The controllability matrix for this system is 10 ù 1 -2 ù é 10 -10 é0 (A) ê-10 (B) ê1 -1 0 -20 ú 1ú ê ú ê ú 40 úû 4 úû êë 10 -20 êë1 -2 10 ù é 10 -10 (C) ê-10 0 20 ú ê ú êë 10 -20 -40 úû
1
-2
10
Fig. P6.7.34-36
Consider the system shown in fig. P6.7.31-33
u
x1
-2
Statement for Q.31–33:
1 s
1 s
1
10
eramp ( ¥)
(A) 0
x2
-1
1
1 0ù é 0 é0 ù ê ú 30. x& = -5 -9 7 x + ê0 ú r, y = [1 0 0 ]x ê ú ê ú êë -1 0 0 úû êë1 ûú estep ( ¥)
Control & System
x1
1
y
1 -1ù é0 (D) ê1 6 -1ú ê ú êë1 -4 -4 úû
35. The observability matrix is 10 ù 1 -2 ù é 10 -10 é0 (A) ê-10 (B) ê1 -1 0 -20 ú 1ú ê ú ê ú 40 úû 4 úû êë 10 -10 êë1 -2 10 ù é 10 -10 (C) ê-10 0 20 ú ê ú êë 10 -10 -40 úû
1 2ù é0 (D) ê1 -1 1ú ê ú êë1 -2 4 úû
-5
36. The system is
-6
Fig. P6.7.31-33
31. The controllability matrix is é1 0 ù é 1 -2 ù (A) ê (B) ê ú 4 úû ë0 1û ë-2 é1 0 ù (C) ê ú ë0 -1û
é 1 2ù (D) ê ú ë-2 -4 û
(A) Controllable and observable (B) Controllable only (C) Observable only (D) None of the above Statement for Q.37–38: A state flow graph is shown in fig. P6.7.37-38
32. The observability matrix is é1 0 ù é 1 -2 ù (A) ê (B) ê ú 0 1 4 úû ë û ë-2 é1 0 ù (C) ê ú ë0 -1û
(A) Controllable and observable
(C) Observable only
u
1 s x2
1 s
x1
5
y
-5
Fig. P6.7.37-38
37. The state and output equation for this system is éx ù é x& ù é0 -1ù é x ù é0 ù (A) ê 1 ú = ê 21 ú ê 1 ú + ê ú u, y = [5 4 ]ê 1 ú 5 & x x 1 ë 2 û êë ë x2 û 4 úû ë 2 û ë û
(D) None of the above Page 380
1
-21 4
é 1 2ù (D) ê ú ë-2 -4 û
33. The system is (B) Controllable only
4
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Chap 6.7
év& ù é-1 -1ù év1 ù é1 ù év ù (B) ê &1 ú = ê + ê ú vi , iR = [ 4 1]ê 1 ú ê ú ú ë i3 û ë-3 -1û ë i3 û ë0 û ë i3 û
1ù é x ù é0 ù é x& ù é 0 éx ù 21 ú ê 1 ú + ê ú u, y = [5 4 ]ê 1 ú (B) ê 1 ú = ê 5 & x 1 ë x2 û êë ë x2 û 4 úû ë 2 û ë û
é v& ù é1 -3ù é v1 ù é 1 ù év ù (C) ê 1 ú = ê + ê ú vi , iR = [1 4 ]ê 1 ú ê ú ú & ëv2 û ë1 6 û ëv2 û ë-1û ëv2 û
éx ù é x& ù é0 -1ù é x ù é1ù (C) ê 1 ú = ê 21 ú ê 1 ú + ê ú u, y = [ 4 5 ]ê 1 ú 5 x 1 ë x2 û ë x& 2 û êë 4 úû ë 2 û ë û
é v& ù é 1 3ù é v1 ù é 1 ù é v1 ù (D) ê 1 ú = ê ú êv ú + ê-1ú vi , iR = [1 4 ]êv ú & v 1 6 û ë 2û ë û ë 2û ë ë 2û
1ù é x ù é1ù é x& ù é 0 éx ù 21 ú ê 1 ú + ê ú u, y = [ 4 5 ]ê 1 ú (D) ê 1 ú = ê 5 & x x 1 ë 2 û êë ë x2 û 4 úû ë 2 û ë û
Statement for Q.41–43:
38. The system is
Consider the network shown in fig. P6.7.41-43. This system may be represented in state space representation x& = A × x + B × u
(A) Observable and controllable (B) Controllable only (C) Observable only
iC
(D) None of the above iR1
39. Consider the network shown in fig. P6.7.39. The state-space representation for this network is iL
iL
iR2
1 2F
1W
is
1W
4H
vs
2W
Fig. P6.7.41-43
iC
iR
41. The state variable may be
1F
Fig. P6.7.39
év& ù é-0.25 1ù évC ù é 1 ù évC ù (A) ê & C ú = ê ú ê i ú + ê0.25 ú vs , iR = [0.5 0 ]ê i ú i 0 5 0 . û ûë Lû ë ë Lû ë ë Lû
(A) iR1 , iR 2
(B) iL , iC
(C) vC , iL
(D) None of the above
42. If state variable are chosen as in previous question, then the matrix A is é 1 -1ù (A) ê ú ë-1 3û
é 1 -3ù (B) ê ú ë-1 1û
é-1 3ù (C) ê ú ë 1 -1û
é 3 -1ù (D) ê ú ë-1 1û
év& ù é-1 0.25 ù évC ù é0.25 ù év ù (D) ê & C ú = ê vs , iR = [0.5 0 ]ê C ú +ê ê ú ú ú ë iL û ë 0 -0.5 û ë iL û ë 0 û ë iL û
43. The matrix B is é 3ù (A) ê ú ë-1û
é-1ù (B) ê ú ë 3û
40. For the network shown in fig. P6.7.40. The output is
é-3ù (C) ê ú ë 1û
é 1ù (D) ê ú ë-3û
év& ù é -0.5 1ù évC ù é0.25 ù évC ù (B) ê & C ú = ê ú ê i ú + ê 1 ú vs , iR = [0.5 0 ]ê i ú i 0 25 0 . û ûë Lû ë ë Lû ë ë Lû év& ù é1 0.25 ù évC ù é0.25 ù év ù (C) ê & C ú = ê +ê vs , iR = [0.5 0 ]ê C ú ê ú ú ú ë iL û ë0 0.5 û ë iL û ë 0 û ë iL û
i1
1W
1H
v1
i3
v2
iR
i2 vi
1F
4vL
Statement for Q.44–47: 4v1
1W
Consider the network shown in fig. P6.7.44-47 i1
Fig. P6.7.40
iR ( t). The state space representation is
vi
év& ù é 1 -1ù év1 ù é1 ù év ù (A) ê &1 ú = ê + ê ú vi , iR = [ 4 1]ê 1 ú ê ú ú ë i3 û ë-3 1û ë i1 û ë0 û ë i3 û
1W
i3
1W
i5
i2
i4
1H
1H
1W
1F
+ vo -
Fig. P6.7.44-47
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Control & System
44. The state variable may be (A) i2 , i4
(B) i2 , i4 , vo
(C) i1 , i3
(D) i1 , i3 , i5
45. In state space representation matrix A is 1 1ù é 2 ê- 3 - 3 3ú ê 1 2 2ú (A) êú 3 3ú ê 3 ê- 1 - 2 - 1 ú êë 3 3 3 úû
1 2ù é 1 ê 3 - 3 - 3ú ê 2 2 1ú (B) ê - ú 3 3ú ê 3 ê- 1 - 2 - 1 ú êë 3 3 3 úû
1 1ù é 2 ê 3 - 3 - 3ú ê 1 2 2ú (C) ê ú 3 3ú ê 3 1ú ê- 1 - 2 êë 3 3 3 úû
1 2ù é 1 ê- 3 - 3 3ú ê 2 2 1ú (D) êú 3 3ú ê 3 ê- 1 - 2 - 1 ú êë 3 3 3 úû
46. The matrix B is é 2ù ê 3ú ê 1ú (A) ê- ú ê 3ú ê- 1 ú êë 3 úû
é2 ù ê3ú ê1 ú (B) ê ú ê3ú ê1 ú êë 3 úû
é 1ù ê- 3 ú ê 1ú (C) ê- ú ê 3ú ê 1ú êë 3 úû
é2 ù ê3ú ê1 ú (D) ê ú ê3ú ê2 ú êë 3 úû
47. If output is vo , then matrix C is (A) [-1 0 0]
(B) [1 0 0]
(C) [0 0 -1]
(D) [0 0 1]
SOLUTIONS 1. (B) From the SFG x& 3 = -3 x1 - 2 x2 x x2 = 3 Þ s x x1 = 2 Þ s
- x3 + u x& 2 = x3 x& 1 = x2
2. (A) x& 1 = - 2 x1 + x3 , x& 2 = - 2 x2 + u , x& 3 = -3 x3 + u 1 0 ù é x1 ù é0 ù é x& 1 ù é-2 ê x& ú = ê 0 -2 0 ú ê x ú + ê0 ú u ê 2ú ê ú ê 2ú ê ú & x 0 0 3 êë 3 úû êë ûú êë x3 úû êë1 ûú 3. (C) x& 1 = - 2 x1 + x3 , x& 2 = - 2 x2 + u , x& 3 = -3 x3 + u y = 5 x1 + 5 x2 + 5 x3 0 1ù é x1 ù é0 ù é x& 1 ù é-2 ê x& ú = ê 0 -2 0 ú ê x ú + ê1 ú u ê 2ú ê ú ê 2ú ê ú 0 -3úû êë x3 úû êë1 úû êë x& 3 úû êë 0 4. (C) x& 1 = -4 x1 + x2 , x& 2 = x3 + 2 u , x& 3 = - 3 x3 + u é x& 1 ù é-4 1 1ù é x1 ù é0 ù ê x& ú = ê 0 0 1ú ê x2 ú + ê2 ú u ê 2ú ê úê ú ê ú êë x& 3 úû êë 0 0 -3úû êë x3 úû êë1 úû 5. (B) Ds = |sI - A| = s 2 + s + 2 = 0 Þ s = - 0.5 ± j1.323 0 és + 3 6. (A) ( sI - A) = ê s+ ë 0 ( sI - A)
-1
é 1 ê = ês + 3 ê 0 ë
ù 0 ú 1 ú ú s + 3û
ée -3t F( t ) = L-1 {( sI - A )} == ê ë0 ************************
ù 1 , |sI - A|= 3úû ( s + 3) 2
0 ù ú e -3t û
é s -2 ù 2 7. (A) ( sI - A) = ê ú, |sI - A|= s + 4 ë2 s û ( sI - A)
-1
é s 1 é s 2 ù ê s2 + 4 = 2 = s + 4 êë-2 s úû ê -2 ê ë s2 + 4
2 ù s2 + 4 ú s ú ú 2 s + 4û
é cos 2 t sin 2 t ù F( t ) = L-1 {( sI - A )} = ê ú ë- sin 2 t cos 2 t û é s -2 ù 2 8. (A) ( sI - A) = ê ú, Ds =|sI - A|= s + 4 ë2 s û
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( sI - A)
-1
é s 1 é s 2 ù ê s2 + 4 = 2 = s + 4 êë-2 s úû ê -2 ê ë s2 + 4
é1 ù é1ù 15. (D) T( s) = ê ú( sI - A) -1 ê ú ë0 û ë1û
2 ù s + 4ú s ú ú 2 s + 4û 2
( sI - A)
é cos 2 t sin 2 t ù F( t ) = L-1 {( sI - A )} = ê ú ë- sin 2 t cos 2 t û
ì 1 é s 2 ù é0 ù é1ù 1 ü é2 ù ü 1 -1 ì = L-1 í 2 ý =L í 2 ê ú ê ú ê ú ê úý î s + 4 ë-2 s û ë1 û ë1û s þ î s( s + 4) ë s û þ 2 é ù ê s( s 2 + 4) ú ê ú 1 ê 2 ú ë ( s + 4) û
10. (C) (A) is not a state-transition matrix, since F(0) ¹ I (C) is a state-transition matrix since F(0) = I and [ F( t)]-1 = F( -t) 0 ù é s -1 ê 11. (C) ( sI - A) = 0 s -1 ú ê ú êë1 2 s + 3úû
|sI - A| = s3 + 3s2 + 2 s + 1, Þ
s = - 2.325, - 0.338 ± j0.562 X ( s) = ( sI - A) -1 B U ( s)
12. (B)
és 3 + 3s + 2 s+3 1 ù é0 ù 1 ê ú s( s + 3) s ê0 ú = 3 -1 úê ú s + 3s 2 + 2 s + 1 ê 2 -s -2 s - 1 s úû êë1 úû êë é1 ù 1 êsú = 3 s + 3s 2 + 2 s + 1 ê 2 ú êës úû 13. (C)
Y ( s) CX ( s) = U ( s) U ( s)
1 ù é ê s 3 + 3s 2 + 2 s + 1 ú ú ê s 1 = [1 0 0 ]ê 3 ú= 3 2 2 ê s + 3s 2+ 2 s + 1 ú s + 3s + 2 s + 1 s ú ê êë s 3 + 3s 2 + 2 s + 1 úû 14. (D)
Y ( s) = C ( sI - A) -1 B U ( s)
B = [1 1]
1 ù é0 ù s+2 1 és + 1 = ê s + 1úû êë1 úû ( s + 1) 2 Ds ë 0
é 1 ê = ês + 1 ê 0 ë
ù 0 ú 1 ú ú s + 2û ù 0 ú é1 ù 1 = 1 ú êë0 úû s + 1 ú s + 2û
16. (C) x& = A ×x + B× u, y = C × x + Du
ü ïï é0.5(1 - cos 2 t) ù ý= ê ú ï ë 0.5 sin 2 t û ïþ
(B) is not a state-transition matrix since F(0) ¹ I
-1
é 1 é1ù ê s + 1 T( s) = ê ú ê ë1û ê 0 ë
9. (C) q( t) = L-1 {( sI - A) -1 BR( s)}
ì ïï = L-1 í ï ïî
Chap 6.7
1 0ù é 0 é10 ù A =ê 0 0 1ú, B = ê 0 ú, C = [1 0 0 ], D = 0 ê ú ê ú êë-1 -2 -3úû êë 0 úû Y ( s) T( s) = = C ( sI - A ) -1 B + D U ( s) ( sI - A)
-1
és 3 + 3s + 2 s+3 1ù 1 ê ú = 3 s ( s + ) s 1 3 ú s + 3s 2 + 2 s + 1 ê -s -2 s - 1 s 2 úû êë
Substituting the all values, T( s) =
10(2 s 2 + 3s + 2) s 3 + 3s 2 + 2 s + 1
17. (A)
C( s) b0 24 = = R( s) s 3 + a2 s 2 + a1 s + a0 ( s 3 + 9 s 2 + 26 s + 24)
( s 3 + a2 s 2 + a1 s + a0 ) C( s) = b0 R( s) Taking the inverse Laplace transform assuming zero initial conditions &&& c + a2 &&c + a1 c& + a0 c = b0 r x1 = c = y, x2 = c& , x3 = &&c x& 1 = c& = x2 , x& 2 = &&c = x3 x& 3 = &&& c = b0 r - a2 &&c - a1 c& - a0 c = - a0 x1 - a1 x2 - a2 x3 + b0 r , 1 0 ù é x1 ù é 0 ù é x& 1 ù é 0 ê x& ú = ê 0 0 1 ú ê x2 ú + ê 0 ú r ê 2ú ê úê ú ê ú êë x& 3 úû êë-a0 -a1 -a1 úû êë x3 úû êëb0 úû a0 = 24 , a1 = 26, a2 = 9, b0 = 24 1 0 ù é x1 ù é 0 ù é x& 1 ù é 0 ê x& ú = ê 0 0 1 ú ê x2 ú + ê 0 ú r ê 2ú ê úê ú ê ú êë x& 3 úû êë-24 -26 -9 úû êë x3 úû êë24 úû é x1 ù y = [1 0 0 ] ê x2 ú ê ú êë x3 úû 18. (B) Fourth order hence four state variable
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é 0 ê 0 x& = ê ê 0 ê -a ë 0
1 0
0 1
0
0
-a1
-a 2
0 ù é0 ù ú ê0 ú 0 ú x = ê ú r, y = [1 0 0 0 ]x 1 ú ê0 ú ú ê ú -a 3 û ëb0 û
a0 = 100, a1 = 7, a2 = 10,
a3 = 20 ,
=
b0 = 100
Y ( s) = [0 1] X ( s) =
é sin 2 t ù cos 2 t ú F( t ) = L-1 {( sI - A)} = ê 2 ú ê êë- 2 sin 2 t cos 2 t úû
Þ
20. (C) Find the transfer function of option Y ( s) 1 For (A) , , = U ( s) s - 2
y( t) =
0 ù é1ù és + 2 Y ( s) 1 = [0 1] s + 1úû êë1úû U ( s) ( s + 1)( s + 2) êë 2
|sI - A| = s
- 8 s - 11s + 8
Þ
-1
æ é2 ù é1ù 3 ö çç ê ú + ê ú 2 ÷÷ è ë1 û ë1û s + 9 ø
é2 s + 4 s + 21s + 45 ù 1 2 ( s + 5)( s + 9) êë s 3 - 7 s 2 + 12 s - 7 úû 3
1 1 -2 t - e 2 2
A
-1
0.05 -0.05 ù é-0.4 ê = -1 -0.25 -0.25 ú ê ú -0.5 ûú 15 . êë -2
1ù é -2 - ú 27. (A) estep ( ¥) = 1 + CA -1B, A -1 = ê 3 êë 1 0 úû
s = 9.11, 0.53, - 1.64
23. (D) X ( s) = ( sI - A) -1 (x(0) + B × u) és - 1 -2 ù =ê s + 1úû ë 3
1 s( s + 2)
= 1 - 0.2 = 0.8
s = -5.79, - 121 .
-2 -3 ù és 22. (B) ( sI - A) = ê 0 s - 6 -5 ú ê ú êë-1 -4 s - 2 úû 2
é0 ù é1 ù ö ê0 ú + ê0 ú 1 ÷ ê ú ê ú s÷ êë0 úû êë0 úû ÷ø
1 é ù ê ú s( s + 2) ê ú 1 ú =ê 2 ê s ( s + 2) ú 1 ê ú êë s 2 ( s + 1)( s + 2) úû
é-0.4 0.05 -0.05 ù é0 ù estep ( ¥) = 1 + [ -1 1 0 ]ê -1 0.25 -0.25 ú ê0 ú ê úê ú 15 . -0.5 úû êë1 úû êë -2
é-2 -1ù 21. (C) A = ê ú, ë -3 -5 û
3
æ ç ç ç è
X ( s) =
1 0ù é- 5 ê A= 0 -2 1ú, ê ú êë20 -10 1úû
So (C) is correct option.
Þ
-1
26. (D) For a unit step input estep ( ¥) = 1 + CA -1B
és + 2 ù s+3 1 = [0 1] = ( s + 1)( s + 2) êës + 3úû ( s + 1)( s + 3)
|sI - A| = s2 + 7 s + 7
1 1 - e - t + e -2 t 2 2
Y ( s) = [1 0 0 ],
Y ( s) 1 For (B) , = U ( s) s - 2 For (C),
y( t) =
1 s( s + 1)( s + 2)
0 ù és + 2 -1 ê = 0 -1 ú s ê ú 0 s + 1úû êë 0
sin 2 t
2
é1 ù é1ù 1 ö ê0 ú + ê1ú s ÷÷ ë û ë û ø
25. (B) X ( s) = ( sI - A) -1 (x(0) + B × u)
é sin 2 t ù cos 2 t + ê ú x( t) = F( t )x(0) = 2 ê ú êë- 2 sin 2 t + cos 2 t úû 3
0 ùæ és + 1 1 ç ( s + 1)( s + 2) êë -1 s + 2 úû çè
( s + 1) é ù ê s( s + 2) ú =ê ú 1 ê ú ë s( s + 1)( s + 2) û
é 0 1ù 1 é s 1ù 19. (B) A = ê , ( sI - A) -1 = 2 ú s + 2 êë-2 súû ë-2 0 û
y = x1 - x2 =
Control & System
2
1ù é 1 2 -2 - ú é0 ù estep ( ¥) = 1 + [1 1] ê 3 ê ú =1 - = 3 3 êë 1 0 úû ë1 û 28. (C) eramp ( ¥) = lim [(1 + CA -1B) t + C( A -1 ) 2 B] t ®¥
2 1 + CA -1B = , 3
é2 ù eramp ( ¥) = lim ê t + C( A -1 ) 2 Bú = ¥ t ®¥ ë 3 û
2
4 s 3 - 10 s 2 + 45 s - 105 Y ( s) = [1 2 ] X ( s) = ( s 2 + 5)( s 2 + 9) 24. (B) X ( s) = ( sI - A) -1 (x(0) + B × u) Page 384
29. (B) estep ( ¥) = 1 + CA -1B -4 -2 ù é-5 é1 ù A = ê -3 -10 0 ú, B = ê1 ú, C = [ -1 2 1] ê ú ê ú 1 -5 úû êë -1 êë0 úû
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The State-Variable Analysis
A
-1
é-1 1 0 ù A = ê 0 -1 0 ú, C = [10 - 10 10 ], ê ú êë 0 0 -2 ûú
0.134 0.122 ù é-0.305 ê = 0.091 -0.140 -0.037 ú ê ú êë -0.079 -0.055 -0.232 úû
é-1 1 0 ù CA = [10 - 10 10 ] ê 0 -1 0 ú = [10 ê ú êë 0 0 -2 úû
. estep ( ¥) = 1 + 0.0976 = 10976 -1
-1 2
eramp ( ¥) = lim [(1 + CA B) t + C( A ) B] = ¥ t ®¥
-1 ù 0 é 0 30. (B) A -1 = ê 1 0 0 ú ê ú êë1.286 0.143 -0.714 úû
CA 2 = [10
0 -1 ù é0 ù é 0 ê 0 0 ú ê0 ú = 0 estep ( ¥) = 1 + [1 0 0 ] 1 ê úê ú êë1.286 0.143 -0.714 úû êë1 úû . -0.143 0.714 ù é -1286 ê 0 0 -1 ú (A ) = ú ê ëê-0.776 -0.102 -0.776 úû
nonsingular and system is controllable
1ù é 0 é 1ù A =ê , B=ê ú ú ë-6 -5 û ë-2 û
10 ù é 10 -10 det O M = detê-10 0 -10 ú = -3000 ê ú 40 úû êë 10 -20
1ù é 1ù ù é 1 -2 ù é 0 ê-6 -5 ú ê-2 ú ú = ê-2 4 úû ë ûë ûû ë
The rank of O M is 3. Hence system is observable.
32. (A) y = x1 , y = [1 0 ]x , 0 ], CA = [1
37. (B) x& 2 = - 5 x1 -
é C ù é1 0 ù 0 ] , OM = ê ú =ê ú ëCA û ë0 1 û
21 x2 + u , x& 1 = x2 , y = 5 x1 + 4 x2 4
1ù é x ù é0 ù é x& 1 ù é 0 é x1 ù 1 ê x& ú = ê-5 - 21 ú ê x ú + ê1 ú u, y = [5 4 ]ê x ú ë 2 û êë ë 2û 4 úû ë 2 û ë û
33. (C) det CM = 0. Hence system is not controllable. det O M = 1. Hence system is observable.
é C ù é 5 4ù 38. (B) O M = ê ú =ê ú ëCA û ë-20 1 û
34. (B) x& 1 = - x1 + x2 , x& 2 = - x2 + u , x& 3 = - 2 x3 + u
det O M = 0. Thus system is not observable
é-1 1 0 ù é0 ù é-1 1 0 ù é0 ù &x = ê 0 -1 0 ú x + ê1 ú u, A = ê 0 -1 0 ú , B = ê1 ú ê ú ê ú ê ú ê ú êë 0 0 -2 úû êë1 úû êë 0 0 -2 úû êë1 úû
1ù é0 21 ú CM = [B AB] = ê 1 êë 4 úû
é-1 1 0 ù é0 ù é 1ù AB = ê 0 -1 0 ú ê1 ú = ê -1ú ê úê ú ê ú êë 0 0 -2 úû êë1 úû êë-2 úû
det CM = -1. Thus system is controllable. 39. (B)
é-1 1 0 ù é 1ù é-2 ù A 2B = ê 0 -1 0 ú ê -1ú = ê 1ú ê úê ú ê ú êë 0 0 -2 úû êë-2 úû êë 4 úû 1 -2 ù é0 ê 1ú Cm = [B AB A B] = 1 -1 ê ú 4 úû êë1 -2 2
35. (A) y = 10 x1 - 10 x2 + 10 x3 , y = [10 - 10
40 ]
Since the determinant is not zero, the 3 ´ 3 matrix is
31. (B) x& 1 = x2 + u , x& 2 = - 5 x2 - 6 x1 - 2 u
C = [1
é-1 1 0 ù 0 - 20 ] ê 0 -1 0 ú = [10 - 10 ê ú êë 0 0 -2 úû
1 -2 ù é0 36. (A) det Cm = det ê1 -1 1ú = -1, ê ú 4 úû êë1 -2
C( A -1 ) 2 B = 0.714, eramp ( ¥) = 0.714
éé 1ù CM = [B AB] = êê ú ëë-2 û
0 - 20 ]
10 ù é C ù é 10 -10 ê ú ê O M = CA = -10 0 -20 ú ê ú ê ú 2 40 ûú êëCA úû êë 10 -10
-1 2
1ù é 0 é 1ù x& = ê x + ê ú u, ú ë-6 -5 û ë-2 û
Chap 6.7
10 ]x
dvc di v = ic , L = L = 0.25 vL dt dt 4
vC and iL are state variable. v iL = iC + iR , iC = iL - iR = iL - C , vL = vs - vC 2 dvL v Hence equations are = iL - C = - 0.5 vC + iL dt 2 diL = 0.25( vs - vC ) = - 0.25 vC + 0.25 vs dt év& C ù é -0.5 1ù évC ù é0.25 ù ê i& ú = ê-0.25 0 ú ê i ú + ê 1 ú vs , û ûë Lû ë ë Lû ë
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GATE EC BY RK Kanodia
UNIT 6
iR =
év ù iR = [0.5 0 ] ê C ú ë iL û
vC = 0.5 vC , 2
45. (A)
Control & System
di2 di dvo = v2 , 4 = v4 , = i5 dt dt dt i1
40. (B)
dv1 di = i2 , 3 = vL dt dt
1W
vi
Hence v1 and i3 are state variable.
1W
v2 i3
v4 i5
i2
i4
1H
1H
1W
1F
+ vo -
i2 = i1 - i3 = ( vi - v1 ) - i3 , i2 = - v1 - i3 + vi vL = v1 - v2 = v1 - iR , = v1 - ( i3 + 4 v1 ) = -3v1 - i3 dv1 di = - v1 - i3 + vi , 3 = - 3v1 - i3, y = iR = 4 v1 + i3 dt dt év&1 ù é-1 -1ù év1 ù é1 ù év1 ù ê i& ú = ê-3 -1ú ê i ú + ê0 ú vi , iR = [ 4 1]ê i ú û ë 3û ë û ë 3û ë ë 3û 41. (C) Energy storage elements are capacitor and inductor. vC and iL are available in differential form and linearly independent. Hence vC and iL are suitable for state-variable. 1 dvC dvC = iC Þ = 2 iC 2 dt dt 1 diL diL = vL Þ = 2 vL 2 dt dt
42. (B)
iC iR1 is
1W
+
iL
+ vC 1 2F
vL
iR2 + vR2
-
1W
4vL
-
Fig. S6.7.42
vL = vC + vR 2 = vC + iR2 , iC + 4 vL = iR 2 vL = vC + iC + 4 vL , -3vL = vC + iC v iC = is - iR1 - iL , iC = is - L - iL 1
...(i) ...(ii)
Solving equation (i) and (ii) -3 ( is - iL - iC ) = vC + iC , 2 iC = vC - 3iL + 3is
Fig. S6.7.45
Now obtain v2 , v4 and i5 in terms of the state variable -vi + i1 + i3 + i5 + vo = 0 But i3 = i1 - i2 and i5 = i3 - i4 -vi + i1 + ( i1 - i2 ) + ( i3 - i4 ) + vo = 0 2 1 1 1 i1 = i2 + i4 - vo + vi 3 3 3 3 2 1 1 2 v2 = vi - i1 = - i2 - i4 + vo + vi 3 3 3 3 1 1 1 1 i3 = i1 - i2 = - i2 + i4 - vo + vi 3 3 3 3 1 2 1 1 i5 = i3 - i4 = - i2 - i4 - vo + vi 3 3 3 3 1 2 2 1 v4 = i5 + vo = - i2 - i4 + vo + vi 3 3 3 3 1 1ù é2 ù é 2 é i&2 ù ê 3 3 3 ú é i2 ù ê 3 ú ê& ú ê 1 2 2 ú ê ú ê1 ú ú i4 + ê ú vi ê i4 ú = ê- 3 - 3 3úê ú ê3ú êv& o ú ê 1 2 1 êëvo úû ê 1 ú ë û ê- ú êë 3 êë 3 úû 3 3 úû 1 1ù é 2 ê- 3 - 3 3ú ê 1 2 2ú A = êú, 3 3ú ê 3 ê- 1 - 2 - 1 ú êë 3 3 3 úû
47. (D) vo is state variable
-3vL = vC + is - vL - iL , 2vL = - vC + iL - is dvC di = vC - 3iL + 3is , L = - vC + iL - is dt dt év& C ù é 1 -3ù évC ù é 3 ù ê i& ú = ê-1 1ú ê i ú + ê-1ú is ûë Lû ë û ë Lû ë
y = vo ,
é i2 ù y = [0 0 1] = ê i4 ú ê ú êëvo úû
********
é 3ù 43. (A) B = ê ú ë-1û 44. (B) There are three energy storage elements, hence 3 variable. i2 , i4 and vo are available in differentiated form hence these are state variable.
Page 386
é2 ù ê3ú ê1 ú 46. (B) B = ê ú ê3ú ê1 ú êë 3 úû
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GATE EC BY RK Kanodia
UNIT 7
10. If machine is not properly adjusted, the product
Statement for Question 15-16:
resistance change to the case where ax = 1050 W. Now the rejected fraction is
Communication System
The life time of a system expressed in weeks is a Rayleigh random variable X for which
(A) 5046% (C) 2.18%
(B) 10.57% (D) 6.43%
ì x -x e 400 ïï 200 f X ( x) = í ï ïî 0 2
11. Cannon shell impact position, as measured along the line of fire from the target point is described by a
0£x x f0 - W2 > 2Wm
(A) 51 %
(B) 11.8 %
(C) 5.1 %
(D) None of the above
Statement for Question 33-36 The figure 6.54-57 shows the positive portion of the envelope of the output of an AM modulator. The message signal is a waveform having zero DC value. m(t)
(B) 10 %
45
(C) 20 %
(D) 33.33 %
30 15
Statement for Question 30-31
t
Consider the system shown in figP7.4.30-31. The modulating signal m( t) has zero mean and its maximum (absolute) value is Am = max m( t) . It has bandwidth Wm . The nonlinear device has a input-output characteristic y( t) = ax( t) + bx 2 ( t). Page 416
Fig.P7.4.33-36
33. The modulation index is (A) 0.5
(B) 0.6
(C) 0.4
(D) 0.8
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GATE EC BY RK Kanodia
Amplitude Modulation
Chap 7.4
34. The modulation efficiency is
41. The lower sideband of the SSB AM signal is
(A) 8.3 %
(B) 14.28 %
(A) -100 cos(2 p( fc - 1000) t) + 200 sin(2 p( fc - 1000) t)
(C) 7.69 %
(D) None of the above
(B) -100 cos(2 p( fc - 1000) t) - 200 sin(2 p( fc - 1000) t)
35. The carrier power is
(C) 100 cos(2 p( fc - 1000) t) - 200 sin(2 p( fc - 1000) t)
(A) 60 W
(B) 450 W
(C) 30 W
(D) 900 W
(D) 100 cos(2 p( fc - 1000) t) + 200 sin(2 p( fc - 1000) t) Statement for Question 42-43
36. The power in sidebands is (A) 85 W
Consider the system shown in figure 6.69-70. The
(B) 42.5 W
(C) 56 W
average value of m( t) is zero and maximum value of
(D) 37.5 W
m( t) is M. The square-law device is defined by 37. In a broadcast transmitter, the RF output is represented as e( t) = 50[1 + 0.89 cos 5000 t + 0.30 sin 9000 t ]cos( 6 ´ 106 t)V
y( t) = 4 x( t) + 10 x( t). S
m(t)
x(t) Square- Law Device
y(t) Filter
AM Signal
What are the sidebands of the signals in radians ? cos wct
(A) 5 ´ 10 3 and 9 ´ 10 3 (B) 5.991 ´ 106 , 5.995 ´ 106 , 6.005 ´ 106 and 6.009 ´ 106 (C) 4 ´ 10 3, 1.4 ´ 10 4 (D) 1 ´ 10 , 11 . ´ 10 , 3 ´ 10 , and 15 . ´ 10 6
7
6
42. The value of M, required to produce modulation index of 0.8, is
7
38. An AM modulator has output x( t) = 40 cos 400 pt + 4 cos 360 pt + 4 cos 440 pt The modulation efficiency is (A) 0.01
(B) 0.02
(C) 0.03
(D) 0.04
Fig. P7.4.42-43
(A) 0.32
(B) 0.26
(C) 0.52
(D) 0.16
43. Let W be the bandwidth of message signal m( t). AM signal would be recovered if
39. An AM modulator has output
(A) fc > W
(B) fc > 2W
(C) fc ³ 3W
(D) fc > 4W
x( t) = A cos 400 pt + B cos 380 pt + B cos 420 pt The carrier power is 100 W and the efficiency is 40%. The value of A and B are (A) 14.14, 8.16
(B) 50, 10
(C) 22.36, 13.46
(D) None of the above
44. A super heterodyne receiver is designed to receive transmitted signals between 5 and 10 MHz. High-side tuning is to be used. The tuning range of the local oscillator for IF frequency 500 kHz would be (A) 4.5 MHz - 9.5 MHz
Statement for Question 40-41
(B) 5.5 MHz - 10.5 MHz
A single side band signal is generated by modulating signal of 900-kHz carrier by the signal
(C) 4.5 MHz - 10.5 MHz
m( t) = cos 200 pt + 2 sin 2000 pt. The amplitude of the
(D) None of the above
carrier is Ac = 100.
(B) - sin(2 p1000 t) + 2 cos(2000 pt)
45. A super heterodyne receiver uses an IF frequency of 455 kHz. The receiver is tuned to a transmitter having a carrier frequency of 2400 kHz. High-side tuning is to be used. The image frequency will be
(C) sin(2 p1000 t) + 2 cos(1000 t)
(A) 2855 kHz
(B) 3310 kHz
(C) 1845 kHz
(D) 1490 kHz
$ ( t) is 40. The signal m (A) - sin(2 p1000 t) - 2 cos(2000 pt)
(D) sin(2 p1000 t) - 2 cos(2 p1000 t)
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GATE EC BY RK Kanodia
UNIT 7
46. In the circuit shown in fig.P7.4.46, the transformers are center tapped and the diodes are connected as shown in a bridge. Between the terminals 1 and 2 an a.c. voltage source of frequency 400 Hz is connected. Another a.c. voltage of 1.0 MHz is connected between 3 and 4. The output between 5 and 6 contains components at
Communication System
49. In fig.P7.4.49 m( t) =
2 sin 2 pt sin 199pt , s( t) = cos 200 pt and n( t) = t t Multiplier
3
LPF 1 Hz
y(t)
s(t)
n(t)
Fig.P7.4.49
The output y( t) will be sin 2pt (A) t
4
6
2
Multiplier
|H(jw)|=1 s(t)
5
1
Adder
S
m(t)
Fig.P7.4.46
(A) 400 Hz, 1.0 MHz, 1000.4 kHz, 999.6 kHz
(B)
sin 2 pt sin pt + cos 3pt t t
(C)
sin 2 pt sin 0.5 pt + cos 15 . pt t t
(D)
sin 2 pt sin pt + cos 0.75 pt t t
(B) 400 Hz, 1000.4 kHz, 999.6 kHz (C) 1 MHz, 1000.4 kHz, 999.6 kHz (D) 1000.4 kHz, 999.6 kHz 47. A superheterodyne receiver is to operate in the frequency range 550 kHz-1650 kHz, with the denote intermediate frequency of 450 kHz. Let R = CCmax min the required capacitance ratio of the local oscillator and I denote the image frequency (in kHz) of the incoming signal. If the receiver is tuned to 700 kHz, then
50. 12 signals each band-limited to 5 kHz are to be transmitted over a single channel by frequency division multiplexing. If AM -SSNB modulation guard band of 1 kHz is used, then the bandwidth of the multiplexed signal will be (A) 51 kHz
(B) 61 kHz (D) 81 kHz
(A) R = 4.41, I = 1600
(B) R = 2.10, I = 1150
(C) 71 kHz
(C) R = 3, I = 1600
(D) R = 9.0, I = 1150
51. Let x( t) be a signal band-limited to 1 kHz. Amplitude modulation is performed to produce signal A proposed demodulation g( t) = x( t) sin 2000pt. technique is illustrated in figure 6.83. The ideal low pass filter has cutoff frequency 1 kHz and pass band gain 2. The y( t) would be
48. Consider a system shown in Figure . Let X ( f ) and Y ( f ) denote the Fourier transforms of x( t) and y( t) respectively. The ideal HPF has the cutoff frequency 10 kHz. The positive frequencies where Y ( f ) has spectral peaks are
(A) 2 y( t) x(t)
HPF 10 kHz
Balanced Modulator
Balanced Modulator
~
~
10 kHz
13 kHz X(f )
-3
-1
1
Fig.P7.4.48
(A) 1 kHz and 24 kHz (B) 2 kHz and 24 kHz (C) 1 kHz and 14 kHz (D) 2 kHz and 14 kHz Page 418
3
f (kHz)
y(t)
(C)
1 2
y( t)
(B) y( t) (D) 0
52. Suppose we wish to transmit the signal x( t) = sin 200 pt + 2 sin 400 pt using a modulation that create the signal g( t) = x( t) sin 400pt. If the product g( t) sin 400pt is passed through an ideal LPF with cutoff frequency 400p and pass band gain of 2, the signal obtained at the output of the LPF is 1 2
(A) sin 200pt
(B)
(C) 2 sin 200pt
(D) 0
sin 200 pt
53. In a AM signal the received signal power is 10 -10 W with a maximum modulating signal of 5 kHz. The noise spectral density at the receiver input is 10 -18 W/Hz. If the noise power is restricted to the message signal
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Amplitude Modulation
bandwidth only, the signals-to-noise ratio at the input to the receiver is (A) 43 dB
(B) 66 dB
(C) 56 dB
(D) 33 dB
Chap 7.4
SOLUTION 1. (C) u( t) = (20 + 4 sin 500 pt) cos(2 p ´ 10 5 t) V = 20(1 + 0.2 sin 500 pt) cos(2 p ´ 10 5 t) V,
a = 0.2
Statement for Question 54-55 Consider the following Amplitude Modulated (AM)
2. (B) Pc =
signal, where fm < B x AM ( t) = 10(1 + 0.5 sin 2 pfm t) cos 2 pfc t.
(B) 12.5
(C) 6.25
(D) 3.125
25 2 N0 B
(D)
4. (B) x( t) = [20 + 2 cos(2 p1500 t) + 10 cos(2 p3000 t)]cos(2 pfc t) 1 1 æ ö = 20ç 1 + cos(2 p1500 t) + cos(2 p3000 t) cos(2 pfc t) ÷ 10 2 è ø
55. The AM signal gets added to a noise with Power Spectra Density Sn ( f ) given in the figure below. The ration of average sideband power to mean noise power would be 25 25 (B) (A) 8 N0 B 4 N0 B (C)
25 N0 B
Statement for Question 56-57 A certain communication channel is characterized by 80 dB attenuation and noise power-spectral density of 10 -10 W/Hz. The transmitter power is 40 kW and the message signal has a bandwidth of 10 kHz. 56. In the case of conventional AM modulation, the predetecion SNR is
This is the form of a conventional AM signal with message signal 1 1 m( t) = cos(2 p1500 t) + cos(2 p3000 t) 10 2 1 1 = cos 2 (2 p1500 t) + cos(2 p1500 t) 10 2 1 1 The minimum of g( z) = z 2 + is achieved for z10 2 1 201 1 and it is min( g( z)) = . Since z = is in z =20 400 20 the range of cos (2 p1500 t), we conclude that the 201 minimum value of m( t) is . Hence, the modulation 400 201 index is a = 400 5. (B) x( t) = 20 cos(2 pfc t) + cos(2 p( fc - 1500) t) + cos(2 p( fc - 1500) t)
(A) 108
(B) 2 ´ 108
= 5 cos(2 p( fc 3000) t) + 5 cos(2 p( fc + 3000) t)
(C) 10 2
(D) 2 ´ 10 2
The power in the sidebands is 1 1 25 25 + = 26 Psidebands = + + 2 2 2 2
57. In case of SSB, the predetecion SNR is (A) 2 ´ 10 2
(B) 4 ´ 10 2
(C) 2 ´ 10 3
(D) 4 ´ 10 3
*************
ö ÷÷ = 204 W ø
3. (A) Psb = Pt - Pc = 204 - 200 = 4 W
54. The average side-band power for the AM signal given above is (A) 25
æ (0.2) 2 Pt = Pc çç 1 + 2 è
20 2 = 200 W, 2
The total power is Ptotal = Pcarrier + Psidebands = 200 + 26 = 226 The ratio of the sidebands power to the total power is Psidebands 26 = Ptotal 226 æ æ 0.9 2 ö a2 ö 6. (B) Pt = Pc çç 1 + ÷÷ = 2000çç 1 + ÷ = 2810 W 2 ÷ø 2 ø è è æ a2 ö 7. (A) Pt = Pc çç 1 + ÷ 2 ÷ø è Pc = 37.88 kW, www.gatehelp.com
or
æ 0.8 2 ö ÷ 50 ´ 10 3 = Pc çç 1 + 2 ÷ø è
Pi = ( Pt - Pc ) = (50 - 37.88) = 12.12 kW Page 419
GATE EC BY RK Kanodia
UNIT 7
1
1
æ æ a 2 ö2 a 2 ö2 8. (A) I t = I c çç 1 + ÷÷ or 20 = 18çç 1 + ÷ or a = 0.68 2 ÷ø 2 ø è è
18. (A) Pt =
Pt =
500 2 2
Pi = 30 - 20 = 10 kW The DC input power =
10 = 12.5 kW. 0.8
m( t) =
1
é 4 ¥ ( -1) n -1 ù x( t) = ê å cos[2 pfm (2 n - 1)]ú sin(2 p1000 ´ 10 3 t) ë p n =1 2 n - 1 û So frequency component (106 ± fm (2 n - 1) will be present
2 P. If carrier is 3
2 suppressed then P or 66% power will be saved. 3
where n = 1, 2, 3, .... For fm = 10 kHz and n = 1 & 2 frequency present is 990, 970, 1030 kHz.
Thus 1020 kHz will be absent.
20. (B) fc + fm = 705 kHz,
ö ÷÷ ø
BW = 2 fm = 10 kHz or fm = 5 kHz fc = 705 - 5 = 700 kHz
= 1125 . kW
21. (C) x( t) = m( t) c( t) = A(sinc ( t) + sinc 2 ( t) cos(2pfc t)
14. (D) x( t) = Ac (1 + a cos 2 pfm t) cos 2 pfc t Thus a =1, therefor modulation
Here Ac (1 + a) = 2 Ac ,
4 ¥ ( -1) n -1 cos[2 pfm (2 n - 1) ] å p n =1 2 n - 1
The modulated output
11. (C) a 2 = a1a + a 22 = 0.32 + 0.4 2 = 0.5 2 or a = 0.5
ö æ 0.32 0.4 2 ÷÷ = 10çç 1 + + 2 2 ø è
fc = 1000 kHz, x( t) = c( t) m( t)
Expressing square wave as modulating signal m( t)
æ æ 0.7 2 ö a 2 ö2 ÷÷ = 2çç 1 + ÷ = 2.23 kW Pt = Pc çç 1 + 2 ÷ø 2 ø è è
æ a2 a2 13. (A) Pt = Pc çç 1 + 1 + 2 2 2 è
é 0.8 2 ù = 165 kW 1 + ê 2 úû ë
19. (C) c( t) = sin 2p fc t,
a = 70% = 0.7
12. (B) In previous solution Pc =
2 a 2 m( t) ù Ac2 é ê1 + ú 2 ê 2 úû ë
Here modulation index a =1. Thus
1ö æ 9. (C) Pt = 20000ç 1 + ÷, Pt = 30 kW, 2ø è
10. (A) Pc = 2 kW,
Communication System
Taking the Fourier transform of both sides, we obtain
index is 1 or 100% modulation.
X( f ) =
15. (A) If modulation index a is 0, then
=
A [ P( f ) + L( f )] * ( d( f - fc ) + d( f + fc )) 2
A [ P( f - fc ) + L( f - fc ) + P( f + fc ) + L( f - fc )] 2
æ 0 2 ö Ac2 ÷= çç 1 + 2 ÷ø 2 è
1 Since P( f - fc ) ¹ 0 for f - fc < , whereas L( f - fc ) ¹ 0 2
If modulation index is 1 then
for f - fc < 1. Hence, the bandwidth of the bandpass
Pt1 =
Pt2 =
Ac2 2
Ac2 2
æ 12 ö 3 2 ÷ = Ac , çç 1 + 2 ÷ø 4 è
filter is 2.
Pt 2 3 = Pt1 2
22. (C) x( t) = m( t) c( t)
Thus Pt2 = 15 . Pt1 and Pt2 is increases by 50%
= 100[cos(2 p1000 t) + 2 cos(2 p2000 t)]cos(2 pfc t) = 100 cos(2 p1000 t) cos(2 pfc t) + 200 cos(2 p2000 t) cos(2 pfc t) 100 = [cos(2 p( fc + 1000) t) + cos(2 p( fc - 1000) t)] 2
16. (A) fm = 10 kHz, R = 1000 W, C = 10000 pF Hence 2pfm RC = 2 p ´ 10 4 ´ 10 3 ´ 10 -8 = 0.628 a max = (1 + (0.628) 2 ) 17. (B)
-
1 2
= 0.847
1 1 , £ RC £ fc BWm
+
200 [cos(2 p( fc + 2000) t) + cos(2 p( fc - 2000) t)] 2
Thus, the upper sideband (USB) signal is
Here fc = 1 MHz
xu ( t) = 50 cos[2 p( fc + 1000) t ] + 100(2 p( fc + 2000) t)
Signal Bandwidth BWm = 2 fm = 2 ´ 2 ´ 10 3 = 4 kHz Thus
1 1 or 10 -6 £ RC £ 250 ms £ RC £ 106 4 ´ 10 3
Thus appropriate value is 20 m sec Page 420
23. (B) When USSB is employed the bandwidth of the modulated signal is the same with the bandwidth of the message signal. Hence WUSSB = W = 10 4 Hz www.gatehelp.com
GATE EC BY RK Kanodia
UNIT 7
37.
(B)
Sidebands
are
( 6 ´ 106 ± 5000)
and
Communication System
43. (C) The filter characteristic is shown in fig.S7.4.43
( 6 ´ 10 ± 9000) 6
H(f )
Thus 6.005 ´ 10 , 5.995 ´ 10 , 5.991 ´ 10 or 5.991 ´ 10 , 6
6
6
6
m(t)
6.005 ´ 106 and 6.009 ´ 106 38. (B) x( t) can be written as
f W
x( t) = ( 40 + 8 cos 40 pt) cos 400 pt 8 modulation index a = = 0.2 40 1 Pc = ( 40) 2 = 800 W 2
fc + W < 2 f or fc > W
1 1 Psb = B 2 + B 2 = 66.67 2 2
the
Hilbert
44. (B) Since High-side tuning is used fLO = fm + f IF = 500 kHz, fLOL = 5 + 0.5 = 5.5 MHz,
45. (B) fimage = fL + 2 f IF = 2400 = 3310 kHz 46. (D) The given circuit is a ring modulator. The output
or B = 8.161
is DSB-SC signal. So it will contain m( t) cos( nwc t) where n = 1, 2, 3...... Therefore there will be only (1 MHz ± 400
transform
of
sin (2 p1000 t) is cos (2 p1000 t)
Hz) frequency component. 47. (A) fmax = 1650 + 450 = 2100 kHz 1
$ ( t)= sin (2 p1000 t) - cos (2 p1000 t) Thus m
fmin = 550 + 450 = 1000 kHz.
41. (D) The expression for the LSSB AM signal is.
frequency is minimum, capacitance will be maximum
x l ( t) = Ac m( t) cos(2 pfc t) + Ac m( t) sin(2 pfc t)
R=
Substituting $ ( t) = sin(2 p1000 t) - 2 cos(2 p1000 t) and m
= 100[cos(2 p1000 t) cos(2 pfc t) + sin(2 p1000 t) sin(2 pfc t)] +200[cos(2 pfc t) sin(2 p1000 t) - sin(2 pfc t) cos(2 p1000 t)] = 100 cos(2 p( fc - 1000) t) - 200 sin(2 p( fc - 1000) t)
= 4m(t) + 4 cos wc t + 10m2 (t) + 20m(t) cos wc t + 5 + 5 cos 2wc t
xc ( t) = 4[1 + 5 Mmn ( t)]cos wc t 5M = 0.8
or
M = 0.16
component of 11 kHz will remain. The output of 2nd modulator will be (13 ± 11) kHz. So Y ( f ) has spectral peak at 2 kHz and 24 kHz. 49. (C) m( t) s( t) = y1 ( t)
42. (D) y( t) = 4( m( t) + cos wc t) + 10( m( t) + cos wc t) 2
m( t) = Mmn ( t)
or R = 4.41
kHz and (10 - 1) kHz. After passing the HPF frequency
+100[sin(2 p1000 t) - 2 cos(2 p1000 t) sin(2 pfc t)]
xc ( t) = 4[1 + 5 m( t)]cos wc t
2p LC
output of first modulator spectral peak will be at (10 + 1)
xl ( t) = 100[cos(2 p1000 t) + 2 sin(2 p1000 t) cos(2 pfc t)]
The AM signal is,
f =
48. (B) Since X ( f ) has spectral peak at 1 kHz so at the
we obtain
= 5 + 4m(t) + 10m2 (t) + 4[1 + 5m(t)] cos wc t + 5 cos 2wc t
2 Cmax fmax = 2 = (2.1) 2 Cmin fmin
or
fi = fc + 2 f IF = 700 + 2( 455) = 1600 kHz
Ac = 100, m( t) = cos(2 p1000 t) + 2 sin(2 p1000 t)
Page 422
Therefore fc > 3W
fLOU = 10 + 0.5 = 10.5 MHz
40. (D) The Hilbert transform of cos (2 p1000 t) is whereas
2fc
fc+W
fc - W > 2W or fc > 3W ,
A2 39. (A) Carrier power Pc = = 100 W, A = 14.14 2 40 Psb Psb or Eeff = = 0.4 = 100 + Psb Pc + Psb 100
sin (2 p1000 t),
fc-W fc
Fig.S7.4.43
The components at 180 Hz and 220 Hz are side band 1 1 Psb = ( 4) 2 + ( 4) 2 = 16 W, 2 2 Psb 16 Eeff = = Pc + Psb 800 + 16
Psb = 66.67 W,
2W
=
2 sin(2 pt) cos(200 pt) sin(202 pt) - sin(198 pt) = t t
y1 ( t) + n( t) = y2 ( t) =
sin 202 pt - sin 198 pt sin 198 pt + t t
y2 ( t) s( t) = y( t) =
[sin 202 pt - sin 198 pt + sin 199 pt ]cos 200 pt t
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GATE EC BY RK Kanodia
Amplitude Modulation
=
Chap 7.4
55. (D) Noise power = Area rendered by the spectrum
1 [sin( 402 pt) + sin(2 pt) - {sin( 398 pt) - sin(2 pt)} 2 + sin( 399pt) - sin( pt)]
After filtering sin(2 pt) + sin(2 pt) - sin( pt) 2t sin(2 pt) + 2 sin(0.5 t) cos(15 . pt) = 2t sin 2 pt sin 0.5 pt = + cos 15 . pt 2t t y( t) =
= N0 B Ratio of average sideband power to mean noise 6.25 25 Power = = N0 B 4 N0 B 56. (C) Since the channel attenuation is 80 db, then P 10 log T = 80 PR or PR = 10 -8 PT = 10 -8 ´ 40 ´ 10 3 = 4 ´ 10 -4 Watts If the noise limiting filter has bandwidth B, then the
50. (D) The total signal bandwidth = 5 ´ 12 = 60 kHz
pre-detection noise power is
There would be 11 guard band between 12 signal. So
fc +
Pn = 2
guard band width = 11 kHz Total band width = 60 + 11 = 71 kHz
ò
fc -
B 2
B 2
N0 df = N 0 B = 2 ´ 10 -10 B Watts 2
In the case of DSB or conventional AM modulation,
51. (D) x1 ( t) = g( t) cos(2000pt)
B = 2W = 2 ´ 10 4
1 = x( t) sin(2000 pt) cos(2000 pt) = x( t) sin( 4000 pt) 2 1 X1 ( jw) = X ( j( w - 4000 p)) - X ( j( w + 4000 p)) 4j
Hz, whereas in SSB modulation
B = W = 10 . Thus, the pre-detection signal to noise 4
ratio in DSB and conventional AM is SNR DSB,AM =
This implies that X1 ( jw) is zero for w £ 2000 p because w < 2 pfm = 2 p1000. When x1 ( t) is passed through a LPF with cutoff frequency 2000p, the output will be zero.
PR 4 ´ 10 -4 = = 10 2 Pn 2 ´ 10 -10 ´ 2 ´ 10 4
57. (A) In SSB modulation B = W = 10 4 SNR SSB =
52. (A) y( t) = g( t) sin( 400 pt) = x( t) sin 2 ( 400 pt) (1 - cos)( 800 pt) = (sin(200 pt) + 2 sin( 400 pt) 2 1 = [sin(200 pt) - sin(200 pt) cos( 800 pt) + 2 sin( 400 pt) 2
4 ´ 10 -4 = 2 ´ 10 2 2 ´ 10 -10 ´ 10 4
***********
-sin( 400 pt) cos( 800 pt) 1 1 = sin(200 pt) - [sin(1000 pt) - sin( 6000 pt)] 2 4 1 + sin( 400 pt) - [sin(1200 pt) - sin( 400 pt)] 4 If this signal is passed through LPF with frequency 400p and gain 2, the output will be sin(200pt) 53. (A) Message signal BW fm = 5 kHz Noise power density is 10 -18 W/Hz Total noise power is 10 -18 ´ 5 ´ 10 3 = 5 ´ 10 -15 W Input signal-to-noise ratio SNR=
10 -10 = 2 ´ 10 4 or 43 dB 5 ´ 10 -15
54. (C) Average side band power is Ac2 a 2 10 2 (0.5) 2 = = 6.25 W 4 4
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GATE EC BY RK Kanodia
CHAPTER
7.6 DIGITAL TRANSMISSION
Statement for Question 1-5
Statement for Question 6-7
Fig. P7.6.1-5 shows fourier spectra of signal x( t) and y(t). Determine the Nyquist sampling rate for the
A signal x( t) is multiplied by rectangular pulse train c( t) shown in fig.P7.6.6-7.. c(t)
given function in question. X( jw)
0.25 ms
Y( jw)
-2´10-3
-10-3
w
w
Fig.P7.6.1-5
10-3
2´10-3
t
Fig.P7.6.6-7
6. x( t) would be recovered form the product. x( t) c( t) by using an ideal LPF if X ( jw) = 0 for
1. x( t) (A) 100 kHz
(B) 200 kHz
(A) w > 2000 p
(B) w > 1000 p
(C) 300 kHz
(D) 50 kHz
(C) w < 1000 p
(D) w < 2000 p
7. If X ( jw) satisfies the constraints required, then the
2. y( t)
pass band gain A of the ideal lowpass filter needed to
(A) 50 kHz
(B) 75 kHz
recover x( t) from e( t) x( t) is
(C) 150 kHz
(D) 300 kHz
(A) 1
(B) 2
(C) 4
(D) 8
3. x 2 ( t) (A) 100 kHz
(B) 150 kHz
(C) 250 kHz
(D) 400 kHz
4. y 3( t) (A) 100 kHz
(B) 300 kHz
(C) 900 kHz
(D) 120 kHz
8. Consider a set of 10 signals xi ( t), i = 1, 2, 3, ...10.. Each signal is band limited to 1 kHz. All 10 signals are to be time-division multiplexed after each is multiplied by a carrier e( t) shown in Figure. If the period T of e( t) is chosen the have the maximum allowable value, the largest value of D would be c(t)
D
5. x( t) y( t) (A) 250 kHz
(B) 500 kHz
(C) 50 kHz
(D) 100 kHz
Page 434
-2T
-T
0
Fig.P7.6.8
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T
2T
t
GATE EC BY RK Kanodia
Digital Transmission
(A) 5 ´ 10 -3 sec -5
(C) 5 ´ 10 sec
(B) 5 ´ 10 -4 sec
Chap 7.6
15. A CD record audio signals digitally using PCM. The audio signal bandwidth is 15 kHz. The Nyquist samples
-6
(D) 5 ´ 10 sec
are quantized into 32678 levels and then binary coded. 9. A compact disc recording system samples a signals
The minimum number of binary digits required to
with a 16-bit analog-to-digital convertor at 44.1 kbits/s.
encode the audio signal
The CD can record an hours worth of music. The
(A) 450 k bits/sec
(B) 900 k bits/sec
approximate capacity of CD is
(C) 980 340 k bits/sec
(D) 490 170, k bits/sec
(A) 705.6 M Bytes
(B) 317.5 M Bytes
(C) 2.54 M Bytes
(D) 5.43 M Bytes
16. The American Standard Code for Information Interchange has 128 characters, which are binary
10. An analog signal is sampled at 36 kHz and
coded. If a certain computer generates 1,000,000
quantized into 256 levels. The time duration of a bit of
character per second, the minimum bandwidth required
the binary coded signal is
to transmit this signal will be
(A) 5.78 ms
(B) 3.47 ms
(A) 1.4 M bits/sec
(B) 14 M bits/sec
(C) 6.43 ms
(D) 7.86 ms
(C) 7 M bits/sec
(D) 0.7 M bits/sec
11. An analog signal is quantized and transmitted using
17. A binary channel with capacity 36 k bits/sec is
a PCM system. The tolerable error in sample amplitude
available for PCM voic transmission. If signal is band
is 0.5% of the peak-to-peak full scale value. The
limited to 3.2 kHz, then the appropriate values of
minimum binary digits required to encode a sample is
quantizing level L and the sampling frequency will be
(A) 5
(B) 6
(A) 32, 3.6 kHz
(B) 64, 7.2 kHz
(C) 7
(D) 8
(C) 64, 3.6 kHz
(D) 32, 7.2 kHz
18. Fig.P7.4.18 shows a PCM signals in which
Statement for Question 12-13. Ten telemetry signals, each of bandwidth 2kHz, are to be transmitted simultaneously by binary PCM. The maximum tolerable error in sample amplitudes is 0.2% of the peak signal amplitude. The signals must be
amplitude level of + 1 volt and - 1 volt are used to represent binary symbol 1 and 0 respectively. The code word used consists of three bits. The sampled version of analog signal from which this PCM signal is derived is
sampled at least 20% above the Nyquist rate. Framing and synchronizing requires an additional 1% extra bits. Fig.P7.4.18
12. The minimum possible data rate must be (A) 272.64 k bits/sec
(B) 436.32 k bits/sec
(A) 4 5 1 2 1 3
(B) 8 4 3 1 2
(C) 936.32 k bits/sec
(D) None of the above
(C) 6 4 3 1 7
(D) 1 2 3 4 5
13. The minimum transmission bandwidth is
19. A PCM system uses a uniform quantizer followed by
(A) 218.16 kHz
(B) 468.32 kHz
a 8-bit encoder. The bit rate of the system is equal to 108
(C) 136.32 kHz
(D) None of the above
bits/s. The maximum message bandwidth for which the system operates satisfactorily is
14. A Television signal is sampled at a rate of 20% above the Nyquist rate. The signal has a bandwidth of 6 MHz. The samples are quantized into 1024 levels. The
(A) 25 MHz
(B) 6.25 MHz
(C) 12.5 MHz
(D) 50 MHz
minimum bandwidth required to transmit this signal
20. Twenty-four voice signals are sampled uniformly at
would be
a rate of 8 kHz and then time-division multiplexed. The
(A) 72 M bits/sec
(B) 144 M bits/sec
(C) 72 k bits/sec
(D) 144 k bits/sec
sampling process uses flat-top samples with 1 ms duration. The multiplexing operating includes provision
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Page 435
GATE EC BY RK Kanodia
UNIT 7
Communication System
for synchronization by adding and extra pulse of 1 ms
quantized into 256 level using a m-low quantizer with
duration. The spacing between successive pulses of the
m = 225.
multiplexed signal is (A) 4 ms
(B) 6 ms
(C) 7.2 ms
(D) 8.4 ms
26. The signal-to-quantization-noise ratio is
21. A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size d is 100 m V. The modulator is tested with a this test signal required to avoid slope overload is (A) 2.04 V
(B) 1.08 V
(C) 4.08 V
(D) 2.16 V
a
linear
DM
(B) 38.06 dB
(C) 42.05 dB
(D) 48.76 dB
27. It was found that a sampling rate 20% above the rate wou7ld be adequate. So the maximum SNR, that can be realized without increasing the transmission bandwidth, would be (A) 60.4 dB
(B) 70.3 dB
(C) 50.1 dB
(D) None of the above
28. For a PCM signal the compression parameter
Statement fo Question 22-23 : Consider
(A) 34.91 dB
m = 100 and the minimum signal to quantization-noise system
designed
to
ratio is 50 dB. The number of bits per sample would be.
accommodate analog message signals limited to bandwidth
(A) 8
(B) 10
of 3.5 kHz. A sinusoidal test signals of amplitude Amax = 1
(C) 12
(D) 14
V and frequency fm = 800 Hz is applied to system. The
29. A sinusoid massage signal m( t) is transmitted by
sampling rate of the system is 64 kHz. 22. The minimum value of the step size to
binary avoid
overload is
PCM
without
compression.
the
signal
to-quantization-noise ratio is required to be at least 48 dB, the minimum number of bits per sample will be
(A) 240 mV
(B) 120 mV
(A) 8
(B) 10
(C) 670 mV
(D) 78.5 mV
(C) 12
(D) 14
23. The granular-noise power would be (A) 1.68 ´ 10 -3 W (C) 2.48 ´ 10
If
-3
W
30. A speech signal has a total duration of 20 sec. It is
(B) 2.86 ´ 10 -4 W (D) 112 . ´ 10
-4
sampled at the rate of 8 kHz and then PCM encoded. The signal-to-quantization noise ratio is required to be
W
40 dB. The minimum storage capacity needed to 24. The SNR will be (A) 298 (C) 4.46 ´ 10
3
accommodate this signal is (B) 1.75´10 -3
(A) 1.12 KBytes
(B) 140 KBytes
(D) 201
(C) 168 KBytes
(D) None of the above
25. The output signal-to-quantization-noise ratio of a 10-bit
31. The input to a linear delta modulator having fa
PCM was found to be 30 dB. The desired SNR is 42 dB. It
step-size D = 0.628 is a sine wave with frequency fm and
can be increased by increasing the number of quantization
peak amplitude Em . If the sampling frequency fs = 40
level.In this way the fractional increase in the transmission
kHz, the combination of the sinc-wave frequency and
bandwidth would be (assume log 2 10 = 0.3)
the peak amplitude, where slope overload will take
(A) 20%
(B) 30%
piace is
(C) 40%
(D) 50%
Em (A) 0.3 V
fm 8 kHz
(B) 1.5 V
4 kHz
A signal has a bandwidth of 1 MHz. It is sampled
(C) 1.5 V
2 kHz
at a rate 50% higher than the Nyquist rate and
(D) 3.0 V
1 kHz
Statement for Question 26-27.
Page 436
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GATE EC BY RK Kanodia
Digital Transmission
Chap 7.6
32. A sinusoidal signal with peak-to-peak amplitude of
38. Four signals g1 ( t), g 2 ( t), g s ( t) and g 4 ( t) are to be
1.536 V is quantized into 128 levels using a mid-rise
multiplexed and transmitted. g1 ( t) and g 4 ( t) have a
uniform quantizer. The quantization-noise power is
bandwidth of 4 kHz, and the remaining two signals
-6
(A) 0.768 V
(B) 48 ´ 10 V
(C) 12 ´ 10 -6 V 2
(D) 3.072 V
have bandwidth of 8 kHz,. Each sample requires 8 bit
2
for encoding. What is the minimum transmission bit rate of the system.
33. A signal is sampled at 8 kHz and is quantized using
(A) 512 kbps
(B) 16 kbps
for a
(C) 192 kbps
(D) 384 kbps
8 bit uniform quantizer. Assuming SNR q
sinusoidal signal, the correct statement for PCM signal 39. Three analog signals, having bandwidths 1200 Hz,
with a bit rate of R is
600 Hz and 600 Hz, are sampled at their respective
(A) R = 32 kbps, SNR q = 25.8 dB
Nyquist rates, encoded with 12 bit words, and time
(B) R = 64 kbps, SNR q = 49.8 dB
division multiplexed. The bit rate for the multiplexed
(C) R = 64 kbps, SNR q = 55.8 dB
signal is
(D) R = 32 kbps, SNR q = 49.8 dB
(A) 115.2 kbps
(B) 28.8 kbps
(C) 57.6 kbps
(D) 38.4 kbps
34. A 1.0 kHz signal is flat-top sampled at the rate of 180 samples sec and the samples are applied to an ideal rectangular LPF with cat-off frequency of 1100 Hz, then the output of the filter contains
40. The minimum sampling frequency (in samples/sec) required to reconstruct the following signal form its samples without distortion would be
(A) only 800 Hz component
3
2
(B) 800 and 900 Hz components
æ sin 2 p1000 t ö æ sin 2 p1000 t ö x( t) = 5ç ÷ + 7ç ÷ pt pt è ø è ø
(C) 800 Hz and 1000 Hz components
(A) 2 ´ 10 3
B) 4 ´ 10 3
(D) 800 Hz, 900 and 1000 Hz components
(C) 6 ´ 10 3
(D) 8 ´ 10 3
35. The Nyquist sampling interval, for the signal sinc (700 t) + sinc (500 t) is 1 (A) sec 350
p (B) sec 350
1 (C) sec 700
p (D) sec 175
41.
The
minimum
step-size
required
for
a
Delta-Modulator operating at32 K samples/sec to track the signal (here u( t) is the nuit function) x( t) = 125 t( u( t) - u( t - 1)) + (250 - 125 t)( u( t - 1) - u( t - 2)s so that slope overload is avoided, would be
36. A signal x( t) = 100 cos(24 p ´ 10 ) t is ideally sampled 3
(A) 2 -10
(B) 2 -8
(C) 2 -6
(D) 2 -4
with a sampling period of 50 m sec and then passed through an ideal lowpass filter with cutoff frequency of
42. Four signals each band limited to 5 kHz are
15 KHz. Which of the following frequencies is/are
sampled at twice the Nyquist rate. The resulting PAM
present at the filter output
samples are transmitted over a single channel after
(A) 12 KHz only
(B) 8 KHz only
time division multiplexing. The theoretical minimum
(C) 12 KHz and 9 KHz
(D) 12 KHz and 8 KHz
transmissions bandwidth of the channel should be equal to.
37. In a PCM system, if the code word length is
(A) 5 kHz
(B) 20 kHz
increased form 6 to 8 bits, the signal to quantization
(C) 40 kHz
(D) 80 kHz
noise ratio improves by the factor. (A) 8/6
(B) 12
(C) 16
(D) 8
43. Four independent messages have bandwidths of 100 Hz, 100 Hz, 200 Hz and 400 Hz respectively. Each is
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GATE EC BY RK Kanodia
UNIT 7
Communication System
sampled at the Nyquist rate, time division multiplexed
50. A speech signal occupying the bandwidth of 300 Hz
and transmitted. The transmitted sample rate, in Hz,
to 3 kHz is converted into PCM format for use in digital
is given by
communication. If the sampling frequency is8 kHz and
(A) 200
(B) 400
each sample is quantized into 256 levels, then the
(C) 800
(D) 1600
output bit the rate will be
44.
The
Nyquist
sampling
rate
for
the
signal
g( t) = 10 cos(50 pt) cos 2 (150 pt). Where ' t ' is in seconds, is
(A) 3 kb/s
(B) 8 kb/s
(C) 64 kb/s
(D) 256 kb/s
(A) 150 samples per second
51. If the number of bits in a PCM system is increased
(B) 200 samples per second
from n to n + 1, the signal-to-quantization noise ratio
(C) 300 samples per second
will increase by a factor.
(D) 350 samples per second
(A)
45. A TDM link has 20 signal channels and each
(C) 2
channel is sampled 8000 times/sec. Each sample is represented by seven binary bits and contains an additional bit for synchronization. The total bit rate for the TDM link is (A) 1180 K bits/sec
(B) 1280 K bits/sec
(C) 1180 M bits/sec
(D) 1280 M bits/sec
46. In a CD player, the sampling rate is 44.1 kHz and the samples are quantized using a 16-bit/sample quantizer. The resulting number of bits for a piece of music with a duration of 50 minutes is (A) 1.39 ´ 10 9
(B) 4.23 ´ 10 9
(C) 8.46 ´ 10 9
(D) 12.23 ´ 10 9
( n + 1) n
(B)
(D) 4
52. In PCM system, if the quantization levels are increased
form
2
to
8,
the
256
quantization
levels.
(A) remain same
(B) be doubled
(C) be tripled
(D) become four times
53. Assuming that the signal is quantized to satisfy the condition of previous question and assuming the approximate bandwidth of the signal is W. The minimum required bandwidth for transmission of a will be. (A) 5 W
(B) 10 W
(C) 20 W
(D) None of the above
The
bit
transmission rate for the time-division multiplexed signal will be (A) 8 kbps (C) 256 kbps
************ (B) 64 kbps (D) 512 kbps
48. Analog data having highest harmonic at 30 kHz generated by a sensor has been digitized using 6 level PCM. What will be the rate of digital signal generated? (A) 120 kbps (C) 240 kbps
(B) 200 kbps (D) 180 kbps
49. In a PCM system, the number of quantization levels is 16 and the maximum lsignal frequency is 4 kHz.; the bit transmission rate is (A) 32 bits/s (C) 32 kbits/s Page 438
(B) 16 bits/s (D) 64 dbits/s
bandwidth
requirement will.
sampled at Nyquist rate are converted into binary PCM using
relative
binary PCM signal based on this quantization scheme
47. Four voice signals. each limited to 4 kHz and signal
( n + 1) 2 n2
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GATE EC BY RK Kanodia
UNIT 7
16. (C) 128= 2 7,. We need 7 bits/character. For 1,000,000 character we need 7 Mbits/second. Thus minimum bandwidth = 7 Mbits/sec.
26. (B)
Communication System
3 L2 So = = 6394 = 38.06 dB N o [ln(m + 1)]2
27. (C) Nyquist Rate = 2 MHz
17. (D) fs > 2 fm = 6400 Hz, nfs £ 63000 36000 36000 n£ = 5.63, n = 5, L = 2 n = 32, fs = = 7.2 kHz. 6400 5
50% higher rate = 3 MHz,
Thus transmission bandwidth is 3 MHz ´ 8 = 24 Mbits/s. New sampling rate is at 20% above the Nyquist rate. Sampling rate= 12 . ´ 2 = 2.4 MHz.
18. (D) The transmitted code word are
bits per second= 0
0
1
1
0
0
0
1
1
1
0
0
1
0
bits 24 M sec = 10 bits 2.4 MHz
1
So 3(1024) 2 = = 102300 = 50.1 dB N o (ln 256) 2
Level = 210 = 1024, Fig.S.7.6.18
In 1st word 001(1) 28. (B)
In 2nd word 010(2) In 3rd word 011(3)
3 L2 So = ³ 50 dB, m = 100 N o [ln(m + 1)]2
3 L2 = 100 000 or [ln 101) 2
In 4th word 100(4) In 5th word 101 (5)
L = 842.6
Because L is power of 2, we select L = 1024 = 210 .
19. (B) Message bandwidth= W, Nyquist rate = 2W
Thus 10 bits are required.
Bandwidth = 2W ´ 8 = 16W bit/s 29. (A) So = m 2 ( t), N o =
108 W= = 6.25 MHz 16
8
16W= 10 , or
1 20. (A) Sampling interval T, = = 125 ms. There are 24 8k channels and 1 sync pulse, so the time allotted to each channel is Tc =
T 25
= 5 ms. The pulse duration is 1 ms. So
the time between pulse is 4 ms. 21. (B) Amax =
22. (D) Amax =
23. (D) N o =
24. (C)
L = 256 = 28
So 3 L2 m 2 ( t) , = 3 L2 No mp2
since signal is sinusoidal
m 2 ( t) 1 = , mp2 2
3 L2 = 48 dB = 63096, L = 205.09 2 Since L is power of 2, so we select L = 256 Hence 256 = 28 ,
dfs 0.1 ´ 68 k = 108 . V = wm 2 p ´ 10 3
3mp2
So 8 bits per sample is required .
30. (B) ( SNR) q = 176 . + 6.02( n) = 40 dB, n = 6.35 We take the n = 7.
df s A w 1 ´ 2 p ´ 800 or d = max m = = 78.5 mV wm fs 50 ´ 10 3
d2 B (0.0785) 2 ´ 3500 = = 1122 . ´ 10 -4 W 3 fs 3 ´ 64000
So 0.5 = 4.46 ´ 10 3 = N o 112 . ´ 10 -4
Capacity = 20 ´ 8 k ´ 7 = 112 . Mbits = 140 Kbytes 31. (B) For slope overload to take place Em ³
df s 2 pfm
This is satisfied with Em = 15 . V and fm = 4 kHz. 32. (C) Step size d=
2 mp L
=
1536 . = 0.012 V 128
quantization noise power S 25. (A) o µ L2 , No
æ S ö L = 2 n çç o ÷÷ = 10 log( C2 2 n ) è N o ødB
= log C + 20 n log 2 = a + 6 n dB.
=
This equation shows
that increasing n by one bits increase the by 6 dB. Hence an increase in the SNR by 12 dB can be accomplished by increasing 9is form 10 to 12, the transmission bandwidth would be increased by 20% Page 440
d2 (0.012) 2 = 12 ´ 10 -6 V 2 = 12 12
33. (B) Bit Rate = 8 k ´ 8 = 64 kbps (SNR) q = 176 . + 6.02 n dB = 176 . + 6.02 ´ 8 = 49.8 dB 34. (B) fs = 1800 samples/sec,
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fm =
1800 = 900 Hz 2
GATE EC BY RK Kanodia
Digital Transmission
Since the sampling rate is 1800 samples/sec the highest frequency that can be recovered is 900 Hz.
The
maximum
frequency
component
will
be
150 + 25 = 175 Hz.
x( t) is band limited with fm = 350 Hz,
Thus Nyquist 1 rate is 2 fm = 700 Hz, Sampling interval = sec 700 1 1 = = 20 kHz, T 50 ´ 10 -6
æ 1 + cos 300 pt ö 44. (D) g( t) = 10 cos 50 ptç ÷ 2 è ø = 5 cos 50 pt + 5 cos 50 pt cos 300 pt
35. (C) x( t) = sinc 700t + sinc 500t 1 = [sin 700 pt + sin 500 pt ] pt
36. (D) fs =
Chap 7.6
Thus fs = 2 ´ 175 = 350 . sample per second. 45. (B) Total sample = 8000 ´ 20 = 160 k sample/sec Bit for each sample = 7 + 1 = 8 Bit Rate = 160 k ´ 8 = 1280 ´ 10 3 bits/sec
fc = 12 kHz
The frequency passed through LPF are fc , fs - fm or 12 kHz, 8 kHz
46. (B) The sampling rate is fs = 44100 meaning that we take 44100 samples per second. Each sample is quantized using 16 bits so the total number of bits per
37. (C) P =
( SNR)1 2 , Here n = code word length, = (SNR)2 2 2 n 1 2n 2
16
n1 = 61 n2 = 8,
Thus rate =
second is 44100´16. For a music piece of duration 50 min = 3000- sec the resulting number of bits per channel
2 = 16 212
(left
= 2.1168 ´ 10
9
and
right)
is 44100 ´ 16 ´ 3000
and the overall number of bits is
2.1168 ´ 10 ´ 2 = 4.2336 ´ 10 9 9
38. (D) Signal g1 ( t), g 2 ( t), g s ( t) and g 4 ( t) will have 8 k, 8 k, 16 k and 16 k sample/sec at Nyquist rate. Thus
47. (C) Nyquist Rate = 2 ´ 4k = 8 kHz
48000 sample/sec bit rate 48000 ´ 8 = 384 kbps
Total sample = 4 ´ 8 = 32 k sample/sec
39. (C) Analog signals, having bandwidth 1200 Hz, 600 Hz and 600 Hz have 2400, 1200 samples/sec at Nyquist
256 = 28 , so that 8 bits are required Bit Rate = 32 k ´ 8 = 256 kbps
rate. Hence 48000 sample/sec
48. (D) Nyquist Rate = 2 ´ 30 k = 60 kHz
bit rate = 48000 sample/sec ´12 = 57.6 kbps
2 n ³ 6 Thus n = 3,
3
æ sin 2 p1000 t ö æ sin 2 p1000 t ö 40. (C) x( t) = 5ç ÷ + 7ç ÷ pt pt è ø è ø
2
49. (C) Nyquist rate= 2 ´ 4 = 8 kHz 2 n = 16 or n = 4,
Maximum frequency component = 3 ´ 1000 = 3 kHz Sampling rate = 2 fm = 6 kHz
50. (C) fs = 8 kHz,
41. (B) Here fs = 32 k sample/sec 1 1 fm = = Em = 125, T 2
51. (D)
For slope-overload to be averted Em ³ D£
Em fm fs
or D £
Bit Rate = 4 ´ 8 = 32 kbits/sec 2 n = 256 Þ n = 8
Bit Rate = 8 ´ 8k = 64 kb/x
1 2
So a 2 2 n , If PCM is increased form n to n + 1, No
the ratio will increase by a factor 4. Which is
fs fm
independent of n. 1 2
125( ) (128)( ) or D £ 2 -8 or D £ 3 32 ´ 10 32 ´ 1024
42. (D) fm = 5 kHz,
Bit Rate = 60 ´ 3 = 18 kHz
Nyquist Rate = 2 ´ 5 = 10 kHz
Since signal are sampled at twice the Nyquist rate so
52. (C) If L = 2, then 2 = 2 n or n = 1 ND If L = 8, then 8 = 2 n or n = 3. So relative bandwidth will be tripled. 53. (B) The minimum bandwidth requirement for transmission of a binary PCM signal is BW= vW. Since
sampling rate = 2 ´ 10 = 20 kHz.
v = 10, we have BW = 10 W
Total transmission bandwidth = 4 ´ 20 = 80 kHz 43. (D) Signal will be sampled 200, 200, 400 and 800 sample/sec thus 1600 sample per second,
*********** www.gatehelp.com
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GATE EC BY RK Kanodia
CHAPTER
7.8 SPREAD SPECTRUM
Statement for Question 1-3 :
5. The Antijam margin is
A pseudo-noise (PN) sequance is generated using a
(A) 47.5 dB
(B) 93.8 dB
feedback shift register of length m = 4. The chip rate is
(C) 86.9 dB
(D) 12.6 dB
7
10 chips per second 6. 1. The PN sequance length is
A slow
FH/MFSK
system
has
the
following
parameters.
(A) 10
(B) 12
Number of bits per MFSK symbol = 4
(C) 15
(D) 18
Number of MFSK symbol per hop = 5 The processing gain of the system is
2. The chip duration is (A) 1ms
(B) 0.1 ms
(C) 0.1 ms
(D) 1 ms
(B) 15 ms
(C) 6.67 ns
(D) 0.67 ns
(B) 37.8 dB
(C) 6 dB
(D) 26 dB
7.
A fast
FH/MFSK
system
has
the
following
parameters.
3. The period of PN sequance is (A) 15 . ms
(A) 13.4 dB
Number of bits per MFSK symbol = 4 Number of pops per MFSK symbol = 4 The processing gain of the system is
Statement for Question 4-5:
(A) 0 dB
(B) 7 dB
(C) 9 dB
(D) 12 dB
A direct sequence spread binary phase-shiftkeying system uses a feedback shift register of Length
Statement for Question 8-9:
19 for the generation of PN sequence . The system is required to have an average probability of symbol error due to externally generated interfering signals that does not exceed 10 -5
A rate 1/2 convolution code with dfrec = 10 is used to encode a data requeence occurring at a rate of 1 kbps. The modulation is binary PSK. The DS spread spectrum sequence has a chip rate of 10 MHz
4. The processing gain of system is
8. The coding gain is
(A) 37 dB
(B) 43 dB
(A) 7 dB
(B) 12 dB
(C) 57 dB
(D) 93 dB
(C) 14 dB
(D) 24 dB
Page 450
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GATE EC BY RK Kanodia
Spread Spectrum
9. The processing gain is
Chap 7.8
Statement for question 16-18 :
(A) 14 dB
(B) 37 dB
(C) 58 dB
(D) 104 dB
A CDMA system consist of 15 equal power user that transmit information at a rate of 10 kbps, each using a DS spread spectrum signal operating at chip
10. A total of 30 equal-power users are to share a
rate of 1 MHz. The modulation scheme is BPSK.
common communication channel by CDM. Each user transmit information at a rate of 10 kbps via DS spread
16. The Processing gain is
spectrum and binary PSK. The minimum chip rate to
(A) 0.01
(B) 100
(C) 0.1
(D) 10
obtain a bit error probability of 10
-5
(A) 1.3 ´ 106 chips/sec
(B) 2.9 ´ 10 5 chips/sec
(C) 19 . ´ 106 chips/sec
(D) 1.3 ´ 10 5 chips/sec
11. A CDMA system is designed based on DS spread
17. The value of e b/J 0 is (A) 8.54 dB
(B) 7.14 dB
(C) 17.08 dB
(D) 14.28 dB
spectrum with a processing gain of 1000 and BPSK modulation scheme. If user has equal power and the
18. How much should the processing gain be increased
desired level of performance of an error probability of
to allow for doubling the number of users with affecting
10 -6 , the number of user will be
the autopad SNR
(A) 89
(B) 117
(A) 1.46 MHz
(B) 2.07 MHz
(C) 147
(D) 216
(C) 4.93 MHz
(D) 2.92 MHz
12. In previous question if processing gain is changed to
19.
500, then number of users will be
processing gain of 500. If the desired error probability is
(A) 27 users
(B) 38 users
10 -5 and ( e b / J 0 ) required to obtain an error probability
(C) 42 users
(D) 45 users
of 10 -5 for binary PSK is 9.5 dB, then the Jamming
A DS/BPSK
spread
spectrum
signal
has
a
margin against a containers tone jammer is Statement for Question 13-15 : A DS spread spectrum system transmit at a rate of
(A) 23.6 dB
(B) 17.5 dB
(C) 117.4 dB
(D) 109.0 dB
1 kbps in the presets of a tone jammer. The jammer power is 20 dB greater then the desired signal, and the required Îb / J 0 to achieve satisfactory performance is 10 dB.
Statement for Question 20-21 : An m = 10 ML shift register is used to generate the pre hdarandlm sequence in a DS spread spectrum
13. The spreading bandwidth required to meet the specifications is
system. The chip duration is Tc = l ms and the bit duration is Tb = NTc , where N is the length (period of the m sequence).
(A) 10 7 Hz
(B) 10 3 Hz
(C) 10 5 Hz
(D) 106 Hz
20. The processing gain of the system is
14. If the jammer is a pulse jammer, then pulse duty cycle that results in worst case jamming is (A) 0.14
(B) 0.05
(C) 0.07
(D) 0.10
(A) 10 dB
(B) 20 dB
(C) 30 dB
(D) 40 dB
21. If the required e b/J 0 is 10 and the jammer is a tone jammer with an average power J av, then jamming
15. The correspond probability of error is
margin is.
(A) 4.9 ´ 10 -3
(B) 6.3 ´ 10 -3
(A) 10 dB
(B) 20 dB
(C) 9.4 ´ 10 -4
(D) 8.3 ´ 10 -3
(C) 30 dB
(D) 40 dB
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GATE EC BY RK Kanodia
UNIT 7
Statement for Question 22-23 : An FH binary orthogonal FSK system employs an m = 15 stage liner feedback shift register that generates an ML sequence. Each state of the shift register selects one of L non over lapping frequency bands in the hopping pattern. The bit rate is 100 bits/s. The demodulator employ non coherent detection. 22. If the hop rate is one per bit, the hopping bandwidth for this channel is
Communication System
(A) 0.4 GHz
(B) 0.6 GHz
(C) 0.7 GHz
(D) 0.9 GHz
28. The probability of error for the worst-case partial band jammer is (A) 0.2996
(B) 0.1496
(C) 0.0368
(D) 0.0298
29. The minimum hop rate for a FH spread spectrum system that will prevent a jammer from operating five onives away from the receiver is
(A) 6.5534 MHz
(B) 9.4369 MHz
(C) 2.6943 MHz
(D) None of the above
(A) 3.2 bHz
(B) 3.2 MHz
(C) 18.6 MHz
(D) 18.6 kHz
23. Suppose the hop rate is increased to 2 hops/bit and the receiver uses square law combining the signal over two hops. The hopping bandwidth for this channel is (A) 3.2767 MHz
(B) 13.1068 MHz
(C) 26.2136 MHz
(D) 1.6384 MHz
***********
Statement for Qquestion 24-25 : In a fast FH spread
spectrum system, the
information is transmitted via FSK with non coherent detection. Suppose there are N = 3 hops/bit with hard decision decoding of the signal in each hop. The channel is AWGN with power spectral density
1 2
N0 and an SNR
20-13 dB (total SNR over the three hops) 24. The probability of error for this system is (A) 0.013
(B) 0.0013
(C) 0.049
(D) 0.0049
25. In case of one hop per bit the probability of error is (A) 196 . ´ 10 -5
(B) 196 . ´ 10 -7
(C) 2.27 ´ 10 -5
(D) 2.27 ´ 10 -7
Statement for Question 26-29 : A slow FH binary FSK system with non coherent detection
operates
at
e b / J 0 = 10,
with
hopping
bandwidth of 2 GHz, and a bit rate of 10 kbps. 26. The processing gain of this system is (A) 23 dB
(B) 43 dB
(C) 43 dB
(D) 53 dB
27. If the jammer operates as a partial band jammar, the bandwidth occupancy for worst case jamming is Page 452
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GATE EC BY RK Kanodia
Spread Spectrum
Chap 7.8
W/R W/R Î = = b J av/Pav N u - 1 J 0
SOLUTION
æÎ W/R = çç b è J0
1. (C) The PN sequence length is N= 2 m - 1 = 2 4 - 1 = 15
æÎ W = Rçç b è J0
2. (B) The chip duration is 1 TC = 7 s = 0.1 ms 10
ö ÷÷( N u - 1) ø
ö ÷÷( N u - 1) ø
where R = 10 4 bps, N u = 30 and Îb / J 0 = 10 Therefore, W = 2.9 ´ 106 Hz The minimum chip rate is 1 / Tc = W = 2.9 ´ 106 chips/sec
3. (A) The period of the PN sequence is T= NTC = 15 ´ 0.1 = 15 . ms
11. (D) To achieve an error probability of 10 -6 , we
4. (C) m= 19
æÎ required çç b è J0
n= 2 m - 1 = 219 - 1 = 219
ö ÷÷ = 10.5 dB ødB
Then, the number of users of the CDMA system is
The processing gain is 10 log10 N= 10 log10 219
W/R 1000 +1= + 1 = 89 users Îb /J 0 11.3
= 190 ´ 0.3 or 57 dB
Nu =
æ Eb ö 5. (A) Antijam margin = (Processing gain) - 10 log10 çç ÷÷ è N0 ø
12. (D) If the processing gain is reduced to W/R = 500,
The probability of error is 1 Pe = erfc 2
æ ç ç è
Eb N0
N u=
ö ÷ ÷ ø
500 + 1 = 45 users 11.3
13. (D) We have a system where ( J av/Pav) dB = 20 dB,
With Pe = 10 -5, we have Eb / N 0 = 9.
R = 1000 bps and (Îb /J 0 ) dB = 10 dB
Hence, Antijam margin = 57 - 10 log10 9 = 57 - 9.5
æÎ æJ ö æW ö Hence, we obtain ç ÷ = çç av ÷÷ + çç b è R ødB è Pav ødB è J 0
or =47.5 dB 6. (D) The precessing gain (PG) is FH Bandwidth W c PG = = 5 ´ 4 = 20 = Symbol Rate Rs
ö ÷÷ = 30 dB ødB
W = 1000 R W = 1000 R = 106 Hz
Hence, expressed in decibels, PG= 10 log10 20 = 26 db 7. (D) The processing gain is PG = 4 ´ 4 = 16 Hence, in decibels,
14. (C) The duty cycle of a pulse jammer of worst-case 0.71 0.7 jamming is a = = = 0.07 Îb /J 0 10 15. (D) The corresponding probability of error for this
PG = 10 log10 16 = 12 dB 8. (A) The coding gain is Rcd min =
then
worst-case jamming is 1 ´ 10 = 5 or 7 dB 2
9. (B) The processing gain
P2 =
0.083 0.083 = = 8.3 ´ 10 -3 e b/J 0 10
16. (B) Precessing gain is
10 7 W = = 5 ´ 10 3 or 37 dB R 2 ´ 10 3
W 106 = = 100 R 10 4
17. (A) We have N u = 15 users transmitting at a rate of
10. (C) We assume that the interference is characterized
10,000 bps each, in a bandwidth of W = 1MHz.
as a zero-mean AWGN process with power spectral
The e b/J 0 is.
-5
density J 0 . To achieve an error probability of 10 , the required Îb /J 0 = 10 we have
e b W/R 106 / 10 4 100 = = = 14 14 J 0 Nu - 1
= 7.14 or 8.54 dB
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GATE EC BY RK Kanodia
UNIT 7
18. (B) With N u = 30 and e b/J 0 = 7.14, the processing
Communication System
ec
gain should be increased to
1 P= e 2 N 0 2
W/R= (7.14)(29) = 207
where e c / N 0 = 20 / 3. The probability of a bit error is
W= 207 ´ 104 = 2.07 MHz
Pb = 1 - (1 - p) 2 = 1 - (1 - 2 p + p2 ) = 2 p - p2
Hence the bandwidth must be increased to 2.07 MHz 19. (B) The processing gain is given as W = 500 or 27 dB R
=e
-
ec 2 N0
ec
-
1 - 2 N0 e = 0.0013 2
25. (C) In the case of one hop per bit, the SNR per bit is ec
The ( e b/J 0 ) required to obtain an error probability of
20, Hence,
Pb =
1 - 2 N 0 1 -10 e = e = 2.27 ´ 10 -5 2 2
10 -5 for binary PSK is 9.5 dB. Hence, the jamming 26. (D) We are given a hopping bandwidth of 2 GHz and
margin is
a bit rate of 10 kbs.
æe ö æ J av ö æW ö ÷÷ = ç ÷ - çç b ÷÷ = 27.95 or 17.5 dB çç R P è ø è J 0 ødB è av ødB dB
Hence,
20. (C) The period of the maximum length shift register sequence is
a* W = 2/( e b/J 0 ) = 0.2
Since Tb = NTc then the processing gain is T N b = 1023 or 30 dB Tc
Hence a* W = 0.4 GHz 28. (C) The probability of error with worst-case partial-band jamming is
21. (B) A Jamming margin is æe ö ö ÷÷ - çç b ÷÷ = 30 - 10 = 20 dB ødB è J 0 ødB
where J av = J 0W » J 0/Tc = J 0 ´ 10
22. (A) The length of the shift-register sequence is L = 2 m - 1215 - 1 = 32767 bits For binary FSK modulation, the minimum frequency separation is 2/T, where 1/T is the symbol (bit) rate. The hop rate is 100 hops/sec. Since the shift register
e -1 e -1 = 3.68 ´ 10 -2 = ( e b/J 0 ) 10
Dd= 2 ´ 8050 = 16100 Dd Dd= x ´ t or t = t Dd 16100 Þ t= = = 5.367 ´ 10 5 x 3 ´ 108 1 1 f= = = 18.63 kHz t 5.367 ´ 10 -5
L = 32767 states and each state utilizes a
bandwidth of 2/T = 200 Hz, then the total bandwidth for the FH signal is 6.5534 MHz. 23. If the hopping rate is 2 hops/bit and the bit rate is 100 bits/sec, then, the hop rate is 200 hops/sec. The minimum
frequency
separation
for
orthogonality
2/T = 400 Hz. Since there are N = 32767 states of the shift register and for each state we select one of two frequencies
separated
by
400
Hz,
the
hopping
bandwidth is 13.1068 MHz. 24. (B) The total SNR for three hops is 20 ~ 13 dB. Therefore the SNR per hop is 20/3. The probability of a chip error with non-coherent detection is Page 454
P2 =
29. (D) d= 5 miles = 8050 meters
6
has
27. (A) The bandwidth of the worst partial-band jammer is a* W, where
N = 210 - 1 = 1023
æW æ J av ö ÷÷ = çç çç P è av ødB è Rb
W 2 ´ 10 9 = = 2 ´ 10 5or 53 dB 10 4 R
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***********
GATE EC BY RK Kanodia
UNIT 8
10. A field is given as G=
Electromagnetics
16. A field is given as
13 ( yu x + 3u y + xu z ) x2 + y2
G =
25 ( xu x + yu y ) x2 + y2
The unit vector in the direction of G at P(3, 4, -2)
The field at point (-2, 3, 4) is (A) 13( -2 u x + 3u y + 4 u z )
(B) -2 u x + 3u y + 4 u z
(C) 13( 3u x + 4 u y - 2 u z )
(D) 3u x + 4 u y - 2 u z
11. A field is given as F = yu x + zu y + xu z The angle between G and u x at point (2, 2, 0) is (A) 45°
(B) 30°
(C) 60°
(D) 90°
is (A) 0.6 u x + 0. 8 u y
(B) 0. 8 u x + 0.6 u y
(C) 0.6 u y + 0. 8 u z
(D) 0.6 u z + 0.6 u x
17. A field is given as F = xyu x + yzu y + zxu x The value of 4 2
the double integral I = ò ò F × u ydzdx in the plane y = 7 is 0 0
12. A vector field is given as G = 12 xyu x + 6( x 2 + 2) u y + 18 z 2 u z
(A) 128
(B) 56
(C) 190
(D) 0
18. Two vector extending from the origin are given as
The equation of the surface M on which |G | = 60 is (A) 4 x 2 y 2 + 4 x 4 + 9 z 4 + 2 x 2 = 96 (B) 2 x 2 y 2 + x 4 + 9 z 4 + 2 x 2 = 96 (C) 2 x 2 y 2 + 4 x 4 + 9 z 4 + 2 x 2 = 96 (D) 4 x 2 y 2 + x 4 + 9 z 4 + 2 x 2 = 96
R1 = 4 u x + 3u y - 2 u z and R 2 = 3u x - 4 u y - 6 u z . The area of the triangle defined by R1 and R 2 is (A) 12.47
(B) 20.15
(C) 10.87
(D) 15.46
19. The four vertices of a regular tetrahedron are located at O (0, 0, 0), A(0, 1, 0), B(0.5
13. A vector field is given by
(
E = 4 zy u z + 2 y sin 2 x u y + y sin 2 x u z 2
(C) Plane x =
3p 2
, 0.5,
2 3
). The unit vector perpendicular (outward) to
(A) 0.41u x + 0.71u y + 0.29 u z (B) 0.47 u x + 0.82 u y + 0.33u z
(B) Plane x = 0
(C) -0.47 u x - 0.82 u y - 0.33u z
(D) all
(D) -0.41u x - 0.71u y - 0.29 u z 20. The two vector are R AM = 20 u x + 18 u y - 18 u z and
14. The vector field E is given by E = 6 zy 2 cos 2 x u x + 4 xy sin 2 x u y + y 2 sin 2 x u z The region in which E = 0 is
R AN = - 10 u x + 8 u y + 15 u z . The unit vector in the plane of the triangle that bisects the interior angle at A is (A) 0.168 u x + 0.915 u y + 0.367 u z
(A) y = 0
(B) x = 0
(B) 0.729 u x + 0.134 u y - 0.672 u z
(C) z = 0
np (D) x = 2
(C) 0.729 u x + 0.134 u y + 0.672 u z (D) 0.168 u x + 0.915 u y - 0.367 u z
15. Two vector fields are F = -10 u x + 20 x( y - 1) u y and G = 2 x 2 yu x - 4 u y + 2 u z . At point A(2, 3, -4) a unit vector in the direction of F - G is
21. Two points in cylindrical coordinates are A( r = 5, f = 70 ° , z = -3) and B(r = 2, f = 30 ° , z = 1). A unit vector at A towards B is
(A) 0.18 u x + 0.98 u y - 0.05 u z
(A) 0.03u x - 0.82 u y + 0.57 u z
(B) -0.18 u x - 0.98 u y + 0.05 u z
(B) 0.03u x + 0.82 u y + 0.57 u z
(C) -0.37 u x + 0.92 u y + 0.02 u z
(C) -0.82 u x + 0.003u y + 0.57 u z
(D) 0.37 u x - 0.92 u y - 0.02 u z Page 458
3, 0.5, 0) and C
the face ABC is
2
The surface on which E y = 0 is (A) Plane y = 0
0 .5 3
(D) 0.003u x - 0.82 u y + 0.57 u z www.gatehelp.com
GATE EC BY RK Kanodia
Vector Analysis
22. A field in cartesian form is given as D = xu x +
29. The vector
yu y x +y 2
B=
2
10 u r + r cos q u q + u f r
in cartesian coordinates at (-3, 4, 0) is
In cylindrical form it will be u u uf (B) D = r + (A) D = r r r cos f (C) D = ru r
Chap 8.1
(D) D = ru r + cos f u f
(A) u x - 2 u y
(B) - 2u x + u y
(C) 1.36 u x + 2.72 u y
(D) -2.72 u x + 1.36 u x
30. The two point have been given A (20, 30 ° , 45 ° ) and
23. A vector extends from A(r = 4, f = 40 ° , z = -2) to B(
B ( 30, 115 ° , 160 ° ). The |R AB | is
r = 5, f = -110 ° , z = 1). The vector R AB is
(A) 22.2
(B) 44.4
(A) 4.77 u x + 7.30 u y + 4 u z
(C) 11.1
(D) 33.3
(B) -4.77 u x - 7.30 u y + 4 u z (C) -7.30 u x - 4.77 u y + 4 u z
31. The surface r = 2 and 4, q = 30 ° and 60°, f = 20 ° and
(D) 7.30 u x + 4.77 u y + 4 u z
80° identify a closed surface. The enclosed volume is
24. The surface r = 3, r = 5, f = 100 °, f = 130 °, z = 3 and
(A) 11.45
(B) 7.15
(C) 6.14
(D) 8.26
z = 4.5 define a closed surface. The enclosed volume is (A) 480
(B) 5.46
32. The surface r = 2 and 4, q = 30 ° and 50° and f = 20 °
(C) 360
(D) 6.28
and 60° identify a closed surface. The total area of the enclosing surface is
25. The surface r = 2, r = 4, f = 45 °, f = 135 °, z = 3 and z = 4 define a closed surface. The total area of the enclosing surface is
(A) 6.31
(B) 18.91
(C) 25.22
(D) 12.61
(A) 34.29
(B) 20.7
33. At point P(r = 4, q = 0.2 p , f = 0.8 p), u r in cartesian
(C) 32.27
(D) 16.4
component is
26. The surface r = 3, r = 5, f = 100 °, f = 130 °, z = 3 and
(A) 0.48 u x + 0.35 u y + 0.81u z
z = 4.5 define a closed volume. The length of the longest
(B) 0.48 u x - 0.35 u y - 0.81u z
straight line that lies entirely within the volume is
(C) -0.48 u x + 0.35 u y + 0.81u z
(A) 3.21
(B) 3.13
(D) 0.48 u x - 0.35 u y - 0.81u z
(C) 4.26
(D) 4.21 34. The expression for u y in spherical coordinates at P(
27. A vector field H is
r = 4, q = 0.2 p, f = 0.8 p) is
æ fö H = rz 2 sin f u r + e - z sin ç ÷ u f + r 3u z è2 ø
(A) 0.48 u r + 0.35 u q - 0.81u f (B) 0.35 u r + 0.48 u q - 0.81u f
p æ ö At point ç 2, , 0 ÷ the value of H × u x is 3 è ø (A) 0.25
(B) 0.433
(C) -0.433
(D) -0.25
(C) -0.48 u r + 0.35 u q - 0.81u f (D) -0.35 u r + 0.48 u q - 0.81u f 35. Given a vector field
28. A vector is A = yu x + ( x + z) u y. At point P(-2, 6, 3)
D = r sin f u r -
A in cylindrical coordinate is (A) -0.949 u r - 6.008 u f
(B) 0.949 u r - 6.008 u f
(C) -6.008 u r - 0.949 u f
(D) 6.008 u r + 0.949 u f
1 sin q cos f u q + r 2 u f r
The component of D tangential to the spherical surface r = 10 at P(10, 150°, 330°) is
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GATE EC BY RK Kanodia
UNIT 8
Electromagnetics
(A) 0 .043u q + 100 u f
40. The gradient of the functionG = r 3 sin 2 q sin 2 f sin q
(B) -0 .043u q - 100 u f
at point P ( 12 ,
(C) 110 u q + 0.043u f
(A) 1.41u r + 3u z
(B) u x + u y + u z
(D) 0 .043u q - 100 u f
(C) 3.46 u r + 9.3u q
(D) All
36. The circulation of F = x 2 u x - xzu y - y 2 u z around the
41.
path shown in fig. P8.1.36 is
The
1 2
1 2
,
) is
directional
derivative
of
function
F = xy + yz + zx at point P(3, -3, - 3) in the direction toward point Q(4, -1, -1) is
z 1
(A) -3
(B) 1
(C) -2
(D) 0
42. The temperature in a auditorium is given by T = 2 x 2 + y 2 - 2 z 2 . A mosquito located at (2, 2, 1) in the y
auditorium desires to fly in such a direction that it will
1
get warm as soon as possible. The direction, in that it
1
x
must fly is (A) 8 u x + 8 u y - 4 u z
Fig. P8.1.36
(A) -
1 3
(B)
1 6
(B) 2 u x + 2 u y - u z
(C) -
1 6
(D)
1 3
(D) -(2 u x + 2 u y - u z )
(C) 4 u x + 4 u y - 4 u z
37. The circulation of A = r cos f u r + z sin f u z around the edge L of the wedge shown in Fig. P8.1.37 is
43. The angle between the normal to the surface x2 y + z = 3
x ln z - y 2 = -4
and
at
the
point
of
intersection (-1, 2, 1) is
y
(A) 73.4°
(B) 36.3°
(C) 16.6°
(D) 53.7°
L
44. The divergence of vector A = yzu x + 4 xyu y + yu z at 60
point P(1, -2, 3) is
o
x
0
2
Fig. P8.1.37
(A) 2
(B) -2 (D) 4
(A) 1
(B) -1
(C) 0
(C) 0
(D) 3
45.
The
divergence
of
the
38. The gradient of field f = y 2 x + xyz is (A) y( y + z) u x + x(2 y + z) u y + xyu z (B) y(2 x + z) u x + x( x + z) u y + xyu z (C) y 2 u x + 2 yxu y + xyu z
(A) 2.6
(B) 1.5
(C) 4.5
(D) -4.5
46. The divergence of the vector
(D) y(2 y + z) u x + x(2 y + z) u y + xyu z
A = rz 2 cos f u r + z sin 2 f u z is
39. The gradient of the field f = r z cos 2 f at point 2
(1, 45 ° , 2) is (A) 4u f
(B) 4 2u f
(C) -4u f
(D) -4 2u f
Page 460
vector
A = 2 r cos q cos f u r + r u f at point P(1, 30°, 60°) is 1 2
(A) 2rz 2 cos f u r + sin 2 f u z (B) 2rz 2 cos f u r + sin 2 f u z (C) 2 z 2 cos f u r + sin 2 f u z (D) z sin 2
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f u r + 2rz cos f u f + z 2 sin f u z r
GATE EC BY RK Kanodia
Vector Analysis
Chap 8.1
47. The flux of D = r 2 cos 2 f u r + 3 sin f u f over the
54. If V = x 2 y 2 z 2 , the laplacian of the field V is
closed surface of the cylinder 0 £ z < 3,
(A) 2( xy 2 + yz 2 + zx 2 )
r = 3 is
(A) 324
(B) 81p
(B) 2( x 2 y 2 + y 2 z 2 + z 2 x 2 )
(C) 81
(D) 64p
(C) ( x 2 y 2 + y 2 z 2 + z 2 x 2 )
48. The curl of vector A = e xyu x + sin xy u y + cos 2 xz u z is
(D) 0 55. The value of Ñ 2 V at point P(3, 60°, -2) is if
(A) y e xyu x + x cos xy u y - 2 x sin 2 xz u z
V = r 2 z(cos f + sin f)
(B) z sin 2 xy u y + ( y cos xy - xe xy) u z (C) z sin 2 xy u y + ( x cos xy - xe xy) u z (D) xy e xyu x + xy cos xy u y - 2 xz sin 2 xz u z
(A) -8.2
(B) 12.3
(C) -12.3
(D) 0
56. If the scalar field V = r 2 (1 + cos q sin f) then Ñ 2 V is
49. The curl of vector field A = rz sin f u r + 3rz 2 cos f u r at point (5, 90°, 1) is (A) 0
(B) 12u q
(C) 6u r
(D) 5u f
(A) 1 + 2(1 - r 2 ) cos q sin f (B) 6 + 4 cos q sin f - cot q cosec q sin f (C) 2 + 2(1 - r 2 ) cos q sin f (D) 0
50. The curl of vector field 1 A = r cos q u r - sin q u q + 2 r 2 sin q u f is r 1 (A) cos q u r - cos q u q r
57. Ñ ln r is equal to
æ 1 ö (B) 2 r 2 cos q u r - 4 r sin q u q + ç 2 sin q - r sin q ÷u f r è ø
(A) Ñ ´ ( fu z )
(B) Ñ ´ ( zu f)
(C) Ñ ´ (ru f)
(D) Ñ ´ (ru z )
58. If r = xu x + yu y + zu x then ( r × Ñ ) r 2 is equal to (A) 2 r 2
(B) 3r 2
(D) 0
(C) 4 r 2
(D) 0
51. If A = ( 3 y 2 - 2 z) u x - 2 x 2 z u y + ( x + 2 y) u z , the value
59. If r = xu x + yu y + zu x is the position vector of point
of Ñ ´ Ñ ´ A at P(-2, 3, -1) is
P( x, y, z) and r =|r| then Ñ × r n r is equal to
(C) 4 r cos q u r - 6 r sin q u q + sin q u f
(A) -( 6 u x + 4 u y)
(B) 8 ( u x + u y)
(A) nr n
(B) ( n + 3) r n
(C) -8( u x + u y)
(D) 0
(C) ( n + 2) r n
(D) 0
52. The grad × Ñ ´ A of a vector field
60. If F = x 2 yu x - yu y, the circulation of vector field F
A = x 2 yu x + y 2 zu y - 2 xzu z is
around closed path shown in fig. P8.1.60 is,
(A) 2 xy + 2 yz - 2 x
y
(B) x 2 y + y 2 z - 2 xz
1
(C) 2 x 2 y + 2 y 2 z - 2 xz S
(D) 0 53. If V = xy - x 2 y + y 2 z 2 , the value of the div grad V
0
is
L
2
1
x
Fig. P8.1.60
(A) 0 (B) z + x 2 + 2 y 2 z (C) 2 y( z 2 - yz - x) (D) 2( z - y - y) 2
2
(A)
7 3
(B) -
7 6
(C)
7 6
(D) -
7 3
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GATE EC BY RK Kanodia
UNIT 8
Electromagnetics
61. If A = r sin f u r + r 2 u r , and L is the contour of fig.
SOLUTIONS
ò A. dL is
P8.1.61, then circulation
L
1. (D) d = ( x1 - x2 ) 2 + ( y1 - y2 ) 2 + ( z1 - z 2 ) 2
y
2
= ( 4 - 2) 2 + ( -6 - 3) 2 + ( 3 - ( -1)) 2 = 4 + 81 + 16 = 101 L
1
2. (A) R AB = R B - R A = ( 3u x + 0 u y + 2 u z ) - (5 u x - u y + 0 u z ) -2
-1
1
2
= -2 u x + u y + 2 u z
x
|R AB | =
Fig. P8.1.61
(A) 7 p + 2
(B) 7 p - 2
(C) 7p
(D) 0
uR = -
22 + 1 + 22 = 3 2 1 2 ux + u y + uz 3 3 3
3. (C) The component of F parallel to G is
62. The surface integral of vector F = 2r 2 z 2 u r + r cos 2 f u z
=
over the region defined by 2 £ r £ 5, -1 < z < 1,
F ×G (10, - 6, 5) × (0.1, 0.2, 0.3) G = (0.1, 0.2, 0.3) 2 G 0.12 + 0.2 2 + 0.32
= 9.3(0.1, 0.2, 0.3) = (0.93, 1.86, 2.79)
0 < f < 2 p is (A) 44
(B) 176
(C) 88
(D) 352
63. If D = xyu x + yzu y + zxu z , then the value of òò A × dS
is, where S is the surface of the cube defined by 0 £ x £ 1, 0 £ y £ 1, 0 £ z £ 1 (A) 0.5
(B) 3
(C) 0
(D) 1.5
4. (C) The vector component of F perpendicular to G is F ×G ( 3, 2, 1) × ( 4, 4, - 2) =F G = ( 3, 2, 1) ( 4, 4, - 2) G2 42 + 42 + 22 = (3, 2, 1)- (2, 2, - 1) = (1, 0, 2) = u x + 2 u z 5. (C) R = 3u x + 4 M - N = 3u x + 4(2 u x + 3u y - 4 u z ) -( -4 u x + 4 u y + 3u z ) = 15 u x + 8 u y - 19 u z
64. If D = 2rzu r + 3z sin f u r - 4r cos f u z and S is the
|R| =
15 2 + 8 2 + 19 2 = 25.5 = 25.5
surface of the wedge 0 < r < 2, q < f < 45 °, 0 < z < 5, then 6. (B) R = - M + 2N
the surface integral of D is (A) 24.89
(B) 131.57
= - ( 8 u x + 4 u y - 8 u z ) + 2( 8 u x + 6 u y - 2 u z )
(C) 63.26
(D) 0
= 8u x + 8u y + 4u z 8u x + 8u y + 4u z uR = 82 + 82 + 42
65. If the vector field F = ( axy + bz 3) u x + ( 3 x 2 - gz) u y + ( 3 xz 2 - y) u z is irrotational, the value of a , b and g is
=
2 2 1 æ2 2 1ö ux + u y + uz = ç , , ÷ 3 3 3 è 3 3 3ø
(A) a = b = g = 1
(B) a = b = 1, g = 0
-M + 2N = -( 8, 4, - 8) + 2( 8, 6, - 2) = (8, 8, 4)
(C) a = 0,
(D) a = b = g = 0
uR =
b = g =1
**************
uR =
æ2 2 1ö =ç , , ÷ è 3 3 3ø 8 +8 +4 ( 8, 8, 4)
2
2
2
2 2 1 ux + u y + uz 3 3 3
æ 1 + 7 -6 - 2 4 + 0 ö 7. (C) Mid point is ç , , ÷ = (4, -4, 2) 2 2 ø è 2` uR = Page 462
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( 4, - 4, 2) 4 + 4 +2 2
2
2
2 1ö æ2 =ç , - , ÷ 3 3 3ø è
GATE EC BY RK Kanodia
Vector Analysis
=
2 2 1 ux - u y + uz 3 3 3
8. (A) G = 24(1)(2) u x + 12(1 + 2) u y + 18( -1) 2 u z = 48 u x + 36 u y + 18 u z 2 1ö æ2 9. (A) A = ( 6, - 2, - 4), B = k ç , - , ÷ 3 3ø è3 |B - A |= 10 2
2
2
2 ö 2 ö 1 ö æ æ æ ç 6 - k ÷ + ç -2 + k ÷ + ç -4 - k ÷ = 100 3 3 3 ø è ø è ø è k2 - 8 k - 44 = 0
Þ
k = 1175 . ,
Chap 8.1
= -34 u x + 84 u y - 2 u z -34 u x + 84 u y - u z uR = 34 2 + 84 2 + 2 2 = - 0.37 u x + 0.92 u y - 0.02 u z 16. (A) At P(3, 4, -2) 25 G= 2 ( 3u x + 4 u y) = 3u x + 4 u y 3 + 42 3u x + 4 u y = 0.6 u x + 0.8 u y uG = 32 + 4 2 17. (B) F × u y = Fy = yz
2 1ö æ2 B = 1175 . ç ,- , ÷ 3 3ø è3
4 2
I=
ò ò yzdzdx = 0 0
= (7.83, -7.83, -3.92)
æ2 ò0 çç ò0 yzdz è 4
4 ö ÷dx = ò 2 ydx = 2( 4) y = 8 y ÷ 0 ø
At y = 7, I = 8(7) = 56
13 10. (D) G = ( 3u x + 4 u y - 2 u z ) 2 ( -2) + ( 3) 2 = 3u x + 4 u y - 2 u z
éu x u y u z ù 1 1ê 18. (B) Area = R1 ´ R 2 = 4 3 -2 ú ú 2 2ê êë 3 -4 -6 úû
11. (A) Let q be the angle between F and u x ,
= u x ( -18 - 8) - u y( -24 + 6) + u z ( -16 - 9)
Magnitude of F is |F|= y + z + x
= -26 u x + 18 u y - 25 u z
2
F × u x = (F)(1) cos q = y y cos q = = 2 y + z 2 + x2
2
2
R1 ´ R 2 = 26 2 + 18 2 + 25 2 = 40.31 2 2 +2 2
2
=
1
area =
2
q = 45 °
40.31 = 20.15 2
19. (B) R BA = (0, 1, 0) - (0.5 3 , 0.5, 0) = ( -0.5 3 , 0.5, 0) æ 0.5 R BC = çç , 0.5, è 3
12. (D) |G|= 60 (12 xy) 2 + ( 6( x 2 + 2)) 2 + (18 z 2 ) 2 = 60 Þ
144 x 2 y 2 + 36( x 4 + 4 x 2 + 4) + 324 z 4 = 3600
Þ
4 x y + ( x + 4 x + 4) + 9 z = 100
Þ
4 x 2 y 2 + x 4 + 9 z 4 + 4 x 2 = 96
2
2
4
2
4
R BA ´ R BC
13. (D) For E y = 0, 2 y sin 2 x = 0
Þ
sin 2 x = 0, Þ
3p x = 0, 2
2 x = 0, p , 3p , Þ
y =0
Hence (D) is correct. 14. (A) E = y( 6 zy cos 2 x u x + 4 x sin 2 x u y + y sin 2 x u z ) Hence in plane y = 0,
E = 0.
15. (C) R = F - G = ( -10 u x + 20 x( y - 1) u y) - (2 x 2 yu x - 4 u y + 2 u z ) At P(2, 3, - 4) , R = F - G = ( -10 u x + 80 u y) - (24 u x - 4 u y + 2 u z )
= u x 0.5
2 3
ö ÷ - (0.5 3 , 0.5, 0) ÷ ø
é uy ê ux = ê0.5 3 0.5 ê ê 1 0 êë 3
æ 1 = çç , 0, 3 è
2 3
ö ÷ ÷ ø
ù uz ú 0 ú ú 2ú 3 úû
2 0.5 - u y (0.5 2 ) + u z 3 3
= 0.41u x - 0.71u y + 0.29 u z The required unit vector is 0.41u x + 0.71u y + 0.29 u z = 0.412 + 0.712 + 0.29 2 = 0.47 u x + 0.81u y + 0.33u z 20. (A) The non-unit vector in the required direction is 1 = ( u AN + u AM ) 2 ( -10, 8, 15) u AN = = ( -0.507, 0.406, 0.761) 100 + 64 + 225
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GATE EC BY RK Kanodia
UNIT 8
(20, 18, - 10)
u AM =
400 + 324 + 100
Electromagnetics
26. (A) A(r = 3, f = 100 ° , z = 3) = A( -0.52, 2.95, 3)
= (0.697, 0.627, -0.348)
B(r = 5, f = 130 ° , z = 4.5) = B( -321 . , 3.83, 4.5)
1 ( u AM + u AN ) 2 1 = [(0.697, 0.627, - 0.348) + ( -0.507, 0.406, 0.761) ] 2
length = |B - A| B - A = ( -321 . , 3.83, 4.5) - ( -0.52, 2.95, 3) = ( -2.69, 0.88, 15 . )
= (0.095, 0.516, 0.207) (0.095, 0.516, 0.207) = (0.168, 0.915, 0.367) u bis = 0.095 2 + 0.516 2 + 0.2612 Hence (A) is correct.
|B - A| = |( -2.69,
0.88, 15 . )| = 2.69 2 + 0.88 2 + 15 . 2 = 3.21
æ p ö 27. (C) At P ç 2, , 0 ÷, H = 0.5 u f + 8 u z è 3 ø u x = cos f u r - sin f u f =
21. (D) In cartesian coordinates A (5 cos 70 ° , 5 sin 70 ° , - 3) = A(171 . , 4.70, - 3)
1 ( u r - 3u f) 2
æ 3 ö ÷ = -0.433 H × u x = (0.5)çç ÷ 2 è ø
B (2 cos ( -30 ° ), 2 sin ( -30 ° ), - 3) = B(173 . , - 1, 1) R AB = R B - R A = (173 . , - 1, 1) - (171 . , 4.70, - 3)
é Ar ù é cos f sin f 0 ù é Ax ù 28. (A) ê Af ú = ê- sin f cos f 0 ú ê A y ú ê ú ê úê ú 0 1 úû êë Az úû êë Az úû êë 0
= (0.02, - 5.70, 4) (0.02, - 5.70, 4) = (0.003, -0.82, 0.57) u AB = 0.02 2 + 5.70 2 + 4 2 22. (A) x = r cos f ,
At P (-2, 6, 3)
y = r sin f
æ 6 ö A = 6 u x + u y , f = tan -1 ç ÷ = 108.43° è -2 ø
r cos f u x + r sin f u y 1 D= 2 = (cos f u x + sin f u y) r cos 2 f + r 2 sin 2 f r
cos f = - 0.316, sin f = 0.948
1 Dr = D × u r = [cos f ( u x × u r) + sin f ( u y × u r)] r =
1 1 [cos 2 f + sin 2 f] = r r
Df = D × u f = =
0.948 0 ù é6 ù é Ar ù é-0.316 ê A ú = ê -0.948 -0.316 0 ú ê1 ú ê fú ê úê ú 0 1 úû êë0 úû êë Az úû êë 0 Ar = 6 ( -0.316) + 0.948 = -0.949,
1 [cos f ( u x × u f) + sin f ( u y × uf)] r
Af = 6( -0.948) - 0.316 = -6.008, Az = 0 Hence (A) is correct option.
1 [cos f ( - sin f) + sin f (cos f)] = 0 r
Therefore D =
29.(B) At P (-3, 4, 0)
1 ur r
r = x 2 + y 2 + z 2 = 32 + 4 2 + 0 2 = 5
23. (B) A( 4 cos 40 ° , 4 sin 40 ° , - 2) = A( 306 . , 2.57, - 2) B (5 cos ( -110 ° ), 5 sin ( -110 ° ), 2) = B ( -171 . , - 4.7, 2) R AB = R B - R A = ( -171 . , - 4.7, 2) - ( 306 . , 2.57, - 2)
4 .5 130 ° 5
ò ò ò rdrdfdz
éBx ù ésin q cos f cos q cos f - sin q ù éBr ù êB ú = êsin q sin f cos q cos f cos f ú êB ú ê yú ê ú ê qú - sin q 1 úû êëBf úû êë Bz úû êë cos q
= 6.28
3 100 ° 3
25. (C) Area is =2
135° 4
ò
ò rdrdf +
45° 2
4
4 135°
ò
ò 2 dfdz +
3 45°
4 135°
4 4
3 45°
3 2
ò ò 4 dfdz + 2 ò ò drdz
ér ù æ p ö æ pö = 2 ê ú ç ÷ + (2)(1)ç ÷ = 32.27 è2 ø ë 2 û2è 2 ø 2
Page 464
f = tan -1
x2 + y2 p = z 2 y 4 = tan -1 = 126.87 ° x -3
B = 2u r + u f
= ( -4.77, - 7.3, 4) 24. (D) Vol =
q = tan -1
é-0.6 0 -0.8 ù é2 ù = ê 0.8 0 -0.6 ú ê0 ú ê úê ú -1 0 úû êë1 úû êë 0 Bx = 2( -0.6) + 0.8 = -2 B y = 2(0.2) - 0.6 = 1 Bz = 0
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UNIT 8
2
C3 = ò r cos f dr
Along 3,
0
ò A × dL
= f=60 °
r2 = -1 2
Electromagnetics
44. (D) Ñ × A =
¶ Ax ¶ A y ¶ Az =0 + 4x + 0 + + ¶x ¶y ¶z
At P (1, -2, 3), Ñ × A = 4
= C1 + C2 + C3 = 1
L
45. (A) 38. (A) Ñf = u x
¶f ¶f ¶f + uy + uz ¶x ¶y ¶z
Ñ× A =
= y( y + z) u x + x(2 y + z) u y + xyu z 39. (C) Ñf = u r
=
¶f 1 ¶f ¶f + uf + uz ¶x r ¶y ¶z
1 ( 6 r 2 cos q cos f) + 0 + 0 r2
At P (1, 30°, 60°), Ñ × A = 6(1)(cos 30 ° )(cos 60 ° ) = 2.6
= 2 r 2 z cos 2 f u r - 2rz sin 2 f u f + r 2 cos 2 f u z
46. (C) Ñ × A =
At P (1, 45°, 2), Ñf = - 4u f 40. (B) r sin q cos f = x , r sin q sin f = y , r cos q = z G = r 3 sin 2 q sin 2 f sin q
=
1 ¶(rrz 2 cos f) ¶( z sin 2 f) = 2 z 2 cos f + sin 2 f + r ¶r ¶z
ò D × dS
¶( 4 xyz) ¶( 4 xyz) ¶( 4 xyz) ÑG = u x + uy + uz ¶x ¶x ¶z
S
Ñ ×D =
= 4 yzu x + 4 xzu y + 4 xyu z æ1 1 1ö At P ç , , ÷ , ÑG = u x + u y + u z è2 2 2 ø
ÑF = -6u x , R PQ = ( 4, - 1, - 1) - ( 3, - 3, - 3) = (1, 2, 2)
¶T ¶T ¶T + uy + uz ¶x ¶y ¶z
= 2 xu x + 2 yu z - 4 z u z
v
1 ¶(rr 2 cos 2 f) 1 ¶( z sin f) 3 + = 3r cos 2 f + sin f r r r ¶r ¶r 2
f+
v
z sin r
ö f ÷÷rdfdzdr ø
3
3
2p
3
3
2p
0
0
0
0
0
0
= ò dz ò zr 2 dr ò cos 2 f df + ò zdz ò dr ò sin f df = 81p
At P(3, -3, - 3),
= -2
ò Ñ × D dv
æ
41. (C) ÑF = ( y + z) u x + ( x + z) u y + ( x + y) u z
( -6 u x ) × ( u x + 2 u y + 2 u z )
=
ò Ñ × D dv = ò ççè 3r cos v
42. (C) ÑT = u x
By divergence theorem
S
= 4( r sin q cos f)( r sin q sin f)( r cos q) = 4 xyz
3
1 ¶(rAr) 1 ¶( Af) ¶( Az ) + + r ¶r r ¶q ¶z
47. (B) The flux is ò D × dS ,
= r 3(2 sin q cos q)(2 sin f cos f) sin q
ÑF × u R =
1 ¶(r 2 Ar ) 1 ¶(sin q Aq) 1 ¶(sin q Af) + + r2 r sin q r sin q ¶f ¶r ¶q
éu x ê¶ 48. (B) Ñ ´ A = ê ê ¶x êë Ax
uy ¶ ¶y Ay
u z ù éu x uy uz ù ¶ú ê¶ ¶ ¶ ú ú =ê ú ¶z ú ê ¶x ¶y ¶z ú Az úû êë e xy sin xy cos 2 xz úû
u x (0 - 0) - u y(2 cos xz ( - sin xz) z) + u z ( y cos xy - xexy) = z sin 2 xy u y + ( y cos xy - xe xy) u z
At P(2, 2, 1), ÑT = 4 u x + 4 u y - 4 u z 43. (A) Let f = x 2 y + z - 3, g = x ln z - y 2 + 4 Ñf = 2 xyu x + x 2 u y + u z Ñg = ln z u x - 2 yu y +
x uz z
At point P (-1, 2, 1) uf =
-4 u x + u y + u z 18
cos q = ± u f × u g = ± -1
q = cos 0.28 = 73.4 ° Page 466
, ug = -5 18 ´ 17
-4 u y - u z 17 = 0.286
éu r 1ê ¶ 49. (D) Ñ ´ A = ê r ê ¶r ëê Ar
uf uz ù ¶ ¶ú ú ¶f ¶z ú Af Az úû
é ur uf uz ù ê ¶ ¶ ¶ú 1 = ê ú r ê ¶r ¶f ¶z ú êërz sin f 2r 2 z 2 cos f 0 úû =
1 1 u r (-6r 2 z cos f) - u f (-r sin f) u z (6rz 2 cos f - rz cos f) r r
At point P(5, 90°, 1), Ñ ´ A = 5 u f www.gatehelp.com
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Vector Analysis
éu r ê¶ 1 50. (C) Ñ ´ A = 2 ê r sin q ê ¶r êë Ar
56.(B)
r sin qu f ù ú ¶ ú ¶f ú r sin q Af úû
ru q ¶ ¶q rAq
Ñ2 V =
1 r 2 sin q
u r ( 2r 3 sin q - 0) -
¶2 V 1 ¶ æ 2 ¶V ö 1 ¶ æ ¶V ö 1 çr ÷+ 2 ç sin q ÷+ 2 2 2 r ¶r è ¶r ø r sin q ¶q è ¶q ø r sin q ¶f2
= 6 + 4 cos q sin f - cot q cosec q sin f 57. (A) r = x 2 + y 2
1 1 u q ( 6r 2 sin 2 q - 0) + u f ( 0 + r sin q) r sin q r
Ñ ln r = u x
= 4 r cos q u r - 6 r sin q u q + sin q u f ux uy uz ù é ê ú ¶ ¶ ¶ 51. (A) Ñ ´ A = ê ú ¶x ¶y ¶z ú ê 2 2 êë( 3 y - 2 z) ( -2 x z) ( x + 2 y) úû ux uy ¶ ¶ ¶x ¶y + 2 x2 ) 3
=
uz ù ú ¶ ú ¶z ú -( 4 xz + 6 y) úû
Ñ ´ Ñ ´ A = -6 u x - 4 u y = -( 6 u x + 4 u y) uz ù ¶ ú ú ¶z ú -2 xz úû
n
-1 æ nö Ñ × r n r = 2 x 2 ç ÷( x 2 + y 2 + z 2 ) 2 è2 ø n
2
n
= n( x 2 + y 2 + z 2 )( x 2 + y 2 + z 2 ) 2 2
2
2
L
2
¶ 55. (A) Ñ V = ¶r 2
2
æ ¶V çç r è ¶r
2
2
S
Ñ ´ F = -x uz dS = dxdy ( -u z )
ò (Ñ ´ F) × dS = - òò ( - x ) dxdy 2
= 2( y z + x z + x y ) = 2( x y + y z + z x ) 2
ò (Ñ ´ F) × dS
2
¶2V ¶2V ¶2V 54. (B) Ñ 2 V = + + ¶x 2 ¶y 2 ¶z 2 2
+ 3r n
60. (C) By Stokes theorem ò F × dL =
= -2 y + 2 z - 2 y = 2( z - y - y) 2
-1
= nr n + 3r n = ( n + 3) r n
¶( z - 2 xy) ¶(2 yz - x ) ¶( x - 2 y z) + + ¶x ¶y ¶z 2
2
n
+ rn + rn + rn
= ( z - 2 xy) u x + (2 yz - x ) u y + ( x - 2 y z) u z
2
¶( xr n ) ¶( yr n ) ¶( zr n ) + + ¶x ¶y ¶z
-1 -1 æ nö æ nö + 2 y 2 ç ÷( x 2 + y 2 + z 2 ) 2 + 2 z 2 ç ÷( x 2 + y 2 + z 2 ) 2 è2 ø è2 ø
2
2
x y x y ux + 2 u y = ux + u y 2 2 r r x +y x +y
n
¶V ¶V ¶V 53. (D) ÑV = u x + uy + uz ¶x ¶y ¶z
2
tan -1
where r n = ( x 2 + y 2 + z 2 ) 2
Ñ (Ñ ´ A) = 0
2
0
ù ú ú ú ú yú x úû
uz ¶ ¶z
2
59. (B) Ñ × r n r =
= - y 2 u x + 2 zu y - x 2 u z
Ñ × (ÑV ) =
uy ¶ ¶y
= x(2 x) + y(2 y) + z(2 z) = 2( x 2 + y 2 + z 2 ) = 2 r 2
At P(-2, 3, -1),
uy ¶ ¶y y2 z
é êu x ê¶ y uz = ê x ê ¶x ê 0 êë
æ ¶ ¶ ¶ ö 2 58. (A) ( r × Ñ ) r 2 = çç x +y +z ÷÷( x + y 2 + z 2 ) x y z ¶ ¶ ¶ è ø
= u x ( -6) - u y( -4 z) + u z (0) = -6 u x + 4 zu y
é ux ê ¶ 52. (D) Ñ ´ A = ê ê ¶x 2 ëê x y
¶ ln r ¶ ln r ¶ ln r x y + uy + uz = 2 ux + 2 u y r r ¶x ¶y ¶z
Ñ ´ f u z = Ñ ´ tan -1
= u x (2 + 2 x 2 ) - u y(1 - ( -2)) + u z ( -4 xz - 6 y) é ê Ñ ´Ñ ´ A =ê ê êë(2
2 sin q cos q sin f cos q sin f sin q sin 2 q
= 6(1 + cos q sin f) -
é ur ru q r sin q u f ù ú ê ¶ ¶ ¶ 1 = 2 ê ú r sin q ê ¶r ¶q ¶f ú 3 2 ëêr cos q - sin q 2 r sin q úû =
Chap 8.1
2
2
2
S
2
ö 1 ¶2V ¶2V + ÷÷ + 2 2 ¶z 2 ø r ¶f
ò ò x dydx + ò ò x dydx 2
0 0
2
1
1
= 4 z(cos f + sin f) - z(cos f + sin f) + 0 = 3z(cos f + sin f) æ1 3ö ÷= -8.2 At P(3, 60°, -2), Ñ 2 V = 3( -2)çç + 2 ÷ø è2
2 -x + 2
1 x
=
0
1
=
2
ò x dx + ò x (2 - x) dx 3
0
2
1
2
é x4 ù 1 ù é2 = ê ú + ê x3 - x4 ú 4 3 4 û1 ë ë û0 =
1 é16 1 7 ù é2 1 ù 1 14 + - 4ú - ê - ú = + -4+ = 4 êë 3 3 4 6 û ë3 4 û 4
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UNIT 8
æ
ò A × dL = ççè ò
61. (C)
C
ab
+
ò
+
bc
ò
ò
+
cd
da
ò D × dS = ò (Ñ × D)dv
ö ÷÷A × dL ø
S
2
2
= 4 ò rdr
L
1
0
a
b
d
-2
-1
1
2
=0
p
c
ò A × dL
ò r df
=
3
b
A × dL = r 3df
0
= r 3 ò df = (1) 3( -p) = - p p
62. (B) Using divergence theorem
Ñ ×F =
ò
=
Ñ × Fdv
v
1 ¶ (2r 2 z 2 ) = 4 z 2 = 4 z 2 r ¶r 1
ò Ñ × F dv = òòò 4 z rdrdfdz = 4 ò z 2
2
0
¶( xy) ¶( yz) ¶( zx) + + x =y+z+ x ¶x ¶y ¶z
S
æ1 = 3çç ò xdx è0
1
1
1
0
0
0
( y + z + x) dxdydz
1 ö æ1ö dy ò0 ò0 dz ÷÷ = 3çè 2 ÷ø = 15. ø 1
64. (B) Ñ × D =
Page 468
2p
v
ò A × dS = ò ò ò
= 4z +
5
dz ò rdr ò df = 176
ò A × dS = ò (Ñ × A)dv S
Ñ ×A =
2
-1
v
63. (D)
0
0
0
uy ¶ ¶y 3 x 2 - gz
*******
=0
ò A × dL = 0 + 8 p + 0 - p = 7 p ò F × dS
0
f = p, A × dL = 0,
Along da, dr = 0,
d
0
0
c
ò A × dL
45°
i.e. a = 1 = b = g.
= (2) 3 p = 8 p
d
a
5
If F is irrotational, Ñ ´ F = 0
A × dL = r 3df
Along cd, df = 0,
ò A × dL
2
uz ¶ ¶z 3 xz 2 -
= ( -1 + g) u x + ( 3bz 2 - 3z 2 ) u y + ( 6 x - ax) u z
a
Along bc, dr = 0,
45°
é ux ê ¶ 65. (A) Ñ ´ F = ê ¶ x ê a x bz 2 + êë
f = 0,
ò A × dL
5
ò zdz ò df + 3ò dr ò zdz ò cos f df
x
b
A × dL = 0,
æ ö 3z cos f ÷÷ rdrdfdz çç 4 z + r è ø
æ 4 öæ 25 öæ p ö æ 25 öæ 1 ö = 4ç ÷ç . ÷ = 13157 ÷ç ÷ + 3(2)ç ÷ç è 2 øè 2 øè 4 ø è 2 øè 2 ø
Fig. S8.1.61
Along ab, df = 0,
v
ò D × dS = ò ò ò
y
c
Electromagnetics
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CHAPTER
8.2 ELECTROSTATICS
1. Let Q1 = 4 mC
be located at P1 (3, 11, 8) while
Q2 = -5 mC is at P2 (6, 15, 8). The force F2 on Q2 will be
6. A 2 mC point charge is located at A (4, 3, 5) in free space. The electric field at P(8, 12, 2) is
(A) -( 4.32 u x + 5.76 u y) N
(B) 4.32 u x + 5.76 u y N
(A) 131.1u r + 159.7 u f - 49.4 u z
(C) -( 4.32 u x + 5.76 u y) mN
(D) 4.32 u x + 5.76 u y mN
(B) 159.7 u r + 27.4 u f - 49.4 u z
2. Four 5 nC positive charge are located in the z = 0 plane at the corners of a square 8 mm on a side. A fifth
(C) 1311 . u r + 27.4 u f - 49.4 u z (D) 159.7 u r + 137.1u f - 49.4 u z
5 nC positive charge is located at a point 8 mm distant from each of the other charge. The magnitude of the
7. A point charge of -10 nC is located at P1 (0, 0, 0.5)
total force on this fifth charge is
and a charge of 2 mC at the origin. The E at P(0, 2, 1) is
(A) 2 ´ 10 -4 N (C) 0.014 N
(B) 4 ´ 10 -4 N (D) 0.01 N
3. Four 40 nC are located at A(1, 0, 0), B(-1, 0, 0), C(0,
(A) 68.83u r + 14.85 u f
(B) 68.83u r + 64.01u f
(C) 68.83u r - 14.85 u f
(D) 68.83u r - 64.01u f
8. Charges of 20 nC and -20nC are located at (3, 0, 0)
1, 0) and D(0, -1, 0) in free space. The total force on the
and (-3, 0, 0) and (-3, 0, 0), respectively. The magnitude
charge at A is
of E at y axis is
(A) 24.6u x mN (C) -13.6u x mN
(B) -24.6u x mN (D) 13.76u x mN
4. Let a point charge 41 nC be located at P1 (4, -2, 7) and a charge 45 nC be at P2 (–3, 4, –2). The electric field E at P3(1, 2, 3) will be
1080 (9 + y 2 ) 3 2
(B)
1080 (9 + y 2 ) 3
(C)
108 (9 + y 2 ) 3 2
(D)
108 (9 + y 2 ) 3
9. A charge Q0 located at the origin in free space
(A) 0.13u x + 0.33u y + 0.12 u z
produces a field for which E2 = 1 kV m at point P(–2, 2,
(B) -0.13u x - 0.33u y - 0.12 u z
–1). The charge Q0 is
(C) 115 . u x + 2.93u y + 109 . uz (D) -115 . u x - 2.93u y - 109 . uz 5. Let a point charge 25 nC be located at P1 (4, -2, 7) and a charge 60 nC be at P2 (-3, 4, -2). The point, at which on the y axis, is Ex = 0, is (A) -7.46 (C) -6.89
(A)
(B) -22.11 (D) (B) and (C)
(A) 2 mC
(B) -3 mC
(C) 3 mC
(D) -2 mC
10. The volume charge density r n = r oe -|x|-| y|-|z| exist over all free space. The total charge present is (A) 2r o
(B) 4r o
(C) 8r o
(D) 3r o
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UNIT 8
Electromagnetics
11. A uniform volume charge density of 0.2 mC m 2 is
18. Two identical uniform charges with r l = 80 nC m are
present throughout the spherical shell extending from
located in free space at x = 0, y = ± 3 m. The force per
r = 3 cm to r = 5 cm. If r = 0 elsewhere, the total charge
unit length acting on the line at positive y arising from
present throughout the shell will be
the charge at negative y is
(A) 41.05 pC
(B) 257.92 pC
(A) 9.375u y mN
(B) 37.5u y mN
(C) 82.1 pC
(D) 129.0 pC
(C) 19.17u y mN
(D) 75u y mN
12.
If r v =
1 z 2 + 10
5 e -0 .1 r( p-|f|) mC m 3 in
the region
19. A uniform surface charge density of 10 nC m 2 is
0 £ r £ 10, -p < f < p and all z, and r v = 0 elsewhere,
present in the region x = 0, - 2 < y < 2 and all z if e = e o
the total charge present is
, the electric field at P(3, 0, 0) has
(A) 1.29 mC
(B) 2.58 mC
(C) 0.645 mC
(D) 0
(A) x component only (B) y component only
13. The region in which 4 < r < 5, 0 < q < 25 °, and
(C) x and y component
0.9 p < f < 11 . p contains the volume charge density of
(D) x, y and z component
f 2
r v = 10 ( r - 4) ( r - 5) sin q sin . Outside the region, r v = 0. The charge within the region is
20. The surface charge density is r s = 5 nC m 2 , in the
(A) 0.57 C
(B) 0.68 C
region r < 0.2 , z = 0, and is zero elsewhere. The electric
(C) 0.46 C
(D) 0.23 C
field E at A(r = 0, z = 0.5) is
14. A uniform line charge of 5 nC m is located along the
(A) 5.4 V m
(B) 10.1 V m
(C) 10.5 V m
(D) 20.2 V m
line defined by y = -2, z = 5. The electric field E at P(1, 21. Three infinite charge sheet are positioned as
2, 3) is (A) -9 u y + 4.5 u z
(B) 9 u y - 4.5 u z
follows: 10 nC m 2 at x = -3, - 40 nC m 2 at y = 4 and 50
(C) -18 u y + 9 u z
(D) 18 u y - 9 u z
nC m 2 at z = 2. The E at (4, 3, -2) is
15. A uniform line charge of 6.25 nC m is located along the line defined by y = -2, z = 5. The E at that point in the z = 0 plane where the direction of E is given by
(
1 3
u y - 23 u z ), is
(A) 4.5 u y + 9 u z
(B) 4.5 u y - 9 u z
(C) 9 u y - 18 u z
(D) 18 u y - 36 u z
(C) -48u y V m
(D) 24u y V m
(D) -0.56 u x - 2.23u y + 2.8 u z kV m Let E = 5 x 3u x - 15 x 2 yu y . The equation of the
(A) y =
m
respectively. The E at P(6, 0, 6) will be (B) 48u y V m
(C) 0.56 u x + 2.23u y + 2.8 u z kV m
stream line that passes through P(4, 2, 1) is
and y = -3
(A) -24u y V m
(B) 0.56 u x - 2.23u y + 2.8 u z kV m
22.
16. Uniform line charge of 20 nC m and -20 nC m are located in the x = 0 plane at y = 3
(A) 0.56 u x + 2.23u y - 2.8 u z kV m
128 x3
(B) x =
128 y3
64 x2
(D) x =
64 y2
(C) y =
23. A point charge 10 nC is located at origin. Four
17. Uniform line charges of 100 nC m lie along the
uniform line charge are located in the x = 0 plane as
entire extent of the three coordinate axes. The E at
follows : 40 nC m
at y = 1 and -5 m, -60 nC m
P(-3, 2, -1) is
y = -2 and - 4 m. The D at P(0, -3, 4) is
(A) -192 . u x + 2 u y - 108 . u z kV m
(A) -19.1u y + 25.5 u z pC m 2
(B) -0.96 u x + u y - 0.54 u z kV m
(B) 19.1u y - 25.5 u z pC m 2
(C) 0.96 u x - u y + 0.54 u z kV m
(C) -16.4 u y + 219 . u z pC m 2
(D) 192 . u x - 2 u y + 108 . u z kV m
(D) 16.4 u y - 219 . u z pC m 2
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Electrostatics
24. A point charge 20 nC
Chap 8.2
is located at origin. Four
31. A spherical surface of radius of 3 mm is centered at
uniform line charge are located as follows 40 nC m at
P(4, 1, 5) in free space. If D = xu x C m 2 the net electric
y = ± 1 and 50 nC m at y = ± 2. The electric flux that
flux leaving the spherical surface is
leaves the surface of a sphere, 4 m in radius, centered at origin is (A) 1.33 nC
(B) 1.89 mC
(C) 1.33 mC
(D) 1.89 mC
(A) 113.1 mC
(B) 339.3 nC
(C) 113.1 nC
(D) 452.4 nC
32. The electric flux density is
25. The cylindrical surface r = 8 C contains the surface charge density, r s = 5 e
-20|z|
D=
2
nC m . The flux that leaves
the surface r = 8 cm, 1 cm < z < 5 cm
30 ° < f < 90 ° is
(A) 270.7 nC
(B) 9.45 nC
(C) 270.7 pC
(D) 9.45 pC
1 [10 xyzu x + 5 x 2 zu y + (2 z 3 - 5 x 2 y) u z ] z2
The volume charge density r v at (-2, 3, 5) is (A) 6.43
(B) 8.96
(C) 10.4
(D) 7.86
26. Let D = 4 xy u x + 2( x 2 + z 2 ) u y + 4 yzu z C m 2 . The total charge enclosed in the rectangular parallelepiped
33. If D = 2ru r C m 2 , the total electric flux leaving the
0 < x < 2, 0 < y < 3, 0 < z < 5 m is
surface of the cube, 0 < x , y, z < 0.4 is
(A) 360 C
(B) 180 C
(A) 0.32
(B) 0.34
(C) 100 C
(D) 560 C
(C) 0.38
(D) 0.36
27. Volume charge density is located in free space as rv = 2e
-1000 r
nC m
for 0 < r < 1 mm
3
and
rv = 0
34. If E = 4 u x - 3u y + 5 u z in the neighborhood of point P(6, 2, -3). The incremental work done in moving 5 C
elsewhere. The value of Dr on the surface r = 1 mm is
charge a distance of 2 m in the direction u x + u y + u z is
(A) 1.28 pC m 2
(B) 0.28 pC m 2
(A) -60 J
(B) 34.64 J
(C) 0.78 pC m 2
(D) 0.32 pC m 2
(C) -34.64 J
(D) 60 JJ
and 4 carry uniform
35. If E = 100 u r V m , the incremental amount of work
charge densities of 20 nC m and -4 nC m . The Dr at
done in moving a 60 mC charge a distance of 2 mm from
2 < r < 4 is
P(1, 2, 3) toward Q(2, 1, 4) is
28. Spherical surfaces at r = 2 2
(A) -
16 nC m 2 r2
2
(B)
80 (C) 2 nC m 2 r
16 nC m 2 r2
80 (D) - 2 nC m 2 r
+ 6 z 3u z C m 2 . The total charge enclosed in the volume 0 < x, y , z < a is 5 4 a 3
(B) a 5 + 6 a 4
(C) 6 a 5 + a 4
(D)
(B) 3.1 mJ
(C) -31 . mJ
(D) 0
36. A 10 C charge is moved from the origin to P(3, 1, -1)
29. Given the electric flux density, D = 2 xyu x + x 2 u y
(A) 6 a 5 +
(A) -5.4 mJ
5 5 a + 6a4 3
in the field E = 2 xu x - 3 y 2 u y + 4 u z V m along the straight line path x = -3 y, y = x + 2 z. The amount of energy required is (A) -40 J
(B) 20 J
(C) -20 J
(D) 40 J
37. A uniform surface charge density of 30 nC m 2 is present on the spherical surface r = 6 mm in free space.
30. Let D = 5 x y z u y . The flux enclosed by volume
The V AB between A ( r = 2 cm, q = 35 ° , f = 55 ° )
x = 3 and 3.1, y = 1 and 1.1, and z = 2 and 2.1 is
( r = 3 cm, q = 40 ° , f = 90 ° ) is
(A) 49.6
(B) 24.8
(A) 2.03 V
(B) 10.17 V
(C) 35.4
(D) 36.4
(C) 4.07 mV
(D) -10.17 V
4
4
4
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UNIT 8
Electromagnetics
38. A point charge is located at the origin in free space.
(A) -4.94 nC
(B) -4.86 nC
The work done in carrying a charge 10 C from point A
(C) -5.56 nC
(D) -3.68 nC
( r = 4, q = p 6 , f = p 4) to B( r = 4, q = p 3 , p 6) is (A) 0.45 mJ
(B) 0.32 mJ
(C) -0.45 mJ
(D) 0
46. A dipole having Qd = 100 V × m 2 4 pe o
39. Let a uniform surface charge density of 5 nC m 2 be present at the z = 0 plane, a uniform line charge density of 8 nC m be located at x = 0, z = 4 and a point charge of 2 mC be present at P(2, 0, 0). If V = 0 at A(0, 0, 5), the V at B(1, 2, 3) is (A) 10.46 kV
(B) 1.98 kV
(C) 0.96 kV
(D) 3.78 kV
is located at the origin in free space and aligned so that its moment is in the u z direction. The E at point ( r = 1, 45 ° , f = 0) is (A) 158.11 V m
(B) 194.21 V m
(C) 146.21 V m
(D) 167.37 V m
47. A dipole located at the origin in free space has a 40. A non uniform linear charge density, r L = 6 ( z + 1)
moment p = 2 ´ 10 -9 u z C.m. The points at which | E|q = 1
nC m lies along the z axis. The potential at P(r = 1, 0, 0)
mV m on line y = z, x = 0 are
in free space is ( V¥ = 0)
(A) y = z = ± 23.35
(B) y = z = ± 16.5, x = 0
(C) y = z = 16.5
(D) y = 0, z = 23.35, x = 0
2
(A) 0 V
(B) 216 V
(C) 144 V
(D) 108 V 48. A dipole having a moment p = 3u x - 5 u y + 10 u z
41. The annular surface, 1 cm < r < 3 cm carries the
nC.m is located at P(1, 2, -4) in free space. The V at Q
nonuniform surface charge density r s = 5r nC m . The V
(2, 3, 4) is
2
at P(0, 0, 2 cm) is (A) 81 mV
(B) 90 mV
(C) 63 mV
(D) 76 mV
42. If V = 2 xy 2 z 3 + 3 ln( x 2 + 2 y 2 + 3z 2 ) in free space the magnitude of electric field E at P (3, 2, -1) is (A) 72.6 V/m
(B) 79.6 V/m
(C) 75 V/m
(D) 70.4 V/m
V. The volume charge density at r = 0.6 m is (B) -1.79 nC m 3
(C) 1.22 nC m 3
(D) -1.22 nC m 3
(B) 1.26 V
(C) 2.62 V
(D) 2.52 V
49. A potential field in free space is expressed as V = 40 xyz . The total energy stored within the cube 1 < x, y, z < 2 is
43. It is known that the potential is given by V = 70 r 0 .6 (A) 1.79 nC m 3
(A) 1.31 V
(A) 1548 pJ
(B) 0
(C) 774 pJ
(D) 387 pJ
50. Four 1.2 nC point charge are located in free space at the corners of a square 4 cm on a side. The total potential energy stored is
44. The potential field V = 80 r 2 cos q V. The volume
(A) 1.75 mJ
(B) 2 mJ
(C) 3.5 mJ
(D) 0
charge density at point P(2.5, q = 30 °, f = 60 °) in free 51. Given the current density
space is (A) -2.45 nC m 3 (C) -1.42 nC m
3
(B) 1.42 nC m 3 (D) 2.45 nC m
J = 10 5[sin (2 x) e -2 yu x + cos (2 x) e -2 yu y ] kA m 2
3
The total current crossing the plane y = 1 in the
45. Within the cylinder r = 2, 0 < z < 1 the potential is
u y direction in the region 0 < x < 1, 0 < z < 2 is
given by V = 100 + 50r + 150r sin f V. The charge lies
(A) 0
(B) 12.3 mA
within the cylinder is
(C) 24.6 mA
(D) 6.15 mA
Page 472
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GATE EC BY RK Kanodia
UNIT 8
Electromagnetics
64. In a certain region where the relative permitivity is
69. A potential field exists in a region where e = f ( x). If
2.4, D = 2 u x - 4 u y + 5 u z nC m 2 . The polarization is
r v = 0, the Ñ 2 V is 1 dF ¶V (A) f ( x) dx ¶x
(A) 2.8 u x - 5.6 u y + 7 u z nC m 2 (B) 3.4 u x - 6.9 u y + 8.6 u z nC m 2 (C) 1.2 u x - 2.3u y + 2.9 u z nC m 2
(C)
(D) 3.89 u x - 6.43u y + 8.96 u z nC m 2 65. Medium 1 has the electrical permitivity e1 = 15 . eo and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permitivity e 2 = 2.5 e o and occupies the region to the right of x = 0 plane. If E1 in medium 1 is E1 = (2 u x - 3u y + 1u z ) V m
then E2
in
(B) f ( x)
1 df ¶V f ( x) dx ¶x
(D) -f ( x)
where r v = 0. It is know that both Ex and V are zero at the origin. The V ( x, y) is (A) 3( x 2 - y 2 )
(B) 3( y 2 - x 2 )
(C) 4 x 2 - 3 y 2
(D) 4 y 2 - 3 x 2
(A) (2.0 u x - 1.8 u x + 0.6 u z ) V m (B) (1.67 u x - 3u y + u z ) V m (C) (1.2 u x - 3u y + u z ) V m (D) (1.2 u x - 1.8 u y + 0.6 u z ) V m
*********
66. Two perfect dielectrics have relative permitivities e r1 = 2 and e r 2 = 8. The planner interface between them is the surface x - y + 2 z = 5. The origin lies in region 1. If E1 = 100 u x + 200 u y - 50 u z V m then E2 is (A) 400 u x + 800 u y - 200 u z V m (B) 400 u x + 200 u y - 50 u z V m (C) 25 u x + 200 u y - 50 u z V m (D) 125 u x + 175 u z V m 67. The two spherical surfaces r = 4 cm and r = 9 cm are for
u r then E2 4 < r < 6 and e r 2 = 5 for 6 < r < 9. If E1 = 1000 r2 is (A)
5000 ur V m r2
(B)
400 ur V m r2
(C)
2500 ur V m r2
(D)
2000 ur V m r2
68. The surface x = 0 separate two perfect dielectric. For x > 0,
let
e r1 = 3,
while
er 2 = 5
where
x < 0.
If
E1 = 80 u x - 60 u y - 40 u z V m then E2 is (A) (133.3u x - 100 u z - 66.7 u z ) V m (B) (133.3u x - 60 u z - 40 u z ) V m (C) ( 48 u x - 36 u y - 24 u z ) V m (D) ( 48 u x - 60 u y - 40 u z ) V m Page 474
df ¶V dx ¶x
70. If V ( x, y) = 4 e 2 x + f ( x) - 3 y 2 in a region of free space
medium 2 is
separated by two perfect dielectric shells, e r1 = 2
df ¶V dx ¶x
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GATE EC BY RK Kanodia
Electrostatics
SOLUTIONS 1. (C) F2 = =
=
f = tan -1
= 65.9 cos 56.3° + 148.3 sin 56.3 = 159.7
(4, 4), (-4, 4), (4, -4), (-4, 4), the fifth charge will be on the z-axis at location z = 4 2. By symmetry, the force on the fifth charge will be z
directed, and will be four
times the z component
=
2
q2 4 pe o d 2
(5 ´ 10 -9) 2 ´ = 10 -2 N 4 p ´ ( 8.85 ´ 10 -12 )( 8 ´ 10 -3) 2 2
= -65.9 sin 56.3° + 148.3 cos 56.3° = 27.4 , Ez = -49.4 7. (C) Ep =
2 ´ 10 -8 4 pe o
é R1 2R 2 ù ê- |R |3 + |R |3 ú ë 1 2 û
R 2 = (0, 2, 1) - (0, 0, 0) =(0, 2, 1) é - (2 u y + 0.5 u z ) 2(2 u y + u z ) ù Ep = 9 ´ 10 9 ´ 10 -8 ê + ú ( 4.25) 3 2 ( 5)3 2 û ë E p = 54.9 u y + 44.1u z
3. (D) The force will be ( 40 ´ 10 -9) 2 é R CA R DA R BA ù F= + + 3 3 3ú ê 4 pe o ë|R CA| |R DA| |R BA| û where R CA = u x - u y , R DA = u x + u y , R BA = 2 u x
F=
Ef = E r × u f = 65.9( u x × u f) + 148.3 ( u y × u f)
R1 = (0, 2, 1) -(0, 0, 0.5) =(0, 2, 0.5)
4
|R CA|=|R DA|= 2,
12 = 56.3°, and z = 2 8
Er = Ep × u p = 65.9( u x × u r) + 148.3 ( u y × u r)
2. (D) Arranging the charge in the xy plane at location
´
é 4 u x + 9 u y - 3u z ù ê ú (106) 3 / 2 ë û
Then at point P, r = 8 2 + 12 2 = 14.4,
= ( 4.32 u x + 5.76 u y) mN
4
2 ´ 10 -6 4 pe o
= 65.9 u x + 148.3u y - 49.4 u z
Q1Q2 R12 4 pe o |R12|3
( 4 ´ 10 -6 )( -5 ´ 10 -6 ) ( 3u x + 4 u y) ´ 4 pe o 53
F=
Chap 8.2
|R BA|= 2
éu x - u y u x + u y 2u x ù ( 40 ´ 10 -9) 2 + + -9 ê 4 p ´ ( 8.85 ´ 10 ) ë 2 2 8 úû 2 2
At P, r = 5 , q = cos -1
1 5
= 63.4 ° and f = 90 °
So Er = E p × u r = 54.9[ u y × u r ] + 44.1[ u z × u r ] = 54.9 sin q sin f + 44.1 cos q = 68.83 Eq = E r × u q = 54.9[ u y × u q ] + 44.1[ u z × u q ] = 54.9 cos q sin f + 44.1 ( - sin q) = -14.85 Ef = Er × u f = 54.9( u y × u f) + 44.1( u z × u f) =54.9 cos f = 0
= 1376 . u x mN
8. (A) Let a point on y axis be P(0, y, 0)
10 -9 é 41R13 45R 23 ù 4. (C) E = + 2 ê |R 23|2 úû 4 pe o ë |R13|
Ep =
R13 = -3u x + 4 u y - 4 u z ,
R 23 = 4 u x - 2 u y + 5 u z
é 41( -3u x + 4u y - 4u z 45( 4u x + -2u y + 5u z ù E = 9 ´ 10 9 ´ 10 -9 ´ ê + ú ( 41) 3 / 2 ( 45) 3 / 2 ë û
= 1152 . u x + 2.93u y + 1089 . uz 5. (D) The point is P3(0, y, 0) R13 = -4 u x + ( y + 2) u y - 7 u z , R12 = 3u x + ( y - 4) u y + 2 u z Ex =
ù 10 -9 é 25 ´ ( -4) 60 ´ 3 + 4 pe o êë[ 65 + ( y + 2) 2 ]3 2 [13 + ( y - 4) 2 ]3 2 úû
To obtain Ex = 0,
0.48 y 2 + 1392 . y + 7312 . =0
which yields the two value y = -6.89, - 22.11 6. (B) Ep =
2 ´ 10 -6 R AP 4 pe o |R AP|3
20 ´ 10 -8 4 pe o
é R1 R2 ù ê-|R |3 - |R |3 ú ë 1 2 û
R1 = (0, y, 0) - (3, 0, 0) =(-3, y, 0) R 2 = (0, y 0) - (-3, 0, 0) = (3, y, 0),
|R1 | = |R 2 | =
9 + y2
é -3u x + yu y 3u x + yu y ù + Ep = 20 ´ 10 -9 ´ 9 ´ 10 9 ê ú 2 3 ( 9 + y 2 ) 3 úû êë ( 9 + y ) -1080 u x 1080 , |E| = = = 2 32 (9 + y ) (9 + y 2 ) 3 2 9. (B) The field at P will be Q é -2 u x + 2 u y - u z ù Ep = 0 ê ú, Ez = 1 kV m 4 pe o ë 93 2 û Q0 = -4 pe o ´ 9 3 2 ´ 10 3 = -3 m C 10. (C) This will be 8 times the integral of r n over the first octant
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GATE EC BY RK Kanodia
UNIT 8
¥
Q = 8ò
¥
¥
0
0
ò ò
0
é 6 u x - 3u y 6 u x + 3u y ù EP = 20nC ´ 2 ´ 9 ´ 10 9 ê 36 + 9 úû ë 36 + 9
r oe - x - y- z dxdydz = 8ro
11. (C) Q =
2p
p
0
0
= -48u y V m
0 .05
ò ò ò
0.2 r sin q drd qdf 2
0 .03
é r3 ù = ê4 p(0.2) ú = 82.1 pC 3 û 0 .03 ë 12. (A) Q =
òòò ¥ -p 0
ée =5 ê ë
R xp = ( -3, 0, - 1, ) - ( -3, 0, 0, ) = (0, 2, -1) Similarly R yp = (-3, 0, -1), R zp = (-3, 2, 0)
5 e -0 .1 r( p-|f|) z 2 + 10 10
-0 .1 r
é R xp R yp R zp ù + + ê 3 3 3ú ë|R xp| |R yp| |R zp| û
rL 2 pe o
17. (B) EP =
0 .05
-¥ p 10
Electromagnetics
ù ( -0.1 - 1) ( -0.101) ú ´ (0.1) 2 û0
Ep = 100 ´ 10 -9 ´ 2 ´ 9 ´ 10 -9 ¥
é 2u y - u z -3u x - u z -3ux + 2u y ù + + E p = 100 ´ 10 -9 ´ 2 ´ 9 ´ 10 -9 ´ ê ú 5 10 13 ë û
2p
( p - f) dz 2 + 10 0 z
ò 2ò
-¥
= -0.96 u x + u y - 0.54 u z kV m
¥
p2 dz -¥ z + 10
Q = 5 ´ 26.4 ò
18. (C) At y = 4, E =
2
¥
z ù é 1 = 5 ´ 26.4 ´ p2 ê tan -1 = 129 . mC 10 úû -¥ ë 10
dF = dqE = r L dzE r 2L dzu y
1
F=
ò
2 pe o
0
13. (A) f = 10
1 .1 p 25° 5
f
ò ò ò ( r - 4)( r - 5) sin q × sin 2 r
0 .9p 0 4
25°
5
1 .1 p
é r 5 9 r 4 20 r 3 ù é 1 1 fù é ù = 10 ê + ú ê2 q - 4 sin 2 q ú ´ ê-2 cos 2 ú 5 4 3 ë û 0 .9p ë û ë û4 0 = 10 [ -3.39 ][0.0266 ][0.626 ] = 0.57 C
( 6) = 18.75 u y mN r s dS R - R¢ 4 pe o |R - R¢|3
ò ò
19. (A) E =
sin q drdqdf
2
rL uy , 2 pe o
where R = 3u x and R¢ = yu y + zu z , ¥ 2 3u x - yu y - zuz r dydz EPA = L ò ò 4 pe o -¥ -2 (9 + y 2 + z 2 ) 3 2 Due to odd function
Rp r , 14. (D) Ep = L 2 pe o |R p|2
EPA =
R p = (1, 2, 3) - (1, -2, 5) = (0, 4, -2)
So there is only x component.
|R | p
2
= 20,
5 ´ 10 -9 Ep = 2 pe o
é4u y - 2u z ù ê ú = 18 u y - 9 u z 20 ë û
15. (C) With z = 0, the general field is Ez = 0 =
r L é( y + 2) u y - 5 u z ù 2 pe o êë ( y + 2) 2 + 25 úû
we require |E2 | = |2E y| So 2 ( y + 2) = 5 E=
Þ
y=
1 2
6.25 é2.5 u y - 5 u z ù = 9 u y - 18 u z 2 pe o êë 6.25 + 25 úû
rL 4 pe o
¥
2
òò
-¥ -2
3u x dydz (9 + y 2 + z 2 ) 3 2
20. (D) There will be z component of E only R = zu z , R¢ = ru r , Ez , Pa =
rs 4 pe o
2 p 0 .2
R - R¢ = zu z - ru z
zrdrdf 2 + r 2)3 2
ò ò (z 0
0
0 .2
ù ù 2pr s z é rzé 1 1 1 = ú = s ê 2 ê 2 ú 2 2 eo ê z 4 pe o ê z + r ú z + 0.04 úû û0 ë ë Ez =
rs eo
é z ê1 2 z + 0.04 êë
ù ú, at z = 0.5, Ez = 20.2 V m úû
21. (A) Since charge sheet are infinite, the field magnitude associated with each one will be r s 2 e o , which is position independent. The field direction will
r 16. (C) Ep = L 2 pe o
é R+ Q R- Q ù ê 2 2ú ë|R + Q| |R - Q| û
R+ Q = (6, 0, 6) - (0, 3, 6) = (6, -3, 0) R- Q = (6, 0, 6) - (0, -3, 6) = (6, 3, 0) Page 476
depend on which side of a given sheet one is positioned. é10 ´ 10 -9 40 ´ 10 -9 50 ´ 10 -9 ù EA = ê ux uy uzú 2 eo 2 eo ë 2 eo û = 0.56 u x + 2.23u y - 2.8 u z www.gatehelp.com
GATE EC BY RK Kanodia
Electrostatics
22. (A)
dy E y -15 x 2 y -3 y = = = 5 x3 dx Ex x
dy 3dx =y 3
Þ
At P, 2 =
C1 43
Þ
a a
ln y = -3 ln x ln C y = C1 = 128
Þ
Chap 8.2
Þ
y =
C1 , x3
+
128 x3
ò ò -x
a a
= a4 + 0 -
a a + + 0 + 6a5 = a4 + 6a5 3 3
of
¶D y ¶y
= 20 x 4 y 3z 4
cube
=(3.05
1.05
2.05)
and
Volume
Ñ V = (0.1) 3 = 0.001 f = 20 ( 305 . ) 4 (105 . ) 3 (2.05) 3 (2.05) 4 0.001 =35.4
24. (C) h1 = 2 4 2 - 1 = 7.75,
31. (C) F = (Ñ × D ) Dv
h2 = 2 4 2 - 1 = 6.93
=
QT = 2 ´ 7.75 ´ 40n + 2 ´ 6.93 ´ 50 + 20n =1.33 mC
ò ò
Top
4
30. (C) Ñ × D = Center
Botoom
Right
4
D = -19.1u y + 25.5 u z pC m 2
a a
dxdz + ò ò x 2 dxdz + ò ò -6(0) 3 dxdy + ò ò 6a 3dxdy 0 0 0 0 0 0 0 0 1 4 4244 3 1 4243 1 442443 1 4 4244 3 Left
at that point. Thus D arises from the point charge alone 10 ´ 10 -9 ( -3u y + 4 u z ) , D= ( 32 + 4 2 )1 .5 4p
a a
2
arrangement of line charges, whose field will all cancel
0 .05 90 °
Back
Front
a a
23. (A) This point lies in the center of a symmetric
25. (D) Q =
a a
29. (C) Q = ò D × n dS = ò ò 2 aydydz + ò ò -2(0) dydz S 0 0 0 0 1 4 24 3 1 44244 3
¶( x) æ 4 3ö . nC ç p(0.003) ÷ = 1131 ¶x è 3 ø
é10 y æ 2 + 10 x 2 y öù =8.96 32. (B) r v = Ñ × D = ê ÷÷ú + 0 + çç z3 øû ( -2 , 3, 5) è ë z
5 e -20 z (0.08) dfdz nC
0 .01 30 0 .05
æp pö æ 1 ö -20 z = ç - ÷(5)(0.08)ç ÷e è2 6 ø è 20 ø 0 .01
33. (C) Ñ × D =
= 9.45 ´ 10 -3 nC = 9.45 pC
ò Ñ × D dv = 6 ´ (0.4)
26. (A) Out of the 6 surface only 2 will contribute to the net outward flux. The y component of D will penetrate the surface y = 0 and y = z and net flux will be zero. At x = 0 plane Dx = 0 and at z = 0 plane Dz = 0. This leaves the 2 remaining surfaces at x = 2 and z = 5. The net outward flux become 5 3
f = ò ò D x = 2 × u x dydz + 0 0
3
3
0
0
2p p
ò òD
x =2
× u z dzdy
0 0
é -r e Q = 8 pê êë 1000
2 -1000 r
-1000 r
0
2 e -1000 r ( -1000 r - 1) 1000(1000) 2 0
Q = 4 ´ 10 -9nC, Dr =
4.0 ´ 10 -9 Q = 0.32 C m 2 = 2 4 pr 4 p ´ (0.001) 2
28. (C) 4 pr Dr = 4 p(2 ) 20 ´ 10 2
Dr =
80 nC m 2 r2
2
= -5( 4 u x - 3u y + 5 u z ) × =-
10 3
( u x + u y + u z )(2) 3
( 4 - 3 + 5) = -34.64 J
æ ( u x - u y + u z )(2 ´ 10 -3) ö ÷ = -( 60 ´ 10 -6 )ç 100 u r × ç ÷ 3 è ø
r 2 sin q drdqdf 0 .001
+
34. (C) dW = - qE × dL
(2, 1, 4) - (7, 2, 3) = (1, -1, 1) ux - u y + uz , dW = - qE × dL u PQ = 3
0
0 .001
= 0.38
V
35. (B) The vector in this direction is
0 0
0 .001
ò ò ò 2e
3
3 2
= 5 ò 4(2) ydy + 2 ò 4(5) ydy = 360 C.
27. (D) Q =
1 d 2 ( r ´ 2 r) = 6 , r 2 dr
-9
2
Cm ,
ù ú úû
= - 12 ´
10 -6 3
( u r × u x - u r × u y)
æ2 ö At r , f = tan -1 ç ÷ = 63.4 ° è1ø u r × u x = cos 63.4 ° = 0.447, u r × u y = sin 63.4 ° = 0.894 dW = 31 . mJ 36. (A) W = - q ò E × dL
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UNIT 8
Electromagnetics
= -10 ò (2 xu x - 3 y 2 u y + 4 u z ) ( dxu x + dyu y + dzu z ) 3
1
-1
0
0
0
é ù 18 z -ê6 xy 2 z 2 + 2 uz 2 2 ú x + y + z 2 3 ë û
= -10 ò 2 xdx + 10 ò 3 y 2 dy - 10 ò 4 dz = -40 J
. u z V m, Ep = 7.1u x + 22.8 u y - 711 |E|= 75 V m
37. (D) rA = 2 cm, rB = 3 cm V ( r) =
43. (D) E = -ÑV = -
Q 4 pa 2r s ( 6 ´ 10 -3) 2 ´ 30 ´ 10 -9 = = 4 pe o r 4 pe o r 8.85 ´ 10 -12 ´ r
= -42 r -0 .4 u r V m,
0.122 0.122 0.122 , V AB = V A - VB = V ( r) = = = 2.03 r 0.02 0.03 38. (D) W = -
Q1Q2 4 pe o
rB
ò
rA
dr Q1Q2 = r 2 4 pe o
D = e oE = -42 r -0 .4 e o u r C m 2 1 d 2 1 d e rv = Ñ × D = 2 ( r Dr ) = 2 ( -42 e o r 1 .6 )= - 67.2 1o.4 r dr r dr r
é1 1ù êr - r ú ë B Aû
At r = 0.6 m, rv = -
rA = rB , W = 0 39. (B) VP ( s) =
Q , 4pe o r
r L dr r + C1 = - L ln r + C1 2 pe or 2 pe o
Vs ( z) = - ò
rs rz dz + C2 = s + C2 2 eo 2 eo
D = e oE = -80 e o (2 r cos q u r - r sin q u q) é1 2 2 ù 1 ¶ rv = Ñ × D = ê 2 ( r Dr ) + ( Dr sin q) ú r sin q ¶q ë r ¶r û
Here r , r , z are the scalar distance from the charge. r = 2 2 + 5 2 = 29 , r = (5 - 4) 2 = 1 , z = 5 By putting these value. C = -193 . ´ 10 3 2
r = 12 ` + 12 = 2 , z = 3, 40. (D) Vr =
¥
-¥
2
é1 ¶ ù 1 ¶ r v = eo ê (rEr) + Ef ú r ¶f û ër ¶r é (50 + 150 sin f) 150 sin f ù 50e o = e o ê+ =C m2 ú r r r ë û 1 2p 2
Q =ò
R = zu z , R¢ = ru r, dS = rdrdf, Vr =
ò ò 0
0 .01
(5 ´ 10 -9)r 2 drdf 4 pe o r 2 + z 2
0
ù z2 5 ´ 10 -9 ér 2 2 r + z ln (r + r 2 + z 2 ) ú ê 2 eo 2 ë2 û 0 .01
At z = 0.02, Vp = 0.081 V
0 0
50 e o rdrdfdz = - 2 p(50) e o 2 = -5.56 nC r
46. (A) V =
42. (C) E = -ÑV é ù 6x = - ê2 y 2 z 2 + 2 ux 2 2ú x + 2 y + 3z û ë é ù 12 y - ê4 xyz 3 + 2 uy 2 2 ú x + 2 y + 3z û ë
Qd cos q 100 cos q , = 4 pe o r 2 r2
1 ¶V æ ¶V ö E = -ÑV = -ç ur + uq ÷ r r ¶ ¶q è ø =
Page 478
òò
, 0 .03
Vp =
1 ¶V ¶V ur uf ¶r r ¶f
D = e o E, r v = Ñ × D = e oÑ × E
¥
r s dS 41. (A) Vr = òò , 4 pe o|R - R¢| 2 p 0 .03
r V = -320 e o cos q = -2.45 nC m 3
= -(50 + 150 sin f) u r - (150 cos f) u f ,
VB = 198 . kV
6 ´ 10 -9 dz r L dz = ò = 108 V 4 ( 2 1) 3 2 oR -¥ pe o z +
ò 4 pe
é1 12 sin q cos q ù r v = -80 e o ê 2 ´ 3r 2 cos q ú r sin q ër û
45. (C) E = -ÑV = -
At point N, r = (2 - 1) + 2 + z = 14 2
¶V 1 ¶V ur uq ¶r r ¶q
= -160 r cos q u r + 80 r sin q u q V m
Q r r - L ln r - s z + C 4 pe o r 2 pe o 2 pe o
V =
67.2 ´ 8.85 ´ 10 -12 = -1.22 nC m 3 (0.6)1 .4
44. (A) E = -ÑV = -
Vl (r) = - ò
dV u r = -(0.6)(70) r -0 .4 u r dr
100 (2 cos q u r + sin q u q), r3
|E| = 100( 4 cos 2 q + sin 2 q)1 2 47. (A) E = www.gatehelp.com
= 100 ´
5 = 158.11 V m 2
2 ´ 10 -9 [2 cos q u r + sin q u q ], 4 pe o r 3
GATE EC BY RK Kanodia
Electrostatics
55. (A) I = òò J × n z = 0 .2 dS
y = z lies at q = 45° 2 ´ 10 -9 1 = 10 -3 Eq = 4 pe o r 3 2
S
(required) =
r 3 = 12.73 ´ 10 3, r = 23.35
0
56. (B) So Eal = Est =
Solving
4 pe o (1 + 1 + 8 2 )1 .5
é 1 ù 1 1 49. (A) E = -ÑV = 40 ê 2 u x + uy + uzú 2 2 x yz xy z xyz ë û
òòò E × Edv 2
2
2
ò ò
1
1
1
Þ
J al =
sac J st sst
J st = 3.2 ´ 10 5 A m 2
57. (B) J =
We = 800 e o ò
J al J st = sal sst
I = p(2 ´ 10 -3) J st + p[( 4 ´ 10 -3) 2 - (2 ´ 10 -3) 2 ]J al = 80
( 3u x - 5 u y + 10 u z ) × ( u x + u y + 8 u z ) ´ 10 -9
=1.31 V
eo 2
ò ò
0 .4
40 40 rdrdf = log[r 2 + (0.1) 2 ] (2 p) 2 2 r + (0.1) 2 0
= 40 p log 17 = 356 A
where R - R¢ = Q - P = (1, 1, 8)
We =
2 p 0 .4
0
P × (R - R¢) 48. (A) V = 4 pe o|R - R¢|3
So VP =
Chap 8.2
E=
4 u r A m2, 2prl
4 12.73 J ur V m = = s 2 prls rl 3
5
12.73 12.73 5 6.51 V u r × u rdr = ln = l r l 3 l 3
V = - ò E × dL = ò
é 1 1 1 ù ê x 4 y 2 z 2 + x 2 y 4 z 2 + x 2 y 2 z 4 ú dxdydz ë û
5
V 6.51 1.63 R= = = W I 4l r
= 1548 pJ æ ¶V ö 1 ¶V ¶V 58. (D) E = -ÑV = -çç ur + uf + u z ÷÷ r ¶f ¶z è ¶r ø (r + 1) 2 = -z 2 cos f u r + z sin f u f - 2(r + 1) z cos f u z r
1 4 50. (A) W = å qn Vn 2 n =1 V1 = V21 + V31 + V41 =
q é 1 1 1 ù + + ê 4 pe o ë0.04 0.04 0.04 2 úû
E = -1.82 u f + 14.5 u f - 2.67 u z V m E ×E r s = eo E × n s = eo |E|
V1 = V2 = V3 = V4 W =
2 ´ (1.2 ´ 10 -9) 2 1 ( 4) q1 V1 = 4 pe o (0.04) 2
51. (B) I = òò J × n =
ò ò 10
5
r s = e o 1.82 2 + 14.5 2 + 2.67 2 = 1315 . pC m 2
2 1
y =1
S
2 1
1 ù é . mJ êë2 + 2 úû = 175
dS = ò ò J × u y 0 0
y =1
dxdz 59. (C) V =
cos (2 x) 2 -2 y dxdz = 12307 kA = 12.3 MA
p p sin 3 2 = 2.5 V 23
40 cos
So the equation of the surface is 40 cos q sin f = 2.5, 16 cos q sin f = r 3 r2
0 0
52. (C) I = òò J × n dS S
=
60. (A) E = -ÑV, D = eE = - e oÑV
2 p 0 .3p
ò 0
800 sin q (0.80) 2 sin q dqdf =154.8 A ò 2 + ( 0 . 80 ) 4 0 .1 p
53. (D) F = ma = qE, a=
qE ( -1.602 ´ 10 -19)( -4 ´ 106 ) = u z = 7.0 ´ 1017 u z m s 2 , m 9.11 ´ 10 -3
v = at = 7.0 ´ 10 tu z m s 17
54. (D) =
¶r V 1 ¶ ¶J z = -Ñ × J = (rJ r) + ¶t r ¶r ¶z
1 ¶ ¶ æ -20 ö (25) + ç ÷ =0 r ¶r ¶z çè r 2 + 0.01 ÷ø
é ù x 200 x ¶ ux + 2 u z ú C m2 = -e o ê200 z 2 ( ) x x + 4 x + 4 ¶ ë û 200 e o x D (z= 0 ) = - 2 u z C m2, x +4 r s = D × u z z= 0 = Q=
0
2
-3
0
ò ò
- 200 e o x C m2, x2 + 4 2
1 - 200 e o x dxdy = -( 3)(200) e o ln[ x 2 + 4 ] 2 x2 + 4 0
= - 300 ln 2 = - 1.84 nC
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Page 479
GATE EC BY RK Kanodia
UNIT 8
61. (B) E = -ÑV ,
-3 xy 2 z 3 satisfy this equation.
62. (D) The plane can be replaced by -60 nC at Q (2, 5, -6).
Electromagnetics
Since E is normal to the surface e 2 1000 400 En 2 = r1 En 1 = ´ ur = 2 ur V m r2 r er 2 5
R = (5, 3, 1) - (2, 4, 6) = (3, -1, -5)
68. (D) D n 1 = D n 2
R¢ = (5, 3, 1)-(2, 4, -6) = (3, -1, 7),
D n 1 = 80 ´ 3e o = D n 2 = 5 e oEn 1 ,
|R |= 35 , |R¢|= 59
E2 = 48 u x - 60 u y - 40 u z
VP =
q q q é 1 1 ù = ê 4 pe o R 4 pe o R¢ 4 pe o ë 35 59 úû -9
Vp = 60 ´ 10 ´ 9 ´ 10
-9
rL 2p eo
0 = Ñ × ( - f ( x)ÑV ) é dF ¶V ¶2V ¶2V ¶2V ù =-ê + f ( x) 2 + f ( x) 2 + f ( x) 2 ú ¶x ¶y ¶z û ë dx ¶x
é Final Distance from the charge ù rL ln ê 2 pe o ëInitial Distance from the charge úû
é 1 12 + (2 - 1) 2 12 + (2 - (-2)) 2 12 + 32 ù - ln - ln êln + ln ú 1 1 1 êë 2 úû
= 2.40 kV 64. (C) P = D - e oE = D -
Þ
. e o = 2.5 e o En 2 = 2 ´ 15 Et1 = Et 2 ,
Þ
D D = ( e r - 1) er er
e1E N1 = e 2E N 2
En 1 = - 817 .
1 6
6
f ( x1 ) = -4 e 2 x + 3 x 2 + C, f (0, 1) = -4 + C2 V (0, 0) = 0 = 4 + f (0) f ( x) = -4 e
C2 = 0. f (0) = -4
+ 3x , 2x
- 4 e2 x + 3 x 2 - 3 y 2 = 3 ( x 2 - y 2 )
[u x - u y + 2u z ] *********
Et1 = Et 2 and D n 1 = D n 2 e r1 e oEn 1 = e r 2 e oEn 2 e 1 1 = r1 En 1 = E n 1 , E2 = Et 2 + En 1 er 2 4 4
= 133.3 u x + 166.7 u y + 16.67 u z - 8.33 u x + 8.33 u y - 16.67 u z
= 125 u x + 175 u y V m 67. (B) D n 1 = D n 2 , Page 480
Þ
2
[100 - 200 - 100 ] = -817 . Vm
Et1 = E1 - En 1 = 133.3u x + 166.7 u z + 16.67 u z
En 2
2x
V ( x, y) = 4 e
= -33.33u x + 33.33u y - 66.67 u z V m
Þ
Ñ 2 V = 0,
Integrating again
66. (D) The unit vector that is normal to the surface is u x - u y + 2u z ÑF , uN = = |ÑF| 6 1
Þ
It follow that C1 = 0
En 2 = 12 .
E2 = 1.2 u x - 3u y + 1u z
En 1 = E1 × u N =
70. (A) r v = 0
¶2 f - 6 =0 ¶x 2 ¶f ¶f = -16 e 2 x + 6 Þ = - 8 e 2 x + 6 x + C1 ¶x ¶x ¶V ¶f Ex = = 8 e2 x + ¶x ¶x ¶f ¶f Ex (0) = 8 + =0 Þ = -8 ¶x x = 0 ¶x x = 0
P = 12 . u x - 2.3u y + 2.9 u z nC m 2 65. (C) D n 1 = D n 2
é dF ¶V ù =-ê + f ( x)Ñ 2 V ú ë dx ¶x û 1 dF ¶V Þ Ñ 2V = f ( x) dx ¶x
Ñ 2 V = 16 e2 x +
(2.4 - 1) ´ 10 -9 2.4
P = (2 u x - 4 u y + 5 u z )
e r1 e oEn 1 = e r 2 e oEn 2
En1 = 48
Ñ × D = r v = 0 = Ñ × ( - f ( x)ÑV )
1 ù é 1 êë 35 - 59 úû = 21 V
V0 = 0, Vp = -
D = eE
69. (A) D = eE = - f ( x)ÑV ,
63. (A) Using method of images Vp - V0 = -
and Et1 = Et 2 ,
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UNIT 8
Electromagnetics
Statement for Q.15–16:
Statement for Q.9–11: An infinite filament on the z-axis carries 10 mA in
In the cylindrical region
the u z direction. Three uniform cylindrical current
2 r + for r £ 0.6 r 2 3 for r > 0.6 Hf = r Hf =
sheets are also present at 400 mA m at r = 1 cm, -250 mA m at r = 2 cm and 300 mA m at r = 3 cm. 9. The magnetic field H f at r = 0.5 cm is (A) 0.32 A m
(B) 0.64 A m
15. The current density J for r < 0.6 mm is
(C) 1.36 mA m
(D) 0
(A) 2u z A m
(B) -2u z A m
(C) u z A m
(D) 0
10. The magnetic field H f at r = 15 . cm is (A) 1.63 A m
(B) 0.37 A m
16. The current density J for r > 0.6 mm is
(C) 2.64 A m
(D) 0
(A) 2u z A m
(B) -3u z A m
(C) 3u z A m
(D) 0
11. The magnetic field H f at r = 3.5 cm is (A) 0.14 A m (C) 0.27 A m
(B) 0.56 A m
17.
(D) 0.96 A m
v = ( 3u x + 12 u y - 4 u z ) ´ 10
An
electron 5
with
velocity
m s experiences no net
forces at a point in a magnetic field B = u x + 2 u y + 3u z Statement for Q.12–14:
mWb m 2 . The electric field E at that point is
In the fig. P8.3.12–14 The region 0 £ z £ 2 is filled
(A) -4.4 u x + 1.3u y + 0.6 u z kV m
with an infinite slab of magnetic material (m r = 2.5). The
(B) 4.4 u x - 1.3u y - 0.6 u z kV m
surface of the slab at z = 0 and z = 2, respectively, carry
(C) -4.4 u x + 1.3u y + 0.6 u z kV m
surface current 30u x A m and -40u x as shown in fig.
(D) 4.4 u x - 1.3u y - 0.6 u z kV m
z
18. A point charge of 2 ´ 10 -16 C and 5 ´ 10 -26 kg is
mo
z=2
-40ux A/m
moving in the combined fields B = - 3u x + 2 u y - u z mT and E = 100 u x - 200 u y + 300 u z V m. If the charge velocity at t = 0 is v(0) = (2 u x - 3u y - 4 u z ) 10 5 m s, the
mr = 2.5 mo
z=0
30ux A/m
Fig. P8.3.12–14
12. In the region 0 < z < 2 the H is (A) -35u y A m
(B) 35u y A m
(C) -5u y A m
(D) 5u y A m
x
acceleration of charge at t = 0 is (A) 600[ 3u x + 2 u y - 3u z ]10 9 m s 2 (B) 400[ 6 u x + 6 u y - 3u z ]10 9 m s 2 (C) 400[ 6 u x - 6 u y + 3u z ]10 9 m s 2 (D) 800[ 6 u x + 6 u y - u z ]10 9 m s 2 19. An electron is moving at velocity v = 4.5 ´ 10 7 u y
13. In the region z < 0 the H is (A) 5u y A m
(B) -5u y A m
(C) 10u y A m
(D) -10u y A m
14. In the region z > 2 the H is (A) 5u y A m
(B) -5u y A m
(C) 35u y A m
(D) -35u y A m
Page 482
m s along the negative y-axis. At the origin, it encounters the uniform magnetic field B = 2 .5 u z mT, and remains in it up to y = 2.5 cm. If we assume that the electron remains on the y –axis while it is in the magnetic field, at y = 50 cm the x and z coordinate are respectively (A) 1.23 m, 0.23 m
(B) -1.23 m, -0.23 m
(C) -11.7 cm, 0
(D) 11.7 cm, 0
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Magnetostatics
Statement for Q.20–22:
Chap 8.3
25. If the current filament is located at y = 0.5, z = 0,
A rectangular loop of wire in free space joins points
and u L = u x , then F is
A(1, 0, 1) to B(3, 0, 1) to C(3, 0, 4) to D(1, 0, 4) to A. The
(A) 35 .2 u y nN m
(B) 68.3u x nN m
wire carries a current of 6 mA flowing in the u z
(C) 105.6 u z nN m
(D) 0
direction from B to C. A filamentary current of 15 A flows along the entire z, axis in the u z directions. 20. The force on side BC is (A) -18u x nN
(B) 18u x nN
(C) 3.6 u x nN
(D) -3.6 u x nN
21. The force on side AB is (A) 23.4 u z mN
(B) 16.4 u z mN
(C) 19.8 u z nN
(D) 26.3u z nN
(B) -36u x nN
(C) 54u x nN
(D) -54u x nN
fig.
P8.3.23.
The
magnetic
B = 6 xu x - 9 yu y + 3zu z Wb m 2 .
flux
The
(C) 4u x mN m
(D) -4u x mN m
27. The force exerted on the filament by the current
density total
(B) -0.4 u x N m
A conducting current strip carrying K = 6 u z A m lies in the x = 0 plane between y = 0.5 and y = 15 . m. There is also a current filament of I = 5 A in the u z direction on the z –axis.
23. Consider the rectangular loop on z = 0 plane shown in
(A) 0.4 u x N m
Statement for Q.27–28:
22. The total force on the loop is (A) 36u x nN
26. Two infinitely long parallel filaments each carry 100 A in the u z direction. If the filaments lie in the plane y = 0 at x = 0 and x = 5 mm, the force on the filament passing through the origin is
is
force
experienced by the rectangular loop is
strip is (A) 12.2 u y mN m
(B) 6.6 u y mN m
(C) -12.2 u y mN m
(D) -6.6 u y mN m
28. The force exerted on the strip by the filament is
y 2 5A
(A) -6.6 u y mN m
(B) 6.6 u y mN m
(C) 2.4 u x mN m
(D) -2.4 u x mN m
Statement for Q.29–32:
1
In a certain material for which m r = 6.5, 0
1
2
3
x
H = 10 u x + 25 u y - 40 u z A m
Fig. P8.3.23
(A) 30u z N
(B) -30u z N
(C) 36u z N
(D) -36u z N
29. The magnetic susceptibility c m of the material is (A) 5.5 (C) 7.5
(B) 6.5 (D) None of the above
30. The magnetic flux density B is
Statement for Q.24–25: Three uniform current sheets are located in free
(A) 82 u x + 204 u y - 327 u z mWb m 2
space as follows: 8u z A m at y = 0, -4u z A m at y = 1
(B) 82 u x + 204 u y - 327 u z mA m
and -4u z A m at y = -1. Let F be the vector force per
(C) 82 u x + 204 u y - 327 u z mT
meter length exerted on a current filament carrying 7
(D) 82 u x + 204 u y - 327 u z mA m
mA in the u L direction.
31. The magnetization M is
24. If the current filament is located at x = 0, y = 0.5
(A) 75 u x + 187.5 u y - 300 u z A m 2
and u L = u z , then F is
(B) 75 u x + 187.5 u y - 300 u z A m 2
(A) 35 .2 u y nN m
(B) -35 .2 u y nN m
(C) 55 u x + 137.5 u y - 220 u z A m 2
(C) 105.6 u y nN m
(D) 0
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GATE EC BY RK Kanodia
UNIT 8
32. The magnetic energy density is (A) 19 mJ m
2
If H increases from 0 to 210 A m, the energy
(B) 9.5 mJ m
(C) 16.3 mJ m
2
Electromagnetics
stored per unit volume in the alloy is
2
(D) 32.6 mJ m 2
(A) 6.2 MJ m 3
(B) 1.3 MJ m 3
(C) 2.3 kJ m 3
(D) 2.9 kJ m 3
Statement for Q.33–34: For a given material magnetic susceptibility c m = 31 . and within which B = 0.4 yu z T.
(B) 151.6 yu z kA m
(C) 102.7 yu z kA m
(D) 77.6 yu z kA m
cube of size a, the magnetization volume current density is 12 (A) uz a
33. The magnetic field H is (A) 986.8 yu z kA m
40. If magnetization is given by H = 6a ( - yu x + xu y) in a
(C)
34. The magnetization M is
6 uz a
(B)
6 ( x - y) a
(D)
3 ( x - y) a
41. The point P(2, 3, 1) lies on the planner boundary
(A) 241yu z kA m
(B) 318.2 yu z kA m
separating region 1 from region 2. The unit vector
(C) 163yu z kA m
(D) None of the above
u N12 = 0.6 u x + 0.48 u y + 0.64 u z is directed from region 1 to
35. In a material the magnetic field intensity is H = 1200 A m when B = 2 Wb m 2 . When H is reduced to 400
A m,
B = 1.4
Wb m 2 .
The
change
in
the
magnetization M is (A) 164 kA m
(B) 326 kA m
(C) 476 kA m
(D) 238 kA m
region
2.
If
m r1 = 2,
H1 = 100 u x - 300 u y + 200 u z A m, then H 2 is (A) 40.3u x + 48.3u y - 178.9 u z A m (B) 80.2 u x - 315.8 u y + 178.9 u z A m (D) 80.2 u x + 48.3u y + 178.9 u z A m
each atom has a dipole moment of 2.6 ´ 10 -30 u y A × m 2 . The H in material is (m r = 4.2) (A) 2.94 u y A m
(B) 0.22 u y A m
(C) 0.17 u y A m
(D) 2.24 u y A m
42. The plane separates air ( z > 0, m r = 1) from iron ( z £ 0, m r = 20). In air magnetic field intensity is H = 10 u x + 15 u y - 3u z A m. The magnetic flux density in iron is (A) 5.02 u x + 7.5 u y - 0.076 u z mWb m 2 (B) 12.6 u x + 18.9 u y - 75.4 u z mWb m 2
37. In a material magnetic flux density is 0.02 Wb m 2
(C) 251u x + 377 u y - 377 . u z mWb m 2
and the magnetic susceptibility is 0.003. The magnitude
(D) 251u x + 377 u y - 1508 u z mWb m 2
of the magnetization is (B) 23.4 A m
(C) 16.3 A m
(D) 8.4 A m
43. The plane 2 x + 3 y - 4 z = 1 separates two regions. Let m r1 = 2 in region 1 defined by 2 x + 3 y - 4 z > 1, while m r 2 = 5 in region 2 where 2 x + 3 y - 4 z < 1. In region
38. A uniform field H = - 600 u y A m exist in free space.
H1 = 50 u x - 30 u y + 20 u z A m. In region 2, H 2 will be
The total energy stored in spherical region 1 cm in
(A) 63.4 u x + 4318 . u y - 19.4 u z A m
radius centered at the origin in free space is
(B) 52.9 u x - 25.66 u y + 14.2 u z A m
(A) 0.226 J m
3
(B) 1.452 J m
(C) 1.68 J m 3
(C) 48.6 u x - 16.4 u y - 46.3u z A m
3
(D) None of the above
(D) 0.84 J m 3
39. The magnetization curve for an iron alloy is approximately given by B= Page 484
1 H + H 2 mWb m 2 3
and
(C) 40.3u x - 315.8 u y - 178.9 u z A m
36. A particular material has 2.7 ´ 10 29 atoms m 3 and
(A) 47.6 A m
m r2 = 8
********
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Magnetostatics
SOLUTIONS 1. (C) H = =
I 4p
IdL ´ u R 4 pR 2 -¥
ò
[2 + ( 3 - y) ] 2
0
2 32
R
R = (1 + x) + 32 + 2 2 = x 2 + 2 x + 14 2
H=
-u ydy[2 u x + ( 3 - y) u y ]
¥
-(1 + x) u x + 3u y + 2 u z
uR =
¥
ò
Chap 8.3
4 dxu x ´ [ -(1 + x) u x + 3u y + 2 u z ]
¥
ò
4 p ( x 2 + 2 x + 14) 3 2
-¥
=
(12 u z - 8 u y) dx
¥
ò 4 p( x
=
+ 2 x + 14)
2
-¥
¥
3/2
=
2 u z dy I = ò 2 4 p 0 [2 + ( 3 - y) 2 ]3 2
= 0.147 u z - 0.098 u y A m
3 - y = 2 tan q, -dy = 2 sec 2 q,
5. (B) H =
q1 = 56.31°, q 2 = -90 ° H=
I 4p
56 .31 °
ò
-90 °
2 u z dq 2 sec q
=
I u z [sin q ] -5690.31° °= 145.8 u z mA m = 4p 2. (A) H = H y + H z , H z =
Iz uf 2pr
r = ( -3) 2 + ( 4) 2 = 5 3u y + 4 u x = 5 5 24 ( 4 u x + 3u y) Hz = = 0.611u x + 0.458 u y mA m 2 p(5) 5 uf =
Hy =
- u z ´ ( -3u x + 4 u y)
Iy 2pr
2
I uf 2 pr
( -3u x + 5 u z ) 34
=
3u z - 5 u x 34
( -5 u x + 3u z )
12 2 p 34
-a
3. (A) H =
I uf 2 pr
4
ò
-4
Idzu z ´ (ru r - zu z ) 4 p(r 2 + z 2 ) 3 2
I æç a 12 2 pr ç (r + a 2 ) è I At r = 1, H = 2pr
Þ
a
1Þ
1+ a 1
a=
3
=
2p
ò
-a
Iau f 2 pr(r 2 + z 2 ) 3 2
ö ÷u A m ÷ f ø
1 2
= 0.577 m
IdL ´ u R 4 pR 2 Idfu f ´ ( - ur ) I uz A m = 2a 4 pa
K ´ u R dxdy 4 pR 2 ¥ 4 u x ´ ( - xu x - yu y - 3u z ) dx dy
7. (D) H = òò 2
=
ò ò
4 p( x 2 + y 2 + 9) 3 2
-2 -¥
=
ö I æç 4 ÷u 1f 2 pr ç (r 2 + 16) ÷ø è
=-
IdL ´ u R , IdL = 4 dxu x 4 pR 2
=
=
I = 3 A, a = 0.5 m, H = 3u z A m
I I 8 uf uf 2 pr 4 p r (r 2 + 16)
4. (B) H = ò
2
a
6. (A) H = ò
=
H = 0.1u f A m
rIdzu f 4 p(r 2 + z 2 ) 3 2
=
4
f = 60 °, I = 3p,
-a
Idzu z ´ (ru r - zu z ) 4 p(r 2 + z 2 ) 3 2
rIdzu f rIu f z ò- a 4 p(r 2 + z 2 ) 3 2 = 4 p r 2 (r 2 + z 2 ) 3 2
rdz I I uf uf ò 2 2 pr 4 p -4 4 p(r + z 2 ) 3 2
At r = - 3,
a
ò
4 p13
a
0
34
H = H y + H z = 0.331u x + 0.458 u y + 0.168 u z mA m
=
ò
2
= - 0.281u x + 0.168 u z mA m
=
a
u f , r = ( -3) + (5) = 34
uf = u y ´ Hy =
I uf 2 pr
2(12 u z - 8 u y)
2
¥
ò ò
4 ( - yu z - 3u y) dx dy
-2 -¥ 2
¥
4 p( x 2 + y 2 + 9) 3 2 12 u ydx dy
ò ò 4 p( x
-2 -¥
3 uy p
2
ò
-2
2
+ y 2 + 9) 3 2 2 dy y2 + 9 2
-
6 1 æ yö u y tan -1 ç ÷ = -0.75 u y A/m p 3 è 3 ø -2
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GATE EC BY RK Kanodia
UNIT 8
IdL ´ u R 4 pR 2 -Idzu z ´ ( -zu z + u y)
8. (D) H = ò ¥
=ò 0
¥
=ò 0
4 p(1 + z )
2 32
Idzu x + 4 p(1 + z 2 ) 3 2
æ I ç zu x = 4 p çç (1 + z 2 ) è =
0
¥
+
Idxu x ´ ( - xu x + u y)
¥
ò
4 p(1 + x )
2 32
0
¥
ò 0
Idxu z 4 p(1 + x 2 ) 3 2
uy 2 12
uzù 3 ú ´ 10 5 ´ 10 -3 ú -4 úû
= [ u x ( -8 - 36) - u y( -4 - 9) + u z (12 - 6)] ´ 10 2 V m = -4.4 u x + 1.3u y + 0.6 u z kV m 18. (D) v(0) ´ B = (2 u x - 3u y - 4 u z )10 5
ö ÷ + ÷ (1 + x 2 ) ¥ ÷ ø xu z
éu x ê1 ê êë 3
Electromagnetics
0
´ ( -3u x + 2 u y - u z )10 -3
I ( u x + u z ) = 0.8 ( u x + u z ) mA m 4p
= 1100 u x + 1400 u y - 500 u z F(0) = Q [E + v ´ B] = 2 ´ 10 -16 [1200 u x + 1200 u y - 200 u z ]
9. (A) Using Ampere’s circuital law
= 4 ´ 10 -14 [ 6 u x + 6 u y - u z ]
ò H × dL = 2prHf = I encl
F = ma
At r = 0.5 cm,
I encl = 10 mA
-3
2 p(5 ´ 10 ) H f = 10, H f = 0.32 A m 10. (B) At r = 15 . cm enclosed current I encl = 10 + 2 p(0.01)( 400) = 35.13 mA 2 p(0.015) H f = 35.13 ´ 10
-3
Þ
H f = 0.37 A m
11. (C) The enclosed current is I encl = 10 + 2 p(0.01) 400 - 2 p(0.02)250 + 2 p(0.03) 300 = 60.3 mA m 2 p(0.035) H f = 60.3 M 12. (A) H =
Þ
H f = 0.27 A m
1 ( -30 - 40) u x ´ ( -u z ) = -35 u y A m 2
Þ
a=
F 4 ´ 10 -14 = [6u x + 6u y - u z ] m 5 ´ 10 -26
= 800[ 6 u x + 6 u y - u z ] 10 9 m s 2 19. (C) F = e v ´ B = - (1.6 ´ 10 -19)( 4.5 ´ 10 7 u y)(2.5 ´ 10 -3 u z ) = -1.8 ´ 10 -14 u x N This force will be constant during the time the electron travels the field. It establishes a negative x –directed velocity as it leaves the field, given by the acceleration times the transit time tt , Ftt æ -1.8 ´ 10 -14 öæ 2.5 ´ 10 -2 ÷ç =ç m çè 9.1 ´ 10 -31 ÷øçè 4.5 ´ 10 7 0.5 - 0.025 = = 106 . ´ 10 -8 s 4.5 ´ 10 7
vx = t50
ö ÷÷ = -11 . ´ 10 7 m s ø
In that time, the electron moves to an x coordinate 1 13. (B) H = K ´ u n 2 1 = ( 30 - 40) u x ´ ( -u z ) = -5 u y A m 2
given by . ´ 10 7)(106 . ´ 10 -8 ) = -0.117 m x = vx t50 = -(11 x = -117 . cm, z = 0 C
1 14. (A) H = ( -30 + 40) u x + ( -u z ) = 5 u y A m 2 15. (C) J = Ñ ´ H = =
1 ¶ æ r2 çç 2 + 2 r ¶r è
1 ¶(rH f) uz r ¶r
ö ÷÷u z = u z A m ø
20. (A) FBC = ò I loop dL ´ Bfrom 4
= ò ( 6 ´ 10 -3) dzu z ´ 1
1 ¶ æ 3ö çr ÷ = 0 r ¶r çè r ÷ø
position along the loop segment. 3
ò ( 6 ´ 10 1
=
-3
) dxu x ´
15m o uy 2 px
45 ´ 10 -3 m o ln 3u z = 19.8 u z nN p
17. (A) F = e(E + v ´ B) If F = 0, E = -v ´ B = B ´ v Page 486
15m o u y = -1.8 ´ 10 -8 u x = -18 u x nN 2 p( 3)
21. (C) The field from the long wire now varies with
FAB = 16. (D) J =
wire at BC
B
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GATE EC BY RK Kanodia
Magnetostatics
22. (A) This will be the vector sum of the forces on the four sides. By symmetry, the forces on sides AB and CD will be equal and opposite, and so will cancel. This leaves the sum of forces on side BC and DA 4
FDA =
ò -( 6 ´ 10
-3
) dxu z ´
1
15m o u y = 54 u x nN 2 p(1)
Chap 8.3
28. (A) F =
ò K ´ BdS =
area
1 1 .5
ò ò 6u
z
´
0 0 .5
-5m o u x dy 2 py
15m o æ 15 . ö =ln ç ÷u y = -6.6 u y mN m p è 0.5 ø 29. (A) c m + 1 = m r , c m + 1 = 6.5 , c m = 5.5
Ftotal = FDA + FBC = (54 - 18) u x = 36 u x nN 30. (A) B = mH = m om rH
23. (A) F = ò IdL ´ R
= 4 p ´ 10 -7 ´ 6.5(10 u x + 25 u y - 40 u z )
2
2
3
1
1
1
1
2
= I ò dxu x ´ B + I ò dyu y ´ B + I ò dxu x ´ B + I ò dyu y ´ B
= 82 u x + 204 u y - 327 u z mWb m 2
u x ´ B = u x [ 6 xu x - 9 yu y + 3zu z ] = 3zu y - 9 yu z
31. (C) M = c m H = 55 u x + 137.5 u y - 220 u z A m
u y ´ B = u y[ 6 xu x - 9 yu y + 3zu z ] = 3zu x - 6 xu z z = 0 for all element 3
2
1
1
1
1
3
2
32. (B) W =
F = I ò dx (-9 yu z) y=1 + I ò dy(-6 xu z) x=3 + I ò dx (-9 yu z) y=2 + I ò dy(-6 xu z) x=1
=
= I ( -18 - 18 + 36 + 6) u z = 5 ´ 6 u z = 30 u z N 24. (B) Within the region -1 < y < 1, the magnetic fields from the two outer sheets (carrying -4u z A m) cancel, leaving only the field from the center sheet. Therefore
1 1 H × B = mH 2 2 2
1 ´ 6.5 ´ 4 p ´ 10 -7 (100 + 625 + 1600) = 9.5 mJ m 2 2
33. (D) m r = c m + 1 = 31 . + 1 = 4.1, m = m om r = 4.1m o H=
0.4 yu z B = = 77.6 yu z kA m m 4.1 ´ 4 p ´ 10 -7
H = -4 u x A m (0 < y < 1) and H = 4 u x A m ( -1 < y < 0). Outside (y > 1 and y < - 1) the fields from all three sheet cancel, leaving H = 0 ( y > 1, y < - 1). So at x = 0, y = 0.5 F = Iu z ´ B = (7 ´ 10 -3) u z ´ - 4m o u x = -35.2 u y nN m m
0
0
-100 m o u y 2 p (5 ´ 10 -3)
= 0.4 u x N m
m r1 =
27. (B) The field from the current strip at the filament location 3m o . ö 6m o u x æ 15 dy = ln ç ÷u x y p 0 2 p è .5 ø 0 .5
1 .5
B=
ò
= 1.32 ´ 10 -6 u x Wb m 2 1
F = ò IdL ´ B 0
1
1 1 m = ´ = 1326.3 m o 600 4 p ´ 10 -7
M1 = c m H1 = 1590 . ´ 106 A m 1.4 B For case 2, m = 2 = H 2 400
ò IdL ´ B
= ò 100 dzu z ´
2 B1 = H 1200
c m = m r - 1 = 1325.3
1
1
35. (C) For case 1, m = m r1 =
F 25. (D) = Iu x ´ ( -4m o u x ) = 0 m 26. (A) F =
34. (A) M = c m H = ( 31 . )(77.6) yu z = 241 yu z kA m
= ò 5 dzu z ´ 1.32 ´ 10 -6 u x dz = 6.6 u y mN m
1.4 = 2785.2 400 ´ 4 p ´ 10 -7
c m = 2784.2 2784.2 M 2 = c m H 2 = 1114 . ´ 106 A m DM = (1590 . - 1114 . ) ´ 106 = 476 kA m 36. (B) M = Nm = (2.7 ´ 10 29)(2.6 ´ 10 -30 u y) = 0.7 u y A m H=
0.7 u y M = = 0.22 u y A m m r - 1 4.2 - 1
0
37. (A) M = www.gatehelp.com
B mo
ö æ 1 çç + 1 ÷÷ ø è cm
-1
Page 487
GATE EC BY RK Kanodia
UNIT 8
-1
=
Electromagnetics
The normal component of H1 is
0.02 æ 1 ö + 1 ÷ = 47.6 A m ç 4 p ´ 10 -7 è 0.003 ø
H N1 = (H1 × u N 21 ) u N 21 H1 × u N 21 = (50 u x - 30 u y + 20 u z ) × (0.37 u x + 0.56 u y - 0.74u z )
= 18.5 - 16.8 - 14.8 = -131 .
1 1 38. (A) W = H × B = m o H 2 2 2 1 = ( 4 p ´ 10 -7)( 600) 2 = 0.226 J m 3 2
=
Ho
0
0
= -4.83u x - 7.24 u y + 9.66 u z A m Tangential component of H1 at the boundary
æ1
HT1 = H1 - H N1
ö
ò H. dB = ò Hçè 3 + 2 H ÷ødH
39. (A) W = 2 o
Ho
(H1 × u N 21 ) u N 21 = ( -131 . )(0.37 u x + 0.56 u y - 0.74 u z )
= (50 u x - 30 u y + 20 u z ) - ( -4.83u x - 7.24 u y + 9.66 u z ) = 54.83u x - 22.76 u y + 10.34 u z A m
3 o
2H H + = 6.2 MJ m 3 6 3
40. (A) J b = Ñ ´ M =
H T 2 = H T1 m 2 H N 2 = r1 H N1 = ( -4.83u x - 7.24 u y + 9.66 u z ) m r2 5
12 uz a
= -193 . u x - 2.90 u y + 3.86 u z A m H 2 = H T 2 + H N 2 = (54.83u x - 22.76 u y + 10.34 u z )
41. (B) B1 = 200m o u x - 600m o u y + 400m o u z
+ ( -193 . u x - 2.9 u y + 3.86 u z ) = 52.9 u x - 25.66 u y + 14.2 u z
Its normal component at the boundary is B1 N = (B1 × u N12 ) u N12
********
= (52.8 u x + 42.24 u y + 56.32 u z ) m o = B2 N B Þ H 2 N = 2 N = 6.60 u x + 5.28 u y + 7.04 u z 8m o H1 N =
B1 N = 26.40 u x + 2112 . u y + 28.16 u z 12m o
H1 T = H1 - H1 N = (100 u x - 300 u y + 200 u z ) -(26.4 u x + 2112 . u y + 28.16 u z ) = 73.6 u x - 32112 . u y + 171.84 u z H1 T = H 2 T H 2 = H 2 N + H 2 T = 80.2 u x - 315.8 u y + 178.9 u z A m 42. (C) H N1 = - 3u z , H T1 = 10 u x + 15 u y H T 2 = H T1 = 10 u x + 15 u y m 1 ( -3u z ) = 0.15 u z H N 2 = 1 H N1 = m2 20 H 2 = H N 2 + H T 2 = 10 u x + 15 u y - 0.15 u z B2 = m 2H 2 = 20 ´ 4 p ´ 10 -7(10 u x + 15 u y - 0.15 u z ) = 251u x + 377 u y - 377 . u z mWb m 2 43. (B) At the boundary normal unit vector 2 u x + 3u y - 4 u z Ñ (2 x + 3 y - 4 z) un = = |Ñ (2 x + 3 y - 4 z)| 29 = 0.37 u x + 0.56 u y - 0.74 u z Since this vector is found through the gradient, it will point
in
the
direction
of
increasing
values
of
2 x + 3 y - 4 z, and so will be directed into region 1. Thus u n = u n 21 . Page 488
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GATE EC BY RK Kanodia
UNIT 8
Statement for Q.8–9:
Electromagnetics
Statement for Q.12–13:
The location of the sliding bar in fig. P8.4.8–9 is
Consider the fig. P8.4.12–13. The rails have a
given by x = 5 t + 4 t . The separation of the two rails is
resistance of 2 W m. The bar moves to the right at a
30 cm. Let B = x u z T.
constant speed of 9 m s in a uniform magnetic field of
3
2
y
0.8 T. The bar is at x = 2 m at t = 0.
B
B
z
B
a
VM
v
0.2 cm I
v
x
b
Fig. P8.4.8–9.
8. The voltmeter reading at t = 0.5 s is
16 cm
(A) -21.6 V
(B) 21.6 V
Fig. P8.4–12–14
(C) -6.3 V
(D) 6.3 V
12. If 6 W resistor is present across the left-end with the
9. The voltmeter reading at x = 0.6 m is
right end open-circuited, then at t = 0.5 sec the current
(A) -1.68 V
(B) 1.68 V
I is
(C) -0.933 V
(D) 0.933 V
(A) -45 mA
(B) 45 mA
(C) -60 mA
(D)60 mA
Statement for Q.10–11: A perfectly conducting filament containing a 250W
13. If 6 W resistor is present across each end, then I at
resistor is formed into a square as shown in fig.
0.5 sec is
P8.4.10-11.
(A) -12.3 mA
(B) 12.3 mA
(C) -7.77 mA
(D) 77.7 mA
y I(t)
Statement for Q.14–15: 0.5 cm
B
The internal dimension of a coaxial capacitor is
250 W
a = 1.2 cm, b = 4 cm and c = 40 cm. The homogeneous material inside the capacitor has the parameter
x
e = 10 -11 F m, m = 10 -5 H m and s = 10 -5 S m.The electric
Fig. P8.4.10–11
field intensity is E = 10r cos (10 5 t) u p V m. 7
10. If B = 6 cos (120 pt - 30 ° ) u z T, then the value of I ( t) is
14. The current density J is
(A) 2.26 sin (120 pt - 30 ° ) A (B) 2.26 cos (120 pt - 30 ° ) A (C) -2.26 sin (120 pt - 30 ° ) A (D) -2.26 cos (120 pt - 30 ° ) A 11. If B = 2 cos p( ct - y) u z mT, where c is the velocity of light, then I ( t) is
200 sin (10 5 t) u r A m 2 r
(B)
400 sin (10 5 t) u r A m 2 r
(C)
100 cos (10 5 t) u r A m 2 r
(A) 1.2(cos pct - sin pct) mA
(D) None of the above
(B) 1.2(cos pct - sin pct) mA
15. The quality factor of the capacitor is
(C) 1.2(sin pct - sin pct) mA
(A) 0.1
(B) 10
(C) 0.2
(D) 20
(D) 1.2(sin pct - sin pct) mA Page 490
(A)
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GATE EC BY RK Kanodia
Maxwell’s Equations
16. The following fields exist in charge free regions
(B) -4 a sin (15 . ´ 108 t - ax) u z mA m
P = 60 sin ( wt + 10 x) u z
(C) 4 a sin (15 . ´ 108 t - ax) u z m A m
Q = 10r cos ( wt - 2r) u f
(D) 4 a sin (15 . ´ 108 t - ax) u z mA m
R = 3r 2 cot f u r + 1r cos fu f
21. The value of a is
S=
(A) 4.3
(B) 2.25
(C) 5
(D) 6
1 r
sin q sin ( wt - 6 r) u q
The possible electromagnetic fields are (A) P, Q
(B) R, S
(C) P, R
(D) Q, S
Statement for Q.22–23: Let H = 2 cos (1010 t - bx) u z A m, m = 3 ´ 10 -5 H m,
17. A parallel-plate capacitor with plate area of 5 cm 2 and plate separation of 3 mm has a voltage 50 sin (10 3 t) V
Chap 8.4
applied to its plates. If e r = 2,
the displacement
current is
e = 1.2 ´ 10 -10 F m and s = 0 everywhere. 22. The electric flux density D is (A) 120 cos (1010 t - bx) nC m 2
(A) 148 cos (1010 t) nA
(B) 261 cos (1010 t) mA
(B) -120 cos (1010 t - bx) nC m 2
(C) 261 cos (1010 t) nA
(D) 148 cos (1010 t) mA
(C) 120 cos (1010 t + bx) nC m 2 (D) None of the above
18. In a coaxial transmission line ( e r = 1), the electric 23. The magnetic flux density B is
field intensity is given by E=
(A) 6.67 ´ 10 4 cos (1010 t + bx) T
100 cos (10 9 t - 6 z) u r V m. r
(B) 6.67 ´ 10 4 cos (1010 t - bx) (C) 6 ´ 10 -5 cos (1010 t + bx) T
The displacement current density is 100 (A) sin (10 9 t - 6 z) u r A m 2 r (B)
(D) 6 ´ 10 -5 cos (1010 t - bx) T Statement for Q.24–25:
116 sin (10 9 t - 6 z) u r A m 2 r
(C) -
0.9 sin (10 9 t - 6 z) u r A m 2 r
(D) -
216 cos (10 9 t - 6 z) u r A m 2 r
A material has s = 0 and e r = 1. The magnetic field intensity is H = 4 cos (106 t - 0.01z) u y A m. 24. The electric field intensity E is (A) 4.52 sin (106 t - 0.01z) kV m (B) 4.52 sin (106 t - 0.01z) V m
Statement for Q.19–21:
(C) 4.52 cos (106 t - 0.01z) V m
Consider the region defined by |x|,|y| and |z|< 1. Let e = 5 e o , m = 4m o , and s = 0 the displacement current
(D) 4.52 cos (106 t - 0.01z) kV m
density J d = 20 cos (15 . ´ 108 t - ax) u y m A m 2 . Assume
25. The value of m r is
no DC fields are present.
(A) 2
(B) 3
(C) 4
(D) 16
19. The electric field intensity E is (A) 6 sin (15 . ´ 108 t - ax) u y mV m
26. The surface r = 3 and 10 mm, and z = 0 and 25 cm
(B) 6 cos (15 . ´ 10 t - ax) u y mV m
are perfect conductors. The region enclosed by these
(C) 3 cos (15 . ´ 108 t - ax) u y mV m
surface has m = 2.5 ´ 10 -6 H m, e = 4 ´ 10 -11 F m and
8
(D) 3 sin (15 . ´ 108 t - ax) u y mV m
s = 0. If H = 2r cos 8 pz cos wt u f A m, then the value of w is
20. The magnetic field intensity is
(A) 2 p ´ 106 rad s
(B) 8 p ´ 106 rad s
(A) -4 a sin (15 . ´ 108 t - ax) u z m A m
(C) 2 p ´ 108 rad s
(D) 8 p ´ 108 rad s
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Page 491
GATE EC BY RK Kanodia
UNIT 8
Electromagnetics
27. For distilled water m = m o , e = 81e o , and s = 2 ´ 10 -3
SOLUTIONS
S m, the ratio of conduction current density to displacement current density at 1 GHz is (A) 111 . ´ 10 -5
(B) 4.44 ´ 10 -4
(C) 2.68 ´ 10 -6
(D) 1.68 ´ 10 -7
1. (B) emf = -
= 2 p(0.2) 2 (20)( 377) sin 377 t mV = 0.95 cos 377 t V
28. A conductor with cross-sectional area of 10 cm 2 carrier
a
conductor
current
2 sin (10 9 t)
mA.
If
2. (A) emf =
s = 2.5 ´ 106 S m and e r = 4.6, the magnitude of the (A) 48.4 m A m
(B) 8.11 nA m
2
F = p(0.1) 2 B = 0.31 cos (120 pt)
(D) 16.4 m A m 2
(C) 32.6 nA m 2
emf = Vba ( t) = -
29. In a certain region
dF dt
= 0.31(120 p) sin (120 pt) = 118.43 sin(120 pt)
J = ( 4 yu x + 2 xzu y + z u z ) sin (10 t) A m 3
4
Vab = - 118.43 sin (120 pt)
If volume charge density r v in z = 0 plane is zero, 4. (D) I =
then r v is (A) 3z 2 cos (10 4 t) mC m 3 (B) 0.3z 2 cos (10 4 t) mC m 3 (C) -3z cos (10 t) mC m 2
1 1 Bo wL2 = ( 4)(2)(2) 2 = 16 V 2 2
3. (C) Since B is constant over the loop area, the flux is
displacement current density is 2
dF d B × dS =dt dt ò
4
Vab 118.43 sin (120 pt) = = 0.47 sin (120 pt) R 250
5. (A) emf = -
3
(D) -0.3z 2 cos (10 4 t) mC m 3
=
30. In a charge-free region (s = 0, e = e o e r , m = m o ) magnetic field intensity is
H = 10 cos (10 t - 4 y) u z 11
dF d =dt dt
òò B × u dz z
loop area
d ( 3)( 4)( 6) cos 5000 t = -360000 sin 5000 t dt
I=
emf 360000 sin 5000 t = - 0.4 sin 5000 t A =R 900 ´ 10 3
A m. The displacement current density is 1 1
(A) -40 sin (10 9 t - 4 y) u y A m
6. (C) F =
(B) 40 sin (10 9 t - 4 y) u y A m
ò ò 20m
o
cos ( 3 ´ 108 t - y) dx dy
0 0
= [20m o sin ( 3 ´ 108 t - y)]10
(C) -40 sin(10 3 t - 4 y) u x A m
= 20 m o [sin ( 3 ´ 108 t - 1) - sin ( 3 ´ 108 t)] Wb
(D) 40 sin (10 9 t - 4 y) u x A m 31. In a nonmagnetic medium ( e r = 6.25) the magnetic field of an EM wave is H = 6 cos bx cos (108 t) u z A m. The corresponding electric field is (A) 903 sin (0.83 x) sin (108 t) V m
Emf = -
dF dt
= - 20 ´ (4p ´ 10 -7)(3 ´ 108 ) ´ [cos (3 ´ 108 t - 1) - cos (3 ´ 108 t)]
= 7540[cos ( 3 ´ 108 t) - cos ( 3 ´ 108 t - 1)] V 7. (D) In this case F = [20m o (2 p) sin ( 3 ´ 108 t - y)]20 p = 0
8
(B) 903 sin (12 . x) sin (10 t) V m (C) 903 sin (0.83 x) cos (108 t) V m
8. (A) F =
(D) 903 sin (12 . x) cos (108 t) V m
òò B × dS =
area
E = 5 cos(10 9 t - 8 x) u x + 4 sin(10 9 t - 8 x) u z V m.
(A) 3.39
(B) 1.84
(C) 5.76
(D) 2.4 ***********
Page 492
ò ò t dtdy 2
0
0
= 0.1 x 3 = 0.1(5 t + 4 t 3) 3 Wb
32. In a nonmagnetic medium
The dielectric constant of the medium is
0 .3 x
emf = -
dF = -0.1( 3)(5 t + 4 t 3) 2 (5 + 12 t 2 ) dt
At t = 0.5 s, emf = -0.1( 3)(2.5 + 0.5) 2 (5 + 3) = -21.6 V 9. (C) At x = 0.6 m, 0.6 = 5 t + 4 t 3 At t = 0.119 s, www.gatehelp.com
emf = -0.933 V
Þ
t = 0.119 s
GATE EC BY RK Kanodia
Maxwell’s Equations
10. (A) F =
òò B × dS = 6(0.5)
2
Chap 8.4
I d = 2 prlJ d = - 2 pl(10) sin (10 5 t) = - 8 psin (10 5 t) A
cos (120 pt - 30 ° ) Wb
area
Quality factor
dF emf = = 6(0.5) 2 (120 p) sin (120 pt - 30 ° ) dt The current is
emf 6(0.5) 2 (120 p) = sin (120 pt - 30 ° ) A R 250
= 2.26 sin (120 pt - 30 ° ) A
òò B × dS = (0.5)(2) ò cos( py - pct) dy
area
0
1é æ pö ù 1 sinç pct - ÷ - sin pct ú = [ - cos pct - sin pct ] mWb p êë è 2ø û p dF emf = = c [cos pct - sin pct ] mV dt =
I ( t) =
=
Ic
8 = 0.1 80
16. (A) Ñ × P = 0, Ñ ´ P = -
emf 3 ´ 108 = [cos pct - sin pct ] mA 250 R
Q
is a possible EM field sin f 1 ¶ ( 3r 2 cot f) × Ñ ×R = ¹ 0, R is not an EM field. r ¶r r Ñ ×S =
¶(sin 2 f) 1 sin ( wt - 6 r) ¹ 0 r sin q ¶r 2
S is not an EM field. Hence (A) is correct.
= 12 . [cos pct - sin pct ] A
17. (A) D = eE = e
12. (A) The flux in the left-hand closed loop is Id = J × S =
Fl = B ´ area = (0.8)(0.2)(2 + 9t) dF L = - (0.16)(9) = -1.44 V emfl = dt
V d
Þ
increasing with time, Rl = 6 + 2[2 (2 + 9 t)]W, At t = 0.5, Rl = 32 W 1.44 Il = = - 45 mA 32
18. (C) ¶D ¶E 100 =e Jd = = eo [ - sin (10 9 t - 6 z)] 10 9 u r A m 2 ¶t ¶t r =-
current from the right loop, which is now closed. The flux in the right loop, whose area decreases with time, is Fr = (0.8)(0.2)(16 - 2 - 9 t) dF R = 1.44 V emfR = dt
0.9 sin (10 9 t - 6 z) u r A m 2 r
19. (D) D = ò J d dt + C1 =
20 ´ 10 -6 sin (15 . ´ 108 - ax) u y 15 . ´ 108
. ´ 108 t - ax) u y C m 3 = 1.33 ´ 10 -13 sin (15 C1 is set to zero since no DC fields are present. E=
Rr = 44 W
The contribution to the current from the right loop -144 Ir = = 032.7 mA 44 The total current = -32.7 - 45 = -77.7 mA 14. (C) J = s E =
dD e dV = dt d dt
eS dV 2 e o 5 ´ 10 -4 = 10 3 ´ 50 cos (10 3 t) 3 ´ 10 -3 d dt
13. (C) In this case, there will be contribution to the
At 0.5 s,
Jd =
= 148 cos (1010 t) nA
While the bar in motion, the loop resistance is
Rr = 6 + 2 (2 (14 - 9 t)),
¶Pz uy ¹ 0 ¶x
P is a possible EM field 1 ¶ Ñ × Q = 0, Ñ ´ Q = [10 cos ( wt - 2r)]u z ¹ 0 r ¶r
0 .5
11. (D) F =
Id
100 cos (10 5 t) u r A m 2 r
15. (A) Total conduction current 100 cos (10 5 t) u r A m 2 I C = òò J × dS = 2prlJ = 2 prl r = 80 pcos (10 5 t) A ¶D ¶eE 10 = sin (10 5 t) A m 2 Jd = =¶t ¶t r
D 1.33 ´ 10 -13 = sin (15 . ´ 108 - ax) u y e 5 eo
= 3 ´ 10 -3 sin (15 . ´ 108 t - ax) V m 20. (D) Ñ ´ E =
¶E y ¶x
uz = -
¶B ¶t
= -a( 3 ´ 10 -3) cos (15 . ´ 108 t - ax) u z = -
¶B ¶t
B=
a( 3 ´ 10 -3) sin (15 . ´ 108 t - ax) u z 15 . ´ 108
H=
2 ´ 10 -11 B = sin (15 . ´ 108 t - ax) u z A m m 4 ´ 4 p ´ 10 -7
. ´ 108 t - ax) u z mA m = 4 ´ 10 -6 a sin (15 21. (B) Ñ ´ H = www.gatehelp.com
¶H z u y = Jd ¶x Page 493
GATE EC BY RK Kanodia
UNIT 8
= a 2 ( 4 ´ 10 -6 ) cos (15 . ´ 108 t - ax) = J D
Electromagnetics
27.(B) At high frequency
Comparing the result a 2 4 ´ 10 -6 = 20 ´ 10 -6 , a = 5 = 2.25 22. (B) Ñ ´ H = -
=
¶H z ¶D uy = ¶x ¶t
Þ
2b cos (1010 t - bx) u y C m 2 D =1010
Þ
b=
w . ´ 10 -10 = 600 = 1010 me = 1010 3 ´ 10 -5 ´ 12 v
Þ
¶z
Jd =
E=
Ic sS
4.6 e o (10 9) 2 cos (10 9 t) 10 -3 2.5 ´ 106 ´ 10 ´ 10 -4
29. (B) Ñ × J = (0 + 0 + 3z 2 ) sin (10 4 t) = -
23. (D) B = mH = 6 ´ 10 -5 cos (1010 t - bx) u z T ¶H y
Ic = sE Þ S ¶E e ¶I c Jd = e = ¶t sS ¶t
|J d | = 32.6 nA m 2
D = -120 cos (1010 t - bx) u y nC m 2
24. (A) Ñ ´ H = -
2 ´ 10 -3 = 4.44 ´ 10 -4 2 p ´ 10 9 ´ 81 ´ e o
28. (C) J c =
¶D = 2b sin (1010 t - bx) u y ¶t
Jc sE s = = J d weE we
3z 2 cos (10 4 t) + C1 10 4
rv =
ux
8r ¶t
At z = 0, r v = 0, C1 = 0 r v = 0.3z 2 cos (10 4 t) mC m 3
¶E Ñ ´ H = 0.04 cos (10 t - 0.01z) u x = e o ¶t 6
30. (D) J d = Ñ ´ H =
0.04 sin (106 t - 0.01z) u x E= 106 e o
¶H z ux ¶y
= 40 sin (10 9 t - 4 y) u y A m
= 4.52 sin (10 t - 0.01z) u x kV m 6
31. (A) Ñ ´ H = -
¶Ex ¶H 25. (B) Ñ ´ E = uy = -m ¶z ¶t
= 6b sin (bx) cos (108 t) u y 1 E = ò 6b sin (bx) cos (108 t) u y dt e 6b sin (bx) sin (108 t) u y = e1010
-0.04(0.01) ¶H cos (106 t - 0.01z) u y = - m rm o 106 e o ¶t H=
0.04(0.01) sin (106 t - 0.013) u y (106 )(106 )m rm o e o
0.04(0.01) =4 1012 m rm o e o
Þ
mr =
(0.04)(0.01) ´ ( 3 ´ 108 ) 2 = 9 4(1012 )
H=
¶Er (16 p)( 8 p) ¶H uf = cos ( 8 pz) sin ( wt) u f = -m ¶z rew ¶t
6(0.833) sin (bx) sin (108 t) u y V m 6.25 e o ´ 108
b=
w w = er , v c
8=
10 9 ´ 108 e r 3
128 p2 cos ( 8 pz) cos ( wt) u f rew2
w=
8p me
=
8p 4 ´ 10
-11
´ 2.5 ´ 10 -6
= 8 p ´ 108 rad s Page 494
w = 10 9, b = 8, Þ
e r = 5.76
**********
This result must be equal to the given H field. Thus Þ
6.25 = 0.833
32. (C) For nonmagnetic medium m r = 1
16 p sin ( 8 pz) sin ( wt) u r rew
128 p2 2 = remw2 r
w w 108 = er = v c 3 ´ 108
= 903 sin (0.83 x) sin (108 t) u y V m
16 p ¶E = sin ( 8 pz) cos ( wt) u r = e r ¶t
Ñ ´E=
e = 6.25 e o , b = E=
¶H f 26. (D) Ñ ´ H = ur ¶z
E=
¶H z ¶E uz = e ¶y ¶t
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GATE EC BY RK Kanodia
CHAPTER
8.5 ELECTROMAGNETIC WAVE PROPAGATION
Statement for Q.1–3: A y-polarized uniform plane wave with a frequency
(A) 6.43 ´ 106 m s
(B) 2.2 ´ 10 7 m s
(C) 1.4 ´ 108 m s
(D) None of the above
of 100 MHz propagates in air in the + x direction and impinges normally on a perfectly conducting plane at
Statement for Q.5–6:
x = 0. The amplitude of incident E-field is 6 mV m. 1. The phasor H s of the incident wave in air is (A) 16 e (C) 16 e
-j
2p x 3
-j
2p x 3
uz m A m
(B) -16 e
ux m A m
-j
2p x 3
-j
2p x 3
(D) -16 e
A uniform plane wave in free space has electric field Es = (2 u z + 3u y) e - jbx V m.
uz m A m
5. The magnetic field phasor H s is
ux m A m
(B) (5.3u y - 8 u z ) e - jbx m A m
(A) ( -5.3u y - 8 u z ) e - jbx m A m (C) ( -5.3u y + 8 u z ) e - jbx m A m
2. The E-field of total wave in air is
(D) (5.3u y + 8 u z ) e - jbx m A m
æ 2p ö (A) j12 sin ç x ÷ u y mV m è 3 ø
6. The average power density in the wave is
æ 2p ö (B) - j12 sin ç x ÷ u y mV m è 3 ø
(A) 34 mW m 2
(B) 17 mW m 2
(C) 22 mW m 2
(D) 44 mW m 2
æ 2p ö (C) 12 cos ç x ÷ u y mV m è 3 ø
7. The electric field of a uniform plane wave in free
æ 2p ö (D) -12 cos ç x ÷ u y mV m è 3 ø
field phasor H s is
space is given by Es = 12 p( u y + ju z ) e - j15x . The magnetic
3. The location in air nearest to the conducting plane, where total E-field is zero, is (A) x = 15 . m
(B) x = -15 . m
(C) x = 3 m
(D) x = - 3 m
vp is
( -u z + ju y) e - j15x
(B)
12 ho
( u z + ju y) e - j15x
(C)
12 ho
( -u z - ju y) e - j15x
(D)
12 ho
( u z - ju y) e - j15x
A lossy material has m = 5m o , e = 2 e o . The phase
uniform plane wave propagating in a certain lossless material is ( 6 u y - j5 u z ) e
12 ho
Statement for Q.8–9:
4. The phasor magnetic field intensity for a 400 MHz - j18 x
(A)
A m . The phase velocity
constant is 10 rad m at 5 MHz. 8. The loss tangent is (A) 2913
(B) 1823
(C) 2468
(D) 1374
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Page 495
GATE EC BY RK Kanodia
UNIT 8
Electromagnetics
9. The attenuation constant a is
16.
(A) 4.43
(B) 9.99
m r = 1, e 2 = 1) pipe with inner and outer radii 9 mm and
(C) 5.57
(D) None of the above
12 mm carries a total current of 16 sin (106 pt) A. The
A
60
m
long
aluminium
( s = 35 . ´ 10 7 S m,
effective resistance of the pipe is Statement for Q.10–11: At
50
MHz
a
lossy
dielectric
material
is
characterized by m = 2.1m o , e = 3.6 e o and s = 0.08 S m.
(A) 0.19 W
(B) 3.48 W
(C) 1.46 W
(D) 2.43 W
The electric field is Es = 6 e - jgx u z V m.
17. Silver plated brass wave guide is operating at 12
10. The propagation constant g is
m r = e r = 1) is 5d, the minimum thickness required for
(A) 7.43 + j2.46 per meter
wave-guide is
(B) 2.46 + j7.43 per meter
(A) 6.41 mm
(B) 3.86 mm
(C) 6.13 + j5.41 per meter
(C) 5.21 mm
(D) 2.94 mm
GHz. If at least the thickness of silver ( s = 6.1 ´ 10 7 S m,
(D) 5.41 + j 6.13 per meter
Statement for Q.18–19:
11. The impedance h is
A uniform plane wave in a lossy nonmagnetic
(A) 101.4 W
(B) 167.4 W
(C) 98.3 W
(D) 67.3 W
media has Es = (5 u x + 12 u y) e - gz , g = 0.2 + j 3.4 m -1
Statement for Q.12–13: A
non
magnetic
18. The magnitude of the wave at z = 4 m and t = T 8 medium
has
an
intrinsic
impedance 360 Ð30 ° W.
is (A) 10.34
(B) 5.66
(C) 4.36
(D) 12.60
12. The loss tangent is (A) 0.866
(B) 0.5
19. The loss suffered by the wave in the interval
(C) 1.732
(D) 0.577
0 < z < 3 m is
13. The Dielectric constant is (A) 1.634
(B) 1.234
(C) 0.936
(D) 0.548
(A) 4.12 dB
(B) 8.24 dB
(C) 10.42 dB
(D) 5.21 dB
Statement for Q.20–22: The plane wave E = 42 cos ( wt - z) u x V m in air
Statement for Q.14–15:
normally hits a lossless medium (m r = 1, e r = 4) at z = 0.
The amplitude of a wave traveling through a lossy nonmagnetic medium reduces by 18% every meter. The
20. The SWR s is
wave operates at 10 MHz and the electric field leads the
(A) 2
magnetic field by 24°.
(C)
14. The propagation constant is (A) 0.198 + j0.448 per meter (B) 0.346 + j0.713 per meter (C) 0.448 + j0.198 per meter
1 3
(D) 3
22. The reflected electric field is
15. The skin depth is (A) 2.52 m
(B) 5.05 m
(C) 8.46 m
(D) 4.23 m
Page 496
(D) None of the above
21. The transmission coefficient t is 2 4 (A) (B) 3 3 (C)
(D) 0.713 + j0.346 per meter
1 2
(B) 1
(A) -14 cos ( wt - z) u x V m (B) -14 cos ( wt + z) u x V m www.gatehelp.com
GATE EC BY RK Kanodia
UNIT 8
Electromagnetics
33. The region z < 0 is characterized by e r = m r = 1 and
Region 2 (0 < z < 6 cm): m 2 = 2 mH m, e 2 = 25 pF m
s = 0.
Region 3 (z > 6 cm):
The
total
electric
field
E s = 150 e - j10 z u x + 50 Ð20 ° e j10 z u x
here
is
given
39. The lowest frequency, at which a uniform plane
impedance of the region z > 0 is (A) 692 + j176 W (C) 176 + j 692 W
(B) 193 - j 49 W
wave incident from region 1 onto the boundary at z = 0
(D) 49 - j193 W
will have no reflection, is
Statement for Q.34–35: Region 1, z < 0 and region 2, z > 0, are both perfect dielectrics. A uniform plane wave traveling in the u z direction
has
a
frequency
m 3 = 4 mH m, e 3 = 10 pF m
V m. The intrinsic
of
3 ´ 1010
rad s.
Its
wavelength in the two region are l1 = 5 cm and l2 = 3 cm.
(A) 2.96 GHz
(B) 4.38 GHz
(C) 1.18 GHz
(D) 590 MHz
40. If frequency is 50 MHz, the SWR in region 1 is (A) 0.64
(B) 1.27
(C) 2.38
(D) 4.16
41. A uniform plane wave in air is normally incident
34. On the boundary the reflected energy is
onto a lossless dielectric plate of thickness l 8 , and of
(A) 6.25%
(B) 12.5%
intrinsic impedance h = 260 W. The SWR in front of the
(C) 25%
(D) 50%
plate is
35. The SWR is (A) 1.67
(B) 0.6
(C) 2
(D) 1.16
(A) 1.12
(B) 1.34
(C) 1.70
(D) 1.93
42. The E-field of a uniform plane wave propagating in a dielectric medium is given by
36. A uniform plane wave is incident from region 1 (m r = 1, s = 0) to free space. If the amplitude of incident
z ö z ö æ æ 8 E = 2 cos ç 108 t ÷ u x - sin ç 10 t ÷ uy V m 3ø 3ø è è
wave is one-half that of reflected wave in region, then the value of e r is (A) 4
(B) 3
(C) 16
(D) 9
37. A 150 MHz uniform plane wave is normally incident from air onto a material. Measurements yield a SWR of 3 and the appearance of an electric field minimum at 0.3l in front of the interface. The impedance of material is (A) 502 - j 641 W
(B) 641 - j502 W
(C) 641 + j502 W
(D) 502 + j 641 W
The dielectric constant of medium is (A) 3
(B) 9
(C) 6
(D)
43. An electromagnetic wave from an under water source with perpendicular polarization is incident on a water-air interface at angle 20° with normal to surface. For water assume e r = 81, m r = 1. The critical angle q c is (A) 83.62°
(B) 6.38°
(C) 42.6°
(D) None of the above ***********
38. A plane wave is normally incident from air onto a semi-infinite slab of perfect dielectric ( e r = 3.45). The fraction of transmitted power is (A) 0.91
(B) 0.3
(C) 0.7
(D) 0.49
Statement for Q.39–40: Consider three lossless region : Region 1 (z < 0): Page 498
6
m 1 = 4 mH m, e1 = 10 pF m www.gatehelp.com
GATE EC BY RK Kanodia
Electromagnetic Wave Propagation
SOLUTIONS 1. (A) w = 2 p ´ 10
8
8. (B) Loss tangent
rad s
w 2 p ´ 10 2p = = rad m c 3 ´ 108 3
Es = 6 e
2p -j x 3
u y mV m
uE ´ u H = ux , Hs =
6 e 120 p
-j
2p x 3
u y ´ u H = ux , uz
= 16 e
-j
2p x 3
u H = uz
Þ
10 =
Þ
x=
uz m A m
Er = -6 e
j
2p x 3
5´2 2
[ 1+ x
s = 1823 we 1 + x2 - 1 1 + x2 + 1
2 ù me é æ s ö ê 1+ç ÷ - 1ú 2 ê è we ø úû ë
æ 2p ö = - j12 sin ç x ÷ u y mV m è 3 ø
10. (D) a = w
3. (B) The electric field vanish at the surface of the
s 0.08 = =8 we 3.6 ´ 50 ´ 106 ´ 2 pe o
conducting plane at x = 0. In air the first null occur at 3 l p x =- 1 ==- m 2 2 b1 w 2 p ´ 400 ´ 10 = = 1.4 ´ 108 m s b 18
5. (C) The wave is propagating in forward x direction.
2 p ´ 50 ´ 106 3 ´ 108
=
(2.1)( 3.6) ( 65 - 1) = 5.41 2
2 ù me é æ s ö ê 1+ç ÷ + 1ú 2 ê è we ø úû ë
2 p ´ 50 ´ 106 3 ´ 108
(2.1)( 3.6) ( 65 + 1) = 6.13 2
g = a + jb = 5.41 + j 6.13 per meter.
Therefore u E ´ u H = u x . Þ
uH = - uy
For u E = u y , u y ´ u H = u x Þ u H = u z 1 Hs = ( -2 u y + 3u z ) e - jbx = ( -5.3u y + 8 u z ) e - jbx mA m 120 p 6. (B) Pavg =
a=
b=w
6
For u E = u z , u z ´ u H = u x
]
+1
a = 10 ´ 0.999 = 9.99
2p -j x æ - j 2 px ö E = Ei + Er = çç 6 e 3 u y - 6 e 3 u y ÷÷ mV m è ø
4. (C) vp =
2
a 1822 = b 1824
Þ
u y mV m,
2 p ´ 5 ´ 106 3 ´ 108
a = b
9. (B)
2. (B) For conducting plane G = -1,
s =x we
2 ù me é æ s ö ê 1+ç ÷ + 1ú 2 ê è we ø úû ë
b=w
8
b=
Chap 8.5
11. (A) |h| =
1 Re {Es ´ H *s } 2 12. (C)
1 {(5.3) u x + 3( 8) u x } ´ 10 -3 = 17.3u x mW m 2 2 7. (D) Since Pointing vector is in the positive x direction, therefore u E ´ u H = u x . For u E = u y , u y ´ u H = u x
Þ
u H = uz
For u E = u z , u z ´ u H = u x 12 Hs = ( u z - ju y) e - j 15x ho
Þ
u H = -u y ,
m e æ ç 1 + æç s ö÷ ç è we ø è
ö ÷ ÷ ø
=
1 4
2.1 3.6 = 101.4 1
64 4
s = tan 2 q n = tan 60 ° = 1732 . we
13. (D) |h| =
Þ
2
120 p
360 =
m e
1
2 4 æ ö ç 1 + æç s ö÷ ÷ ç è we ø ÷ø è 120p er Þ 1
e r = 0.548
2 4 (1 + 1732 . )
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Page 499
GATE EC BY RK Kanodia
UNIT 8
14. (A) |E| = Eo e - az
ho - ho 1 h2 - h1 G= = 2 =ho 3 h2 + h1 + ho 2 1 1 + |G| 1 + 3 s= = =2 1 - |G| 1 - 1 3
Eo e - a1 = (1 - 0.18) Eo e - a1 = 0.82 q n = 24 °
Þ Þ
1 = 0.198 0.82 s tan 2 q n = = 1111 . we a = ln
2
æ s ö 1+ç ÷ -1 è we ø
a = b
h 2 o 2h2 2 =2 21. (A) t = = h2 + h1 ho + 2 3 2
2
æ s ö 1+ç ÷ +1 è we ø
0.198 = b
234 - 1
Þ
234 + 1
b = 0.448
22. (A) Eor = GEoi = -
g = a + jb = 0.198 + j0.448 15. (B) d =
d=
1 pfsm
Rac =
=
Þ
23. (C) h1 = ho , h2 =
1 p ´ 5 ´ 10 ´ 35 . ´ 10 7 ´ m o
= 120 mm
l sdw
Since d is very small, w = 2pr outer 60 = 0.19 W Rac = 7 . ´ 10 ´ 120 ´ 10 -6 ´ 2 p ´ 12 ´ 10 -3 35
=
5 p ´ 12 ´ 10 ´ m o ´ 6.1 ´ 10 7
= 2.94 mm
h1 = ho
18. (B) E = Re{E s e jwt } = (5 u x + 12 u y) e -0 .2 z cos ( wt - 3.4 z) T 8
Þ
377 h - 377 = 2 3000 h 2 + 377 Þ
e r = 12.5,
Þ
æp ö |E|= 13e -0 .8 cos ç - 13.6 ÷ = 5.66 è4 ø
G=
19. (D) Loss = aDz = 0.2 ´ 3 = 0.6 Np
Eor = -
20. (A) h1 = ho , h2 = ho
Page 500
mr h = o 2 er
æ h - h1 ö 18 ÷÷ h1 = çç 2 -3 è h2 + h1 ø 6 ´ 10
h2 = 485.37 = ho
mr er
m r = 20.75
25. (A) h1 = ho , h2 = ho
0.6 Np = 5. 21 dB.
Þ
æ h - ho ö ÷÷3000 ho = çç 2 è h2 + ho ø
æp ö E = (5 u x + 12 u y) e -0 .8 cos ç - 13.6 ÷ è4 ø
1 Np = 8.686 DB,
mr mr = ho 12.5 er
Eor h - h1 =G = 2 Eoi h2 + h1 æ h - h1 ö ÷÷ Eoi h1 H or = çç 2 è h2 + h1 ø
pfms
At z = 4 m, t =
24. (D) h1 = ho , h2 = ho
But Eor = h1 H or = GEoi
5
9
m ho ho = = 2 e er
ho - ho 1 h2 - h1 G= = 2 =ho 3 h2 + h1 1 + ho 2
f = 5 ´ 10 5 Hz, 5
17. (D) t = 5 d =
1 ( 42) = -14 3
Er = - 14 cos ( wt - z) u x V m
1 1 = = 5.05 a 0.198
16. (A) w = p106
Electromagnetics
m r ho = 2 er
1 h2 - h1 =3 h2 + h1
1 10 (10) = 3 3 10 Eor H or = = = 8.8 ´ 10 -3 ho 3 ´ 377 uE ´ u H = uk ,
-u y ´ u H = -u z
H r = -8.8 cos ( wt - z) u x mA m w w b =1 = = m r er v c
www.gatehelp.com
Þ
u H = -u x
GATE EC BY RK Kanodia
Electromagnetic Wave Propagation
w=
3 ´ 108
= 0.5 ´ 108 rad s.
12 ´ 3
26. (D) h1 = ho , h2 = ho
33.(A) G = G=
m r ho = = 0.58ho er 3
h - h1 0.58ho - ho G= 2 = = -0. 266 h2 + h1 0.58h o + h o
27. (B) ETotal = Ei + Er ,
Eor = GEoi = -2.66
ho G=
Þ
1 sin 45 ° = 0.333 4.5
sin q t1 =
q t1 = 19.47 °
q t2 = sin -1 0.47 = 28 °
But cos q t =
Þ
G=
h1 h cos q B = o cos 58.2 ° = 2.6 cos 58.2 ° ho h2 2.6
q t = 31.8 ° mr h = o = 0.447ho er 5
Ei Er
=
ho -
Þ
h - h1 G= 2 = - 0.38, t = 1 + G = 0.62 h2 + h1
G=
Þ Þ
h2 - h1 = h2 + h1
m r2 m r1 + ho er 2 e r1
ho
=
m r 2 1 + 0.447 = = 0.382, 2.62 m r1 1 ± 0.447 e r1 æ m r 2 =ç e r 2 çè m r1
3
ö ÷÷ = 0.056, 17.9 ø
m r1 m r2 m r32 m r31 m r2 m r1 + m r32 m r31
2
e r1 -1 er 2 e r1 er 2
l2 -1 l = 1 l2 +1 +1 l1
mr h = o er er
1 h2 - h1 = 2 h2 + h1 ho
er 1 = ho 2 ho + er
Þ
37. (C) At minimum
G = ± 0.447
m r2 m r1 - ho er 2 e r1
ho
ö ÷÷ ø
1+
Et = tEi = 92.7 cos ( wt - 8 y) u z V m 32. (B) |G| = 0.2 ,
e r1 æ l2 =ç e r 2 çè l1
1 l2 - l1 3 - 5 = =4 l2 + l1 3 + 5
36. (D) h2 = ho , h1 = ho
G=
2
Þ
1 4 = 5 35. (A) s = = 1 1 - |G| 3 14
2.6 e o = 2.6 eo
31. (A) h1 = ho , h2 = ho
ho
er 2 e r1 h2 - h1 = = ho ho h2 + h1 + er 2 e r1
1 + |G|
30. (A) Since both media are non magnetic e1 = e2
Þ
-
2
The fraction of the incident energy that is reflected is 1 = 6.25%. G2 = 16
e 4.5 29. (B) sin q t 2 = 1 sin q t1 = (0.333) = 0.47 e2 2.25
tan q B =
ö ÷ ÷ = 692 + j176 W ÷ ÷ ø
e j 20 3 e j 20 3
2
28. (B) m o = m 1 = m 2 eo sin q i e1
æ ç1 + ö ÷÷ = 377ç ç ø ç1è
æ 2 pc ö æ 2 pc ö 34. (A) e r1 = çç ÷÷ , e r 2 = çç ÷÷ è l1 w ø è l2 w ø
ETotal = 10 cos ( wt - z) u y - 2.66 cos ( wt + z) u y V m
Þ
h2 - h1 , h1 = ho , h2 + h1
æ1 + G h2 = ho çç è1-G
Et = 7.34 cos ( wt - z) u y V m
Þ
Er 50 Ð20 ° e j 20 = = Ei 150 3
Eot = tEoi = 7.34
t = 1 + G = 0.734,
sin q t1 =
Chap 8.5
er = 9
( f + p) = 0.3l , 2b
2p Þ f = 0.2 p l s -1 3 -1 1 = = |G| = s+1 3+1 2 h - ho G = 0.5 e j 0 .2 p = 2 h2 + ho b=
=
m r1 - m r 2 m r1 + m r 2
Þ
æ 1 + 0.5 e j 0 .2 p ö ÷ = 641 + j502 W h2 = ho çç j 0 .2 p ÷ è 1 - 0.5 e ø
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GATE EC BY RK Kanodia
UNIT 8
ho 3.45 ho 3.45
- ho + ho
= -0.3
43. (B) q c = sin -1
39. (C) This frequency gives the condition b2 d = p Where d = 6 cm, b2 = w m 2 e 2
Þ
f =
p 0.06 1
2 ´ 0.06 2 ´ 10 -6 ´ 25 ´ 10 -12
= 118 . GHz
40. (B) At 50 MHz, b2 = w m 2 e2 = 2 p ´ 50 ´ 10 6 2 ´ 10 -6 ´ 25 ´ 10 -12 = 2. 2 b2 d = 2.22(0.06) = 0.133 h1 =
m1 4 ´ 10 -6 = = 632 W e1 10 -11
h3 = 632 W h2 =
m2 2 ´ 10 -6 = = 283 W e2 25 ´ 10 -12
The input impedance at the first interface is æ h + jh2 tan (b2 d) ö æ 632 + j283(0.134) ö ÷÷ = 283çç hin = h2 çç 3 ÷÷ è 283 + j 632(0.134) ø è h2 + jh3 tan (b2 d) ø = 590 - j138 h - h1 590 - j138 - 632 G = in = = 0.12 Ð - 100.5° h in + h1 590 - j138 + 632 s=
1 + |G| 1 - |G|
=
1 + 0.12 = 1.27 1 - 0.12
41. (C) bd =
2p l p p × = , tan =1 l 8 4 4
h2 = 260, h1 = h3 = ho æ h + jh2 tan (b2 d) ö æ 377 + j260 ö ÷÷ = 260çç hin = h2 çç 3 ÷÷ + j tan ( d ) h h b è 260 + j 377 ø ø è 2 3 2 = 243 - j92 W h - ho 243 - j92 - 377 G = in = = 0.26 Ð - 137 ° hin + ho 243 - j92 + 377 s=
1 + |G| 1 - |G|
=
126 . = 170 . 0.74
42. (A) w = 108 rad s, b = Page 502
er
=
1
3
Þ
e r = 3.
er 2 = sin -1 e r1
1 = 6.38 ° 81
******** 2
w m 2 e2 =
108
Þ
The transmitted fraction is 1 - |G| = 1 - 0.09 = 0.91.
Þ
3 ´ 108
mr ho = er 3.45
38. (A) h1 = ho , h2 = ho h - h1 G= 2 = h2 + h1
Electromagnetics
1 3
rad m, v =
c er
=
w b www.gatehelp.com
GATE EC BY RK Kanodia
CHAPTER
8.7 WAVEGUIDES
Statement for Q.1–3:
Statement for Q.6–7:
A 2 cm by 3 cm rectangular waveguide is filled
In an air-filled rectangular waveguide the cutoff
with a dielectric material with e r = 6. The waveguide is
frequencies for TM11 and TE03 modes are both equal to
operating at 20 GHz with TM11 mode.
12 GHz.
1. The cutoff frequency is
6. The dominant mode is
(A) 3.68 GHz
(B) 22.09 GHz
(A) TM10
(B) TM 01
(C) 9.02 GHz
(D) 16.04 GHz
(C) TE01
(D) TE10
2. The phase constant is (A) 816 rad m
(B) 412 rad m
(C) 1009 rad m
(D) 168 rad m
7. At dominant mode the cutoff frequency is (A) 11.4 GHz
(B) 4 GHz
(C) 5 GHz
(D) 8 GHz
3. The phase velocity is (A) 1.24 ´ 108 m s
(B) 154 . ´ 106 m s
(C) 305 . ´ 10 m s
(D) 7.48 ´ 10 m s
8
8. For an air-filled rectangular waveguide given that
8
4. In an an-filled rectangular wave guide, the cutoff frequency of a TE10 mode is 5 GHz where as that of TE01
æ 2 px ö æ 3py ö 12 Ez = 10 sin ç ÷ sin ç ÷ cos (10 t - bz) V m è a ø è b ø If the waveguide has cross-sectional dimension
mode is 12 GHz. The dimensions of the guide is
a = 6 cm and b = 3 cm, then the intrinsic impedance of
(A) 3 cm by 1.25 cm
(B) 1.25 cm by 3 cm
this mode is
(C) 6 cm by 2.5 cm
(D) 2.5 cm by 6 cm
(A) 373.2 W
(B) 378.9 W
(C) 375.1 W
(D) 380.0 W
5. Consider a 150 m long air-filled hollow rectangular waveguide with cutoff frequency 6.5 GHz. If a short pulse of 7.2 GHz is introduced into the input end of the guide, the time taken by the pulse to return the input end is
Statement for Q.9–10: In an air-filled waveguide, a TE mode operating at 6 GHz has
(A) 920 ns
(B) 460 ns
(C) 230 ns
(D) 430 ns
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æ 2 px ö æ py ö E y = 15 sin ç ÷ cos ç ÷ sin ( wt - 12 z) V m è a ø è b ø Page 511
GATE EC BY RK Kanodia
UNIT 8
9. The cutoff frequency is
Electromagnetics
16. The cross section of a waveguide is shown in fig.
(A) 4.189 GHz
(B) 5.973 GHz
P8.7.16. It has dielectric discontinuity as shown in fig.
(C) 8.438 GHz
(D) 7.946 GHz
P8.7.16. If the guide operate at 8 GHz in the dominant mode, the standing wave ratio is y
10. The intrinsic impedance is (A) 35.72 W
(B) 3978 W
(C) 1989 W
(D) 71.44 W
2.5 cm mo, eo
Statement for Q.11–12.
x
mo, 2.25eo z
5 cm
Fig. P8.7.16
Consider an air-filled rectangular wave guide with cm. The y-component of the a = 2.286 cm and b = 1016 .
(A) -3.911
(B) 2.468
TE mode is
(C) 1.564
(D) 4.389
æ 2 px ö æ 3py ö 10 E y = 12 sin ç ÷ cos ç ÷ sin (10 p ´ 10 t - bz) V m è a ø è b ø
Statement for Q.17–19: Consider the rectangular cavity as shown in fig.
11. The propagation constant g is
P8.7.17–19.
(A) j4094.2
(B) j400.7
(C) j2733.3
(D) j276.4
y
z
12. The intrinsic impedance is (A) 743 W
(B) 168 W
(C) 986 W
(D) 144 W
b
x
Consider a air-filled waveguide operating in the TE12 mode at a frequency 20% higher than the cutoff frequency. 13. The phase velocity is (A) 1.66 ´ 108 m s
(B) 5.42 ´ 108 m s
(C) 2.46 ´ 108 m s
(D) 9.43 ´ 108 m s
(A) 1.66 ´ 108 m s
(B) 4.42 ´ 108 m s
(C) 2.46 ´ 108 m s
(D) 9.43 ´ 108 m s
rectangular
waveguide
is
filled
17. If a < b < c, the dominant mode is (A) TE011
(B) TM110
(C) TE101
(D) TM101
18. If a > b > c, then the dominant mode is
14. The group velocity is
A
0
Fig. P8.7.17–19
Statement for Q.13–14:
15.
a
c
(A) TE011
(B) TM110
(C) TE101
(D) TM101
19. If a = c > b, then the dominant mode is
with
(A) TE011
(B) TM110
(C) TE101
(D) TM101
a
polyethylene ( e r = 2.25) and operates at 24 GHz. The
20. The air filled cavity resonator has dimension a = 3
cutoff frequency of a certain mode is 16 GHz. The
cm, b = 2 cm, c = 4 cm. The resonant frequency for the
intrinsic impedance of this mode is
TM110 mode is
(A) 2248 W
(B) 337.2 W
(A) 5 GHz
(B) 6.4 GHz
(C) 421.4 W
(D) 632.2 W
(C) 16.2 GHz
(D) 9 GHz
Page 512
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GATE EC BY RK Kanodia
UNIT 8
Electromagnetics
SOLUTIONS
frequency, the TM1 mode propagates through the guide without suffering any reflective loss at the dielectric interface. This frequency is
1. (A) fc = er2 = 2.1
er1 = 4 Incident wave
1 cm z
=
Fig. P8.7.34
(A) 8.6 GHz
(B) 12.8 GHz
(C) 4.3 GHz
(D) 7.5 GHz
2
2
2
2
v æmö æ nö ç ÷ +ç ÷ 2 è aø è bø
3 ´ 108 2 6 ´ 10 -2
æ1ö æ1ö ç ÷ + ç ÷ = 3.68 GHz è2 ø è 3ø 2
æf ö æf ö w 2. (C) bp = b 1 - çç c ÷÷ = 1 - çç c ÷÷ f v è f ø è ø
Statement for Q.35–36:
2
2
A 6 cm ´ 4 cm rectangular wave guide is filled with
Þ
bp =
2 p ´ 20 ´ 10 9 6 æ 3.68 ö 1 -ç ÷ = 1009 rad m 3 ´ 108 è 20 ø
dielectric of refractive index 1.25. 35. The range of frequencies over which single mode
3. (A) vp =
w 2 p ´ 20 ´ 10 9 = = 1.24 ´ 108 m s bp 1009
operation will occur is (A) 2.24 GHz < f < 3.33 GHz
4. (A) For TE10 mode fc =
(B) 2 GHz < f < 3 GHz (C) 4.48 GHz < f > a b c
if a < b < c, then
fr1 =
2
2
377 = 251.33 W . 15 251.33 æ 16 ö 1 -ç ÷ è 24 ø
2
v æ1ö æ1ö ç ÷ +ç ÷ 2 èaø è bø
2
The lowest TE mode is TE011 with
= 337.2 W
16. (C) Since a > b, the dominant mode is TE10 . In free space fc =
2
For TE mode to z
= 5. 42 ´ 108 m s
2
er
2
v æmö æ nö æ pö ç ÷ +ç ÷ +ç ÷ 2 è aø b è ø ècø
The lowest TM mode is TM110 with
3 ´ 10
=
= 259.23 W
p = 0, 1, 2 ......
æ f ö æf ö 14. (A) v g = v 1 - çç c ÷÷ = c 1 - çç c ÷÷ = 1.66 ´ 108 m s . fc ø è f ø è 12 377
2
n = 1, 2, 3...... ,
2
fr 2 =
15. (B) h =
æ2 ö 1 -ç ÷ è8ø
m = 1, 2, 3.....,
8
=
2
251.33
G=
2
v
= 2 GHz
2 ´ 0.05 2.25
m = 1, 2, 3...... ,
13. (A) v = c, f = 1.2 fc vp =
= 406.7 W
2
where for TM mode to z
2 p ´ 50 ´ 10 9 = 3 ´ 108
377
ho
æ 3ö 1 -ç ÷ è8ø
2
= 400.7 m -1 , g = jbp = j 400.7 12. (C) hTE =
2 a er
=
17. (A) fr =
10 p ´ 1010 = 50 GHz 2p
æf ö w 1 - çç c ÷÷ bp = v è f ø
c
=
2
= 46.2 GHz f =
æf ö 1 - çç c ÷÷ è f ø
2
c 3 ´ 108 = = 3 GHz 2 a 2 ´ 0.05
v æ1ö æ1ö ç ÷ +ç ÷ 2 è bø è cø
2
fr 2 > fr1 , Hence the dominant mode is TE011 18. (B) If a > b > c then
1 1 1 < < a b c
The lowest TM mode is TM110 with 2
fr1 =
v æ1ö æ1ö ç ÷ +ç ÷ 2 èaø è bø
2
The lowest TE mode is TE101 with 2
fr 2 =
v æ1ö æ1ö ç ÷ +ç ÷ 2 èaø è cø
fr 2 > fr1 www.gatehelp.com
2
Hence the dominant mode is TM110 . Page 515
GATE EC BY RK Kanodia
UNIT 8
1 1 1 = < a c b
19. (C) If a = c > b, then
The lowest TM mode is TM110 with 2
fr1 =
v æ1ö æ1ö ç ÷ +ç ÷ 2 èaø è bø
fr 2 =
v æ1ö æ1ö ç ÷ +ç ÷ 2 èaø è cø
2
2
2
v æmö æ nö æ pö ç ÷ +ç ÷ +ç ÷ 2 è aø è bø ècø 3 ´ 108 2 2 a
m<
mc
Þ
=
2 b er
f >
,
Þ
a = 10.6 cm
2 ´ 3 ´ 108 2 ´ 0.01 e r
mc 2 b er
Þ
15 . ´ 108 a
12 . ´ 15 . ´ 108 a
Þ
3 ´ 10 9 <
9
Þ
er = 9
m<
15 . ´ 108 , b
0.8 ´ 15 . ´ 108 b
b < 4 cm, Thus (C) is correct option.
27. (C) fc =
= 10 ´ 10
Þ
c 3 ´ 108 = = 2.3 GHz 2 a 2 ´ 0.065
c
vp =
æf ö 1 - çç c ÷÷ è f ø
2
=
3 ´ 108 æ 2.3 ö 1 -ç ÷ è 3 ø
2
= 4.7 ´ 108 m s
28. (B) For TE10 mode
2 fb e r c
Rs =
m < 316 .
p ´ 9 ´ 10 9 ´ 4 p ´ 10 -7 = 0.0568 . ´ 10 7 11
1 pfm = = sc d sc
2 2 æ æ 2 b æ fc ö ö÷ 2 ´ 15 . æ 3.876 ö ö÷ Rs ç 1 + çç ÷÷ 0.0568ç 1 + ç ÷ ç ç a è f ø ÷ 2.4 è 9 ø ÷ø ø= è ac = è 2 2 æf ö æ 3.876 ö 15 . ´ 10 -2 ´ 233.8 1 - ç bh 1 - çç c ÷÷ ÷ è 9 ø è f ø
= 0.022 ,
fc 2 = 2 fc1 = 15 GHz 29. (B)
f =
c 3 ´ 108 = = 20 GHz l 0.015
vg2
æf ö æ 15 ö 8 = c 1 - çç c ÷÷ = 3 ´ 108 1 - ç ÷ = 2 ´ 10 m s è 20 ø è f ø
2
2
s
2
Page 516
3 ´ 10 9 >
m 2 + k2 n2
a > 6 cm
Þ
Thus total 7 modes.
25. (A) fc =
Þ
3 GHz < 0.8 fc 01
The maximum allowed m is 3. The propagating mode
2 b er
2
c æmö 15 . ´ 108 æ nö ç ÷ +ç ÷ = 2 è aø a è bø
The next higher mode is TE01 , fc 01 =
will be TM1 , TE1 , TM 2 , TE2 , TM 3 , TE3 and TEM
24. (B) fcm =
7.21
2
2 ´ 30 ´ 10 9 ´ 0.01 2.5 3 ´ 108
mc
6.15
3 GHz > 12 . fc
23. (A) For a propagating mode f > fcm , fcm =
6.8
2
2
2 b er
14.4
Dominant mode is TE10 , fc10 =
21. (A) m = n = 1, p = 0, a = b = c, fr = 2 Ghz,
22. (A) fc =
lc (cm)
2
v æmö æ nö æ pö ç ÷ +ç ÷ +ç ÷ 2 è aø è bø ècø
mc
TE20
fcmn =
2
2 ´ 10 9 =
TE11
26. (C) Let a = kb , 1 < k < 2
3 ´ 108 æ 1 ö æ1ö ç ÷ + ç ÷ = 9 GHz 2 ´ 0.01 è 3 ø è2 ø
fr =
TE01
2
2
=
TE10
l > lc . Hence TE10 mode can be used.
fr2 < fr1 Hence the dominant mode is TE101 . 20. (D) fr =
Mode
2
The lowest TE mode is TE101 with 2
Electromagnetics
sd 10 -15 10 -15 = = we 2 p ´ 9 ´ 10 9 ´ 2.6 ´ 8.85 ´ 10 -12 1.3
sd 3.
45. (A) Let the required point be P ( x, y). Then,
f ( x) is increasing in ] -¥, 2 [ È ] 3, ¥ [.
perpendicular distance of P ( x, y) from y - 3 x + 3 = 0 is
37. (B) f ¢ ( x) =
p=
( x + 1) - 2 x 1- x = 2 ( x 2 + 1) 2 ( x + 1) 2 2
2
2
Clearly, ( x 2 + 1) 2 > 0 for all x. So, f ¢ ( x) > 0 Þ
Þ
=
y - 3x + 3 10
x2 + 4 x + 5
(1 - x 2 ) > 0
10
= =
x2 + 7 x + 2 - 3 x + 3 10 ( x + 2) 2 + 1 10
or p =
( x + 2) 2 + 1 10
dp 2 ( x + 2) d p 2 and = = 2 dx dx 10 10 2
(1 - x) (1 + x) > 0
So,
This happens when -1 < x < 1.
æ d2 p ö ÷ > 0. x = -2, Also, çç 2 ÷ è dx øx = -2
So, f ( x) is increasing in ] -1, 1 [.
dp =0 dx
38. (A) f ¢ ( x) = 4 x 3 - 4 x = 4 x( x - 1)( x + 1).
So, x = -2 is a point of minima.
Clearly, f ¢ ( x) < 0 when x < - 1 and also when x > 1.
When x = -2, we get y = ( -2) 2 + 7 ´ ( -2) + 2 = -8.
Sol. f ( x) is decreasing in ] -¥, -1 [ È ] 1, ¥ [.
The required point is ( -2, - 8).
39.(C) f ¢ ( x) = 9 x8 + 21 x6 > 0 for all non-zero real values
46. (C) Let A (0, c) be the given point and P ( x, y) be any
of x.
point on y = x 2 .
Þ
D = x 2 + ( y - c) 2 is shortest when E = x 2 + ( y - c) 2 is 40. (C) f ¢ ( x) = 3kx - 18 x + 9 = 3 [ kx - 6 x + 3]
shortest.
This is positive when k > 0 and 36 - 12 k < 0 or k > 3.
Now,
2
2
41. (A) f ( x) = ( e ax + e - ax ) = 2 cosh ax. f ¢( x) = 2 a sinh ax < 0 When x > 0 because a < 0 42. (D) f ¢( x) = - x 2 e - x + 2 xe - x = e - x x(2 - x). Clearly, f ¢( x) > 0 when x > 0 and x < 2. 43. (B) f ¢( x) = (2 x + a) 1 < x 0. ê dx 2 ú ë û x = cos x
p é pù in ê0, ú. 4 ë 2û
dz =0 dx
Þ
Sign of
dz changes from -ve to +ve when x passes dx
x=
p f= . 2
dz = - sin 2 f + 8 cos f Þ df
2
Þ
x2 y2 + = 1 which is an 5 4
z = 5 cos 2 f + 4 (1 + sin f) 2 dz Þ = -10 cos f sin f + 8(1 + sin f) cos f df
49. (D) f ¢( x) = (2 cos x - 1)(cos x + 1) and
tan x = 1
1 and a = 2. 2
when z = D 2 is maximum
250 ö æ Thus minimum value = ç 25 + ÷ = 75. 5 ø è
=
Thus z has a minima and therefore y has a maxima at x = cos x.
through the point p 4. So, z is minimum at x = p 4 and ************
therefore, f ( x) is maximum at x = p 4. Maximum value = Page 544
a + 2 b = 1....(i)
D = ( 5 cos f - 0) 2 + (2 sin f + 2) 2 is maximum
x = 5.
f ¢¢( x) = - sin x(1 + 4 cos x). 1 or cos x = -1 Þ f ¢( x) = 0 Þ cos x = 2
Þ
Let the required point be ( 5 cos f , 2 sin f). Then,
f ¢¢(5) = 6 > 0. So, x = 5 is a point of minima.
50. (C) f ( x) =
Þ
- a - 2b + 1 = 0
< 0.
1 is a point of maxima. Maximum value = e1 e . e
2x -
Þ
Solving (i) and (ii) we get b = -
eø
Þ
dy a = + 2 bx + 1 dx x
é dy ù =0 êë dx úû ( x = -1 )
dy = - x - x (1 + log x) dx
Þ
Engineering Mathematics
2 2 = 1. [sec( p 4) + cosec ( p 4)] www.gatehelp.com
GATE EC BY RK Kanodia
CHAPTER
9.3 INTEGRAL CALCULUS
1.
òx
(A)
2
x dx is equal to +1
1 log ( x 2 + 1) 2
(C) tan -1
(B) log ( x 2 + 1)
x 2
2. If F ( a) =
5.
(D) 2 tan -1 x 1 , log a
to 1 (A) ( a x - a a + 1) log a (C)
3.
1 ( a x + a a + 1) log a
1 (B) (ax - aa) log a (D)
1 ( a x + a a - 1) log a
(B) cot x + cosec x + c
(C) tan x - sec x + c
(D) tan x + sec x + c
4.
( 3 x + 1) dx is equal to 2 -2x + 3
ò 2x
æ2x -1ö 3 5 ÷ (A) log (2 x 2 - 2 x + 3) + tan -1 çç ÷ 4 2 è 5 ø (B)
æ2x -1ö 4 ÷ log (2 x 2 - 2 x + 3) + 5 tan -1 çç ÷ 3 è 5 ø
æ2x -1ö 4 2 ÷ (C) log (2 x 2 - 2 x + 3) + tan -1 çç ÷ 3 5 è 5 ø æ2x -1ö 3 2 ÷ (D) log (2 x 2 - 2 x + 3) + tan -1 çç ÷ 4 5 è 5 ø
is equal to
tan -1 (tan x)
(B) 2 tan -1 (tan x)
(C)
1 2
tan -1 (2 tan x)
(D) 2 tan -1 ( 12 tan x)
2 sin x + 3 cos x
ò 3 sin x + 4 cos x dx is equal to
(A)
9 1 x+ log( 3 sin x + 4 cos x) 25 25
(B)
18 2 x+ log( 3 sin x + 4 cos x) 25 25
(C)
18 1 x+ log( 3 sin x + 4 cos x) 25 25
(D) None of these
ò
(A) (A) - cot x + cosec x + c
x
1 2
7.
dx ò 1 + sin x is equal to
2
(A)
6.
a > 1 and F ( x) = ò a 2 dx + K is equal
dx
ò 1 + 3 sin
(B) (C)
3 + 8 x - 3 x 2 dx is equal to 3x - 4 3 3
3 + 8 x - 3 x2 -
æ 3x - 4 ö sin -1 ç ÷ 18 3 è 5 ø 25
3x - 4 25 3 æ 3x - 4 ö sin -1 ç 3 + 8 x - 3 x2 + ÷ 6 18 è 5 ø 3x - 4 6 3
3 + 8 x - 3 x2 -
æ 3x - 4 ö sin -1 ç ÷ 18 3 è 5 ø 25
(D) None of these 8.
dx
ò
(A) (C)
2 x + 3x + 4 2
1 2 1 2
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sin -1
is equal to
4x + 3
cosh -1
23 4x + 3 23
(B)
1 2
sinh -1
4x + 3 23
(D) None of these
Page 545
GATE EC BY RK Kanodia
UNIT 9
9.
2x + 3
ò
x + x+1 2
dx is equal to
(A) 2 x 2 + x + 1 + 2 sinh -1
(C) 2 x 2 + x + 1 + sinh -1 (D) 2 x 2 + x + 1 - sinh -1
10. (A)
ò1+ x + x
(A)
1 2
é ( x + 1) 2 -1 êlog x 2 + 1 + tan ë
(B)
1 4
é ( x + 1) 2 log + 2 tan -1 ê x2 + 1 ë
ù xú û
2x + 1
(C)
1 2
é ( x + 1) 2 -1 êlog x 2 + 1 - 2 tan ë
3
ù xú û
2x + 1
+ x3
is equal to ù xú û
(D) None of these
is equal to
x - x2
2
3 16.
dx
ò
3
dx
15.
3
2x + 1
x 2 + x + 1 + 2 sinh -1
(B)
2x + 1
Engineering Mathematics
x - x2 + c
(C) log (2 x - 1) + c
sin x
ò 1 - sin x dx
is equal to
(A) - x + sec x + tan x + k
(B) - x + sec x + tan x
(B) sin -1 (2 x - 1) + c
(C) - x + sec x - tan x
(D) - x - sec x - tan x
(D) tan -1 (2 x - 1) + c
17.
ò e { f ( x) + f ¢( x)} dx is equal to x
(A) e f ¢( x)
(B) e x f ( x)
(C) e x + f ( x)
(D) None of these
x
1
11.
ò ( x + 1)
(A)
æ 2 ö ÷ 2 cosh -1 çç ÷ è1 + xø
1 - 2 x - x2
dx is equal to
(B)
æ 2 ö ÷ (C) - 2 cosh -1 çç ÷ è1 + xø 12. (A)
(C)
13.
dx
ò sin x + cos x 1 2
æ 2 ö ÷ cosh -1 çç ÷ 2 è1 + xø
1
æ 2 ö ÷ cosh -1 çç ÷ 2 è1 + xø
1
(D) -
is equal to
pö æ log tanç x + ÷ 4 è ø
æ x pö log tanç + ÷ 2 è2 8ø dx
ò sin( x - a) sin( x - b)
(D)
1 2
æ x pö log tanç + ÷ è2 6ø
æ x pö log tanç + ÷ 2 è4 4ø
1
is equal to
(A) sin( x - a) log sin( x - b)
ì sin( x - a) ü (C) sin( a - b) log í ý î sin( x - b) þ
dx ò ex - 1 is equal to
(C) log ( e Page 546
-x
- 1)
x
x +c 2
(B) e x cot
(C) e x tan x + c
òx
x +c 2
(D) e x cot x + c
x3 dx is equal to +1
2
(A) x 2 + log ( x 2 + 1) + c (B) log ( x 2 + 1) - x 2 + c (C)
1 2 1 x - log ( x 2 + 1) + c 2 2
(D)
1 2 1 x + log( x 2 + 1) + c 2 2
(A) x sin -1 x + 1 - x 2 + c
(B) x sin -1 x - 1 - x 2 + c
(C) x sin -1 x + 1 + x 2 + c
(D) x sin -1 x - 1 - x 2 + c
21.
ì sin( x - a) ü 1 (D) log í ý sin( a - b) î sin( x - b) þ
(A) log ( e x - 1)
æ 1 + sin x ö
ò e ççè 1 + cos x ÷÷ødx is
20. ò sin -1 x dx is equal to
æ x -aö (B) log sinç ÷ è x -bø
14.
(A) e x tan
19. (B)
1
18. The value of
ò
sin x + cos x 1 + sin 2 x
dx is equal to
(A) sin x
(B) x
(C) cos x
(D) tan x 1
22. The value of (B) log (1 - e x ) (D) log (1 - e ) x
(A) -1/2 (C) 1/2
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ò 5 x - 3 dx is 0
(B) 13/10 (D) 23/10
GATE EC BY RK Kanodia
UNIT 9
1
39.
x
ò ò(x
46. The area bounded by the curve r = q cos q and the p lines q = 0 and q = is given by 2 ö ö p æ p2 p æ p2 (B) (A) çç çç - 1 ÷÷ - 1 ÷÷ 4 è 16 16 è 6 ø ø
+ y 2 ) dydx is equal to
2
0 x
(A)
7 60
(B)
(C)
4 49
(D) None of these
3 35
(C) 1
1 + x2
0
0
ò ò dy dx is
40. The value of p log ( 2 + 1) 4
(B)
(C)
p log ( 2 + 1) 2
(D) None of these
p log ( 2 - 1) 4
(C)
A
(B)
36 5
(D) None of these
x + y = a and x + y = a in the first quadrant is given 2
2
by a2 - x 2
a
(A)
(C) 4 ò
p2
ò ò
a cos 2 q
0
rdrdq
rdrdq
(B) 2 ò
p2
ò
0
(D) 2 ò
cos 2 q
a
0
p a cos 2 q
ò
0 0
rdrdq
rdrdq
48. The area of the region bounded by the curve 1
x2 4
0
y= 0
(A)
ò ò
(C)
ò ò
2
0
dxdy
3x ( x 2 + 2)
y= x 2 4
dydx
x2 4
y= 0
1
ò
ò ò
(D)
ò
y= 0
dydx
3x ( x 2 + 2)
y= x 2 4
dxdy
below by z = 0 and bounded above by z = h is given by (B) pa 2 h
1 pa 3h 3
(D) None of these
50.
òòò
0
1
0
(B)
49. The volume of the cylinder x 2 + y 2 = a 2 bounded
(C)
0
cos 2 q
a
0
ò ò dxdy
(B)
a-x
0
p4
(A) pah
a2 - x 2
a
ò ò dxdy
(D) None of these
y ( x 2 + 2) = 3 x and 4 y = x 2 is given by
42. The area of the region bounded by the curves 2
(A) 4 ò
0
òò ydxdy is equal to
32 5
ö æ p2 çç - 1 ÷÷ ø è 16
0
41. If A is the region bounded by the parabolas y 2 = 4 x and x = 4 y, then 48 (A) 5
p 16
47. The area of the lemniscate r 2 = a 2 cos 2 q is given by
(A)
2
Engineering Mathematics
(D) None of these
a 2 - y2 a
ò ò dxdy
(C)
a-x
0
43. The area bounded by the curves y = 2 x , y = - x, x = 1 and x = 4 is given by 33 (B) 2
(A) 25 47 (C) 4
44. The area bounded by the curves y 2 = 9 x, x - y + 2 = 0
e x + y+ z dxdydz is equal to (B)
(C) ( e - 1) 2
(D) None of these
1
z
x+ z
-1
0
x -z
ò ò ò
( x + y + z) dy dx dz is equal to
(A) 4
(B) -4
(C) 0
(D) None of these
is given by (A) 1
(B)
1 2
3 2
(D)
5 4
(C)
*************
45. The area of the cardioid r = a (1 + cos q) is given by (A) 2 ò
p
ò
a (1 + cos q)
ò
a (1 + cos q)
q= 0 r = 0
(C) 2 ò
p2
0
Page 548
r=0
rdrdq
(B) 2 ò
rdrdq
(D) 2 ò
p
0 p4
0
ò
a (1 + cos q)
r=a
ò
a (1 + cos q)
r=0
3 ( e - 1) 2
(A) ( e - 1) 3
51.
101 (D) 6
1 1 1
0 0 0
rdrdq rdrdq www.gatehelp.com
GATE EC BY RK Kanodia
Integral calculus
SOLUTIONS 1. (A)
ò
Þ
F ( a) =
ax +K log a
aa +K log a
1 aa 1 - aa K = = log a log a log a F ( x) =
3. (C)
a 1-a 1 + = [ a x - a a + 1] log a log a log a x
a
dx ò 1 + sin x
dx x x x x ö 2 + cos 2 ÷ + 2 sin cos ç sin 2 2ø 2 2 è x sec 2 dx 2 =ò =ò dx 2 2 x xö xö æ æ ç 1 + tan ÷ ç cos + sin ÷ 2 2ø 2ø è è x Put 1 + tan = t 2 x 2 dt 2 Þ sec2 dx = 2 dt Þ ò 2 dt = - + K 2 t t x -2 cos -2 2 +K = +K = x x x cos + sin 1 + tan 2 2 2 x x x -2 cos cos - sin 2 2 2 = ´ +K x x x x cos + sin cos - sin 2 2 2 2 x x 2 x -2 cos + 2 sin cos 2 2 2 +K = 2 x 2 x - sin cos 2 2 -(1 + cos x) + sin x = + k = tan x - sec x - 1 + K cos x
òæ
= tan x - sec x + c
3x + 1 dx 2x -2x + 3 2
=
p=
Þ
5 dx 3 log (2 x 2 - 2 x + 3) + ò 2 2 4 æ 4 æ 5ö 1ö ÷ ç x - ÷ + çç ÷ 2ø è è 2 ø
3 5 1 = log (2 x 2 - 2 x + 3) + tan -1 4 4æ 5ö ç ÷ ç 2 ÷ è ø 5. (C) Let I = ò =ò
1 2 5 2
x-
dx 1 + 3 sin 2 x
cosec2 x dx cosec2 x dx = cosec 2 x + 3 ò (1 + cot2 x) + 3
Put cot x = t Þ I =ò =
3 5 , q = 4 2
3 4x -2 5 dx dx + ò 2 4 ò 2 x2 - 2 x + 3 2 2x -2x + 3
I=
1 1 log t = log ( x 2 + 1) 2 2
2. (A) F ( x) = ò a x dx + K =
=
4. (A) Let I = ò
Let 3 x + 1 = p( 4 x - 2) + q
x dx 2 x +1
Put x 2 + 1 = t Þ 2 xdx = dt 1 1 x ò x 2 + 1 dx = ò 2 × t dt =
Chap 9.3
- cosec2 x dx = dt
1 -dt t 1 æ cot x ö = cot-1 = cot-1 ç ÷ 4 + t2 2 2 2 è 2 ø
1 tan -1 (2 tan x) 2
6. (C) Let I = ò
2 sin x + 3 cos x dx 3 sin x + 4 cos x
Let (2 sin x + 3 cos x) = p( 3 cos x - 4 sin x) + q( 3 sin x + 4 cos x) p=
1 18 , q= 25 25
I=
1 25
=
3 cos x - 4 sin x
3 sin x + 4 cos x
18
ò 3 sin x + 4 cos x dx + 25 ò 3 sin x + 4 cos x dx
1 18 log ( 3 sin x + 4 cos x) + x 25 25 2
ò
1 = 3 2
ì 4 öü æ 2 2 2 ç x - ÷ï ïæ 4 ö æ5 ö æ 4ö æ5 ö -1 3 í ç x - ÷ ç ÷ - ç x - ÷ + ç ÷ sin ç 5 ÷ý 3 3 3 3 è ø è ø è ø è ø çç ÷÷ï ï è 3 øþ î
=
3 + 8 x - 3 x 2 dx = 3
ò
2
4ö æ5 ö æ ç ÷ - ç x - ÷ dx 3 3ø è ø è
7. (B)
3x - 4 25 3 3x - 4 3 + 8 x - 3 x2 + sin -1 6 18 5
8. (B)
www.gatehelp.com
ò
dx 2 x + 3x + 4 2
=
1 2
ò
dx æ 23 ö 3ö æ ÷ ç x + ÷ + çç ÷ 4 è ø è 4 ø 2
2
Page 549
GATE EC BY RK Kanodia
UNIT 9
1
=
sinh
2
æ ç ç è
2x + 3
ò
9. (B)
3 4 = 1 sinh -1 4 x + 3 2 23 23 ö ÷ ÷ 4 ø
x+
-1
=ò
x2 + x + 1
=ò
x2 + x + 1
=
dx +
2x + 1
ò
dx + 2 ò
= 2 x 2 + x + 1 + 2 sinh -1
dx
ò
x 1- x
13. (D)
x2 + x + 1 æ 3ö 1ö æ ÷ ç x + ÷ + çç ÷ 2ø è è 2 ø 2
2
1 x+ 2 3 2
2x + 1
I =ò
( x + 1) 1 - 2 x - x 2
Þ
dx = -
1 dt t2
ò
dt æ 1 t 2 - çç è 2
ö ÷ ÷ ø
2
=-
2
2
ò
1 sin( a - b)
ò sin( x - a) sin( x - b)
=
1 sin ( a - b)
sin [( x - b) - ( x - a)]
dx
sin( x - b) cos( x - a) - cos( x - b) sin( x - a) dx sin( x - a) sin( x - b)
=
1 [cot( x - a) - cot( x - b)]dx sin( a - b) ò
=
1 [log sin ( x - a) - log sin ( x - b)] dx sin ( a - b)
=
ì sin( x - a) ü 1 log í ý sin ( a - b) î sin( x - b) þ
14. (D) Let I =
òe
dx e - x dx =ò 1 - e- x -1
x
dx 15. (B) Let I = ò =ò dt 2 t2 - 1
dx 1 + x + x2 + x3
dx (1 + x) (1 + x 2 ) A Bx + C 1 = + 2 (1 + x)(1 + x ) 1 + x 1 + x 2
Let
1 = A(1 + x 2 ) + ( Bx + C)(1 + x) 1 2
æ 2 ö ÷ cosh -1 çç ÷ 2 è x + 1ø
1
=-ò
sin( a - b) dx
Put 1 - e - x = t Þ e - x dx = dt dt I =ò = log t = log (1 - e - x ) t
1 dt t2
1
12. (C)
Page 550
1
1 ö æ1 ö æ1 1 - 2ç - 1 ÷ - ç - 1 ÷ t t t è ø è ø 2
=-
=
1 t -
1
=-
x+c
dx
ò sin( x - a) sin( x - b)
=
=I
(2 x - 1) + c
Put x + 1 =
2
pö
æ
ò cosecçè x + 4 ÷ødx
ò sin( x - a) sin( x - b)
3
I = ò 2 dq = 2 q + c = 2 sin -1
11. (D) Let I = ò
1
1 sin( a - b)
´ò
Put x = sin q Þ dx = 2 sin q cos q dq 2 sin q cos q 2 sin q cos q I =ò dq = ò dq 2 sin q cos q sin q 1 - sin q
I = sin
pö æ sinç x + ÷ 4ø è
=
=
dx
2
-1
dx
1 é 1æ p öù 1 æ x pö log tan ç + ÷ - log cot ç x + ÷ú = ê 2è 4 øû 2 ë 2 è2 8ø
=
2 dx
( x 2 + x + 1)1 2 + 2 sinh -1 1 2
10. (B)
2
ò
dx
x2 + x + 1
2x + 1
1
=
Engineering Mathematics
dx
ò sin x + cos x dx p p sin x cos + cos x sin 4 4
t
cosh -1 1
Comparing the coefficients of x 2 , x and constant terms, 2
A + B = 0, B + C = 0, C + A = 1 Solving these equations, we get 1 1 1 A = , B=- , C= 2 2 2 1 1 1 x -1 I = dx - ò 2 dx 2 ò1+ x 2 x +1 =
1 1 1 log (1 + x) - log ( x 2 + 1) + tan -1 x 2 2 2
=
1 4
é ( x + 1) 2 log + 2 tan -1 ê x2 + 1 ë
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ù xú û
GATE EC BY RK Kanodia
Integral calculus
16. (B) Let I = ò
sin x dx 1 - sin x
ò
=
=ò
1 - (1 - sin x) dx 1 - sin x
=ò
=ò
1 1 + sin x dx - ò dx = ò dx - x 1 - sin x 1 - sin 2 x
=ò
=ò
1 + sin x dx - x = ò (sec2 x + sec x tan x) dx - x cos 2 x
= tan x + sec x - x
Chap 9.3
sin x + cos x (sin x + cos 2 x) + 2 sin x cos x 2
sin x + cos x (cos x + cos x) 2
dx
sin x + cos x dx = ò dx = x sin x + cos x 35
35
ò 5 x - 3 dx = - ò 5 x - 3 dx +
22. (D)
0
0
= ò e x f ( x) dx + ò e x f ¢( x) dx = { f ( x) e x - ò f ¢( x) ex dx} + ò ex f ¢( x) dx = f ( x) × ex
9ö æ 9 =ç+ ÷+ è 10 5 ø =
æ 1 + sin x ö 18. (A) Let I = ò e çç ÷÷ dx è 1 + cos x ø x
23. (B)
24. (D)
c
dt = [tan -1 t ]1e +1
2
p 4
c
ò x(1 - x) dx = ò ( x - x ) dx 2
0
1 ö 1 æ1 = ç x 2 - x 3 ÷ = c 2 ( 3 - 2 c) 2 3 è ø0 6
x3 x × x2 dx = ò 2 dx x +1 x +1
c
x( x + 1 - 1) x dx = ò xdx - ò 2 dx 2 x +1 x +1 2
20. (A) Let I = ò sin = sin -1 x × x - ò
-1
1 1 - x2 x 1 - x2
x dx = ò sin
-1
x × 1 × dx
× x dx dx
xdx = - tdt = x sin -1 x + ò dt = x sin -1 x + t = x sin -1 x + 1 - x 2 + c sin x + cos x 1 + sin 2 x
dx
ò
x(1 - x) dx = 0
1 2 c ( 3 - 2 c) = 0 6
Þ
0
Þ
c=
3 2
25. (D) Put x 2 + x = t
In second part put 1 - x 2 = t 2
ò
e
òt
e x dx = dt =
Þ
c
1 2 1 x - log ( x 2 + 1) + c 2 2
21.
e x dx 2x +1
0
0
2
ò
1
òe
= tan -1 e - tan -1 1 = tan -1 e -
x = e tan + c 2
x-
dx = + e- x
1
x
= x sin
x
Put e x = t
1ì x x x x ü x x í e × 2 tan - ò e × 2 tan dx ý + ò e tan dx 2î 2 2 2 þ
-1
òe 0
1 x x = ò e x sec 2 dx + ò e x tan dx 2 2 2
=
éæ 5 ö æ 9 9 öù êç 2 - 3 ÷ - ç 10 - 5 ÷ú è ø è øû ë
9 æ 1 9 ö 13 + ç- + ÷= 10 è 2 10 ø 10 1
x xö æ ç 1 + 2 sin cos ÷ x 2 2 ÷ dx =òe ç x çç ÷÷ 2 cos 2 ø è 2
=ò
ò 5 x - 3 dx
35
1
35
19. (C) I = ò
1
ö æ 5 x2 æ 5 ö = ç - x 2 + 3 x ÷ + çç - 3 x ÷÷ è 2 ø0 ø3 5 è 2
17. (B) Let I = ò e { f ( x) + f ¢( x)} dx x
=
dx
1
ò
2x + 1
0
x+ x
26. (A)
2
2
dx = ò 0
dt t
Þ
(2 x + 1) dx = dt
= 2( t1 2 ) 20 = 2 2
p
òx
4
sin 5 xdx
p
Since, f ( - x) = ( - x) 4 sin 5 ( - x) = -x 4 sin 5 x f ( x) is odd function thus p
òx
4
sin 5 x dx = 0
p
27. (A) www.gatehelp.com
p2
p2
0
0
2 ò cos x dx =
ò
1 (cos 2 x + 1) dx 2 Page 551
GATE EC BY RK Kanodia
UNIT 9
Engineering Mathematics
1 æ1 ö ç sin 2 x + x ÷ 2 è2 ø0
pö æp = ò 2 sinç (1 - t) - ÷dt = 2 4 è ø 0
=
1 é1 æp öù (sin p - sin 0) + ç - 0 ÷ú 2 êë2 è2 øû
pö æp = - ò 2 sin ç t - ÷dt = - 1 2 4 è ø 0
=
1 é1 pù p (0 - 0) - 0 + ú = ê 2 ë2 2û 4
2I = 0
1
p2
=
p ö
ò 2 sinçè 4 - 2 t ÷ødt 0
1
æ 3ö æ 1ö Gç ÷ Gç ÷ 1 p p2 p 2 2 Aliter 1. ò cos 2 x dx = è ø è ø = 2 = 2 4 æ4ö 0 2 Gç ÷ è2 ø p2
Aliter 2. Use Walli’s Rule
æp
1
ò cos
2
x =
0
Þ
I =0
31. (C) Let I =
2a
f ( x)
ò f ( x) + f (2 a - x) dx ....(1) 0
I=
f (2 a - x)
2a
ò f (2 a - x) + f ( x) dx....(2) 0
1 p p × = 2 2 4
Adding (1) and (2), we get 2I =
2a
f ( x) + f (2 a - x)
ò f ( x) + f (2 a - x) dx
=
0
a
28. (B) Let I = ò a 2 - x 2 dx
Þ
Put x = a sin q Þ dx = a cos q dq when x = 0, q = 0, p when x = a, q = 2
32. (C) Let I =
2a
ò 1 × dx = [ x ]
2a 0
= 2a
0
I = a
0
Put
0
Þ
p2
= a 2 ò cos 2 q dq = a 2 × 0
1 p (By Walli’s Formula) × 2 2
1 - x2
1 - x2
xdx
1 - x2 = t 1 2 1 - x2
( -2 x) dx = dt
when x = 0, t = 1, when x = 1, t = 0 0
pa 2 4
I = ò -e t dt = -[ e t ]10 = -[ e 0 - e1 ] = e - 1 1
a
Aliter:
ò
e
0
p2
I = ò a 2 - a 2 sin 2 q a cos q dq
=
1
ò
a 2 - x 2 dx
1
0
a
é 1 pa 2 ù pa 2 xù é1 = = ê x a 2 - x 2 + a 2 sin -1 ú = ê0 + 4 2 a û0 ë 4 úû ë2 p2
dx 1 x + x2 0
33. (B) Let I = ò
1
1
=ò 0
29. (D) Let I = ò log (tan x) dx ....(1) 0
dx æ 3ö 1ö æ ÷ ç x - ÷ + çç ÷ 2ø è è 2 ø 2
p2
æp ö I = ò log tan ç - x ÷ dx 2 è ø 0
=
0
=
æ 2 é 1 1 öù 2 æ p p ö -1 ÷ú= - tan -1 çç ç + ÷ êtan 3 è6 6ø 3 ë 3 3 ÷øû è
=
p2
I = ò log (cot x)....(2)
2
é 1ù x- ú 1 ê -1 2ú êtan 3 ê 3 ú 2 êë 2 úû 0
2p 3 3
=
2p 3 9
Adding (1) and (2), we get p2
2 I = ò [log (tan x) + log (cot x)]dx
1
34. (B) Let I =
-1
0
0
p2
= ò log (tan x × cot x) dx 0
p2
= ò log 1 dx = 0
Þ
I =0
=
1
Page 552
x
-x ò-1 x dx + 0
dx =
1
ò -1dx + ò 1 × dx = -[ x ]
0 -1
-1
+ [ x ]10
0
= - [0 - ( -1)] + [1 - 0 ] = 0
0
æ pt p ö 30. (D) Let I = ò 2 sin ç - ÷dt ....(i) è 2 4ø 0
ò
x
35. (C)
100 p
p
0
0
ò |sin x|dx = 100 ò |sin x|dx
[ . .. sin x is periodic with period p] www.gatehelp.com
1
x
ò x dx 0
GATE EC BY RK Kanodia
Integral calculus
p
1
= 100 ò sin x dx = 100( - cos x) 0p
40. (D)
0
1 + x2
ò ò 0
= 100( - cos p + cos 0) = 100(1 + 1) = 200. p
p
0
0
= ( - cos x) m (sin x) n
1 [ x 1 + x 2 + log( x + 1 + x 2 )]10 2 1 = [ 2 + log (1 + 2 )] 2 =
41. (A) Let I = òò ydxdy,
= -cos m x sin n x, if m is odd I = ò cos
dx
0
f ( p - x) = cos m ( p - x) sin n ( p - x)
m
x2
0
1
Where f ( x) = cos m x sin n x
p
0
1
dydx = ò [ y ]01 +
= ò 1 + x 2 dx
36. (C) Let I = ò cos m x sin nx dx = ò f ( x) dx
A
x sin x dx = 0, if m is odd
Solving the given equations y 2 = 4 x and x 2 = 4 y , we get
n
x = 0, x = 4 . The region of integration A is given by
0
2 x
4 é y2 ù = ydydx ò ò ê 2 ú dx û x2 4 0 x2 4 0 ë 4 2 x
p
A =ò
37. (A) Let I = ò xF (sin x) dx ....(1) 0
p
4
é 48 x5 ù 1æ x4 ö ÷÷dx = ê x 2 çç 4 x ú = 5 160 2 10 ø ë û0 è 0 4
= ò ( x - p) F [sin ( p - x)]dx
=ò
0
I =
Chap 9.3
p
ò ( p - x) F(sin x) dx ....(2)
42. (A) The curves are
0
Adding (1) and (2), we get p
2 I = ò pF (sin x) dx
x 2 + y 2 = a 2 ...
....(i)
x + y = a...
....(ii)
0
Þ
I=
The curves (i) and (ii) intersect at A (a, 0) and B (0,a)
p
1 pF (sin x) dx 2 ò0 p2
38. (B) Let I = ò 0
The required area A =
ex æ 2 x xö + 2 tan ÷dx ç sec 2 è 2 2ø
p2
x 1 = ò e xsec 2 dx + 2 2 0
4 2 x
A =ò
x 1 Here, I1 = ò e x sec 2 dx 2 2 0
òe 0
= e - I 2 , I1 + I 2 = e p2
I = I1 + I 2 = e
x ò0 e × 2 tan 2 dx x
tan
ò dydx
y = - x....(ii)
2 x -x
1
4
= ò [2 x + x ]dx 1
æ 32 ö æ 4 1 ö 101 =ç + 8÷ -ç + ÷ = 6 è 3 ø è3 2ø
p2
x
4
ò dydx = ò [ y ]
1 -x
p2
y= a - x
ò
If a figure is drawn then from fig. the required area is
p2
p æ ö = ç e p 2 tan - 0 ÷ 4 è ø
x =0
y = 2 x i.e., y 2 = 4 x....(i)
x ò0 e tan 2 dx = I1 + I 2 x
p2
a2 - x 2
43. (D) The given equations of the curves are
p2
1 xù é1 = ê e x × 2 tan ú 2 û0 2 ë2
a
44. (B) The equations of the given curves are
x dx 2
y 2 = 9 x....(i)
x - y + 2 = 0....(ii)
The curves (i) and (ii) intersect at
p2
A(1, 3) and B(4, 6)
p2
If a figure is drawn then from fig. the required area is 1
39. (B)
ò 0
1
x
x
1 3ù é 2 2 2 òx ( x + y ) dy dx = ò0 êë x y + 3 y úû x dx
1
1 1 ù é = ò ê x 5 2 + x 3 2 - x 3 - x 3 ú dx 3 3 û 0 ë 1
2 52 1 4 ù 3 é2 = ê x7 2 + x - x ú = 15 3 û 0 35 ë7
4 3 x
A =ò
4
ò dydx = ò [ y ]
1 x+ 2
3 x x+ 2
dx
1
4
4
1 é ù = ò [ 3 x - ( x + 2)]dx = ê2 x 3 2 - x 2 - 2 x ú 2 ë û1 1 1 ö 1 æ = (16 - 8 - 8) - ç 2 - - 2 ÷ = 2 è ø 2
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GATE EC BY RK Kanodia
UNIT 9
Thus the equation volume is V = 4 ò zdxdy
45. (A) The equation of the cardioid is r = a (1 + cos q)
A
....(i)
If a figure is drawn then from fig. the required area is p a (1 + cos q)
ò
Required area A = 2
= =
1 2 1 4
cos 2 qdq =
0
2 ò q dq + 0
1 4
1 4
dq
1 1 1
50. (A)
p2
ò q (1 + cos 2 q)dq 2
òòòe
1 1
= ò ò [ ex +
0
2 ò q cos 2 q dq
dxdydz 1 1
] dydz = ò ò [ e1 + 0 0
0
1
= ò [( e 2 + z - e1 + z ) - ( e1 + z - e z )]dz 0
1
= ò ( e 2 + z - 2 e1 + z + e z ) dz = [ e 2 + z - 2 e1 + z + e z ]10 0
=
p3 1 æ -p ö 1 + ç -0÷ 96 4 è 4 ø 8
= ( e 3 - 2 e 2 + e) - ( e 2 - 2 e + 1)
p2
= e 3 - 3e 2 + 3e - 1 = ( e - 1) 3
p2
1
ò cos 2 q dq
51. (C)
0
z x+ z
ò ò ò ( x + y + z) dydxdz
-1 0 x - z p2
p p3 p 1æ1 ö - ç sin 2 q ÷ = 96 16 8 è 2 ø0 16
ö æ p2 çç - 1 ÷÷ 16 ø è
47. (A) The curve is r 2 = a 2 cos 2 q If a figure is drawn then from fig. the required area is p4
a
é1 2 ù òr = 0rdrdq = 4 ò0 êë2 r úû 0
p4
x+ z
=
é( x + y + z) 2 ù ò-1 ò0 êë ú dxdz 2 ûx- y
=
é(2 x + 2 z) 2 æ 2 x ö2 ù -ç ÷ ú dxdz òòê 2 è 2 ø úû -1 0 ê ë
1 z
1 z
z
1 1 3 é ù é( x + z) 3 x 3 ù = 2 ò êò (( x + z) 2 - x 2 ) dx ú dz = 2 ò ê - ú dz 3 3 û0 -1 ë 0 -1 ë û
cos 2 q
dq
1
ésin 2 q ù = 2 ò a cos 2 q dq= 2 a ê = a2 ú 2 ë û0 0
2 2 = ò [(2 z) 3 - z 3 - z ]dz = 3 -1 3
48. (C) The equations of given curves are
æ1 1ö = 4ç - ÷ = 0 è4 4ø
p4
2
y( x 2 + 2) = 3 x....(i)
1
éz4 ù z dz = 6 4 ò ê4ú ë û -1 -1 1
3
and 4 y = x 2 ....(ii)
The curve (i) and (ii) intersect at A (2, 1). If a figure is drawn then from fig. the required area is The required area A =
2
3x ( x 2 + 2 )
x =0
y= x 2 4
ò
ò dxdy
49. (B) The equation of the cylinder is x 2 + y 2 = a 2 The equation of surface CDE is z = h. If a figure is drawn then from fig. the required area is Page 554
- e y + z ]dydz
= ò [ e1 + y + z - e y + z ]10 dz
p3 1 éæ cos 2 q ö æ cos 2 q ö ù - êç -q ÷ - òç÷d q ú 96 4 ëêè 2 ø0 2 ø úû 0è
2
y+ z
1
0
p 4 a cos 2 q
x + y+ z
y+ z 1 0
0 0
p2
=
ò
1 p × = pa 2 h. 2 2
0 0 0
p2
q= 0
dx = a cos q dq,
0
p2 ù p3 1 é = + ê- ò q sin 2 q dq ú 96 4 ë 0 û
A =4
0
0
p2 p2 p2 ù 1 é1 ù 1 éæ sin 2 q ö sin 2 q = ê q 3 ú + êç q 2 dq ú ÷ - ò 2q 4 ë3 û0 4 êëè 2 ø0 2 úû 0
=
a
dx = 4 hò a 2 - x 2 dx
q cos q
p2
p2
p2
Þ
= 4 ha 2 ò cos 2 q dq = 4 ha 2 ×
é1 2 ù òr = 0rdrdq = ò0 êë2 r úû o
2
0
0
p2
p 2 q cos q
òq
a2 - x 2
0
Volume V = 4 h ò a 2 - a 2 sin 2 q × a cos q dq
The required area
ò
a
p2
r = q cos q....(i)
q= 0
a
ò ò hdydx = 4 hò [ y ]
Let x = a sin q,
r=0
46. (C) The equation of the given curve is
A=
=4
a2 - x 2
0
ò rdrdq
q= 0
Engineering Mathematics
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GATE EC BY RK Kanodia
Complex Variables
Chap 9.5
10. The integration of f ( z) = x 2 + ixy from A(1, 1) to B(2,
17. The value of f ( 3) is
4) along the straight line AB joining the two points is
(A) 6
(B) 4i
-29 (A) + i11 3
(C) -4i
(D) 0
(C)
29 (B) - i11 3
23 + i6 5
(D)
18. The value of f ¢(1 - i) is
23 - i6 5
11.
e2 z òc ( z + 1) 4 dz = ? where c is the circle of z = 3
(A)
4 pi -3 e 9
(B)
4 pi 3 e 9
(C)
4 pi -1 e 3
(D)
8 pi -2 e 3
12.
1 - 2z òc z( z - 1)( z - 2) dz = ? where c is the circle z = 15.
(A) 2 + i 6 p
(B) 4 + i 3p
(C) 1 + ip
(D) i3p
(A) 7 ( p + i2)
(B) 6 (2 + ip)
(C) 2 p (5 + i13)
(D) 0
Statement for 19–21: Expand the given function in Taylor’s series. 19. f ( z) =
(A) 1 + 2( z + z 2 + z 3......)
(B) -1 - 2( z - z 2 + z 3......)
(C) -1 + 2( z - z 2 + z 3......)
(D) None of the above
20. f ( z) =
13. ò ( z - z 2 ) dz = ? where c is the upper half of the circle
(B)
-3 (D) 2
3 (C) 2 14.
2 3
-1 é 1 1 ù 1 - ( z - 1) + 2 ( z - 1) 2 .......ú ê 2 ë 2 2 û
(B)
1é 1 1 ù 1 - ( z - 1) + 2 ( z - 1) 2 .......ú 2 êë 2 2 û
(C)
1é 1 1 ù 1 + ( z - 1) + 2 ( z - 1) 2 .......ú 2 êë 2 2 û
cos pz dz = ? where c is the circle z = 3 c z -1 (B) - i2 p
(C) i6 p2
(D) - i6 p2
15.
(D) None of the above
ò
(A) i2 p
21. f ( z) = sin z about z =
sin pz 2 òc ( z - 2)( z - 1) dz = ? where c is the circle z = 3
(A) i6p
(B) i2p
(C) i4p
(D) 0
16. The value of
1 cos pz dz around a rectangle with 2 pi òc z 2 - 1
vertices at 2 ± i , -2 ± i is (A) 6
(B) i2 e
(C) 8
(D) 0
2 ù 1 é pö 1 æ pö æ + 1 z z ç ÷ ç ÷ - .......ú ê 4 ø 2 !è 4ø 2 êë è úû
(B)
2 ù 1 é pö 1 æ pö æ ê1 + ç z - ÷ + ç z - ÷ + .......ú 4 ø 2 !è 4ø 2 êë è úû
(C)
2 ù 1 é æ pö 1 æ pö ê1 - ç z - ÷ - ç z - ÷ - .......ú 4 ø 2 !è 4ø 2 êë è úû
(D) None of the above 22. If z + 1 < 1, then z -2 is equal to (A) 1 +
x + y = 4.
¥
å ( n + 1)( z + 1)
n -1
n =1
(B) 1 +
¥
å ( n + 1)( z + 1)
n +1
n =1
3z 2 + 7 z + 1 f ( z0 ) = ò dz , where c is the circle ( z - z0 ) c 2
p 4
(A)
Statement for Q. 17–18:
2
1 about z = 1 z +1
(A)
c
z =1 -2 (A) 3
z -1 about the points z = 0 z +1
(C) 1 +
¥
å n( z + 1)
n
n =1
(D) 1 +
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¥
å ( n + 1)( z + 1)
n
n =1
Page 565
GATE EC BY RK Kanodia
UNIT 9
Statement for Q. 23–25.
28. The Laurent’s series of f ( z) = 1 in Laurent’s ( z - 1)( z - 2)
Expand the function
series for the condition given in question. 23. 1 < z < 2 1 2 3 (A) + 2 + 3 + ....... z z z (B) K - z -3 - z -2 - z -1 (C)
Engineering Mathematics
1 1 1 1 3 - z - z2 z -K 2 4 8 18
where z < 1 1 5 3 21 5 (A) z z + z .......... 4 16 64 (B)
1 1 2 5 4 21 6 + z + z + z .......... 2 4 16 64
(C)
1 3 15 5 z - z3 + z .......... 2 4 8
(D)
1 1 2 3 4 15 6 + z + z + z .......... 2 2 4 8
1 3 7 + 2 + 4 ........... 2 z z z
1 - e Zz at its pole is z4 -4 (B) 3
(D) None of the above
29. The residue of the function
24. z > 2
(A)
4 3
(C)
-2 3
(A)
6 13 20 + + 3 + ........ z z2 z
(B)
1 8 13 + + + ......... z z2 z3
(C)
1 3 7 + 3 + 4 + ......... 2 z z z
(D)
2 3 4 - 3 + 4 - ........ 2 z z z
25. z < 1 (A) 1 + 3z
+7 2 15 2 z + z ..... 2 4
(B)
1 3 7 15 3 + z + z2 + z ... 2 4 8 16
(C)
1 3 z2 z3 + + + ....... 4 4 8 16
26. If z - 1 < 1 , the Laurent’s series for
30. The residue of z cos
1 is z( z - 1)( z - 2)
( z - 1) 3 ( z - 1) 5 - ......... 2! 5!
(D) - ( z - 1)
-1
1 at z = 0 is z -1 (B) 2
1 2
(C)
1 3
31.
ò z(1 - z)( z - 2) dz = ? where c is
(D) 1 - 2z
-1 3 z = 15 .
(A) - i3p
(B) i3 p
(C) 2
(D) -2
32.
z cos z dz = ? where c is z - 1 = 1 pö c z ç ÷ 2ø è
òæ
(A) 6 p
(B) - 6p
(C) i2p
(D) None of the above
(C) - ( z - 1) - ( z - 1) - ( z - 1) - .......... 3
2 3
(A)
( z - 1) 3 ( z - 1) 5 - ........... 2! 5!
(B) - ( z - 1) -1 -
(D)
c
(D) None of the above
(A) - ( z - 1) -
z is, ( z + 1)( z 2 + 4) 2
5
1
- ( z - 1) - ( z - 1) - ( z - 1) - ......... 3
5
33.
2 ò z e z dz = ? where c is z = 1 c
27. The Laurent’s series of
1 for z < 2 is z( e z - 1)
1 1 1 1 2 (A) 2 + + + 6z + z + .......... z 2 z 12 720 1 1 1 1 2 (B) 2 + z + .......... z 2 z 12 720 (C)
1 1 1 1 2 + + z2 + z + .......... z 12 634 720
(D) None of the above Page 566
(B) - i3p
(A) i3p C)
34.
ip 3
(D) None of the above
2p
dq
ò 2 + cos q = ? 0
(A)
-2 p 2
(C) 2 p 2
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(B)
2p 3
(D) -2 p 3
GATE EC BY RK Kanodia
Complex Variables
¥
35.
x2 ò ( 2 + a 2 )( x 2 + b2 ) dx = ? -¥ x
(A)
p ab a+b
(B)
(C)
p a+b
(D) p ( a + b)
36.
¥
dx
ò1+ x
6
p ( a + b) ab
=?
0
(A) (C)
p 6
(B)
2p 3
(D)
p 2
***************
p 3
Chap 9.5
SOLUTIONS 1. (C) Since, f ( z) = u + iv = Þ
u=
x3 - y3 ; x2 + y2
v=
x 3(1 + i) - y 3(1 - i) ; z ¹0 x2 + y2
x3 + y3 x2 + y2
Cauchy Riemann equations are ¶u ¶v ¶u ¶v and = =¶x ¶y ¶y ¶x By differentiation the value of we get
¶u ¶y ¶v ¶v at(0, 0) , , , ¶x ¶y ¶x ¶y
0 , so we apply first principle method. 0
At the origin, ¶u u(0 + h, 0) - u(0, 0) h3 h2 = lim = lim = 1 h ®0 h ®0 ¶x h h ¶u u(0, 0 + k) - u(0, 0) - k3 k2 = lim = lim = -1 k ®0 ¶v h ®0 k k ¶v v(0 + h, 0) - v(0, 0) h3 h2 = lim = lim =1 h ®0 ¶x h ®0 h h ¶v v(0, 0 + k), v(0, 0) k3 k2 = lim = lim =1 k ®0 ¶y k ®0 k k Thus, we see that
¶u ¶v ¶u ¶v and = =¶x ¶y ¶y ¶x
Hence, Cauchy-Riemann equations are satisfied at z = 0. Again, f ¢(0) = lim z ®0
f ( z) - f (0) z
é( x 3 - y 3) + i( x 3 + y 3) 1 ù = lim ê 2 2 z ®0 (x + y ) ( x + iy) úû ë Now let z ® 0 along y = x, then é( x 3 - y 3) + i( x 3 + y 3) 1 ù 2i 1+ i f ¢ (0) = lim ê = = 2 2 ú z ®0 (x + y ) ( x + iy) û 2(1 + i) 2 ë Again let z ® 0 along y = 0, then é x 3 + i( x 3) 1 ù f ¢ (0) = lim ê =1 + i 2 x ®0 x úû ë (x ) So we see that f ¢(0) is not unique. Hence f ¢(0) does not exist. Df df = lim dz Dz ®0 Dz Du + iDv or f ¢( z) = lim Dz ®0 Dx + iDy
2. (A) Since, f ¢( z) =
....(1)
Now, the derivative f ¢( z) exits of the limit in equation (1) is unique i.e. it does not depends on the path along which Dz ® 0. www.gatehelp.com
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UNIT 9
Let Dz ® 0 along a path parallel to real axis Þ
Dy = 0 \ Dz ® 0
Þ
Now let v be the conjugate of u then ¶v ¶v ¶u ¶u dv = dx + dy = dx + dy ¶x ¶y ¶y ¶x
Dx ® 0
Now equation (1) Du + iDv Du Dv f ¢( z) = lim = lim + i lim Dx ®0 Dx ®0 Dx Dx ®0 Dx Dx ¶u ¶v f ¢( z) = +i ¶x ¶x
Engineering Mathematics
(by Cauchy-Riemann equation) Þ ....(2)
dv = 2 x dx + 2(1 - y) dy
On integrating v = x 2 - y 2 + 2 y + C
Again, let Dz ® 0 along a path parallel to imaginary
5. (C) Given f ( z) = u + i v
....(1)
axis, then Dx ® 0 and Dz ® 0 ® Dy ® 0
Þ
....(2)
Thus from equation (1) Dz + iDv Du Dv ¶u ¶v = lim + i lim = + f¢( z) = lim Dy ®0 Dy ®0 iDy Dy ®0 iDz iD y i ¶y ¶y
add equation (1) and (2)
f ¢( z) =
-i ¶ u ¶ v + ¶y ¶y
....(3)
Now, for existence of f ¢( z) R.H.S. of equation (2) and (3) must be same i.e., ¶u ¶v ¶v ¶u +i = -i ¶x ¶x ¶y ¶y
if ( z) = -v + iu
Þ
(1 + i) f ( z) = ( u - v) + i( u + v)
Þ
F ( z) = U + iV
where, F ( z) = (1 + i) f ( z); U = ( u - v); V = u + v Let F ( z) be an analytic function. Now, U = u - v = e x (cos y - sin y) ¶U ¶U and = e x (cos y - sin y) = e x ( - sin y - cos y) ¶x ¶y Now, dV =
¶u ¶v ¶v -¶u and = = ¶x ¶y ¶x ¶y
-¶U ¶U dx + dy....(3) ¶y ¶x
= e x (sin y + cos y) dx + e x (cos y - sin y) dy
¶u ¶u ¶v ¶v f ¢( z) = -i = +i ¶x ¶y ¶y ¶x
= d[ e x (sin y + cos y)] on integrating V = e x (sin y + cos y) + c1
3. (A) Given f ( z) = x 2 + iy 2 since, f ( z) = u + iv
F ( z) = U + iV = e x (cos y - sin y) + ie x (sin y + cos y) + ic1
Here u = x 2 and v = y 2 ¶u ¶u Now, u = x 2 Þ = 2 x and =0 ¶x ¶y
= e x (cos y + i sin y) + ie x (cos y + i sin y) + ic1
and v = y 2
and f ¢( z) =
(1 + i) f ( z) = (1 + i) e z + ic1 ( i + 1) i i(1 - i) Þ f ( z) = e z + c1 = e z + c1 = ez + c1 2 1+ i (1 + i)(1 - i)
¶v ¶v = 0 and =2y ¶x ¶y
Þ
we know that
F ( z) = (1 + i) e x + iy + ic1 = (1 + i) ez + ic1
f ¢( z) =
¶u ¶u -i ¶x ¶y
....(1)
Þ
f ( z) = e z + (1 + i) c
6. (C) u = sinh x cos y
¶v ¶v + i ....(2) ¶y ¶x
Now, equation (1) gives f ¢( z) = 2 x
....(3)
and equation (2) gives f ¢( z) = 2 y
....(4)
Now, for existence of f ¢( z) at any point is necessary that the value of f ¢( z) most be unique at that point, whatever be the path of reaching at that point
¶u = cosh x cos y = f( x, y) ¶x ¶u and = - sinh x sin y = y( x, y) ¶y by Milne’s Method f ¢( z) = f( z, 0) - iy( z, 0) = cosh z - i × 0 = cosh z On integrating f ( z) = sinh z + constant
From equation (3) and (4) 2 x = 2 y Hence, f ¢( z) exists for all points lie on the line x = y.
Þ
f ( z) = w = sinh z + ic
¶u ¶2u 4. (B) = 2(1 - y) ; =0 ¶x ¶x 2
....(1)
the function x and hence i.e. in w).
¶u ¶2u = -2 x ; =0 ¶y ¶y 2
....(2)
7. (A)
(As u does not contain any constant, the constant c is in
Þ Page 568
¶2u ¶2u + = 0, Thus u is harmonic. ¶x 2 ¶y 2
¶v ¶v = 2 y = h( x, y), = 2 x = g( x, y) ¶x ¶y
by Milne’s Method f ¢( z) = g( z, 0) + ih( z, 0) = 2 z + i 0 = 2 z On integrating f ( z) = z 2 + c www.gatehelp.com
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Complex Variables
8. (D)
¶v -( x 2 + y 2 ) - ( x - y)2 y = ¶y ( x2 + y2 )2
f ¢¢( z o) =
y - x - 2 xy = g( x, y) ( x2 + y2 )2 2
=
2
Chap 9.5
n! 2 pi
f ( z) dz
ò (z - z ) o
c
Taking n = 3,
or
n +1
c
f ( z) dz
ò (z - z )
4
=
o
c
f ( z) dz
ò (z - z ) pi f ¢¢( z o) 3
n +1
=
o
2pi n f ( z o) n!
....(1)
e 2 z dz e 2 z dz = ( z + 1) 4 òc [ z - ( -1)]4
¶v ( x 2 + y 2 ) - ( x - y)2 x y 2 - x 2 + 2 xy = = = h( x, y) ¶x ( x2 + y2 )2 ( x2 + y2 )2
Given fc
By Milne’s Method
Taking f ( z) = e 2 z , and z o = -1 in (1), we have
f ¢( z) = g( z, 0) + ih( z, 0) = -
e 2 z dz
1 1 æ 1 ö + iç - 2 ÷ = - (1 + i) 2 z2 z z è ø
ò ( z + 1)
=
c
pi f ¢¢¢( -1)....(2) 3
Þ
f ¢¢¢( -1) = 8 e
Þ
e 2 z dz
ò ( z + 1)
4
Since, f ( z) is analytic within and on z = 3 8 pi - z e 2 z dz = e ò 4 3 |z |= 3 ( z + 1)
By Milne’s Method
12. (D) Since,
f ¢( z) = f( z, 0) - iy( z, 0) 2 cos 2 z - 2 -2 = - i(0) = = - cosec2 z (1 - cos 2 z) 2 1 - cos 2 z
1 - 2z
c
Formula 1 I1 = ò dz = 2 pi c z
On A, z = 1 + i and On B, z = 2 + 4 i Let z = 1 + i corresponds to t = 0
é 1 ê f ( z o) = 2pi ë
x = b, y = d
b = 1, d = 1
ò f ( z) dz = ò ( x c
2
x = a + b, y = c + d a = 1, c = 3
+ ixy)( dx + idy)
ò [( t + 1)
everywhere in c i.e. z = 15 . , hence by Cauchy’s integral
+ i( t + 1)( 3t + 1)][ dt + 3i dt ]
t= 0
theorem 1 I3 = ò dz = 0....(4) -2 z c
1
= ò [( t 2 + 2 t + 1) + i( 3t 2 + 4 t + 1)](1 + 3i) dt 0
1
é t3 ù 29 = (1 + 3i) ê + t 2 + t + i( t 3 + 2 t 2 + t) ú = + 1 1i 3 3 ë û0
using equations (2), (3), (4) in (1), we get 1 - 2z 1 3 òc z( z - 1)( z - 2) dz = 2 (2 pi) + 2 pi - 2 (0) = 3pi
11. (D) We know by the derivative of an analytic function that
ò
the circle z = 15 . , so the function f ( z) is analytic
1
=
....(2)
inside z = 15 . , therefore I 2 = 2 pi....(3) 1 For I 3 = ò dz, the singular point z = 2 lies outside -2 z c
dx = dt ; dy = 3 dt
c
2
and it
f ( z) dz ù ú [Here f ( z) = 1 = f ( z o) and z o = 0] c z - zo û 1 Similarly, for I 2 = ò dz, the singular point z = 1 lies -1 z c
and z = 2 + 4 i corresponding to t = 1
2 = a + 1, 4 = c + 1 Þ
1
ò z dz
z = 15 . , therefore by Cauchy’s integral
lies inside
AB is , y = 3t + 1 Þ
1 3 I1 + I 2 - I 3....(1) 2 2
c
10. x = at + b, y = ct + d
Þ
=
Since, z = 0 is the only singularity for I1 =
f ( z) = - ò cosec2 z dz + ic = cot z + ic
Þ
1 - 2z 1 1 3 = + z( z - 1)( z - 2) 2 z z - 1 2( z - 2)
ò z( z - 1)( z - 2) dz
On integrating
and t = 1
....(3)
If is the circle z = 3
¶u 2 sin 2 x sinh 2 y = = y( x, y) ¶y (cosh 2 y - cos 2 x) 2
Þ
8 pi -2 e 3
=
c
2 cos 2 x cosh 2 y - 2 = f( x, y) (cosh 2 y - cos 2 y) 2
Þ
-2
equation (2) have
¶u 2 cos 2 x (cosh 2 y - cos 2 x) - 2 sin 2 2 x 9. (A) = ¶x (cosh 2 y - cos 2 x) 2
then, t = 0
f ¢¢¢( z) = 8 e 2 z
Þ
Now, f ( z) = e 2 z
On integrating 1 1 f ( z) = (1 + i) ò 2 dz + c = (1 + i) + c z z
=
4
13. (B) Given contour c is the circle z = 1
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UNIT 9
Þ
z = e iq
3z 2 + 7 z + 1 òc z - zo dz = 2 pif( zo)
dz = ieiqdq
Þ
Now, for upper half of the circle, 0 £ q £ p 2 ò ( z - z ) dz = c
p
ò (e
iq
Þ
- e 2 iq) ie iqdq
Þ
f ¢( z o) = 2pif¢( z o)
f¢ ( z) = 6 z + 7 and f¢¢( z) = 6
f ¢(1 - i) = 2 pi[ 6(1 - i) + 7 ] = 2 p (5 + 13i)
1 é1 1 ù 2 × ( e 2 pi - 1) - ( e 3px - 1) ú = i êë2 3 û 3
19. (C) f ( z) =
14. (B) Let f ( z) = cos pz then f ( z) is analytic within and on z = 3, now by Cauchy’s integral formula 1 f ( z) dz f ( z) f ( z o) = dz Þ ò = 2pif ( z o) 2pi òc z - z o c z - zo
Þ Þ
take f ( z) = cos pz, z o = 1, we have cos pz òz = 3 z - 1 dz = 2 pif (1) = 2pi cos p = -2pi
z -1 2 =1 z +1 z +1
f (0) = -1, f (1) = 0 2 Þ f ¢( z) = ( z + 1) 2
f ¢¢( z) =
-4 ( z - 1) 3
f ¢¢¢( z) =
12 ( z + 1) 4
Þ Þ
f ¢(0) = 2;
f ¢¢(0) = -4; f ¢¢¢(0) = 12; and so on.
Now, Taylor series is given by
sin pz 2 15. (D) ò dz c ( z - 1)( z - 2)
f ( z) = f ( z 0 ) + ( z - z 0 ) f ¢( z 0 ) +
( z - z0 ) 2 f ¢¢( z 0 ) + 2!
sin pz 2 sin pz 2 dz - ò dz z -2 z -1 c c
( z - z0 ) 3 f ¢¢¢( z 0 ) + ..... 3!
ò
= 2 pif (2) - 2 pif (1) since, f ( z) = sin pz 2
about z = 0
Þ
f ( z) = -1 + z(2) +
f (2) = sin 4 p = 0 and f (1) = sin p = 0
16. (D) Let, I = =
Þ
since, f( z) = 3z 2 + 7 z + 1
p é e 2 iq e 3iq ù = i ò ( e 2 iq - e 3iq)dq = i ê 3i úû 0 ë 2i 0
=
f ( z o) = 2pif( z o)
and f ¢¢( z o) = 2pi f¢¢( z o)
q= 0
p
=i×
Engineering Mathematics
1 2 pi
òz c
2
1 cos pz dz -1
z2 z3 ( -4) + (12) + .... 2! 3!
= -1 + 2 z - 2 z 2 + 2 z 3.... f ( z) = -1 + 2( z - z 2 + z 3 ....)
1 1 ö æ 1 ç ÷ cos pz dz ò 2 × 2 pi c è z - 1 z + 1 ø
20. (B) f ( z) =
1 z +1
f ¢( z) =
-1 ( z + 1) 2
Þ
f ¢(1) =
-1 4
3z 2 + 7 z + 1 òc z - 3 dz , since zo = 3 is the only
f ¢¢( z) =
2 ( z + 1) 3
Þ
f ¢¢(1) =
1 4
3z 2 + 7 z + 1 singular point of and it lies outside the z -3
f ¢¢¢( z) =
-6 ( z + 1) 4
Þ
f ¢¢¢(1) = -
Or I =
1 æ cos nz cos nz ö ç ÷dz z +1 ø 4 pi òc è z - 1
17. (D) f ( 3) =
circle x 2 + y 2 = 4 i.e., z = 2, therefore
3z 2 + 7 z + 1 is z -3
Þ
f (1) =
f ( z) = f ( z 0 ) + ( z - z 0 ) f ¢( z 0 ) +
( z - z0 ) 2 f ¢¢( z 0 ) 2! +
Hence by Cauchy’s theorem—
c
3z 2 + 7 z + 1 dz = 0 z -3
( z - z0 ) 3 f ¢¢¢( z 0 ) + K 3!
about z = 1 f ( z) =
18. (C) The point (1 - i) lies within circle z = 2 ( . .. the distance of 1 - i i.e., (1, 1) from the origin is 2 which is less than 2, the radius of the circle). Let f( z) = 3z 2 + 7 z + 1 then by Cauchy’s integral formula Page 570
3 and so on. 8
Taylor series is
analytic everywhere within c. f ( 3) = ò
1 2
=
2 1 æ -1 ö ( z - 1) + ( z - 1)ç ÷+ 2 2! è 4 ø
3 æ 1 ö ( z - 1) æ 3 ö ç - ÷+K ç ÷+ 3! è 8 ø è4ø
1 1 1 1 ( z - 1) + 3 ( z - 1) 2 - 4 ( z - 1) 3 +.... 2 22 2 2
or f ( z) =
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1 1 1é 1 ù 1 - ( z - 1) + 2 ( z - 1) 2 - 3 ( z - 1) 3 + ....ú 2 êë 2 2 2 û
GATE EC BY RK Kanodia
Complex Variables
21. (A) f ( z) = sin z f ¢( z) = cos z
p 1 æ pö f ç ÷ = sin = 4 2 è4ø
Þ
ö 1æ 1æ 1 1 1 z z2 z3 ö çç 1 + + + + .. ÷÷ - ç 1 + + 2 + 3 + K÷ 2è 2 4 9 z z z z è ø ø 1 1 1 1 3 or f ( z) = K-z -4 - z -2 - z -1 - - z - z 2 z -K 2 4 8 18 f ( z) = -
1 æ pö f ¢ç ÷ = 2 è4ø
Þ
f ¢¢( z) = - sin z
Þ
1 æ pö f ¢¢ç ÷ = 4 2 è ø
f ¢¢¢( z) = - cos z
Þ
1 æ pö and so on. f ¢¢¢ç ÷ = 4 2 è ø
24. (C)
( z - z0 ) f ¢¢( z 0 ) 2!
about z =
( z - z0 ) 3 f ¢¢¢( z 0 ) + .... 3!
p 4 2
æ 1 ö ÷ çç 2 ÷ø è
pö æ çz - ÷ 4ø +è 3!
22. (D) Let f ( z) = z -2 =
1 1 = 2 z [1 - (1 + z)]2
=
-1
=
1æ 2 4 8 ö ç 1 + + 2 + 3 + .... ÷ zè z z z ø
f ( z) =
1 3 7 + + +K z2 z3 z4
25. (B) z < 1,
zö 1 1 1æ = - ç1 - ÷ z -2 z -1 2è 2ø
1 1 1 1 = + z( z - 1)( z - 2) 2 z z - 1 2( z - 2)
Since, 1 + z < 1, so by expanding R.H.S. by binomial
For z - 1 < 1 Let z - 1 = u
theorem, we get
Þ
+ ( n + 1)(1 + z) n + K or f ( z) = z -2 = 1 +
å ( n + 1)( z + 1)
z = u + 1 and u < 1 1 1 1 1 = + z( z - 1)( z - 2) 2 z z - 1 2( z - 2)
=
n
n =1
1 1 1 23. (B) Here f ( z) = ....(1) = ( z - 1)( z - 2) z - 2 z - 1 Since, z > 1
Þ
1 < 1 and z < 2 z
1 1 1æ 1ö = = ç1 - ÷ 1ö zè z -1 zø æ zç 1 - ÷ zø è =
Þ
z 2
2 sx s y i.e. if ( s y - sx ) 2 > 0, which is true. 46. (A) r = 1.6 ´ 0.4 = .64 = 0.8 byx = r × m1 =
sy sx
Þ
sy sx
=
byx r
1 sy 1 5 , × = ´2= r sx 0.8 2
æ m - m2 tan q = çç 1 è 1 + m1 m2
=
1.6 =2 0.8 m2 = r ×
sy sx
= 0.8 ´ 2 = 1.6.
ö æ 2.5 - 1.6 ö 0.9 ÷÷ = çç = 0.18. ÷÷ = ø è 1 + 2.5 ´ 1.6 ø 5
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GATE EC BY RK Kanodia
(B)
x2 x5 x8 x11 + + + 2 20 160 4400
(C)
x2 x5 x8 x11 + + + 2 20 160 2400 2
(D)
5
8
Chap 9.7
Statement for Q. 18–19: dy For = 1 + y 2 given that dx x:
0
0.2
0.4
0.6
y:
0
0.2027
0.4228
0.6841
11
x x x x + + + 2 40 480 2400
12. For dy dx = xy given that y = 1 at x = 0. Using Euler method taking the step size 0.1, the y at x = 0.4 is
Using Milne’s method determine the value of y for x given in question.
(A) 1.0611
(B) 2.4680
18. y (0.8) = ?
(C) 1.6321
(D) 2.4189
(A) 1.0293
(B) 0.4228
(C) 0.6065
(D) 1.4396
Statement for Q. 13–15. For dy dx = x 2 + y 2 given that y = 1 at x = 0.
19. y (10 . ) =?
Determine the value of y at given x in question using
(A) 1.9428
(B) 1.3428
modified method of Euler. Take the step size 0.02.
(C) 1.5555
(D) 2.168
13. y at x = 0.02 is
Statement for Q.20–22:
(A) 1.0468
(B) 1.0204
(C) 1.0346
(D) 1.0348
Apply Runge Kutta fourth order method to obtain y (0.2), y (0.4) and y (0.6) from dy dx = 1 + y 2 , with y = 0 at x = 0. Take step size h = 0.2.
14. y at x = 0.04 is (A) 1.0316
(B) 1.0301
(C) 1.403
(D) 1.0416
20. y (0.2) = ?
15. y at x = 0.06 is (A) 1.0348
(B) 1.0539
(C) 1.0638
(D) 1.0796
(A) 0.2027
(B) 0.4396
(C) 0.3846
(D) 0.9341
21. y (0.4) = ? (A) 0.1649
(B) 0.8397
(C) 0.4227
(D) 0.1934
16. For dy dx = x + y given that y = 1 at x = 0. Using
22. y (0.6) = ?
modified Euler’s method taking step size 0.2, the value
(A) 0.9348
(B) 0.2935
(C) 0.6841
(D) 0.563
of y at x = 1 is (A) 3.401638
(B) 3.405417
(C) 9.164396
(D) 9.168238
23. For dy dx = x + y 2 , given that y = 1 at x = 0. Using Runge Kutta fourth order method the value of y
17. For the differential equation dy dx = x - y 2 given
x = 0.2 is (h = 0.2)
that
(A) 1.2735
(B) 2.1635
(C) 1.9356
(D) 2.9468
x:
0
0.2
0.4
0.6
y:
0
0.02
0.0795
0.1762
24. For dy dx = x + y given that y = 1 at x = 0. Using Runge Kutta fourth order method the value of y
Using Milne predictor–correction method, the y at next value of x is (A) 0.2498
(B) 0.3046
(C) 0.4648
(D) 0.5114
at
x = 0.2 is
at
(h = 0.2)
(A) 1.1384
(B) 1.9438
(C) 1.2428
(D) 1.6389 *********
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GATE EC BY RK Kanodia
UNIT 9
SOLUTIONS
x2 = x0 = 35 . -
1. (B) Let f ( x) = x 3 - 4 x - 9
Engineering Mathematics
x1 - x0 f ( x0 ) f ( x1 ) - f ( x0 )
0.5 ( - 0.5441) = 37888 . 0.3979 + 0.5441
Since f (2) is negative and f ( 3) is positive, a root lies
Since f ( 37888 . ) = - 0.0009 and f ( 4) = 0.3979, therefore
between 2 and 3.
the root lies between 3.7888 and 4.
First approximation to the root is 1 x1 = (2 + 3) = 2.5. 2
Taking x0 = 37888 . , x1 = 4, we obtain 0.2112 x3 = 37888 . ( - .009) = 37893 . 0.3988
Then f ( x1 ) = 2.5 3 - 4(2.5) - 9 = - 3.375
Hence the required root correct to three places of
i.e. negative\The root lies between x1 and 3. Thus the
decimal is 3.789.
second approximation 1 x2 = ( x1 + 3) = 2.75. 2
4. (D) Let f ( x) = xe x - 2, Then f (0) = - 2, and
to
the
root
is
Then f ( x2 ) = (2.75) 3 - 4(2.75) - 9 = 0.7969 i.e. positive. The root lies between x1 and x2 . Thus the third 1 approximation to the root is x3 = ( x1 + x2 ) = 2.625. 2 Then
f ( x3) = (2.625) 3 - 4(2.625) - 9 = - 1.4121
i.e.
negative. The root lies between x2 and x3 . Thus the fourth 1 approximation to the root is x4 = ( x2 + x3) = 2.6875. 2 Hence the root is 2.6875 approximately. 2. (B) Let f ( x) = x 3 - 2 x - 5
f (1) = e - 2 = 0.7183 So a root of (i ) lies between 0 and 1. It is nearer to 1. Let us take x0 = 1. Also f ¢( x) = xe x + e x and f ¢(1) = e + e = 5.4366 By Newton’s rule, the first approximation x1 is f ( x0 ) 0.7183 x1 = x0 =1 = 0.8679 f ¢( x0 ) 5.4366 f ( x1 ) = 0.0672, f ¢( x1 ) = 4.4491. Thus the second approximation x2 is f ( x1 ) 0.0672 x2 = x1 = 0.8679 = 0.8528 f ( x1 ) 4.4491 Hence the required root is 0.853 correct to 3 decimal
So that f (2) = - 1 and f ( 3) = 16
places.
i.e. a root lies between 2 and 3. Taking
x0 = 2, x1 = 3, f ( x0 ) = - 1, f ( x1 ) = 16,
in
the
method of false position, we get x1 - x0 1 x2 = x0 f ( x0 ) = 2 + = 2.0588 f ( x1 ) - f ( x0 ) 17
5. (B) Let y = x + log10 x - 3.375 To obtain a rough estimate of its root, we draw the graph of (i ) with the help of the following table :
Now, f ( x2 ) = f (2.0588) = - 0.3908 i.e., that root lies between 2.0588 and 3. Taking x0 = 2.0588, x1 = 3, f ( x0 )
x
1
2
3
4
y
-2.375
-1.074
0.102
1.227
= - 0.3908, f ( x1 ) = 16 in (i), we get 0.9412 x3 = 2.0588 ( - 0.3908) = 2.0813 16.3908
Taking 1 unit along either axis = 0.1, The curve crosses
Repeating this process, the successive approxima- tions
approximation to the root.
are
Now let us apply Newton–Raphson method to
x4 = 2.0862, x5 = 2.0915, x6 = 2.0934, x7 = 2.0941,
f ( x) = x + log10 x - 3.375 1 f ¢( x) = 1 + log10 e x
the x–axis at x0 = 2.9, which we take as the initial
x8 = 2.0943 etc. Hence the root is 2.094 correct to 3 decimal places.
Taking x0 = 35 . , x1 = 4, in the method of false position,
f (2.9) = 2.9 + log10 2.9 - 3.375 = - 0.0126 1 f ¢(2.9) = 1 + log10 e = 11497 . 2.9
we get
The first approximation x1 to the root is given by
3. (C) Let f ( x)2 x - log10 x - 7
Page 584
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GATE EC BY RK Kanodia
x1 = x0 -
f ( x0 ) 0.0126 = 2.9 + = 2.9109 ¢ f ( x0 ) 11497 .
at x = 0,
Thus the second approximation x2 is given by
d2 y =1 + 2 = 3 dx 2
at x = 0, y = 1,
f ( x1 ) 0.0001 = 2.9109 + = 2.91099 f ¢( x1 ) 11492 .
2
Hence the desired root, correct to four significant figures, is 2.911
d 3y d2 y æ dy ö y = 2 2 ç ÷ dx 3 dx 2 è dx ø
6. (B) Let x = 28 so that x 2 - 28 = 0 Taking f ( x) = x 2 - 28, Newton’s iterative method gives 28 ö f ( xn ) x 2 - 28 1 æ ÷ = xn - n = çç xn + ¢ 2 xn 2è f ( xn ) xn ÷ø
é dy d 2 y d 3y ù y 3 + ê dx dx 2 dx 3 úû ë
d4 y = -2 dx 4
at x = 0, y = 1
Now since f (5) = - 3, f ( 6) = 8, a root lies between 5 and 6.
d4 y = 34 dx 4
The Taylor series expression gives y( x + h) = y( x) + h
Taking x0 = 5.5, x1 =
1æ 28 ö 1 æ 28 ö ÷÷ = ç 5.5 + çç x0 + ÷ = 5.29545 2è 5.5 ø x0 ø 2 è
x2 =
1æ 28 ö 1 æ 28 ö ÷÷ = çç 5.29545 + çç x1 + ÷ = 5.2915 2è 5.29545 ÷ø x1 ø 2 è
d 3y =-8 dx 3
x = 0, y = 1,
at
xn + 1 = xn -
dy = -1 dx
y = 1,
d2 y dy =1 -2y 2 dx dx
f ( x1 ) = - 0.0001, f ¢( x1 ) = 11492 .
x2 = x1 -
Chap 9.7
dy h2 d 2 y h3 d 3 y h4 d 4 y +K + + + 3 ! dx 3 4 ! dx 4 dx 2 ! dx 2
y(0.1) = 1 + 0.1( -1) +
(0.1) 2 (0.1) 3 (0.1) 4 3+ ( -8) + 34 + ...... 2! 3! 4!
= 1 - 0.1 + 0.015 - 0.001333 + 0.0001417 = 0.9138 9. (C) Here f ( x, y) = x 2 + y 2 , x0 = 0
1æ 28 ö 1 æ 28 ö ÷ = ç 5.2915 + x3 = çç x2 + ÷ = 5.2915 2è 5.2915 ÷ø x2 ÷ø 2 çè
y0 = 0
We have, by Picard’s method
Since x2 = x3 upto 4 decimal places, so we take
y = y0 +
28 = 5.2915.
x
ò f ( x, y) dx
....(1)
x0
The first approximation to y is given by 7. (B) Let h = 0.1, dy = 1 + xy Þ dx 3
given x0 = 0, d2 y dy =x +y dx 2 dx
2
d y d y dy , =x +2 dx 3 dx 2 dx given that Þ
x1 = x0 + h = 0.1
x = 0,
4
3
2
d y d y d y =x +3 dx 4 dx 3 dx 2 y =1
The Taylor series expression gives : dy h2 d 2 y h3 d 3 y + + + 3 ! dx 3 dx 2 ! dx 2
Þ
y (0.1) = 1 + 0.1 ´ 1 +
Þ
y(0.1) = 1 + 0.1 +
(0.1) 2 (0.1) 3 ×1 + 2 +K 2! 3!
0.01 0.001 + +K 2 3
= 1 + 0.1 + 0.005 + 0.000033 ......... 8. (B) Let h = 0.1,
given
dy = x - y2 x1 = x0 + h = 0.1, dx
x
ò f ( x, y ) dx 0
x0
dy d2 y d 3y d4 y , = 1 ; 2 = 1, = 2 = 3 and so on dx dx dx 3 dx 4
y( x + h) = y( x) + h
y (1 ) = y0 +
x0 = 0,
Where y0 = 0 +
x
x
ò f ( x, 0) dx = ò x dx. 2
0
...(2)
0
The second approximation to y is given by x
ò f ( x, y
y ( 2 ) = y0 +
(1 )
) dx = 0 +
0
x0
æ x = 0 + ò çç x 2 + 9 0è x
Now,
6
æ
x3 ö ÷dx 3 ÷ø
ö x x ÷÷dx = + 3 63 ø 3
y (0.4) =
10. (C) Here
x
ò f ççè x,
7
(0.4) 3 (0.4) 7 + = 0.02135 3 63
f ( x, y) = y - x ; x0 = 0, y0 = 2
We have by Picard’s method
= 11053 . y0 = 1
y = y0 +
ò
x
x0
f ( x, y) dx
The first approximation to y is given by y (1 ) = y0 +
x
ò f ( x, y0 ) dx = 2 +
x0
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x
ò f ( x, 2) dx 0
Page 585
GATE EC BY RK Kanodia
UNIT 9
x
= 2 + ò (2 - x) dx
=2 + 2x -
0
x2 2
....(1)
x
ò f ( x, y
(1 )
=2 +
Here
æ
ò f ççè x, 2 + 2 x -
x0 2
= 2 + ò (2 + 2 x 0
x 2
....(2)
æ
ò f ççè x, 2 + 2 x +
=2 +
x0
y4 = y3 + hf ( x3 , y3) = 10302 . + 0.1 f (0.3 , 10302 . )
x x ö ÷dx 2 6 ÷ø 2
= 10302 . + 0.03090
3
. y4 = y( 0 .4 ) = 10611 Hence y( 0 .4 ) = 10611 .
ö æ x x = 2 + ò çç 2 + 2 x + - ÷÷dx 2 6 ø 0è x
2
3
13. (B) The Euler’s modified method gives
x2 x3 x4 + 2 6 24
=2 + 2x +
11. (B) Here
f ( x, y) = x + y 2 , x0 = 0
y1* = y0 + hf ( x0 , y0 ), h y1 = y0 + [ f ( x0 , y0 ) + f ( x1 , y1*)] 2
y0 = 0
Now, here h = 0.02, y0 = 1, x0 = 0
We have, by Picard’s method y = y0 +
. y1* = 1 + 0.02 f (0, 1), y1* = 1 + 0.02 = 102 h Next y1 = y0 + [ f ( x0 , y0 ) + f ( x , y1*)] 2 0.02 =1 + [ f (0, 1) + f (0.02, 102 . )] 2
x
ò f ( x, y0 ) dx
x0
The first approximation to y is given by y (1 ) = y0 +
x
ò f ( x, y ) dx 0
=0 +
x0 x
x
ò f ( x, 0) dx
= 1 + 0.01 [1 + 10204 . ] = 10202 .
0
2
x = 0 + ò xdx = 2 0
So,
x
ò f ( x, y
(1 )
) dx = 0 +
x0
x
æ
ò f ççè x, 0
x2 2
ö ÷÷dx ø
= 10202 . + 0.02 [ f (0.02, 10202 . )] = 10202 . + 0.0204
= 10406 . h Next y2 = y1 + [ f ( x, y) + f ( x2 , y2* )] 2 0.02 y2 = 10202 . + [ f (0.02, 10202 . ) + f (0.04, 10406 . )] 2
æ x x x ö ÷÷dx = + = ò çç x + 2 50 4 ø 0è x
4
2
5
The third approximation is given by y ( 3) = y0 +
x
ò f ( x, y
(2 )
= 10202 . + 0.01 [10206 . + 10422 . ] = 10408 .
) dx
x0
. y2 = y( 0 .04 ) = 10408
æ x2 x5 ö ÷dx = 0 + ò f çç x, + 2 20 ÷ø 0 è x
15. (C) y3* = y2 + hf ( x2 , y2 )
x æ 2 x7 ö x2 x5 x8 x11 x4 x10 ÷÷dx = + + + = ò çç x + + + 2 20 160 4400 4 400 40 ø 0è
12. (A) x: 0 Page 586
0.1
0.2
. y1 = y (0.02) = 10202
14. (D) y2* = y1 + h f ( x1 , y1 )
The second approximation to y is given by y ( 2 ) = y0 +
y3 = y2 + hf ( x2 , y2 ) = 101 . + 0.1 f (0.2 , 101 . ) n = 3 in (1) gives
) dx
x0 x
n = 2 in (1) gives . + 0.0202 = 10302 . y3 = y( 0 .3) = 101
x
(2 )
h = 0.1
Thus y2 = y( 0 .2 ) = 101 .
3
ò f ( x, y
y0 = 1,
= 1 + 0.1 f (0.1 , 1) = 1 + 0.1 (0.1) = 1 + 0.01
The third approximation to y is given by y ( 3) = y0 +
x0 = 0,
n = 0 in (1) gives y2 = y1 + h f ( x1 , y1 )
x x 2 6
=2 + 2x +
....(1)
y1 = 1 + 0.1 f (0, 1) = 1 + 0 = 1
ö ÷÷dx ø
2
x2 - x) dx 2
2
yn + 1 = yn + h( xn , yn ) y1 = y0 + hf ( x0 , y0 )
) dx
x0 x
Euler’s method gives n = 0 in (1) gives
The second approximation to y is given by y ( 2 ) = y0 +
Engineering Mathematics
0.3
0.4
= 10416 . + 0.02 f (0.04, 10416 . ) = 10416 . + 0.0217 = 10633 . h Next y3 = y2 + [ f ( x2 , y2 ) + f ( x3 , y3*)] 2
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UNIT 9
Engineering Mathematics
1 1 ö æ k2 = hf ç x0 + h, y0 + k1 ÷ = (0.2) f (0.1, 0.1) = 0.202 2 2 ø è
h k ö æ k2 = hf ç x0 + , y0 + 1 ÷ 2 2ø è
1 1 ö æ k3 = hf ç x0 + h, y0 + k2 ÷ = (0.2) f (0.1, 0.101) = 0.2020 2 2 è ø
= (0.2) f (0.1, 11 . ) = 0.2(1.31) = 0.262 h k ö æ k3 = hf ç x0 + , y0 + 2 ÷ 2 2 ø è
k4 = hf ( x0 + h, y0 + k3) = 0.2 f (0.2, 0.2020) = 0.20816 1 k = [ k1 + 2 k2 + 2 k3 + k4 ] 6 1 = [0.2 + 2 (.202) + 2 (.20204) + 0.20816 ], 6 k = 0.2027 such that
y1 = y(0.2) = y0 + k = 0 + 0.2027 = 0.2027
21. (C) We now to find y2 = y(0.4), k1 = hf ( x1 , y1 ) = (0.2) f (0.2, 0.2027) = 0.2 (10410 . )
= .2082
= 0.2 f (0.1, 1131 . ) = 0.2758 k4 = hf ( x0 + h, y0 + k3) = (0.2) f (0.2, 12758 . ) = 0.3655 1 k = [ k1 + 2 k2 + 2 k3 + 2 k4 ] 6 1 = [0.2 + 2 (0.262) + 2 (0.2758) + 0.3655 ] = 0.2735 6 Here
y1 = y( 0 .2 ) = y0 + k
= 1 + 0.2735
Þ
12735 .
1 1 ö æ k2 = hf ç x1 + h , y1 + k1 ÷ 2 2 ø è
24. (C) Here f ( x, y) = x + y h = 0.2
= (0.2) f (0.3, 0.3068)
k1 = hf ( x0 , y0 ) = 0.2 f (0, 1) = 0.2 h k ö æ . ) = 0.24 k2 = hf ç x0 + , y0 + 1 ÷ = (0.2) f (0.1, 11 2 2ø è
To find y1 = y( 0 .2 ) ,
= 0.2188
1 1 ö æ k3 = hf ç x1 + h , y1 + k2 ÷ 2 2 è ø
h k ö æ . ) = 0.244 k3 = hf ç x0 + , y0 + 2 ÷ = (0.2) f (0.1, 112 2 2 ø è
= 0.2 f (0.3, 0.3121) = .2194 k4 = hf ( x1 + h, y1 + k3) = 0.2 f (0.4, .4221) = 0.2356 1 k = [ k1 + 2 k2 + 2 k3 + k4 ] 6 1 = [0.2082 + 2(.2188) + 2(.2194) + 0.356 ] = 0.2200 6
k4 = hf ( x0 + h, y0 + k3) = (0.2) f (0.2, 1244 . ) = 0.2888 1 k = [ k1 + 2 k2 + 2 k3 + k4 ] 6 1 = [0.2 + 2(0.24) + 2(0.244) + 0.2888 ] = 0.2428 6
y2 = y( 0 .4 ) = y1 + k = 0.2200 + .2027 = 0.4227
y1 = y( 0 .2 ) = y0 + k = 1 + 0.2428
= 12428 .
22. (C) We now to find y3 = y( 0 .6 ) , k1 = hf ( x2 , y2 ) = (0.2) f (0.4, 0.4228)
= 0.2357
1 1 ö æ k2 = hf ç x2 + h, y2 + k1 ÷ 2 2 è ø
***********
= (0.2) f (0.5, 0.5406) = 0.2584 1 1 ö æ k3 = hf ç x2 + h, y2 + k2 ÷ 2 2 ø è = 0.2 f (0.5, .5520) = 0.2609 1 k4 = [ k1 + 2 k2 + 2 k3 + k4 ] 6 1 = [0.2357 + 2(.2584) + 2(0.2609) + 0.2935 ] 6 1 = [0.2357 + 0.5168 + 0.5218 + 0.2935 ] = 0.2613 6 y3 = y( 0 .6 ) = y2 + k = .4228 + 0.2613 = 0.6841 23. (A) Here given f ( x, y) = x + y
x0 = 0
y0 = 1,
h = 0.2
2
To find y1 = y( 0 .2 ) , k1 = hf ( x0 , y0 ) Page 588
= (0.2) f (0, 1) = (0.2) ´ 1 = 0.2 www.gatehelp.com
GATE EC BY RK Kanodia
CHAPTER
10.5 EC-07
1. If E Denotes expectation, the variance of a random
6. For the function e- x , the linear approximation around
variable X is given by
x = 2 is
(A) E[ X 2 ] - E 2 [ X ]
(B) E[ X 2 ] + E 2 [ X ]
2
2
(C) E[ X ]
(D) E [ X ]
2. The following plot shows a function y which varies 2
linearly with X. The value of the integral I = ò y dx is y
1
(B) 1 - x
(C) [ 3 + 2 2 - 1(1 + 2 x ]e -2
(D) e -2
7. An independent voltage source in series with an impedance Z s = Rs + jX s delivers a maximum average power to a load impedance Z L when
3 2 1 -1
(A) ( 3 - x) e -2
1
2
x
3
(A) Z L = RS + jX S
(B) Z L = RS
(C) Z L = jX S
(D) Z L = RS - jX S
8. The RC circuit shown in the figure is R
(A) 1.0
(B) 2.5
(C) 4.0
(D) 5.0
C +
+
Vi
R
C
Vo
3. For x > ni ) the electron concentration per
5. Which of the following functions is strictly bounded ?
(C) x 2
(B) a high-pass filter
semiconductor are ni per cm 3 at 300 K. Now, if acceptor
(B) 1
1 x2
(A) a low-pas filter
9. The electron and hole concentrations in an intrinsic
(A) 0.5
(A)
-
-
(B) x 2
(A) x
(B) e x (D) e - x
2
cm 3 at 300 K will be (A) ni
(B) ni + N A
(C) N A - ni
(D)
www.gatehelp.com
ni2 NA Page 639
GATE EC BY RK Kanodia
UNIT 10
10. In a
p+ n junction diode under reverse biased the
Previous year Papers
15. If closed-loop transfer function of a control system is
magnitude of electric field is maximum at
given as T( s) =
(A) the edge of the depletion region on the p -side
(A) an unstable system
(B) the edge of the depletion region on the n -side
(B) an uncontrollable system
(C) the p+ n junction
(C) a minimum phase system
(D) the center of the depletion region on the n-side
(D) a non-minimum phase system
11. The correct full wave rectifier circuit is
16. If the Laplace transform of a signal y( t) is
Output
(B)
Input
Output
(A)
Input
Y ( s) =
1 s ( s -1 )
s-5 ( s + 2 )( s + 3)
then It is
, then its final value is
(A) -1
(B) 0
(C) 1
(D) unbounded
17. If R( t) is the auto correlation function of a real, wide-sense stationary random process, then which of (A) R( t) = R( -t)
Output
(D)
Input
Output
C
Input
the following is NOT true (B) R( t) £ R(0) (C) R( t) = - R( -t) (D) The mean square value of the process is R(0) 12. In a trans-conductance amplifier, it is desirable to
18. If S( f )is the power spectral density of a real,
have
wide-sense stationary random process, then which of the following is ALWAYS true?
(A) a large input resistance and a large output resistance (B) a large input resistance and a small output resistance
(A) S(0) £ S( f )
(B) S( f ) ³ 0
(C) S( - f ) = -S( f )
(D)
¥
ò S( f )df
=0
-¥
(C) a small input resistance and a large output resistance
19. A plane wave of wavelength l is traveling in a direction making an angle 30 o with positive x -axis. The
(D) a small input resistance and a small output resistance
®
E field of the plane wave can be represented as (E0 is constant)
13. X = 01110 and Y = 11001 are two 5-bit binary numbers represented in two's complement format. The sum of X and Y represented in two's complement format using 6 bits is (A) 100111
(B) 0010000
(C) 000111
(D) 101001
®
æ 3p p ÷ö j çç wt x z l l ÷ø è
®
æ 3p p ö÷ j çç wt + x z l l ÷ø è
(A) E = yE0 e (C) E = yE0 e
®
æ p 3p ÷ö j çç wt - x z l l ø÷ è
®
æ p 3p ö÷ j çç wt - x + z l l ÷ø è
(B) E = yE0 e (D) E = yE0 e
20. If C is close curve enclosing a surface S, then the ®
®
the electric flux density D are related by 14. The Boolean function Y = AB + CD is to be realized using only 2-input NAND gates. The minimum number of gates required is (A) 2
(B) 3
(C) 4
(D) 5
Page 640
®
magnetic field intensity H , the current density j and
® æ® ö ¶D÷ ® ç (A) ò ò H . d s = ò j + .d l ç ¶t ÷ s c è ø ®
®
® æ® ö ¶D ÷ ® ç (B) ò H . d l = òò j + d .d s ç ¶t ÷ s s è ø ®
www.gatehelp.com
®
GATE EC BY RK Kanodia
UNIT 10
Previous year Papers
29. For the circuit shown in the figure, the Thevenin
33. Group I lists four types of p - n junction diodes.
voltage and resistance looking into X - Y are
match each device in Group I with one of the option in
1W X
Group II to indicate the bias condition of the device in its normal mode of operation. Group-I (P) Zener Diode (Q) Solar cell (R) LASER diode (S) Avalanche Photodiode
2W
2A
1W
2i
Y
(A) (C)
4 V, 2W 3
2 (B) 4 V, W 3
4 2 V, W 3 3
(D) 4 V, 2W
Group-II (1) Forward bias (2) Reverse bias
(A) P - 1 Q - 2 R - 1 S - 2 (B) P - 2 Q - 1 R - 1 S - 2 (C) P - 2 Q - 2 R - 1 S - 2
30. In the circuit shown, Vc is 0 volts at t = 0 sec. for t > 0, the capacitor current ic ( t), where t is in seconds, is given by
(D) P - 2 Q - 1 R - 2 S - 2 34. The DC current gain (b) of a BJT is 50. Assuming that the emitter injection efficiency is 0.995, the base
iC
20 kW
4 mF
20 kW
10 V
transport factor is + VC -
(A) 0.980
(B) 0.985
(C) 0.990
(D) 0.995
35. group I lists four different semiconductor devices. (A) 0.50 exp( -25 t) mA
match each device in Group I with its characteristic
(B) 0.25 exp( -25 t) mA
property in Group II.
(C) 0.50 exp( -25 t) mA (D) 0.25 exp( -25 t) mA 31. In the AC network shown in the figure, the phasor
(B) P - 1 Q - 4 R - 3 S - 2
A
~
Group-II (1) Population inversion (2)Pinch-off voltage (3) Early effect (4) Fat-band voltage
(A) P - 3 Q - 1 R - 4 S - 2
voltage V AB (in volts) is
A
Group-I (P)BJT (Q)MOS capacitor (R) LASER diode (S) JFET
5W
(C) P - 3 Q - 4 R - 1 S - 2
5W
(D) P - 3 Q - 2 R - 1 S - 4 -j3
j3
36. For the Op-Amp circuit shown in the figure, Vo is
B
(A) 0 (C) 12.5 Ð30 o
2 kW
(B) 5 Ð30 o (D) 17 Ð30 o
1 kW vo
1V
32. A p+ n junction has a built-in potential of 0.8 V. The
1 kW 1 kW
depletion layer width at reverse bias of 1.2V is 2 mm. For a reverse bias of 7.2 V, the depletion layer width will be (A) 4 mm
(B) 4.9 mm
(C) 8 mm
(D) 12 mm
(A) -2 V
(B) -1 V
(C) -0.5 V
(D) 0.5 V
37. For the BJT circuit shown, assume that the b of the transistor is very large and VBE = 0.7 V . The mode of operation of the BJT is
Page 642
www.gatehelp.com
GATE EC BY RK Kanodia
EC-07
10 kW
2V
10 V
Chap 10.5
(A) 7.00 to 7.29 V
(B) 7.14 to 7.29 V
(C) 7.14 to 7.43 V
(D) 7.29 to 7.43 V
41.
1 kW
The
Boolean
expression
Y = AB CD + ABCD+ ABCD+ ABCD (A) cut-off
(B) saturation
minimized to
(C) normal active
(D) reverse active
(A) Y = A B CD + A B C + A C D
38. In the Op-Amp circuit shown, assume that the diode current follows the equation I = I s exp(V/VT ). For Vi = 2 V , V0 = V01 ,
and
Vi = 4 V , V0 = V02 .
for
The
relationship between V01 and V02 is
can
be
(B) Y = A B CD + B C D + A B C D (C) Y = AB C D + B C D + AB CD (D) Y = AB C D + B C D + A B C D 42. The circuit diagram of a standard TTL NOT gate is
D
shown in the figure. Vi = 2.5 V , the modes of operation of vi
2 kW
the transistors will be vo
VCC =5 V
(A) V02 = 2 V01
(B) V02 = e V01
(C) V02 = V01 ln 2
(D) V01 - V02 = VT ln 2
100 W
1.4 kW
4 kW 2
Q4 +
39. In the CMOS inverter circuit shown, if the
Q1
Q2
D +
transconductance parameters of the NMOS and PMOS transistors are kn = kp = m n Cox
Wn Ln
= m p Cox
Wp Lp
Q3
= 40 mA/V 2
1 kW
and their threshold voltages are VTHn = VTHp = 1V , the
-
-
current I is 5V
(A) Q1 : revere active;Q2 : normal active; Q3: saturation; Q4 :cut-off
PMOS 2.5 V
I
(B) Q1 : revere active;Q2 : saturation; Q3: saturation; Q4 :cut-off
NMOS
(A) 0 A
(B) 25 mA
(C) 45 mA
(D) 90 mA
(C) Q1 : normal active;Q2 : cut-off; Q3: cut-off; Q4 :saturation (D) Q1 : saturation;Q2 : saturation; Q3: saturation; Q4 :normal active
40. For the Zener diode shown in the figure, the Zener
43. In the following circuit, X is given by
voltage at knee is 7V, the knee current is negligible and 0
the Zener dynamic resistance is 10W. if the input
1 1 0
voltage ( Vi ) range is from 10 to 16V, the output voltage
(V0 ) ranges
from
S1 S0
200 W +
vi
I0 4-to-1 MUX I1 Y I2 I3
A
0 1 1 0
I0 4-to-1 I1 MUX I2 Y I3 S1
B
vo
(A) X = A B C + A B C + A BC + ABC
_
(B) X = ABC + A B C + ABC + ABC www.gatehelp.com
X
S0
C
Page 643
GATE EC BY RK Kanodia
UNIT 10
Previous year Papers
(C) X = AB + BC + AC
47. (A) The 3-dB bandwidth of the low-pas signal e- t u( t),
(D) X = A B + B C + AC 44. The following binary values were applied to the X
where u( t) is the unit step function, is given by 1 1 (A) Hz (B) 2 - 1 Hz 2p 2p
and Y inputs of NAND latch shown in the figure in the
(C) ¥
sequence indicated below
(D) 1 Hz
48. A Hilbert transformer is a
X = 0, Y = 1; X = 0, Y = 0; X = 1, Y = 1.
(A) non-linear system (C) time-varying system
X P
(B) non-causal system (D) low-pass system
49. The frequency response of a linear, time-invariant system is given by H ( f ) = 1 + j510 pf . The step response of
Q Y
the system is
The corresponding stable P , Q outputs will be (A) P = 1, Q = 0; P = 1, Q = 0; P = 1, Q = 0 or P = 0, Q = 1 (B) P = 1, Q = 0; P = 0, Q = 1; or P = 0 Q = 1; P = 0, Q = 1
(C)
(C) P = 1, Q = 0; P = 1, Q = 1; P = 1, Q = 0 or P = 0, Q = 1 (D)P = 1, Q = 0; P = 1, Q = 1; P = 1, Q = 1
t - ö æ (B) 5çç 1 - e 5 ÷÷u( t) è ø
(A) 5( 1 - e -5t ) u( t) 1 (1 - e-5t )u(t) 2
50.
A
5-point
(C)
sequence
t - ö 1æ ç 1 - e 5 ÷u( t) ÷ 5 çè ø
x[ n]
is
given
as
45. For the circuit shown, the counter state (Q1Q0 )
x[ -3] = 1, x[ -2 ] = 1, x[ -1] = 0, x[0 ] = 5, x[1] = 1. Let X ( e jw )
follows the sequence
denote the discrete-time Fourier transform of x[ n]. The p value of
ò X (e
jw
)dw is
-p
(A) 5
(B) 10p (D) 5 + j10 p
(C) 16p Q0
D0
D1
Q1
51. The z-transform x[ z ] of a sequence x[ n] is given by X ( z) = 1 -02.z5-1 . It is given that the region of convergence of x[ n] includes the unit circle. The value of x[0] is
(A) 00, 01, 10, 11, 00
(B) 00, 01, 10, 00, 01
(C) 00, 01, 11, 00, 01
(D) 00, 10, 11, 00, 10
(A) -0.5
(B) 0
(C) 0.25
(D) 0.5
52. A Control system with PD controller is shown in 46.
8085
the figure If the velocity error constant K V = 1000 and
microprocessor system as an I/O mapped I/O as show in
the damping ration z = 0.5, then the value of K P and K D
the figure. The address lines A0 and A1 of the 8085 are
are
An
8255
chip
is
interfaced
to
an
used by the 8255 chip to decode internally its thee ports
R(s)
and the Control register. The address lines A3 to A7 as
C(s)
+
well as the IO/M signal are used for address decoding. The range of addresses for which the 8255 chip would get selected is 8255
(A) K P = 100, K D = 0.09
(B) K P = 100, K D = 0.9
(C) K P = 10, K D = 0.09
(D) K P = 10, K D = 0.9
53. The transfer function of a plant is T( s) = (A) F8H - FBH
(B) F8H - FCH
(C) F8H - FFH
(D) F0H - F7H
Page 644
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5 (s + 5)( s2 + s + 1)
GATE EC BY RK Kanodia
EC-07
The second-order approximation of T( s) using dominant pole concept is 1 (A) (s + 5)(s + 1) (C)
5
(B)
5 s2 + s + 1
(s + 5)(s + 1) 1 s2 + s + 1
(D)
1 s 2 -1
. If the plant is operated in a unity feedback
stabilize this control system is 10( s - 1) 10( s + 4) (A) (B) s+2 s+2 10( s + 2) s + 10
(B)
1 s + 11s + 11
(C)
10 s + 10 s + 11s + 11
(D)
1 s + s + 11
2
(A) (B) (C) (D)
decreasing decreasing decreasing increasing
2
the the the the
step size granular noise sampling rate step size
digital communications. The expression for p( t) with unity roll-off facto is given by
s + 10
p( t) =
55. A unity feedback control system has an open-loop
sin 4 pWt 4 pWt(1 - 16W 2 t 2 )
The value of p( t) at t =
transfer function G( s) =
2
59. The raised cosine pulse p( t) is used for zero ISI in
2(s + 2)
(D)
10 s + 11s + 11 2
can be reduced by
configuration, then the lead compensator that an
(C)
(A)
58. In delta modulation, the slope overload distortion
54. The open-loop transfer function of a plant is given as G( s) =
Chap 10.5
K . s( s 2 + 7 s + 12)
The gain K for which s = 1 + j1 will lie on the root
1 is 4W
(A) -0.5
(B) 0
(C) 0.5
(D) ¥
60. In the following scheme, if the spectrum M ( f ) of
locus of this system is (A) 4
(B) 5.5
(C) 6.5
(D) 10
m( t) is as shown, then the spectrum Y ( f ) of y( t) will be M
m(t)
56. The asymptotic Bode plot of a transfer function is as S
shown in the figure. The transfer function G( s) Hilbert Transform
corresponding to this Bode plot is 0
G(jw)dB 60 dB
-20 dB/dec
(A)
40 dB
(B)
-40 dB/dec
20 dB
w 1
10
20 100
-60 dB/dec
(A)
1 ( s + 1)( s + 20)
(B)
1 s( s + 1)( s + 20)
(C)
100 s( s + 1)( s + 20)
(D)
100 s( s + 1)(1 + 0.05 s)
0
0
(D)
C
0
61.
During
0
transmission
over
a
certain
binary
57. The state space representation of a separately
communication channel, bit errors occurs independently
excited DC servo motor dynamics is given as
with probability p. The probability of AT MOST one bit
é ê ë
dw dt dia dt
1ù é wù é 0 ù ù é-1 ú = ê-1 -10 ú êi ú + ê10 ú u ûë a û ë û û ë
in error in a block of n bits is given by (A) pn (C) np(1 - p) n -1 + (1 - p) n www.gatehelp.com
(B) 1 - pn (D) 1 - (1 - p) n Page 645
GATE EC BY RK Kanodia
UNIT 10
Previous year Papers
62. In a GSM system, 8 channels can co-exist in 200
67. A load of 50W is connected in shunt in a 2-wire
KHz bandwidth using TDMA. A GSM based cellular
transmission line of Z 0 = 50W as shown in the figure.
operator is allocated 5 MHz bandwidth. Assuming a
The 2-port scattering parameter matrix (s-matrix) of
1 5
frequency reuse factor of , i.e. a five-cell repeat pattern,
the shunt element is
the maximum number of simultaneous channels that
é- 1 (A) ê 12 ë 2
can exist in one cell is (A) 200 (C) 25
(B) 40 (D) 5
ù - úû
é0 1 ù (B) ê ú ë1 0 û
1 2 1 2
2 é- 1 3ù (C) ê 23 1ú ë 3 - 3û
63. In a Direct Sequence CDMA system the chip rate is 1.2288 ´ 106 chips per second. If the processing gain is desired to be at Least 100, the data rate (A) must be less than or equal to 12.288 ´ 10 3 bits/sec (B) must be greater than 12.288 ´ 10 3 bits per sec (C) must be exactly equal to 12.288 ´ 10 3 bits per sec
é 1 (D) ê 43 ë- 4
- 43 ù 1ú 4û
68. The parallel branches of a 2-wire transmission line are terminated in 100 W and 200 W resistors as shown in the figure. The characteristic impedance of the line is Z 0 = 50W and each section has a length of 4l . The voltage reflection coefficient G at the input is
(D) can take any value less than 122.88 ´10 3 bits/sec
l/4 l/4
200
64. An air-filled rectangular waveguide has inner
W
dimensions of 3 cm ´ 2 cm. The wave impedance of the TE20 mode of propagation in the waveguide at a frequency
of
30
GHz
is
(free
space
impedance
h0 = 377 W) (A) 308W
(B) 355W
(C) 400W
(D) 461W
200 W
l/4
®
65. The H field (in A/m) of a plane wave propagating in (A) - j
free space is given by ®
H=x
pö 5 3 5 æ cos( wt - bz) + y sinç wt - bz + ÷ h0 h0 2 è ø
The time average power flow density in Watts is h0 100 (B) (A) 100 h0 (C) 50h20
(D)
7 5
(C) j
5 7
69. A
l 2
(B)
-5 7
(D)
5 7
dipole is kept horizontally at a height of
l0 2
above
a perfectly conducting infinite ground plane. The ®
radiation pattern in the lane of the dipole (E plane)
50 h0
looks approximately as (A)
®
y
(B)
y
66. The E field in a rectangular waveguide of inner dimensions a ´ b is given by ®
E=
z
2
wm æ p ö æ 2 px ö ç ÷ H 0 sinç ÷ sin( wt - bz) y h2 è 2 ø è a ø
y
Where H 0 is a constant, and a and b are the dimensions
along
the
x -axis
and
the
(C) TM 20
(D) TE10
Page 646
(D)
y -axis
is (B) TM11
y
C
respectively. The mode of propagation in the waveguide (A) TE20
z
www.gatehelp.com
z
z
GATE EC BY RK Kanodia
EC-07
70. A right circularly polarized (RCP) plane wave is
Common Data for Questions 74, 75 :
incident at an angle of 60 o to the normal, on an air-dielectric interface. If the reflected wave is linearly polarized, the relative dielectric constant xr 2 is
Two 4-ray signal constellations are shown. It is given that f1 and f2 constitute an orthonormal basis for the two constellations. Assume that the four symbols in both the constellations are equiprobable. Let N0 / 2
Linearly Polarized
RCP
Chap 10.5
denote the power spectral density of white Gaussian
air
noise.
Dielectric
(B) 3 (D) 3
(A) 2 (C) 2
0
Common Data for Questions 71, 72, 73: The
figure
shows
the
high-frequency
capacitance-voltage(C-V) characteristics of a Metal/ SiO2/silicon
(MOS)
capacitor
having
an
area
of
1 ´ 10 -4 cm 2 . Assume that the perimitivities ( e 0 e r ) of silicon and Si O2 are 1 ´ 10 -12 F/cm and 35 . ´ 10 -13 F/cm respectively. C
Constellation 1
74. The ratio of the average energy of constellation 1 to the average energy of constellation 2 is (A) 4 a 2
(B) 4
(C) 2
(D) 8
75.
7 pF
Constellation 2
If
these
constellations
are
used
for
digital
communications over an AWGN channel, then which of the following statements is true ? (A) Probability of symbol error for Constellation 1 is lower
1 pF V
0
71. The gate oxide thickness in the MOS capacitor is (A) 50 nm
(B) 143 nm
(C) 350 nm
(D) 1 mm
72. The maximum depletion layer width in silicon is
(B) Probability of symbol error for Constellation 1 is higher (C) Probability of symbol error is equal for both the constellations (D) The value of N 0 will determine which of the two constellations has a lower probability of symbol error,
(A) 0.143 mm
(B) 0.857 mm
Linked Answer Questions: Q. 76 to Q. 85 Carry
(C) 1 mm
(D) 1.143 mm
Two marks Each.
73. Consider the following statements about the C-V characteristics plot:
Statement for Linked Answer Questions 76 & 77: Consider the Op-Amp circuit shown in the figure.
S1: The MOS capacitor has an n-type substrate.
R1
S2: If positive charges are introduced in the oxide, the
R1
C-V plot will shift to the left. vi
Then which of the following is true? (A) Both S1 and S2 are true
vo R C
(B) S1 is true and Se is false (C) S1 is false and S2 is true (D) Both S1 and S2 are false www.gatehelp.com
Page 647
GATE EC BY RK Kanodia
UNIT 10
76. The transfer function V0 ( s)/Vi ( s) is
80. The eigenvalue and eigenvector pairs ( li Vi ) for the
(A)
1 - sRC 1 + sRC
(B)
1 + sRC 1 - sRC
(C)
1 1 - sRC
(D)
1 1 + sRC
system are æ æ é 1ù ö é 1ù ö (A) çç -1, ê ú ÷÷ and çç -2, ê ú ÷÷ ë-1û ø è è ë-2 û ø
77. If Vi = V1 sin( wt) and V0 = V2 sin( wt - f), then the minimum and maximum values of f (in radians) are respectively (A) - 2p and
p 2
(C) -p and 0
(B) 0 and
Previous year Papers
p 2
æ æ é 1ù ö é 1ù ö (B) çç -1, ê ú ÷÷ and çç -2, ê ú ÷÷ ë-1û ø è è ë-2 û ø æ é 1ù ö æ é 1ù ö (C) çç -1ê ú ÷÷ and çç -2, ê ú ÷÷ 1 è ë-2 û ø è ë ûø æ æ é 1ù ö é 1ù ö (D) çç -2, ê ú ÷÷ and çç 1, ê ú ÷÷ 1 ë ûø è è ë-2 û ø
(D) - 2p and 0
Statement for Linked Answer Questions 78 & 79. An 8085 assembly language program is given below.
81. The system matrix A is é 0 1ù (A) ê ú ë-1 1û
1ù é 1 (B) ê ú ë-1 -2 û
1ù é 2 (C) ê ú ë-1 -1û
1ù é 0 (D) ê ú ë-2 -3û
Line 1:
MVI A, B5H
2:
MVI B, OEH
3:
XRI 69H
4:
ADD B
5:
ANI 9BH
6:
CPI 9FH
7:
STA 3010H
density function f ( x) as shown in the figure. Decision
8:
HLT
boundaries of the quantizer are chosen so as t maximize
Statement fo Linked Answer Questions 82 & 83: An input to a 6-level quantizer has the probability
78. The contents of the accumulator just after execution of the ADD instruction in line 4 will be (A) C3H
(B) EAH
(C) DCH
(D) 69H
the entropy of the quantizer output. It is given that 3 consecutive decision boundaries are ' -1' , '0 ' and '1'. f(x) a
b
79. After execution of line 7 of the program, the status -5
of the CY and Z flags will be (A) CY = 0, Z = 0
(B) CY = 0, Z = 1
(C) CY = A, Z = 0
(D) CY = 1, Z = 1
-1
0
1
5
82. The values of a and b are (A) a =
1 1 and b = 6 12
(B) a =
1 3 and b = 5 40
(C) a =
1 1 and b = 4 16
(D) a =
1 1 and b = 3 24
Statement for Linked Answer Questions 80 & 81. Consider a linear system whose state space representation is x( t) = Ax( t). If the initial state vector of é 1ù the system is x(0) = ê ú, then the system response is ë-2 û é e -2 x ù . If the itial state vector of the system x( t) = ê -2 t ú ë-2 e û
83. Assuming that the reconstruction levels of the quantizer are the mid-points of the decision boundaries, the ratio of signal power to quantization noise power is
é 1ù changes to x(0) = ê ú, then the system response ë-2 û
(A)
152 9
(B)
é e- t ù becomes x( t) = ê - t ú ë -e û
(C)
76 3
(D) 28
Page 648
x
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64 3
GATE EC BY RK Kanodia
EC-07
Statement for Linked Answer Questions 84 & 85.
ANSWER
In the digital-to Analog converter circuit shown in the figure below, VR = 10 V and R = 10 kW. R
2R
R
2R
2R
i
R
R
Chap 10.5
1. A
2. B
3. C
4. A
5. D
6. A
7. D
8. C
9. D
10 .C
11. C
12. A
13. C
14. B
15. D
16. A
17. C
18. B
19. A
20. D
21. C
22. A
23. C
24. D
25. B
26. B
27. A
28. D
29. D
30. A
31. D
32. A
33. B
34. B
35. C
2R R vo +
84. The current is (A) 3125 . mA
(B) 62.5mA
36. C
37. B
38. D
39. D
40. C
(C) 125mA
(D) 250mA
41. D
42. B
43. A
44. C
45. B
46. C
47. A
48. A
49. B
50. B
51. D
52. B
53. C
54. A
55. D
56. D
57. A
58. D
59. C
60. A
61. C
62. B
63. A
64. C
65. D
66. A
67. B
68. D
69. C
70. D
71. A
72. B
73. C
74. B
75. B
76. A
77. C
78. B
79. C
80. A
81. D
82. A
83.
84. B
85. C
85. The voltage V0 is (A) -0.781 V
(B) -1.562 V
(C) -3.125 V
(D) -6.250 V
************
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Page 649
GATE EC BY RK Kanodia
CHAPTER
10.1 EC-03
Duration : Three Hours
Maximum Marks : 150
4. The Laplace transform of i( t) is given by I ( s) =
Q.1—30 carry one mark each
As t ® ¥, The value of i( t) tends to
1. The minimum number of equations required to analyze the circuit shown in Fig. Q. 1 is C
C
2 s(1 + s)
(A) 0
(B) 1
(C) 2
(D) ¥
5. The differential equation for the current i( t) in the R
R
~
C
R
circuit of Fig. Q.5 is R
i1(t)
Fig. Q1
(A) 3
(B) 4
(C) 6
(D) 7
sin t
2W
2H
~
1F
Fig. Q5
impedance consisting of 1W resistance in series with 1
d 2i di (A) 2 2 + 2 + i( t) = sin t dt dt
H inductance. The load that will obtain the maximum
(B) 2
d 2i di +2 + 2 i( t) = cos t dt 2 dt
(C) 2
d 2i di +2 + i( t) = cos t 2 dt dt
(D) 2
d 2i di +2 + 2 i( t) = sin t dt 2 dt
2.. A source of angular frequency 1 rad/sec has a source
power transfer is (A) 1 W resistance (B) 1 W resistance in parallel with 1 H inductance (C) 1 W resistance in series with 1 F capacitor (D) 1 W resistance in parallel with 1 F capacitor
6. n-type silicon is obtained by doping silicon with 3. A series RLC circuit has a resonance frequency of
(A) Germanium
(B) Aluminium
1 kHz and a quality factor Q = 100. If each of R, L and C
(C) Boron
(D) Phosphorus
is doubled from its original value, the new Q of the circuit is
7. The bandgap of silicon at 300 K is
(A) 25
(B) 50
(C) 100
(D) 200
(A) 1.36 eV
(B) 1.10 eV
(C) 0.80 eV
(D) 0.67 eV
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GATE EC BY RK Kanodia
EC-03
Chap 10.1
20. A 0 to 6 counter consists of 3 flip flops and a
25. A PD controller is used to compensate a system.
combination circuit of 2 input gate(s). The combination
Compared
circuit consists of
compensated system has
(A) one AND gate
(A) a higher type number
(B) one OR gate
(B) reduced damping
(C) one AND gate and one OR gate
(C) higher noise amplification
(D) two AND gates
(D) larger transient overshoot
21. The Fourier series expansion of a real periodic
26. The input to a coherent detector is DSB-SC signal
signal with fundamental frequency f0 is given by
plus noise. The noise at the detector output is
g p ( t) =
åc
n
n = -¥
e j 2 pf0t . It is given that c3 = 3 + j5. Then c-3 is
to
the
uncompensated
system,
the
(A) the in-phase component
(A) 5 + j 3
(B) -3 - j5
(B) the quadrature component
(C) -5 + j 3
(D) 3 - j5
(C) zero (D) the envelope
22. Let x( t) be the input to a linear, time-invariant system. The required output is 4 x( t - 2). The transfer
27. The noise at the input to an ideal frequency detector
function of the system should be
is white. The detector is operating above threshold. The
(A) 4 e
j 4 pf
(C) 4 e
- j 4 pf
23.
A
(B) 2 e
- j8 pf
power spectral density of the noise at the output is
j8 pf
(A) raised-cosine
(B) flat
(C) parabolic
(D) Gaussian
(D) 2 e sequence
x( n)
with
the
z-transform
X ( z) = z 4 + z 2 - 2 z + 2 - 3z -4 is applied as an input to a
28. At a given probability of error, binary coherent FSK
linear, time-invariant system with the impulse response
is inferior to binary coherent PSK by
h( n) = 2 d( n - 3) where
(A) 6 dB
(B) 3 dB
(C) 2 dB
(D) 0 dB
ì 1, n = 0 d( n) = í î 0, otherwise
29. The unit of Ñ ´ H is
The output at n = 4 is
(A) Ampere
(A) -6
(B) zero
(C) 2
(D) -4
(C) Ampere/meter
(B) Ampere/meter 2
(D) Ampere-meter
30. The depth of penetration of electromagnetic wave in 24. Fig. Q.24 shows the Nyquist plot of the open-loop
a medium having conductivity s at a frequency of 1
transfer function G( s) H ( s) of a system. If G( s) H ( s) has
MHz is 25 cm. The depth of penetration at a frequency
one right-hand pole, the closed-loop system is
of 4 MHz will be
Im GH - plane
w=0
(-1, 0)
Re
(A) 6.25 cm
(B) 12.50 cm
(C) 50.00 cm
(D) 100.00 cm
Q.31—90 carry two marks each.
w is positive
31. Twelve 1 W resistance are used as edges to form a
Fig. Q24
cube. The resistance between two diagonally opposite
(A) always stable (B) unstable with one closed-loop right hand pole (C) unstable with two closed-loop right hand poles (D) unstable with three closed-loop right hand poles
corners of the cube is 5 (A) W 6 (C)
6 5
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(B) 1 W (D)
3 W 2 Page 593
GATE EC BY RK Kanodia
UNIT 10
32. The current flowing through the resistance R in the circuit in Fig. Q.32 has the form P cos 4 t, where P is 1 F 10.24
Previous Year Papers
éV ù ù - Ls ú é I s ù ê ú 1 = s 1 ú êëI 2 s úû ê ú R + Ls + ê0 ú ú Cs û ë û
1 é ê R + Ls + Cs (D) ê - Ls ê ë
M=0.75 H
35. An input voltage 3W
R = 3.92 W V=2cos 4t
v( t) = 10 2 cos ( t + 10) + 10 3 cos (2 + 10 ° ) V
~
is applied to a series combination of resistance R = 1W and an inductance L = 1 H. The resulting steady
Fig. Q32
state current i( t) in ampere is
(A) (0.18 + j 0.72)
(B) (0.46 + j 190 . )
(C) -(0.18 + j 190 . )
(D) -(0.192 + j 0.144)
The circuit for Q.33–34 are given in Fig. Q.33–34. For both the questions, assume that the switch S is in position 1 for a long time and thrown to position 2 at t = 0. 1
(B) 1 - cos ( t + 55 ° ) + 10
3 cos (2 t + 55 ° ) 2
(C) 10 cos ( t - 55 ° ) + 10 cos (2 t + 10 ° - tan -1 2) (D) 1 - cos ( t - 55 ° ) + 10
3 cos (2 t - 35 ° ) 2
C
S
36. The driving-point impedance Z ( s) of a network has
2
the pole-zero locations as shown in Fig. Q.36. If Z(0) = 3, R
V
(A) 10 cos ( t + 55 ° ) + 10 cos (2 t + 10 ° + tan -1 2)
R
i1
L
then Z ( s) is
i2
Im
C 1
Fig. Q33-34
-3
Re
-1
33. At t = 0 + , the current i1 is (A)
-V 2R
-V (C) 4R
(B)
s - plane
-1
-V R
Fig. Q36
(D) zero
34. I1 ( s) and I 2 ( s) are the Laplace transforms of i1 ( t) and
(A)
3( s + 3) s + 2s + 3
(B)
2( s + 3) s + 2s + 2
(C)
3( s - 3) s2 - 2 s - 2
(D)
2( s - 3) s2 - 2 s - 3
2
2
i2 ( t) respectively. The equations for the loop currents I1 ( s) and I 2 ( s) for the circuit shown in Fig. Q.33–34,
37. The impedance parameters Z11 and Z12
after the switch is brought from position 1 to position 2
two-port network in Fig. Q.37 are
at t = 0, are 1 é ê R + Ls + Cs (A)ê - Ls ê ë
2W
1 é ê R + Ls + Cs (B) ê - Ls ê ë
ù é Vù - Ls ú é I s ù ê- ú 1 s = ú 1 ú êëI 2 s úû ê R+ ê 0 ú ú Cs û û ë
1 é ê R + Ls + Cs (C) ê - Ls ê ë
ù é Vù - Ls ú é I s ù ê- ú 1 s = ú 1 ú êëI 2 s úû ê R + Ls + ê 0 ú ú ë û Cs û
Page 594
2W
3W
2
1
éV ù ù - Ls ú é I s ù ê ú 1 = s 1 ú êëI 2 s úû ê ú R+ ê0 ú ú Cs û ë û
1W
1W
2’
1’
Fig. Q37
(A) Z11 = 2.75W and Z12 = 0.25 W (B) Z11 = 3W
and Z12 = 0.5 W
(C) Z11 = 3W
and Z12 = 0.25 W
(D) Z11 = 2.25 W www.gatehelp.com
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and Z12 = 0.5 W
GATE EC BY RK Kanodia
EC-03
Chap 10.1
38. An n-type silicon bar 0.1 cm long and 100 mm 2 in
(A) 2.26 eV
(B) 1.98 eV
cross-sectional
(C) 1.17 eV
(D) 0.74 eV
area
has
a
majority
carrier
concentration of 5 ´ 10 20 / m 3 and the carrier mobility is 0.13 m 2 /V-s at 300 K. If the charge of an electron is 1.5 ´
43..When the gate-to-source voltage ( VGS ) of a MOSFET
10 -19 coulomb, then the resistance of the bar is
with threshold voltage of 400 mV, working in saturation
(A) 106 Ohm
is 900 mV, the drain current is observed to be 1 mA.
(C) 10
-1
(B) 10 4 Ohm
Ohm
(D) 10
-4
Ohm
Neglecting the channel width modulation effect and assuming that the MOSFET is operating at saturation,
39. The electron concentration in a sample of uniformly doped n-type silicon at 300 K varies linearly from 1017 cm 3 at x = 0 to 6 ´ 1016 cm 3 at x = 2 mm. Assume a situation that electrons are supplied to keep this concentration gradient constant with time. If electronic charge is 1.6 ´ 10 -19 coulomb and the diffusion constant Dn = 35 cm 2 s, the current density in the silicon, if no electric field is present, is
the drain current for an applied VGS of 1400 mV is (A) 0.5 mA
(B) 2.0 mA
(C) 3.5 mA
(D) 4.0 mA
44. If P is Passivation, Q is n-well implant, R is metallization and S is source/drain diffusion, then the order in which they are carried out in a standard n-well CMOS fabrication process, is
(A) zero
(B) -112 A cm 2
(C) +1120 A cm 2
(D) -1120 A cm 2
(A) P–Q–R–S
(B) Q–S–R–P
(C) R–P–S–Q
(D) S–R–Q–P
40. Match items in Group 1 with items in Group 2, most suitably.
45. An amplifier without feedback has a voltage gain of
Group 1
Group 2
50, input resistance of 1 kW and output resistance of 2.5
P. LED
1. Heavy doping
kW. The input resistance of the current-shunt negative
Q. Avalanche photo diode
2. Coherent radiation
feedback amplifier using the above amplifier with a
R.Tunnel diode
3.Spontaneous
feedback factor of 0.2, is
S. LASER
4. Current gain
emission
(A) 1/11 kW
(B) 1/5 kW
(C) 5 kW
(D) 11 kW
(A)
(B)
(C)
(D)
P-1
P-2
P-3
P-2
Q-2
Q-3
Q-4
Q-1
46. In the amplifier circuit shown in Fig. Q.46, the
R-4
R-1
R-1
R-4
values of R1 and R2 are such that the transistor is
S-3
S-4
S-2
S-3
operating at VCE = 3 V and I C = 15 . mA when its b is
41. At 300 K, for a diode current of 1 mA, a certain germanium diode requires a forward bias of 0.1435 V,
150. For a transistor with b of 200, the operating point ( VCE , I C ) is VCC = 6 V
whereas a certain silicon diode requires a forward bias R1
of 0.718 V. Under the conditions stated above, the
R2
closest approximation of the ratio of reverse saturation current in germanium diode to that in silicon diode is (A) 1 (C) 4 ´ 10
(B) 5 (D) 8 ´ 10 3
3
Fig. Q46
42. A particular green LED emits light of wavelength 5490 A°. The energy bandgap of the semiconductor material
used
there
is
(Plank’s
(A) (2 V, 2 mA)
(B) (3 V, 2 mA)
(C) (4 V, 2 mA)
(D) (4 V, 1 mA)
constant
= 6.626 ´ 10 -34 J – s) www.gatehelp.com
Page 595
GATE EC BY RK Kanodia
UNIT 10
Previous Year Papers
47. The oscillator circuit shown in Fig. Q.47 has an ideal
51. Three identical amplifiers with each one having a
inverting amplifier. its frequency of oscillation (in Hz) is
voltage gain of 50, input resistance of 1 kW and output resistance of 250 W, are cascaded. The open circuit voltage gain of the combined amplifier is
C
C
C
R
R
R
(A) 49 dB
(B) 51 dB
(C) 98 dB
(D) 102 dB
52. An ideal sawtooth voltage waveform of frequency 500 Hz and amplitude 3 V is generated by charging a capacitor of 2 mF in every cycle. The charging requires
Fig. Q47
1
(A) (2 p 6 RC) 1 (C) ( 6 RC)
1 (B) (2 pRC)
(A) constant voltage source of 3 V for 1 ms
6 (D) (2 pRC)
(C) constant current source of 3 mA for 1 ms
(B) constant voltage source of 3 V for 2 ms
(D) constant current source of 3 mA for 2 ms 48. The output voltage of the regulated power supply 53. The circuit shown in Fig. Q.53 has 4 boxes each
shown in Fig. Q.48 is
described by inputs, P, Q, R and outputs Y, Z with
+
Y = P Å Q Å R, Z = RQ + PR + QP . The circuit acts as
1 kW
a 15 V DC Unregulated Power source
Q Vz = 3 V 40 kW
P
Regulated DC Output
20 kW _
P
Z Y R
Fig. Q48
(A) 3 V (C) 9 V
Q
P
P
Q
Z Y R
(B) 6 V (D) 12 V
Q
Z Y R
P
Q
Z Y R
Output
Fig. Q53
49. The action of a JFET in its equivalent circuit can best be represented as a (A) (B) (C) (D)
(B) 4 bit substracter giving P - Q
Current Controlled Current Source Current Controlled Voltage Source Voltage Controlled Voltage Source Voltage Controlled Current Source
(C) 4 bit substracter giving Q - R (D) 4 bit adder giving P + Q + R
50. If the op-amp in Fig. Q.50 is ideal, the output voltage Vout will be equal to
54. If the functions W , X , Y and Z are as follows W = R + PQ + RS X = PQRS + P Q R S + PQ R S
5 kW
Y = RS + PR + PQ + P Q
1 kW 2V Vout
1 kW 3V 8 kW
Fig. Q50
(A) 1 V
(B) 6 V
(C) 14 V
(D) 17 V
Page 596
(A) 4 bit adder giving P + Q
Z = R + S + PQ + P × Q × R + PQ × S Then (A) W = Z , X = Z
(B) W = Z , X = Y
(C) W = Y
(D) W = Y = Z
55. A 4 bit ripple counter and a 4 bit synchronous counter are made using flip flops having a propagation delay of 10 ns each. If the worst case delay in the ripple
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GATE EC BY RK Kanodia
EC-03
counter and the synchronous counter be R and S respectively, then (A) R =10 n, S =40 ns
(B) R =40 ns,
S =10 ns
(C) R =10 ns, S =30 ns
(D) R =30 ns, S =10 ns
56. The DTL, TTL, ECL and CMOS families of digital ICs are compared in the following 4 columns P
Q
R
S
Fanout is minimum
DTL
DTL
TTL
CMOS
Power consumption is minimum
TTL
CMOS
ECL
DTL
Propagation delay is minimum
CMOS
ECL
TTL
TTL
(A) (B) (C) (D)
Chap 10.1
BCD to binary code Binary to excess -3 code Excess -3 to Gray code Gray to Binary code
59. In the circuit shown in Fig. Q.59, A is a parallel-in, parallel-out 4 bit register, which loads at the rising edge of the clock C. The input lines are connected to a 4 bit bus, W. Its output acts as the input to a 16 ´ 4 ROM whose output is floating when the enable input E is 0. A partial table of the contents of the ROM is as follows MSB
CLK
The correct column is (A) P
(B) Q
(C) R
(D) S
A
1
57. The circuit shown in Fig. Q.57 is a 4 bit DAC. The input bits 0 and 1 are represented by 0 and 5 V respectively. The OP AMP is ideal, but all the resistance and the 5 V inputs have a tolerance of ±10%. The specification (rounded to the nearest multiple of 5%) for the tolerance of the DAC is (A) ±35%
(B) ±20%
(C) ±10%
(D) ±5%
E
ROM
R R
CLK
t
2R t1 4R
Fig. Q59
Vout 8R R
Fig. Q57
58. The circuit shown in Fig. Q.58 converts MSB
+
+
t2
+
Address
Data
0
0011
2
1111
4
0100
6
1010
8
1011
10
1000
12
0010
14
1000
The clock to the register is shown, and the data on the W bus at time t1 is 0110. The data on the bus at time t2 is
MSB
Fig. Q58
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Page 597
GATE EC BY RK Kanodia
UNIT 10
Previous Year Papers
(A) 1111
(B) 1011
Data for Q.65–66 are given below. Solve the
(C) 1000
(D) 0010
problems and choose the correct answers. X ( t) is a random process with a constant mean
60. In an 8085 microprocessor, the instruction CMP B has been executed while the content of the accumulator
value of 2 and the autocorrelation function R X ( t) = 4[ e
is less than that of register B. As a result
-0 .2 t
+ 1].
(A) Carry flag will be set but Zero flag will be reset (B) Carry flag will be reset but Zero flag will be set (C) Both Carry flag and Zero flag will be reset
65. Let X be the Gaussian random variable obtained by sampling the process at t = ti and let ¥
Q ( a) = ò
(D) Both Carry flag and Zero flag will be set
a
61. Let X and Y be two statistically independent random variables uniformly distributed in the ranges ( -1, 1) and (( -2, 1) respectively. Let Z = X + Y . Then the probability that ( Z £ - 2) is (A) zero (C)
1 3
1 2p
e
-
y2 2
dy
The probability that [ x £ 1] is (A) 1 - Q(0.5)
(B) Q(0.5)
(C) Q( 2 1 2 )
(D) 1 - Q( 2 1 2 )
(B)
1 6
66. Let Y and Z be the random variables obtained by
(D)
1 12
sampling X ( t)
at t = 2
and t = 4
respectively. Let
W = Y - Z . The variance of W is 62. Let P be linearity, Q be time-invariance, R be causality and S be stability. A discrete time system has
(A) 13.36
(B) 9.36
(C) 2.64
(D) 8.00
the input-output relationship, 67. Let x( t) = 2 cos ( 800 pt) + cos (1400 pt). x( t) is sampled
n ³1 ì x( n) ï y( n) = í 0, n =0 ï x( n + 1) n £ -1 î
with the rectangular pulse train shown in Fig. Q.67. The only spectral components (in kHz) present in the
where x( n) is the input and y( n) is the output. The
sampled signal in the frequency range 2.5 kHz to 3.5 kHz are
above system has the properties
-3
p(t)
(A) P, S but not Q, R
(B) P, Q, S but not R
(C) P, Q, R, S
(D) Q, R, S but not P
T0 = 10 sec
3
Data for Q.63–64 are given below. Solve the
-T0
t
-T0/6 0 T0/6
problems and choose the correct answers.
T0
Fig. Q67
(A) 2.7, 3.4
(B) 3.3, 3.6
filter (RC-LPF) with R = 1 kW and C = 10 . mF.
(C) 2.6, 2.7, 3.3, 3.4, 3.6
(D) 2.7, 3.3
63. Let H ( f ) denote the frequency response of the
68. The signal flow graph of a system is shown in Fig.
RC-LPF. Let f1 be the highest frequency such that H ( f1 ) 0.95. Then f1 (in Hz) is 0 £ | f | £ f1 H (0)
Q.68. The transfer function
The system under consideration is an RC low-pass
(A) 327.8
(B) 163.9
(C) 52.2
(D) 104.4
R(s)
C(s ) R(s )
1 s
1
1
of the system is 1 s
6 -4
-2
-3
C(s)
Fig. Q68
64. Let t g ( f ) be the group delay function of the given RC-LPF and f2 = 100 Hz. Then t g ( f2 ) in ms, is (A) 0.717
(B) 7.17
(C) 71.7
(D) 4.505
Page 598
6 (A) 2 s + 29 s + 6
(B)
6s s + 29 s + 6
s( s + 2) s + 29 s + 6
(D)
s( s + 27) s + 29 s + 6
(C)
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2
2
2
GATE EC BY RK Kanodia
EC-03
69 The root locus of the system G( s) H ( s) =
Chap 10.1
72. The gain margin and the phase margin of a feedback system with
K s( s + 2)( s + 3)
G( s) H ( s) =
has the break-away point located at
s are ( s + 100) 3
(A) (-0.5, 0)
(B) (-2.548, 0)
(A) - dB, 0°
(B) ¥, ¥
(C) (-4, 0)
(D) (-0.784, 0)
(C) ¥, 0°
(D) 88.5 dB, ¥
70. The approximate Bode magnitude plot of a
73. The zero-input response of a system given by the
minimum phase system is shown in Fig. Q.70. The
state-space equation é x& 1 ù é1 0 ù é x1 ù é x1 (0) ù é1 ù ê x& ú = ê1 1 ú ê x ú and ê x (0) ú = ê0 ú is ûë 2 û ë 2û ë ë 2 û ë û
transfer function of the system is dB 160 140
20
0.1
100
10
( s + 0.1) ( s + 10) 2 ( s + 100)
ée t ù (C) ê t ú ëte û
ét ù (D) ê t ú ëte û
frequency fc = 1 MHz using a nonlinear device with the
( s + 0.1) 3 (B) 10 7 ( s + 10)( s + 100)
2
(C) 108
ée t ù (B) ê ú ët û
74. A DSB-SC signal is to be generated with a carrier
Fig. Q70
( s + 0.1) 3 (A) 108 ( s + 10) 2 ( s + 100)
éte t ù (A) ê ú ët û
( s + 0.1) ( s + 10)( s + 100) 2 3
(D) 10 9
input-output characteristic v0 = a0 v1 + a1 vi3 where a0 and a1
are constants. The output of the nonlinear device
can be filtered by an appropriate band-pass filter. Let vi = Acl cos (2pfcl t) + m( t)
where
m( t) is the message
l c
signal. Then the value of f (in MHz) is 71. A second-order system has the transfer function C (s) 4 = 2 R (s) s + 4s + 4
(A) 1.0
(B) 0.333
(C) 0.5
(D) 3.0
The data for Q.75-76 are given below. Solve the With r ( t) as the unit-step function, the response c( t) of the system is represented by (A) Step Response
Let m( t) = cos [( 4 p ´ 10 3) t ] be the message signal
(B)
and c( t) = 5 cos [(2 p ´ 106 ) t ] be the carrier.
Step Response
2
1
75. c( t) and m( t) are used to generate an AM signal. The Amplitude
Amplitude
1.5 1 0.5 0
0
5
10 15 Time (sec)
20
0
25
modulation index of the generated AM signal is 0.5. Total side band power Then the quantity is Carrier power
0.5
0
(C)
10
5 Time (sec)
(D) 1
Amplitude
1
(A)
1 2
(B)
1 4
(C)
1 3
(D)
1 8
76. c( t) and m( t) are used to generate an FM signal. If
Step Response
Step Response 1.5
Amplitude
problems and choose the correct answers.
the peak frequency deviation of the generated FM is three times the transmission bandwidth of the AM
0.5
signal,
0.5
then
the
coefficient
of
the
term
cos [2 p(1008 ´ 10 t)] in the FM signal (in terms of the 3
0
0 0
2 4 Time (sec)
6
0
2
4 Time (sec)
6
Bessel coefficients) is (A) 5 J 4 ( 3)
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(B)
5 2
J8 ( 3) Page 599
GATE EC BY RK Kanodia
EC-03
Chap 10.1
86. A uniform plane wave traveling in air is incident on
90. Two identical antennas are placed in the q = p
the plane boundary between air and another dielectric
plane as shown in Fig. Q.90. The elements have equal
medium with e r = 4. The reflection coefficient for the
amplitude excitation with 180° polarity difference,
normal incidence, is
operating at wavelength l. The correct value of the
2
magnitude of the far-zone resultant electric field
(A) zero
(B) 0.5 Ð180 °
(C) 0.333 Ð0 °
(D) 0.333Ð180 °
strength normalized with that of a single element, both computed for f = 0, is
87. If the electric field intensity associated with a perfect
dielectric
medium
E( z, t) = 10 cos (2 p ´ 10 t - 0.1pz) 7
is
given
by
then
the
volt/m,
s
uniform plane electromagnetic wave traveling in a f s
velocity of the traveling wave is (A) 3.00 ´ 108 m/sec
(B) 2.00 ´ 108 m/sec
(C) 6.28 ´ 10 7 m/sec
(D) 2.00 ´ 10 7 m/sec
Fig. Q.90
88. A short-circuited stub is shunt connected to a transmission line as shown in Fig. Q.88. If Z 0 = 50 ohm,
æ 2 ps ö (A) 2 cos ç ÷ è l ø
æ 2 ps ö (B) 2 sin ç ÷ è l ø
æ ps ö (C) 2 cos ç ÷ è lø
æ ps ö (D) 2 sin ç ÷ è lø
the admittance Y seen at the junction of the stub and
**************
the transmission line is
l/8
Zo
Zo
Z L 100 W Zo
l/2 Y
Fig. Q.88
(A) (0.01 - j0.02) mho
(B) (0.02 - j0.01) mho
(C) (0.04 - j0.02) mho
(D) (0.02 + j0) mho
89. A rectangular metal wave guide filled with a dielectric material of relative permitivity e r = 4 has the inside dimensions 3.0 cm ´ 1.2 cm. The cut-off frequency for the dominant mode is (A) 2.5 GHz
(B) 5.0 GHz
(C) 10.0 GHz
(D) 12.5 GHz
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Page 601
GATE EC BY RK Kanodia
UNIT 10
ANSWER SHEET
Page 602
1.
(B)
2.
(C)
3.
(B)
4.
(C)
5.
(C)
6.
(D)
7.
(B)
8.
(A)
9.
(C)
10.
(A)
11.
(B)
12.
(D)
13.
(B)
14.
(C)
15.
(A)
16.
(D)
17.
(C)
18.
(B)
19.
(B)
20.
(D)
21.
(D)
22.
(C)
23.
(B)
24.
(A)
25.
(C)
26.
(A)
27.
(A)
28.
(D)
29.
(B)
30.
(B)
31.
(A)
32.
(*)
33.
(D)
34.
(D)
35.
(C)
36.
(B)
37.
(A)
38.
(C)
39.
(C)
40.
(C)
41.
(C)
42.
(A)
43.
(D)
44.
(B)
45.
(A)
46.
(A)
47.
(A)
48.
(C)
49.
(D)
50.
(B)
51.
(D)
52.
(D)
53.
(B)
54.
(A)
55.
(B)
56.
(C)
57.
(A)
58.
(D)
59.
(C)
60.
(A)
61.
(A)
62.
(A)
63.
(C)
64.
(B)
65.
(A)
66.
(C)
67.
(A)
68.
(A)
69
(D)
70
(A)
71.
(B)
72.
(D)
73.
(C)
74
(A)
75
(D)
76.
(D)
77.
(B)
78.
(A)
79
(C)
80
(D)
81.
(B)
82.
(D)
83.
(B)
84.
(C)
85.
(C)
86.
(D)
87.
(B)
88.
(A)
89
(B)
90.
(D)
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Previous Year Papers
GATE EC BY RK Kanodia
UNIT 10
Previous Year Papers
5. For the R-L circuit shown in Fig. Q.5, the input
(C) current controlled current source
voltage vi ( t) = u( t). The current i( t) is
(D) current controlled voltage source
1H
i(t)
10. Voltage series feedback (also called series-shunt feedback) results in vi(t)
2W
(A) increase in both input and output impedances (B) decrease in both input and output impedances (C) increase in input impedance and decrease in output impedance
Fig Q.5
(A)
(B)
i(t)
(D) decrease in input impedance and increase in output impedance
i(t)
0.5 0.31
1 0.63 t(sec)
2
t(sec)
½
(C)
11. The circuit in Fig. Q.11 is a
(D)
i(t)
i(t)
0.5 0.31
vi
1 0.63
R
vo
R
C
C ½
t(sec)
2
t(sec)
6. The impurity commonly used for realizing the base
Fig Q.11
region of a silicon n-p-n transistor is (A) Gallium
(B) Indium
(C) Boron
(D) Phosphorus
7. If for a silicon n-p-n transistor, the base-to-emitter voltage ( VBE ) is 0.7 V and the collector-to-base voltage
(A) low-pass filter
(B) high-pass filter
(C) band-pass filter
(D) band-reject filter
12. Assuming VCEsat = 0.2 V and b = 50, the minimum base current ( I B ) required to drive the transistor in Fig. Q.12 to saturation is
( VCB ) is 0.2 V, then the transistor is operating in the (A) normal active mode
(B) saturation mode
(C) inverse active mode
(D) cutoff mode
3 V IC 1 kW IB
8. Consider the following statements S1 and S2. S1 : The b of a bipolar transistor reduces if the base Fig Q12.
width is increased. S2 : The b of a bipolar transistor increases if the doping concentration in the base is increased. Which one of the following is correct ?
(A) 56 mA
(B) 140 mA
(C) 60 mA
(D) 3 mA
(A) S1 is FALSE and S2 is TRUE
13. A master-slave flip-flop has the characteristic that
(B) Both S1 and S2 are TRUE
(A) change in the input is immediately reflected in the output
(C) Both S1 and S2 are FALSE (D) S1 is TRUE and S2 is FALSE
(B) change in the output occurs when the state of the master is affected
9. An ideal op-amp is an ideal
(C) change in the output occurs when the state of the slave is affected
(A) voltage controlled current source (B) voltage controlled voltage source Page 604
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GATE EC BY RK Kanodia
EC-04
Chap 10.2
14. The range of signed decimal numbers that can be
19. The impulse response h[ n] of a linear time-invariant
represented by 6-bit 1’s complement numbers is
system is given by
(A) -31 to +31
(B) -63 to +64
(C) -64 to +63
(D) -32 to +31
15.
A digital
system is
required
h[ n] = u[ n + 3] + u[ n - 2 ] - 2 u[ n - 7 ] where u[ n] is the unit step sequence. The above
to amplify
a
system is
binary-encoded audio signal. The user should be able to
Vout
control the gain of the amplifier from a minimum to a maximum in 100 increments. The minimum number of bits required to encode, in straight binary, is (A) 8
(B) 6
(C) 5
(D) 7
Vin 0
Fig Q.18
16. Choose the correct one from among the alternatives A, B, C, D after matching an item from Group 1 with the most appropriate item in Group 2.
(A) stable but not causal (B) stable and causal (C) causal but unstable
Group 1
Group 2
P: Shift register
1: Frequency division
Q: Counter
2: Addressing in memory chips
R: Decoder
3: Serial to parallel data conversion
(D) unstable and not causal 20. The distribution function FX ( x) of a random variable X is shown in Fig. Q.20. The probability that X = 1 is Fx(x) 1.0
(A)
(B)
(C)
(D)
P–3
P–3
P–2
P–1
Q–2
Q–1
Q–1
Q–3
R–1
R–2
R–3
R–2
0.55 0.25
-2
AND-OR-Invert (AOI) gate. For the inputs shown in Fig. Q.17, the output Y is
(A) zero
(B) 0.25
(C) 0.55
(D) 0.30
H ( z) = Fig Q.17
(A) 0
(B) 1
(C) AB
(D) AB
x
21. The z-transform of a system is
y
Input are Floating
2
1
Fig Q.20
17. Fig. Q.17 shows the internal schematic of a TTL
A B
0
z z - 0.2
If the ROC is |z |< 0.2, then the impulse response of the system is
18. Fig. Q.18 is the voltage transfer characteristic of (A) an NMOS inverter with enhancement mode transistor as load (B) an NMOS inverter with depletion mode transistor as load
(A) (0.2) n u[ n]
(B) (0.2) n u[ -n - 1]
(C) -(0.2) n u[ n]
(D) -(0.2) n u[ -n - 1]
22. The Fourier transform of a conjugate symmetric function is always
(C) a CMOS inverter
(A) imaginary
(B) conjugate anti-symmetric
(D) a BJT inverter
(C) real
(D) conjugate symmetric
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GATE EC BY RK Kanodia
EC-04
Chap 10.2
33. Consider the Bode magnitude plot shown in Fig.
37. Consider the following statements S1 and S2.
Q.33. The transfer function H ( s) is
S1 : At the resonant frequency the impedance of a series R–L–C circuit is zero.
20 log H(jw) 0
S2 : In a parallel G–L–C circuit, increasing the conductance G results in increase in its Q factor.
20 dB/dec -20 dB/dec
-20
Which one of the following is correct ?
40 dB/dec
(A) S1 is FALSE and S2 is TRUE 1
w
100
10
(B) Both S1 and S2 are TRUE
Fig Q.33
(C) S1 is TRUE and S2 is FALSE
(A)
( s + 10) ( s + 1)( s + 100)
(B)
10( s + 1) ( s + 10)( s + 100)
(D) Both S1 and S2 are FALSE
(C)
10 2 ( s + 1) ( s + 10)( s + 100)
(D)
10 3( s + 100) ( s + 1)( s + 10)
38..
In
an
N A = 9 ´ 10
16
V ( s) of an R–L–C 34. The transfer function H ( s) = o Vi ( s) circuit is given by 6
H ( s) =
10 s + 20 s + 106 2
The Quality factor (Q-factor) of this circuit is (A) 25
(B) 50
(C) 100
(D) 5000
abrupt
concentrations cm
on 3
p–n
the
junction,
p-side
and
the
reverse biased and the total depletion width is 3 mm. The depletion width on the p-side is (A) 2.7 mm
(B) 0.3 mm
(C) 2.25 mm
(D) 0.75 mm
39. The resistivity of a uniformly doped n-type silicon sample is 0.5 W-cm. If the electron mobility (m n ) is 1250 cm 2 /V-sec and the charge of an electron is 1.6 ´ 10 -19 Coulomb, the donor impurity concentration ( N D) in the sample is
conditions
(A) 2 ´ 1016 cm 3
(B) 1 ´ 1016 cm 3
(C) 2.5 ´ 1015 cm 3
(D) 5 ´ 1015 cm 3
zero.
Its
transfer
function
H ( s) = VC ( s) Vi ( s) is 1 (A) 2 s + 10 3 s + 106
10 mH
10 kW
are
respectively. The p-n junction is
35. For the circuit shown in Fig. Q.35, the initial are
doping
n-side
106 (B) 2 s + 10 3 s + 106
40. Consider an abrupt p-n junction. Let Vbi be the built-in potential of this junction and VR be the applied
vi(t)
reverse bias. If the junction capacitance ( C j ) is 1 pF for
vo(t)
100 mF
Vbi + VR = 1 V, then for Vbi + VR = 4 V, C j will be
Fig Q35.
(C)
10 3 s 2 + 10 3 s + 106
(D)
(A) 4 pF
(B) 2 pF
(C) 0.25 pF
(D) 0.5 pF
41. Consider the following statements S1 and S2.
106 s 2 + 106 s + 106
S1 : The threshold voltage ( VT ) of a MOS capacitor decreases with increase in gate oxide thickness.
36. A system described by the following differential equation d2 y dy +3 + 2 y = x( t) dt 2 dt
S2 : The threshold voltage ( VT ) of a MOS capacitor decreases with increase in substrate doping concentration. Which one of the following is correct ?
is initially at rest. For input x( t) = 2 u( t), the output
(A) S1 is FALSE and S2 is TRUE (B) Both S1 and S2 are TRUE
y( t) is -t
-2 t
(A) (1 - 2 e + e ) u( t) -t
-2 t
(C) (0.5 + e + 15 . e ) u( t)
-t
-2 t
(B) (1 + 2 e - e ) u( t) -t
-2 t
(D) (0.5 + 2 e + 2 e ) u( t)
(C) Both S1 and S2 are FALSE (D) S1 is TRUE and S2 is FALSE
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Page 607
GATE EC BY RK Kanodia
UNIT 10
Previous Year Papers
42. The drain of an n-channel MOSFET is shorted to
46. A bipolar transistor is operating in the active region
the gate so that VGS = VDS . The threshold voltage ( VT ) of
with a collector current of 1 mA. Assuming that the b of
the MOSFET is 1 V. If the drain current ( I D) is 1 mA for
the transistor is 100 and the thermal voltage ( VT ) is 25
VGS = 2 V, then for VGS = 3 V, I D is
mV, the transconductance ( g m ) and the input resistance
(A) 2 mA
(B) 3 mA
( rp)
(C) 9 mA
(D) 4 mA
of
the
transistor
in
the
common
emitter
configuration, are (A) g m = 25 mA/V and rp = 15.625 kW
43. The longest wavelength that can be absorbed by
(B) g m = 40 mA/V and rp = 4.0 kW
silicon, which has the bandgap of 1.12 eV, is 1.1 mm. If
(C) g m = 25 mA/V and rp = 2.5 kW
the longest wavelength that can be absorbed by another material is 0.87 mm, then the bandgap of this material is
(D) g m = 40 mA/V and rp = 2.5 kW 47. The value of C required for sinusoidal oscillations of
(A) 1.416 eV
(B) 0.886 eV
(C) 0.854 eV
(D) 0.706 eV
frequency 1 kHz in the circuit of Fig. Q.47 is 2.1 kW
1 kW
44. The neutral base width of a bipolar transistor, biased in the active region, is 0.5 mm. The maximum
C
electron concentration and the diffusion constant in the
1 kW
base are 1014 cm 3 and Dn = 25 cm 2 sec respectively. Assuming negligible recombination in the base, the
C
1 kW
collector current density is (the electron charge is 1.6 ´ 10 -19 Coulomb) (A) 800 A/cm2
(B) 9 A/cm2
(C) 200 A/cm2
(D) 2 A/cm2
Fig Q.47
(A)
45. Assume that the b of the transistor is extremely
1 mF 2p
(C)
large and VBE = 0.7 V, I C and VCE in the circuit shown in Fig. Q.45 are
1 2p 6
(B) 2p mF mF
(D) 2 p 6 mF
48. In the op-amp circuit given in Fig. Q.48, the load
5 V
current iL is IC
R1
2.2 kW
4 kW
vi
+ VEC
R1 vo
1 kW
300 kW R1
R2 iL
RL
Fig Q.45 Fig Q.48
(A) I C = 1 mA, VCE = 4.7 V (B) I C = 0.5 mA, VCE = 375 . V
(A) -
vs R2
(B)
vs R2
(C) -
vs RL
(D)
vs R1
(C) I C = 1 mA, VCE = 2.5 V (D) I C = 0.5 mA, VCE = 39 . V
Page 608
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GATE EC BY RK Kanodia
UNIT 10
Previous Year Papers
57. Consider the sequence of 8085 instructions given
60. A 1 kHz sinusoidal signal is ideally sampled at 1500
below
samples /sec and the sampled signal is passed through
LXI H, 9258
an ideal low-pass filter with cut-off frequency 800 Hz.
MOV A, M
The output signal has the frequency
CMA MOV M , A Which one of the following is performed by this
(A) zero Hz
(B) 0.75 kHz
(C) 0.5 kHz
(D) 0.25 kHz
61. A rectangular pulse train s( t) as shown in Fig. Q.61
sequence?
is convolved with the signal cos 2 ( 4 p ´ 10 3 t). The
(A) Contents of location 9258 are moved to the accumulator
convolved signal will be a s(t)
(B) Contents of location 9258 are compared with the contents of the accumulator
1
(C) Contents of location 8529 are complemented and stored in location 8529 (D) Contents of location 5892 are complemented and stored in location 5892 58. A Boolean function f of two variables x and y is
1 ms
Fig Q.61
(A) DC
(B) 12 kHz sinusoid
(C) 8 kHz sinusoid
(D) 14 kHz sinusoid
62. Consider the sequence
defined as follows :
x[ n] = [ -4 - j5 1 + j2 5 ]
f (0, 0) = f (0, 1) = f (1, 1) = 1; f (1, 0) = 0 Assuming complements of x
and y
are not
available, a minimum cost solution for realizing f using only 2-input NOR gates and 2-input OR gates (each having unit cost) would have a total cost of (A) 1 unit
(B) 4 units
(C) 3 units
(D) 2 units
The conjugate anti-symmetric part of the sequence is (A) [-4 - j2.5
and store the result in the accumulator. The numbers are available in registers B and C respectively. A part of
4 - j2.5]
j2
(B) [- j2.5
1
(C) [- j2.5
j2
(D) [-4
59. It is desired to multiply the numbers 0AH by 0BH
1
j2.5] 0] 4]
63. A causal LTI system is described by the difference equation 2 y[ n] = ay[ n - 2 ] - 2 x[ n] + bx[ n - 1]
the 8085 program for this purpose is given below:
The system is stable only if
MVI A, 00H LOOP:
t
0
––––––––––
(A) a = 2, b < 2
–––––––––––
(B) a > 2, b > 2
–––––––––––
(C) a < 2, any value of b
HLT
(D) b < 2, any value of a
END
64. A causal system having the transfer function
The sequence of instructions to complete the
H ( s) =
program would be (A) JNZ LOOP, ADD B, DCR C
1 s+2
is excited with 10u( t). The time at which the
(B) ADD B, JNZ LOOP, DCR C
output reaches 99% of its steady state value is
(C) DCR C, JNZ LOOP, ADD B
(A) 2.7 sec
(B) 2.5 sec
(C) 2.3 sec
(D) 2.1 sec
(D) ADD B, DCR C, JNZ LOOP Page 610
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GATE EC BY RK Kanodia
EC-04
65. The impulse response h[ n] of a linear time invariant system is given as ì -2 2 ï h[ n] = í 4 2 ï 0 î
n = 2, - 2
sin(2 t) ù ésin( -2 t) (B) ê sin( t ) sin( -3t) úû ë
otherwise
e jpn 4 , then the output is (C) 4 e jpn
(B) 4 2 e - jpn
4
(D) -4 e jpn
4
2ù é-2 69. If A = ê ú, then sin At is ë 1 -3û é sin( -4 t) + 2 sin( -t) - sin( -4 t) + 2 sin( -t) ù (A) ê 2 sin( -4 t) + sin( -t) úû ë - sin( -4 t) + sin( -t)
n = 1, - 1
If the input to the above system is the sequence (A) 4 2 e jpn
Chap 10.2
sin( 4 t) + 2 sin( t) 2 sin( -4 t) - 2 sin( -t) ù - sin( -4 t) + sin( t) 2 sin( 4t) + sin( t) úû
é (C) ê ë
cos( -t) + 2 cos( t) 2 cos( -4 t) - 2 sin( -t) ù é (D) ê -2 cos( 4 t) + cos( t) úû ë - cos( -4 t) + sin( -t)
4
4
66. Let x( t) and y( t) with Fourier transforms F ( f ) and
70. The open-loop transfer function of a unity feedback
Y ( f ) respectively be related as shown in Fig. Q.66. Then
system is
Y ( f ) is
G( s) = x(t)
y(t)
1
-2
0
-2
t
2
0
The range of K for which the system is stable is 21 (A) (B) 13 > K > 0 > K >0 4
t
(C)
-1
Fig Q.66
1 (A) - X ( f 2) e - j 2 pf 2
1 (B) - X ( f 2) e j 2 pf 2
(C) - X ( f 2) e j 2 pf
(D) - X ( f 2) e - j 2 pf
the number of roots which lie in the right half of
phase of the system response at 20 Hz is (B) 0°
(C) 90°
(D) –180°
x3
c
x4
d
(A) 4
(B) 2
(C) 3
(D) 1
x& 1 = -3 x1 - x2 = u, x& 2 = 2 x1 , y = x1 + u
68. Consider the signal flow graph shown in Fig. Q.68. x The gain 5 is x1 b
the s-plane is
72. The state variable equations of a system are :
(A) –90°
x2
(D) -6 < K < ¥
P ( s) = s 5 + s 4 + 2 s 3 + 2 s 2 + 3s + 15
zeros at 5 Hz, 100 Hz and 200 Hz. The approximate
a
21 2, but nonzero for w2 < 2
-¥
of
(A) y(.) is zero for all sampling frequencies ws (C) y(.) is nonzero for ws > 2, but zero for ws < 2
¥
y( t)= ò x( t - t) h(2 t) dt
Which
is sampled at a rate ws rad/s to obtain the final output
four
45. The unit step response of an under-damped second
properties
are
possessed by the system ? BIBO: Bounded input gives a bounded output. Causal: The system is causal,
order system has steady state value of -2. Which one of the following transfer functions has these properties ? -2.24 -3.82 (A) 2 (B) 2 s + 2.59 s + 112 . s + 191 . s + 191 . (C)
LP: The system is low pass.
-2.24 s - 2.59 s + 112 . 2
(D)
-382 s - 191 . s + 191 . 2
LTI: The system is linear and time-invariant.
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Page 667
GATE EC BY RK Kanodia
UNIT 10
46. A discrete random variable X takes values from 1 to 5 with probabilities as shown in the table. A student
®
(D) j =
Previous year Papers
ö 1 B0 z$ æ ç ÷, r ¹ 0 m 0 çè x 2 + y 2 ÷ø
calculates the mean of X as 3.5 and her teacher calculates the variance of X as 1.5. Which of the
50. A transmission line terminates in two branches,
following statements is true ?
each of length l / 4, as shown. The branches are terminated by 50W loads. The lines are lossless and
k
1
2
3
4
5
P ( X = k)
0.1
0.2
0.4
0.2
0.1
have the characteristic impedances shown. Determine the impedance Z i as seen by the source.
(A) Both the student and the teacher are right (B) Both the student and the teacher are wrong (C) The student is wrong but the teacher is right (D) The student is right but the teacher is wrong Q. 47 A message signal given by æ1ö æ1ö 1n( t)= ç ÷ cos w1 t - ç ÷ sin w2 t è2 ø è2 ø
(A) 200W
(B) 100W
(C) 50W
(D) 25W
Common Date Questions
is amplitude-modulated with a carrier of frequency
Common Date for Questions 51 and 52:
wc to generate
Consider s( t)= [1 + m( t)]cos wc t
modulation scheme ? (B) 11.11%
(C) 20%
(D) 25%
the resulting capacity C2 is given by (D) C2 = C1 + 0.3B
®
2 B0 z$ æ çç 2 m 0 è x + y2
ö ÷÷, r ¹ 0 ø
Permittivity of free space = 8.85 ´ 10 -14 F.cm -1 Dielectric constant of silicon = 12 51 The built-in potential of the junction
(D) cannot be estimated from the data given 52. The peak electric field in the device is (A) 0.15 MV . cm -1 , directed from p-region to n-region
[ Hint : The algebra is trivial in cylindrical coordinates.]
(B) j = -
Depletion width on the n-side = 0.1 mm
(C) is 0.82
What current distribution leads to this field ? ö 1 B0 z$ æ ç ÷, r ¹ 0 m 0 çè x 2 + y 2 ÷ø
Doping on the n-side = 1 ´ 1017 cm -3
(B) is 0.76 V
® æ ö x y B= B0 çç 2 y$ - 2 x$ ÷÷ 2 2 x +y ø èx +y
®
(B) 0.15 MV . cm -1 , directed from n-region to p-region (C) 1.80 MV . cm -1 , directed from p-region to n-region (D) 1.80 MV . cm -1 , directed from n-region to p-region Common Data for Questions 53 and 54:
®
(C) j = 0, r ¹ 0 Page 668
room
(A) is 0.70 V
49. A magnetic field in air is measured to be
(A) j =
at
Thermal voltage = 26 mV
capacity C1 . If the SNR is doubled keeping B constant,
(C) C2 = C1 + 2 B
junction
Intrinsic carrier concentration = 1.4 ´ 1010 cm -3
a signal to noise ratio SNR >> 1 and bandwidth B has
(B) C2 = C1 + B
p-n
Depletion width on the p-side = 10 . mm
48. A communication channel with AWGN operating at
(A) C2 = 2 C1
silicon
temperature having the following parameters:
What is the power efficiency achieved by this (A) 8.33%
a
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GATE EC BY RK Kanodia
EC-09
Chap 10.7
The Nyquist plot of a stable transfer function G( s)
Consider the CMOS circuit shown, where the gate
is shown in the figure. We are interested in the stability
voltage VG of the n-MOSFET is increased from zero,
of the closed loop system in the feedback configuration
while the gate voltage of the p-MOSFET is kept
shown.
constant at 3 V. Assume that, for both transistors, the
Im
magnitude of the threshold voltage is 1 V and the product of the trans-conductance parameter and the +
-1-0.5
Re
+
(W/L) ratio, i.e. the quantity mCax (W / L), is 1 mA . V -2 .
G(s) -
-j
53. Which of the following statements is true? (A) G( s) is an all-pass filter (B) G( s) has a zero in the right-half plane (C) G( s) is the impedance of a passive network 57. For small increase in VG beyond 1 V, which of the
(D) G( s) is marginally stable
following gives the correct description of the region of 54. The gain and phase margins of G( s) for closed loop
operation of each MOSFET ?
stability are
(A) Both the MOSFETs are in saturation region
(A) 6 dB and 180°
(B) 3 dB and 180°
(B) Both the MOSFETs are in triode region
(C) 6 dB and 90°
(D) 3 dB and 90°
(C) n-MOSFET is in triode and p-MOSFET is in saturation region
Common Data for Questions 55 and 56: The amplitude of a random signal is uniformly
(D) n-MOSFET is in saturation and p-MOSFET is in triode region.
distributed between -5 V and 5 V. 55. If the signal to quantization noise ratio required in
58. Estimate the output voltage V0 for VG = 15 . V.
uniformly quantizing the signal is 43.5 dB, the step size
[Hint : Use the appropriate current-voltage equation for
of the quantization is approximately
each MOSFET, based on the anser to Q. 57.]
(A) 0.0333 V
(B) 0.05 V
(C) 0.0667 V
(D) 0.10 V
Statement for Linked Answer Questions 59 and 60:
56. If the positive values of the signal are uniformly quantized with a step size of 0.05 V, and the negative
Two products are sold from a vending machine,
values are uniformly quantized with a step size of 0.1 V,
which has two push buttons P1 and P2 . When a button is
the resulting signal to quantization noise ratio is
pressed, the price of the corresponding product is
approximately
displayed in a 7-segment display.
(A) 46 dB
(B) 43.8 dB
(C) 42 dB
(D) 40 dB
If no buttons are pressed, '0' is displayed, signifying 'Rs. 0'. If only P1 is pressed, '2' is displayed, signifying 'Rs.
Linked Answer Questions Statement for Linked Answer Questions 57 and 58
2'. If only P2 is pressed, '5' is displayed, signifying 'Rs.
: 5'.
If both P1 and P2 are pressed, 'E' is displayed, signifying 'Error'.
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