GATEECSP2014

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 1

UNIT 1 NETWORKS

(A) 125/100 and 80/100 (C) 100/100 and 100/100 2013 1.1

1.2

ONE MARK

Three capacitors C1 , C2 and C 3 whose values are 10 mF , 5 mF , and 2 mF respectively, have breakdown voltages of 10 V, 5 V and 2 V respectively. For the interconnection shown below, the maximum safe voltage in Volts that can be applied across the combination, and the corresponding total charge in mC stored in the effective capacitance across the terminals are respectively,

1.6

Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k , k > 0 , the elements of the corresponding star equivalent will be scaled by a factor of

(A) k2

(B) k

(C) 1/k

(D)

The transfer function

(B) 100/100 and 80/100 (D) 80/100 and 80/100

k

(A) 2.8 and 36 (C) 2.8 and 32

V2 ^s h of the circuit shown below is V1 ^s h

(B) 7 and 119 (D) 7 and 80

Common Data For Q. 8 and 9:

A I D

Consider the following figure

(A) 0.5s + 1 s+1 (C) s + 2 s+1 1.3

(B) 3s + 6 s+2 (D) s + 1 s+2

O N

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no w.

d

w

A source vs ^ t h = V cos 100pt has an internal impedance of ^4 + w j3h W . If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in W should be (A) 3 (B) 4 (C) 5 (D) 7

1.7

1.8

2013 1.4

1.5

TWO MARKS

In the circuit shown below, if the source voltage VS = 100+53.13c V then the Thevenin’s equivalent voltage in Volts as seen by the load resistance RL is

(A) 100+90c (C) 800+90c

co ia.

(B) 800+0c (D) 100+60c

1.9

The current IS in Amps in the voltage source, and voltage VS in Volts across the current source respectively, are (A) 13, - 20 (B) 8, - 10 (C) - 8, 20 (D) - 13, 20 The current in the 1W resistor in Amps is (A) 2 (B) 3.33 (C) 10 (D) 12 Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2 . When connected in series, their effective Q factor at the same operating frequency is (A) q1 + q2 (B) ^1/q1h + ^1/q2h (C) ^q1 R1 + q2 R2h / ^R1 + R2h (D) ^q1 R2 + q2 R1h / ^R1 + R2h

2012 1.10

The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage VWX1 = 100 V is applied across WX to get an open circuit voltage VYZ1 across YZ. Next, an ac voltage VYZ2 = 100 V is applied across YZ to get an open circuit voltage VWX2 across WX. Then, VYZ1 /VWX1 , VWX2 /VYZ2 are respectively,

ONE MARK

In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current i (t) for all t is

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) zero (B) a step function (C) an exponentially decaying function (D) an impulse function 1.11

1.12

Page 2 1.15

The average power delivered to an impedance (4 - j3) W by a current 5 cos (100pt + 100) A is (A) 44.2 W (B) 50 W (C) 62.5 W (D) 125 W

1.16

In the circuit shown below, the current through the inductor is

With 10 V dc connected at port A, the current drawn by 7 W connected at port B is (A) 3/7 A (B) 5/7 A (C) 1 A (D) 9/7 A For the same network, with 6 V dc connected at port A, 1 W connected at port B draws 7/3 A. If 8 V dc is connected to port A , the open circuit voltage at port B is (A) 6 V (B) 7 V (C) 8 V (D) 9 V 2011

1.17

(B) - 1 A 1+j

2 A 1+j (C) 1 A 1+j (A)

1.13

TWO MARKS

In the circuit shown below, the value of RL such that the power transferred to RL is maximum is

Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is

O N

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ww

(A) 0.8 W (C) 2 W 1.14

(B) 6.56 - j 7.87 (D) 16 + j 0

A I D 1.18

2012

In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals P and Q is

(A) 6.4 - j 4.8 (C) 10 + j 0

(D) 0 A

ONE MARK

(B) 1.4 W (D) 2.8 W

(A) 5 W (C) 15 W 1.19

(B) 10 W (D) 20 W

The circuit shown below is driven by a sinusoidal input vi = Vp cos (t/RC ). The steady state output vo is

If VA - VB = 6 V then VC - VD is

(A) (Vp /3) cos (t/RC ) (C) (Vp /2) cos (t/RC ) (A) - 5 V (C) 3 V

(B) 2 V (D) 6 V

Common Data For Q. 48 and 49 :

(B) (Vp /3) sin (t/RC ) (D) (Vp /2) sin (t/RC )

2011 1.20

TWO MARKS

In the circuit shown below, the current I is equal to

With 10 V dc connected at port A in the linear nonreciprocal twoport network shown below, the following were observed : (i) 1 W connected at port B draws a current of 3 A (ii) 2.5 W connected at port B draws a current of 2 A

(A) 1.4+0c A (C) 2.8+0c A

(B) 2.0+0c A (D) 3.2+0c A

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 1.21

Page 3

In the circuit shown below, the network N is described by the following Y matrix: 0.1 S - 0.01 S . the voltage gain V2 is Y=> 0.01 S 0.1 SH V1

(A) i (t) = 0.5 - 0.125e-1000t A (C) i (t) = 0.5 - 0.5e-1000t A

The current I in the circuit shown is

1.26

(A) 1/90 (C) –1/99 1.22

(B) –1/90 (D) –1/11

In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with the voltage polarity as indicated. The switch is closed at time t = 0 . The current i (t) at a time t after the switch is closed is

(A) - j1 A (C) 0 A

(A) i (t) = 15 exp (- 2 # 103 t) A (B) i (t) = 5 exp (- 2 # 103 t) A (C) i (t) = 10 exp (- 2 # 103 t) A (D) i (t) =- 5 exp (- 2 # 103 t) A 2010

1.24

-2 S 4H 0.5 S 1H

n (A) o.i 0 W c . ia (C) 10 W

d

(B) 5 W (D) 100 W

GATE 2009 1.28

ONE MARK

In the interconnection of ideal sources shown in the figure, it is known that the 60 V source is absorbing power.

1 - 0.5 (B) > S - 0.5 1H 4 2 (D) > S 2 4H

For parallel RLC circuit, which one of the following statements is NOT correct ? (A) The bandwidth of the circuit decreases if R is increased (B) The bandwidth of the circuit remains same if L is increased (C) At resonance, input impedance is a real quantity (D) At resonance, the magnitude of input impedance attains its minimum value. 2010

1.25

.no

For the two-port network shown below, the short-circuit admittance w parameter matrix is ww

4 (A) > -2 1 (C) > 0.5

A I D

O N ONE MARK

(B) j1 A (D) 20 A

In the circuit shown, the power supplied by the voltage source is

1.27

1.23

(B) i (t) = 1.5 - 0.125e-1000t A (D) i (t) = 0.375e-1000t A

Which of the following can be the value of the current source I ? (A) 10 A (B) 13 A (C) 15 A (D) 18 A 1.29

If the transfer function of the following network is Vo (s) 1 = 2 + sCR Vi (s)

TWO MARKS

In the circuit shown, the switch S is open for a long time and is closed at t = 0 . The current i (t) for t $ 0+ is The value of the load resistance RL is (A) R (B) R 4 2 (C) R

(D) 2R

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 1.30

Page 4

(A) i (t) " V0 R (C) i (t) " V0 (1 + B) R

A fully charged mobile phone with a 12 V battery is good for a 10 minute talk-time. Assume that, during the talk-time the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V as shown in the figure. How much energy does the battery deliver during this talk-time?

(B) i (t) " 2V0 R (D) i (t) " 2V0 (1 + B) R

GATE 2008 1.35

(A) 220 J (C) 13.2 kJ

(A) P = 2, Q = 2 (C) P = 4, Q = 6

TWO MARK

An AC source of RMS voltage 20 V with internal impedance Zs = (1 + 2j) W feeds a load of impedance ZL = (7 + 4j) W in the figure below. The reactive power consumed by the load is

(A) 8 VAR (C) 28 VAR 1.32

In the following graph, the number of trees (P) and the number of cut-set (Q) are

(B) 12 kJ (D) 14.4 J

GATE 2009 1.31

ONE MARK

(B) 16 VAR (D) 32 VAR

1.36

(B) P = 2, Q = 6 (D) P = 4, Q = 10

In the following circuit, the switch S is closed at t = 0 . The rate of change of current di (0+) is given by dt

A I D

(A) 0 (R + Rs) Is (C) L

(B) Rs Is L (D) 3

O N

The switch in the circuit shown was on position a for a long time, n 2008 GATE TWO MARKS o.i c and is move to position b at time t = 0 . The current i (t) for t > 0 . 1.37i d a The Thevenin equivalent impedance Zth between the nodes P and is given by o n Q in the following circuit is w.

ww

(A) 0.2e-125t u (t) mA (C) 0.2e-1250t u (t) mA 1.33

1.34

(B) 20e-1250t u (t) mA (D) 20e-1000t u (t) mA

In the circuit shown, what value of RL maximizes the power delivered to RL ?

(A) 2.4 W

(B) 8 W 3

(C) 4 W

(D) 6 W

The time domain behavior of an RL circuit is represented by L di + Ri = V0 (1 + Be-Rt/L sin t) u (t). dt For an initial current of i (0) = V0 , the steady state value of the R current is given by

(A) 1

1.38

(B) 1 + s + 1 s

2 (C) 2 + s + 1 (D) s2 + s + 1 s s + 2s + 1 The driving point impedance of the following network is given by Z (s) = 2 0.2s s + 0.1s + 2

The component values are (A) L = 5 H, R = 0.5 W, C = 0.1 F (B) L = 0.1 H, R = 0.5 W, C = 5 F (C) L = 5 H, R = 2 W, C = 0.1 F (D) L = 0.1 H, R = 2 W, C = 5 F 1.39

The circuit shown in the figure is used to charge the capacitor C alternately from two current sources as indicated. The switches S1 and S2 are mechanically coupled and connected as follows:

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Page 5

For 2nT # t # (2n + 1) T , (n = 0, 1, 2,..) S1 to P1 and S2 to P2 For (2n + 1) T # t # (2n + 2) T, (n = 0, 1, 2,...) S1 to Q1 and S2 to Q2

The z -parameter matrix for this network is

1.42

1.5 (A) = 4.5 1.5 (C) = 1.5

1.5 (B) = 1.5 4.5 (D) = 1.5

4.5 4.5G 1.5 4.5G

The h -parameter matrix for this network is

1.43

-3 3 (A) = - 1 0.67 G 3 3 (C) = 1 0.67 G

Assume that the capacitor has zero initial charge. Given that u (t) is a unit step function , the voltage vc (t) across the capacitor is given by (A)

1.5 1.5G 4.5 1. 5 G

-3 -1 (B) = 3 0.67 G 3 1 (D) = - 3 - 0.67 G

GATE 2007

3

/ (- 1) n tu (t - nT)

An independent voltage source in series with an impedance Zs = Rs + jXs delivers a maximum average power to a load impedance ZL when (A) ZL = Rs + jXs (B) ZL = Rs (C) ZL = jXs (D) ZL = Rs - jXs

1.44

n=1 3

(B) u (t) + 2 / (- 1) n u (t - nT) n=1 3

(C) tu (t) + 2 / (- 1) n u (t - nT) (t - nT) n=1

(D) / 60.5 - e- (t - 2nT) + 0.5e- (t - 2nT) - T @ 3

ONE MARK

The RC circuit shown in the figure is

1.45

n=1

Common Data For Q. 2.23 & 2.24 : The following series RLC circuit with zero conditions is excited by a unit impulse functions d (t).

1.40

For t > 0 , the output voltage vC ^ t h is

w ww

(A) 1 _e 3 (B) e (C)

3t 2

3 2

(B) a high-pass filter (D) a band-reject filter

.in o c . ia GATE 2007

d .no

-1 (B) 2 te 2 t (A) 2 ^e t - e t h 3 3 1 -1 (C) 2 e 2 t cos c 3 t m (D) 2 e 2 t sin c 3 t m 2 2 3 3 For t > 0 , the voltage across the resistor is -1 2

1.41

O N

A I D

(A) a low-pass filter (C) a band-pass filter

1.46

TWO MARKS

Two series resonant filters are as shown in the figure. Let the 3-dB bandwidth of Filter 1 be B1 and that of Filter 2 be B2 . the value B1 is B2

- e- 2 t i 1

3 1 sin 3 t c 2 mG =cos c 2 t m 3 2 e -21 t sin 3 t c 2 m 3

-1 t 2

(A) 4 (C) 1/2 1.47

-1 (D) 2 e 2 t cos c 3 t m 2 3

(B) 1 (D) 1/4

For the circuit shown in the figure, the Thevenin voltage and resistance looking into X - Y are

Statement for linked Answers Questions 2.25 & 2.26: A two-port network shown below is excited by external DC source. The voltage and the current are measured with voltmeters V1, V2 and ammeters. A1, A2 (all assumed to be ideal), as indicated

(A) (C) 1.48

4 3 4 3

V, 2 W V, 23 W

(B) 4 V, 23 W (D) 4 V, 2 W

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

Under following conditions, the readings obtained are: (1) S1 -open, S2 - closed A1 = 0,V1 = 4.5 V,V2 = 1.5 V, A2 = 1 A (2) S1 -open, S2 - closed A1 = 4 A,V1 = 6 V,V2 = 6 V, A2 = 0

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 6

(C) 1.0 A

In the figure shown below, assume that all the capacitors are initially uncharged. If vi (t) = 10u (t) Volts, vo (t) is given by

1.54

(A) 0.50 exp (- 25t) mA (C) 0.50 exp (- 12.5t) mA 1.49

(D) 0.0 A

(B) 0.25 exp (- 25t) mA (D) 0.25 exp (- 6.25t) mA

In the ac network shown in the figure, the phasor voltage VAB (in Volts) is (A) 8e -t/0.004 Volts (C) 8u (t) Volts

A negative resistance Rneg is connected to a passive network N having driving point impedance as shown below. For Z2 (s) to be positive real,

1.55

(A) 0 (C) 12.5+30c

(B) 5+30c (D) 17+30c

GATE 2006 1.50

TWO MARKS

A two-port network is represented by ABCD parameters given by V1 A B V2 = I G = =C D G=- I G 1 2 If port-2 is terminated by RL , the input impedance seen at port-1 is given by (B) ARL + C (A) A + BRL C + DRL BRL + D (C) DRL + A (D) B + ARL BRL + C D + CRL

1.51

(A) Rneg # Re Z1 (jw), 6w (C) Rneg # Im Z1 (jw), 6w

D O

IA

N

ia

d In the two port network shown in the figure below, Z12 and Z21 and.no w respectively ww

(A) re and br0 (C) 0 and bro

1.53

(B) 0 and - br0 (D) re and - br0

1.57

The first and the last critical frequencies (singularities) of a driving point impedance function of a passive network having two kinds of elements, are a pole and a zero respectively. The above property will be satisfied by (A) RL network only (B) RC network only (C) LC network only (D) RC as well as RL networks A 2 mH inductor with some initial current can be represented as shown below, where s is the Laplace Transform variable. The value of initial current is

(B) 2.0 A

L C L C

The value of x will be (A) n

1.58

such that the step response y (t) in the

L C (C) R $ 2 (D) R = 1 LC The ABCD parameters of an ideal n: 1 transformer shown in the figure are n 0 >0 x H

(C) n2

(A) 0.5 A

ONE MARK

The in condition on R, L and C . o figure has no oscillations, is .c

1.56

(B) Rneg # Z1 (jw) , 6w (D) Rneg # +Z1 (jw), 6w

GATE 2005

(A) R $ 1 2

1.52

(B) 8 (1 - e -t/0.004) Volts (D) 8 Volts

(B) R $

(B) 1 n (D) 12 n

In a series RLC circuit, R = 2 kW , L = 1 H, and C = 1 mF The 400 resonant frequency is (B) 1 # 10 4 Hz (A) 2 # 10 4 Hz p (C) 10 4 Hz

(D) 2p # 10 4 Hz

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 1.59

The maximum power that can be transferred to the load resistor RL from the voltage source in the figure is

(A) 1 W (C) 0.25 W 1.60

(A) 0.238 V (C) - 0.238 V

(B) 10 W (D) 0.5 W

The h parameters of the circuit shown in the figure are

1.65

0. 1 0. 1 (A) = - 0. 1 0. 3 G 30 20 (C) = 20 20G

TWO MARKS

For the circuit shown in the figure, the instantaneous current i1 (t) is

(A) 10 3 90c A 2

(B) 10 3 - 90c A 2

(C) 5 60c A

(D) 5 - 60c A

Impedance Z as shown in the given figure is

A I D

O N w

.in o c . ia (A) 3 V

d .no

ww

1.63

ONE MARK

Consider the network graph shown in the figure. Which one of the following is NOT a ‘tree’ of this graph ?

(B) j9 W (D) j39 W

For the circuit shown in the figure, Thevenin’s voltage and Thevenin’s equivalent resistance at terminals a - b is

(A) 5 V and 2 W (C) 4 V and 2 W 1.64

(C) 4 V

(B) - 3 V (D) - 4 V

GATE 2004 1.67

(A) j29 W (C) j19 W

10 - 1 (B) = 1 0.05G 10 1 (D) = - 1 0.05G

A square pulse of 3 volts amplitude is applied to C - R circuit shown in the figure. The capacitor is initially uncharged. The output voltage V2 at time t = 2 sec is

1.66

1.62

(B) 0.138 V (D) 1 V

The first and the last critical frequency of an RC -driving point impedance function must respectively be (A) a zero and a pole (B) a zero and a zero (C) a pole and a pole (D) a pole and a zero

GATE 2005 1.61

Page 7

(B) 7.5 V and 2.5 W (D) 3 V and 2.5 W

If R1 = R2 = R4 = R and R3 = 1.1R in the bridge circuit shown in the figure, then the reading in the ideal voltmeter connected between a and b is

(A) a (C) c 1.68

(B) b (D) d

The equivalent inductance measured between the terminals 1 and 2 for the circuit shown in the figure is

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) L1 + L2 + M (C) L1 + L2 + 2M 1.69

1.70

(A) sin (10 t - 45c) (C) sin (103 t - 53c)

3

106 s2 + 20s + 106 The Quality factor (Q-factor) of this circuit is (A) 25 (B) 50 n i . (D) 5000 .co(C) 100

1.75

H (s) =

For the circuit shown in the figure, the initial conditions are zero. Its V (s) is transfer function H (s) = c Vi (s)

For the R - L circuit shown in the figure, the input voltage vi (t) = u (t). The current i (t) is

1 s + 106 s + 106 103 (C) 2 s + 103 s + 106 (A)

1.76

GATE 2004 1.72

2

ia

d .no

ww

w

1.74

(B)

2

A I D

O N

(B) sin (10 t + 45c) (D) sin (103 t + 53c)

1-j 1+j (B) = -1 + j 1 - j G 1 + j -1 + j (D) = -1 + j 1 + j G

s+2 s +s+1 (D) 2 1 s +s+1 Vo (s) The transfer function H (s) = of an RLC circuit is given by Vi (s) s s +s+1 (C) 2 s - 2 s +s+1 (A)

For the circuit shown in the figure, the time constant RC = 1 ms. The input voltage is vi (t) = 2 sin 103 t . The output voltage vo (t) is equal to

1+j 1 + jG 1+j 1 - jG

The circuit shown in the figure has initial current iL (0-) = 1 A through the inductor and an initial voltage vC (0-) =- 1 V across the capacitor. For input v (t) = u (t), the Laplace transform of the current i (t) for t $ 0 is

1.73

5 sin (2t + 53.1c) 5 sin (2t - 53.1c) 25 sin (2t + 53.1c) 25 sin (2t - 53.1c)

3

1.71

1-j (A) = 1+j 1+j (C) = 1-j

(B) L1 + L2 - M (D)L1 + L2 - 2M

The circuit shown in the figure, with R = 1 W, L = 1 H and C = 3 F 3 4 has input voltage v (t) = sin 2t . The resulting current i (t) is

(A) (B) (C) (D)

Page 8

TWO MARKS

For the lattice shown in the figure, Za = j2 W and Zb = 2 W . The z11 z12 values of the open circuit impedance parameters 6 z @ = = are z21 z22 G

2

106 s + 103 s + 106 106 (D) 2 s + 106 s + 106 (B)

2

Consider the following statements S1 and S2 S1 : At the resonant frequency the impedance of a series RLC circuit is zero. S2 : In a parallel GLC circuit, increasing the conductance G results in increase in its Q factor. Which one of the following is correct? (A) S1 is FALSE and S2 is TRUE (B) Both S1 and S2 are TRUE (C) S1 is TRUE and S2 is FALSE (D) Both S1 and S2 are FALSE GATE 2003

1.77

ONE MARK

The minimum number of equations required to analyze the circuit shown in the figure is

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

(A) 3 (C) 6 1.78

1.79

1.80

(B) 4 (D) 7

1.84

A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100 . If each of R, L and C is doubled from its original value, the new Q of the circuit is (A) 25 (B) 50 (C) 100 (D) 200

R + Ls + Cs1 - Ls I1 (s) - Vs (B) > = R + Cs1 H=I2 (s)G = 0 G - Ls

R + Ls + Cs1 - Ls I1 (s) - Vs (C) > = R + Ls + Cs1 H=I2 (s)G = 0 G - Ls

The differential equation for the current i (t) in the circuit of the figure is

V R + Ls + Cs1 - Cs I1 (s) s (D) > = R + Ls + Cs1 H=I2 (s)G = 0 G - Ls

O N

A I D

TWO MARKS

.in

co ia.

d

1.86

Twelve 1 W resistance are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is (B) 1 W (A) 5 W 6 (C) 6 W (D) 3 W 5 2

3 (s + 3) 2 (s + 3) (B) 2 s + 2s + 3 s + 2s + 2 3 (s + 3) 2 (s - 3) (C) 2 (D) 2 s + 2s + 2 s - 2s - 3 An input voltage v (t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c) V is applied to a series combination of resistance R = 1 W and an inductance L = 1 H. The resulting steady-state current i (t) in ampere is (A) 10 cos (t + 55c) + 10 cos (2t + 10c + tan-1 2) (A)

2

(B) 10 cos (t + 55c) + 10 23 cos (2t + 55c) (C) 10 cos (t - 35c) + 10 cos (2t + 10c - tan-1 2)

The current flowing through the resistance R in the circuit in the figure has the form P cos 4t where P is

(D) 10 cos (t - 35c) + 1.87

(A) (0.18 + j0.72) (C) - (0.18 + j1.90)

The driving point impedance Z (s) of a network has the pole-zero locations as shown in the figure. If Z (0) = 3 , then Z (s) is

1.85

2 no . (B) d 2i + 2 di + 2i (t) = cos t w dt dt ww 2 d i di (D) 2 + 2 + 2i (t) = sin t dt dt

GATE 2003

1.82

At t = 0+ , the current i1 is (A) - V (B) - V 2R R (C) - V (D) zero 4R I1 (s) and I2 (s) are the Laplace transforms of i1 (t) and i2 (t) respectively. The equations for the loop currents I1 (s) and I2 (s) for the circuit shown in the figure, after the switch is brought from position 1 to position 2 at t = 0 , are V R + Ls + Cs1 - Ls I1 (s) s (A) > == G G 1 H= R + Cs I2 (s) - Ls 0

1.83

A source of angular frequency 1 rad/sec has a source impedance consisting of 1 W resistance in series with 1 H inductance. The load that will obtain the maximum 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

2 (A) 2 d 2i + 2 di + i (t) = sin t dt dt 2 (C) 2 d 2i + 2 di + i (t) = cos t dt dt

1.81

Page 9

3 2

cos (2t - 35c)

The impedance parameters z11 and z12 of the two-port network in the figure are

(B) (0.46 + j1.90) (D) - (0.192 + j0.144)

The circuit for Q. 2.66 & 2.67 is given below. Assume that the switch S is in position 1 for a long time and thrown to position 2 at t = 0 .

(A) (B) (C) (D)

z11 = 2.75 W and z12 = 0.25 W z11 = 3 W and z12 = 0.5 W z11 = 3 W and z12 = 0.25 W z11 = 2.25 W and z12 = 0.5 W

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GATE 2002 1.88

ONE MARK

Page 10

1.93

The dependent current source shown in the figure

1.94

(A) delivers 80 W (C) delivers 40 W 1.89

If each branch of Delta circuit has impedance 3 Z , then each branch of the equivalent Wye circuit has impedance (A) Z (B) 3Z 3 (C) 3 3 Z (D) Z 3 The admittance parameter Y12 in the 2-port network in Figure is

(B) absorbs 80 W (D) absorbs 40 W

In the figure, the switch was closed for a long time before opening at t = 0 . The voltage vx at t = 0+ is

(A) - 0.02 mho (C) - 0.05 mho

(B) 0.1 mho (D) 0.05 mho

GATE 2001 1.95

(A) 25 V (C) - 50 V

The voltage e0 in the figure is

(B) 50 V (D) 0 V

GATE 2002 1.90

TWO MARKS

(A) 48 V (C) 36 V

A I D

TWO MARKS

In the network of the fig, the maximum power is delivered to RL if its value is

O N

1.96

(B) 24 V (D) 28 V

When the angular frequency w in the figure is varied 0 to 3, the locus of the current phasor I2 is given by

.in

no w.

co ia.

d

ww

1.91

(A) 16 W

(B) 40 W 3

(C) 60 W

(D) 20 W

If the 3-phase balanced source in the figure delivers 1500 W at a leading power factor 0.844 then the value of ZL (in ohm) is approximately

(A) 90+32.44c (C) 80+ - 32.44c

(B) 80+32.44c (D) 90+ - 32.44c

GATE 2001 1.92

ONE MARK

The Voltage e0 in the figure is

(A) 2 V (C) 4 V

(B) 4/3 V (D) 8 V

1.97

In the figure, the value of the load resistor RL which maximizes the power delivered to it is

(A) 14.14 W (C) 200 W

(B) 10 W (D) 28.28 W

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 1.98

Page 11

(C) - 4 A

The z parameters z11 and z21 for the 2-port network in the figure are

(D) 4 A

GATE 1999

Identify which of the following is NOT a tree of the graph shown in the given figure is

1.103

(A) z11 = 6 W; z21 = 16 W 11 11 (C) z11 = 6 W; z21 =- 16 W 11 11

(B) z11 = 6 W; z21 = 4 W 11 11 (D) z11 = 4 W; z21 = 4 W 11 11

GATE 2000 1.99

ONE MARK

(A) begh (C) abfg

The circuit of the figure represents a

(A) Low pass filter (C) band pass filter

(B) defg (D) aegh

A 2-port network is shown in the given figure. The parameter h21 for this network can be given by

1.104

1.100

ONE MARK

(B) High pass filter (D) band reject filter (A) - 1/2 (C) - 3/2

In the circuit of the figure, the voltage v (t) is

A I D

(B) + 1/2 (D) + 3/2

GATE 1999

(A) eat - ebt (C) aeat - bebt 1.101

(B) eat + ebt (D) aeat + bebt

O N w

In the circuit of the figure, the value of the voltage source E isww

The Thevenin equivalent voltage VTH appearing between the terminals A and B of the network shown in the given figure is given in . o a.cby

1.105

.no

di

(A) j16 (3 - j4) (C) 16 (3 + j4) 1.106

(A) - 16 V (C) - 6 V

(B) 4 V (D) 16 V

GATE 2000 1.102

(B) j16 (3 + j4) (D) 16 (3 - j4)

The value of R (in ohms) required for maximum power transfer in the network shown in the given figure is

TWO MARKS

Use the data of the figure (a). The current i in the circuit of the figure (b)

(A) 2 (C) 8 1.107

(A) - 2 A

TWO MARK

(B) 2 A

(B) 4 (D) 16

A Delta-connected network with its Wye-equivalent is shown in the given figure. The resistance R1, R2 and R3 (in ohms) are respectively

(A) 1.5, 3 and 9

(B) 3, 9 and 1.5

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(C) 9, 3 and 1.5

Page 12

(C) 4 A

(D) 3, 1.5 and 9 1.116

GATE 1998 1.108

1.109

1.110

1.111

The voltage V in the figure equal to

ONE MARK

A network has 7 nodes and 5 independent loops. The number of branches in the network is (A) 13 (B) 12 (C) 11 (D) 10 The nodal method of circuit analysis is based on (A) KVL and Ohm’s law (B) KCL and Ohm’s law (C) KCL and KVL (D) KCL, KVL and Ohm’s law

(A) 3 V (C) 5 V

Superposition theorem is NOT applicable to networks containing (A) nonlinear elements (B) dependent voltage sources (C) dependent current sources (D) transformers

1.117

The parallel RLC circuit shown in the figure is in resonance. In this circuit

(A) IR < 1 mA (C) IR + IC < 1 mA

1.113

(B) IR + IL > 1 mA (D) IR + IC > 1 mA

(B) 5 V (D) None of the above

The voltage V in the figure is

A I D

O N

The voltage across the terminals a and b in the figure is

.in

no w.

co ia.

d

ww

1.119

(A) 10 V (C) 5 V

(B) 15 V (D) None of the above

In the circuit of the figure is the energy absorbed by the 4 W resistor in the time interval (0, 3) is

(B) 3.0 V (D) 4.0 V

A high-Q quartz crystal exhibits series resonance at the frequency ws and parallel resonance at the frequency wp . Then (A) ws is very close to, but less than wp (B) ws > wp GATE 1997

1.115

1.118

0 - 1/2 The short-circuit admittance matrix a two-port network is > 1/2 0 H The two-port network is (A) non-reciprocal and passive (B) non-reciprocal and active (C) reciprocal and passive (D) reciprocal and active

(A) 0.5 V (C) 3.5 V 1.114

(B) - 3 V (D) None of these

The voltage V in the figure is always equal to

(A) 9 V (C) 1 V

1.112

(D) None or these

(A) 36 Joules (C) 256 Joules 1.120

(B) 16 Joules (D) None of the above

In the circuit of the figure the equivalent impedance seen across terminals a, b, is

ONE MARK

The current i4 in the circuit of the figure is equal to

(A) 12 A

(B) - 12 A

(A) b 16 l W 3 (C) b 8 + 12j l W 3

(B) b 8 l W 3 (D) None of the above

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GATE 1996 1.121

ONE MARK

In the given figure, A1, A2 and A3 are ideal ammeters. If A2 and A3 read 3 A and 4 A respectively, then A1 should read

(A) 1 A (C) 7 A 1.122

(B) 5 A (D) None of these

The number of independent loops for a network with n nodes and b branches is (A) n - 1 (B) b - n (C) b - n + 1 (D) independent of the number of nodes GATE 1996

1.123

Page 13

TWO MARKS

A I D

The voltages VC1, VC2, and VC3 across the capacitors in the circuit in the given figure, under steady state, are respectively.

O N

no w.

(A) 80 V, 32 V, 48 V (C) 20 V, 8 V, 12 V

(B) 80 V, 48 V, 32 V (D) 20 V, 12 V, 8 V

.in

co ia.

d

ww

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 14

SOLUTIONS

j40 100 53.13c j4 + 3 40 90c = 100 53.13c 5 53.13c = 800 90c

VTh = 10 VL1 =

1.5 1.1

Option (B) is correct. In the equivalent star connection, the resistance can be given as Rb Ra RC = Ra + Rb + Rc Ra Rc RB = Ra + Rb + Rc Rb Rc RA = Ra + Rb + Rc So, if the delta connection components Ra , Rb and Rc are scaled by a factor k then ^k Rb h^k Rc h RAl = kRa + kRb + kRc

Option (C) is correct. For the given transformer, we have V = 1.25 1 VWX

Since, So,

2

Rb Rc =k k Ra + Rb + Rc

or, at

= k RA hence, it is also scaled by a factor k

VYZ = 0.8 (attenuation factor) V VYZ = 0.8 1.25 = 1 ^ h^ h VWX VYZ = VWX V VWX = 100 V ; YZ = 100 100 VWX 1

1

1

V at VWZ = 100 V ; WX = 100 100 VYZ Option (D) is correct. For the given capacitance, C = 100mF in the circuit, we have the 1.6 Option (C) is correct. reactance. The quality factor of the inductances are given by 4 1 1 10 XC = = = q 1 = wL 1 s sc s # 100 # 10-6 R1 So, and q 2 = wL 2 10 4 + 10 4 R2 V2 ^s h s = 4 .inin series circuit, the effective quality factor is given by V1 ^s h 10 + 10 4 + 10 4 oSo, c . s s XLeq dia o = = wL 1 + wL 2 Q n s + 1 . R R1 + R 2 eq = w s+2 w w q1 q wL 1 + wL 2 + 2 q R + q2 R2 Option (C) is correct. R R R R 1 R2 1 R2 2 2 = 1 1 = = 1 1 1 1 R1 + R 2 For the purely resistive load, maximum average power is transferred + + R 2 R1 R 2 R1 when 1.7 Option (C) is correct. 2 2 RL = RTh + XTh where RTh + jXTh is the equivalent thevinin (input) impedance of the circuit. Hence, we obtain RL = 42 + 32 5W 2

1.2

2

2

A I D

1.3

1.4

O N

Option (C) is correct. For evaluating the equivalent thevenin voltage seen by the load RL , we open the circuit across it (also if it consist dependent source). The equivalent circuit is shown below

As the circuit open across RL so I2 = 0 or, j40I2 = 0 i.e., the dependent source in loop 1 is short circuited. Therefore, ^ j4h Vs VL1 = j4 + 3

Consider that the voltage across the three capacitors C1 , C2 and C 3 are V1 , V2 and V3 respectively. So, we can write V2 = C 3 V3 C2 ....(1) Since, Voltage is inversely proportional to capacitance Now, given that C1 = 10 mF ; ^V1hmax = 10V C2 = 5 mF ; ^V2hmax = 5 V C 3 = 2 mF ; ^V3hmax = 2V So, from Eq (1) we have V2 = 2 5 V3 for ^V3hmax = 2 We obtain, V2 = 2 # 2 = 0.8 volt < 5 5

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1.8

1.9

1.10

Average power delivered. Pavg. = 1 I 2 Z cos q = 1 # 25 # 5 cos 36.86c = 50 W 2 2

i.e., V2 < ^V2hmax Hence, this is the voltage at C2 . Therefore, V3 = 2 volt V2 = 0.8 volt and V1 = V2 + V3 = 2.8 volt Now, equivalent capacitance across the terminal is Ceq = C2 C 3 + C1 = 5 # 2 + 10 = 80 mF 5+2 7 C2 + C3 Equivalent voltage is (max. value) Vmax = V1 = 2.8 So, charge stored in the effective capacitance is Q = Ceq Vmax = b 80 l # ^2.8h = 32 mC 7 Option (D) is correct.

At the node 1, voltage is given as V1 = 10 volt Applying KCL at node 1 IS + V1 + V1 - 2 = 0 2 1 IS + 10 + 10 - 2 = 0 2 1 IS =- 13 A Also, from the circuit, VS - 5 # 2 = V1 VS = 10 + V1 = 20 volt

Page 15

Alternate method: Z = (4 - j3) W , I = 5 cos (100pt + 100) A 2 Pavg = 1 Re $ I Z . = 1 # Re "(5) 2 # (4 - j3), 2 2 1 = # 100 = 50 W 2 Option (C) is correct

1.12

IA

D O

N

Option (C) is correct. Again from the shown circuit, the current in 1 W resistor is I = V1 = 10 = 10 A 1 1 Option (D) is correct. The s -domain equivalent circuit is shown as below.

Applying nodal analysis at top node. V1 + 1 0c V1 + 1 0c = 1 0c + 1 j1

V1 (j 1 + 1) + j 1 + 1 0c = j1 V1 = - 1 1 + j1 n i o. c . - 1 +1 dia V 1+j o 1 + 1 0c Current = I1 = .n j1 j1 w ww j = = 1 A (1 + j) j 1 + j 1.13 Option (A) is correct. We obtain Thevenin equivalent of circuit B .

Thevenin Impedance :

I (s) =

ZTh = R

vc (0) /s v (0) = c 1 + 1 1 + 1 C1 s C 2 s C1 C 2

Thevenin Voltage :

I (s) = b C1 C2 l (12 V) = 12Ceq vC (0) = 12 V C1 + C 2 Taking inverse Laplace transform for the current in time domain, (Impulse) i (t) = 12Ceq d (t) 1.11

Option (B) is correct. In phasor form,

VTh = 3 0c V Now, circuit becomes as

Z = 4 - j 3 = 5 - 36.86cW I = 5 100c A

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Page 16

When 10 V is connected at port A the network is

I1 = 10 - 3 2+R Power transfer from circuit A to B P = (I 12) 2 R + 3I1 2 P = :10 - 3D R + 3 :10 - 3D 2+R 2+R P = 49R 2 + 21 (2 + R) (2 + R) 49R + 21 (2 + R) P = (2 + R) 2 P = 42 + 70R2 (2 + R) 2 dP = (2 + R) 70 - (42 + 70R) 2 (2 + R) = 0 dR (2 + R) 4 (2 + R) [(2 + R) 70 - (42 + 70R) 2] = 0 140 + 70R - 84 - 140R = 0 56 = 70R R = 0.8 W Current in the circuit,

1.14

Option (A) is correct. In the given circuit

Now, we obtain Thevenin equivalent for the circuit seen at load terminal, let Thevenin voltage is VTh, 10 V with 10 V applied at port A and Thevenin resistance is RTh .

IL =

VTh,10 V RTh + RL

For RL = 1 W , IL = 3 A V 3 = Th,10 V RTh + 1 For RL = 2.5 W , IL = 2 A V = Th,10 V RTh + 2.5 Dividing above two 3 = RTh + 2.5 2 RTh + 1

A I D

O N

...(i) ...(ii)

3RTh + 3 = 2RTh + 5 VA - VB = 6 V n i . RTh = 2 W So current in the branch will be .co a i d Substituting RTh into equation (i) no IAB = 6 = 3 A . 2 VTh,10 V = 3 (2 + 1) = 9 V ww We can see, that the circuit is a one port circuit lookingwfrom Note that it is a non reciprocal two port network. Thevenin voltage terminal BD as shown below seen at port B depends on the voltage connected at port A. Therefore we took subscript VTh,10 V . This is Thevenin voltage only when 10 V source is connected at input port A. If the voltage connected to port A is different, then Thevenin voltage will be different. However, Thevenin’s resistance remains same. Now, the circuit is as shown below :

For a one port network current entering one terminal, equals the current leaving the second terminal. Thus the outgoing current from A to B will be equal to the incoming current from D to C as shown i.e. IDC = IAB = 3 A

For RL = 7 W , 1.16

The total current in the resistor 1 W will be (By writing KCL at node D ) I1 = 2 + IDC = 2+3 = 5A So, VCD = 1 # (- I1) =- 5 V 1.15

Option (C) is correct.

IL =

VTh,10 V = 9 = 1A 2 + RL 2 + 7

Option (B) is correct. Now, when 6 V connected at port A let Thevenin voltage seen at port B is VTh,6 V . Here RL = 1 W and IL = 7 A 3

VTh, 6 V = RTh # 7 + 1 # 7 = 2 # 7 + 7 = 7 V 3 3 3 3

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This is a linear network, so VTh at port B can be written as VTh = V1 a + b where V1 is the input applied at port A. We have V1 = 10 V , VTh,10 V = 9 V ...(i) ` 9 = 10a + b When V1 = 6 V , VTh, 6 V = 9 V ...(ii) ` 7 = 6a + b Solving (i) and (ii) a = 0.5 , b = 4 Thus, with any voltage V1 applied at port A, Thevenin voltage or open circuit voltage at port B will be So, VTh, V = 0.5V1 + 4 For V1 = 8 V VTh,8 V = 0.5 # 8 + 4 = 8 = Voc (open circuit voltage)

Page 17

j j + (1 + j) 2 j = =1 3 (1 + j) 2 + j V = b p l cos (t/RC) 3

Here w = 1 RC

= Vout Vin Thus 1.20

v out

Option (B) is correct. From star delta conversion we have

1

1.17

Thus

Option (A) is correct. Replacing P - Q by short circuit as shown below we have

Ra Rb 6.6 = = 2W Ra + Rb + Rc 6 + 6 + 6

Here R1 = R 2 = R 3 = 2 W Replacing in circuit we have the circuit shown below :

A I D

Using current divider rule the current Isc is 25 ISC = (16 0 ) = (6.4 - j4.8) A 25 + 15 + j30 1.18

R1 =

Option (C) is correct. Power transferred to RL will be maximum when RL is equal to the Thevenin resistance. We determine Thevenin resistance by killing all source as follows :

Now the total impedance of circuit is (2 + j4) (2 - j4) in Z = +2 = 7W . o (2 + j4) (2 - j4) c . a di Current o I = 14+0c = 2+0c n . 7 w

O N ww

1.21

Option (D) is correct. From given admittance matrix we get I1 = 0.1V1 - 0.01V2 and

...(1) ...(2)

I2 = 0.01V1 + 0.1V2 Now, applying KVL in outer loop; RTH = 10 # 10 + 10 = 15 W 10 + 10 1.19

V2 =- 100I2 or I2 =- 0.01V2 From eq (2) and eq (3) we have

Option (A) is correct. The given circuit is shown below

...(3)

- 0.01V2 = 0.01V1 + 0.1V2 - 0.11V2 = 0.01V1 V2 = - 1 11 V1 1.22

For parallel combination of R and C equivalent impedance is R$ 1 jw C R Zp = = 1 1 + j wRC R+ jw C Transfer function can be written as R 1 + jwRC Vout = Z p = Vin R Zs + Zp R+ 1 + jwC 1 + jwRC

Option (A) is correct. Here we take the current flow direction as positive. At t = 0- voltage across capacitor is -3 Q VC (0-) =- =- 2.5 # 10-6 =- 50 V C 50 # 10 + Thus VC (0 ) =- 50 V In steady state capacitor behave as open circuit thus Now,

jwRC = jwRC + (1 + jwRC) 2

V (3) = 100 V VC (t) = VC (3) + (VC (0+) - VC (3)) e-t/RC -t

= 100 + (- 50 - 100) e 10 # 50 # 10

-6

3

= 100 - 150e- (2 # 10 t)

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Page 18

ic (t) = C dV dt

Now

Current is resistor (AB) i (0) = 0.75 = 0.375 A 2

3

= 50 # 10-6 # 150 # 2 # 103 e-2 # 10 t A

Similarly for steady state the circuit is as shown below

3

= 15e-2 # 10 t ic (t) = 15 exp (- 2 # 103 t) A 1.23

Option (A) is correct. Given circuit is as shown below i (3) = 15 = 0.5 A 3 -3 t = L = 15 # 10 = 10-3 sec Req 10 + (10 || 10)

By writing node equation at input port I1 = V1 + V1 - V2 = 4V1 - 2V2 0.5 0.5 By writing node equation at output port I2 = V2 + V2 - V1 =- 2V1 + 4V2 0.5 0.5 From (1) and (2), we have admittance matrix 4 -2 Y => - 2 4H 1.24

Now and So, Hence

...(1)

...(2) 1.26

Option (D) is correct. A parallel RLC circuit is shown below :

i (t) i (0) i (3) B i (t)

At resonance So,

Z in =

1 1 + 1 + jw C R jw L

A I D

1 = wC wL Z in = 1 = R 1/R

O N

co ia.

d

ww

Voltage across Z2 VZ = 2

Z2 : 20 0 = Z1 + Z 2

(maximum at resonance) =c

Thus (D) is not true. Furthermore bandwidth is wB i.e wB \ 1 and is independent of L R , Hence statements A, B, C, are true. 1.25

= 0.5 - 0.125e-1000 t A

Z1 = jwL = j # 103 # 20 # 10-3 = 20j Z2 = R || XC 1 XC = 1 = =- 20j jwC j # 103 # 50 # 10-6 1 (- 20j) Z2 = R = 1W 1 - 20j

.in

no w.

-3

Option (A) is correct. Circuit is redrawn as shown below

Where,

Input impedance

t

= A + Be- 1 # 10 = A + Be-100t = A + B = 0.375 = A = 0.5 = 0.375 - 0.5 =- 0.125

Option (A) is correct. Let the current i (t) = A + Be-t/t t " Time constant When the switch S is open for a long time before t < 0 , the circuit is

- 20j c 1 - 20j m 20j c 20j - 1 - 20j m

(- 20j) : 20 =- j 20j + 400 - 20j m

Current in resistor R is j V I = Z =- =- j A 1 R 2

1.27

Option (A) is correct. The circuit can be redrawn as

At t = 0 , inductor current does not change simultaneously, So the circuit is Applying nodal analysis VA - 10 + 1 + VA - 0 = 0 2 2 Current,

2VA - 10 + 2 = 0 = V4 = 4 V I1 = 10 - 4 = 3 A 2

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: 20

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 19

Current from voltage source is I 2 = I1 - 3 = 0 Since current through voltage source is zero, therefore power delivered is zero. 1.28

Option (A) is correct. Circuit is as shown below

V (0-) = 100 V Thus V (0+) = 100 V + At t = 0 , the circuit is as shown below

Since 60 V source is absorbing power. So, in 60 V source current flows from + to - ve direction So, I + I1 = 12

I (0+) = 100 = 20 mA 5k At steady state i.e. at t = 3 is I (3)= 0

I = 12 - I1 I is always less then 12 A So, only option (A) satisfies this conditions. 1.29

Option (C) is correct. For given network we have (RL XC ) Vi V0 = R + (RL XC ) RL V0 (s) RL + 1 sRL C = = R Vi (s) R RR + L L sC + RL R+ 1 + sRL C =

A I D 1.33

t

t

#0 VIdt = I #0Vdt

O N

(0.5m + 0.3m) 0.2m = 0.16 m F 0.5m + 0.3m + 0.2m 1 1 = = 1250 3 RCeq 5 # 10 # 0.16 # 10-6 i (t) = 20e-1250t u (t) mA

Option (C) is correct. For Pmax the load resistance RL must be equal to thevenin resistance Req i.e. RL = Req . The open circuit and short circuit is as shown n below o.i

w

The open circuit voltage is

= I # Area

From fig

Isc = I1 + I2 = 25 A Rth = Voc = 100 = 4 W Isc 25 Thus for maximum power transfer RL = Req = 4 W

VL = IL ZL The reactive power consumed by load is

1.34

Pr = VL IL* = IL ZL # IL* = ZL IL 2 2 = (7 # 4j) 20+0c = (7 + 4j) = 28 + 16j 8 + 6j Thus average power is 28 and reactive power is 16. Option (B) is correct. At t = 0- , the circuit is as shown in fig below :

Voc = 100 V I1 = 100 = 12.5 A 8 Vx =- 4 # 12.5 =- 50 V I2 = 100 + Vx = 100 - 50 = 12.5 A 4 4

Option (B) is correct. From given circuit the load current is 20+0c = = 20+0c IL = V Zs + ZL (1 + 2j) + (7 + 4j) 8 + 6j = 1 (8 - 6j) = 20+0c = 2+ - f where f = tan - 1 3 10+f 5 4 The voltage across load is

1.32

u (t)

n

. ww

= 2 # 1 (10 + 12) # 600 = 13.2 kJ 2 1.31

t RCeq

.c

ia od

Option (C) is correct. The energy delivered in 10 minutes is E =

i (t) = I (0+) eCeq =

RL 1 = R R + RRL sC + RL 1+ + RsC RL

But we have been given V (s) 1 = T .F. = 0 2 + sCR Vi (s) Comparing, we get 1 + R = 2 & RL = R RL 1.30

Now

1.35

Option (A) is correct. Steady state all transient effect die out and inductor act as short circuits and forced response acts only. It doesn’t depend on initial current state. From the given time domain behavior we get that circuit has only R and L in series with V0 . Thus at steady state i (t) " i (3) = V0 R Option (C) is correct. The given graph is

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Page 20

1 = 0.1 or R = 2 W RC 1 = 2 or L = 0.1 H LC Option (C) is correct. Voltage across capacitor is t Vc = 1 idt C 0

1.39

There can be four possible tree of this graph which are as follows:

#

Here C = 1 F and i = 1 A. Therefore Vc =

There can be 6 different possible cut-set.

t

#0 dt

For 0 < t < T , capacitor will be charged from 0 V Vc =

1.36

t

#0 dt = t

At t = T, Vc = T Volts For T < t < 2T , capacitor will be discharged from T volts as

Option (B) is correct. Initially i (0-) = 0 therefore due to inductor i (0+) = 0 . Thus all current Is will flow in resistor R and voltage across resistor will be Is Rs . The voltage across inductor will be equal to voltage across Rs as no current flow through R.

Vc = T -

t

#T dt = 2T - t

At t = 2T, Vc = 0 volts For 2T < t < 3T , capacitor will be charged from 0 V Vc =

t

#2Tdt = t - 2T

At t = 3T, Vc = T Volts For 3T < t < 4T , capacitor will be discharged from T Volts

Thus but Thus 1.37

A I D

vL (0+) = Is Rs di (0+) vL (0+) = L dt

O N

di (0+) v (0+) Is Rs = L = L L dt

Option (A) is correct. Killing all current source and voltage sources we have,

Vc = T -

t

#3Tdt = 4T - t

At t = 4T, Vc = 0 Volts For 4T < t < 5T , capacitor will be charged from 0 V

.

in co.

ia od

n

. ww

w

Vc =

t

#4Tdt = t - 4T

At t = 5T, Vc = T Volts Thus the output waveform is

Only option C satisfy this waveform. Zth = (1 + s) ( s1 + 1) (1 + s)( s1 + 1) [ s1 + 1 + 1 + s] = = (1 + s) + ( s1 + 1) s + s1 + 1 + 1 Zth = 1

1.40

or Alternative : Here at DC source capacitor act as open circuit and inductor act as short circuit. Thus we can directly calculate thevenin Impedance as 1 W 1.38

s C s RC

+

3

2 Vc (s) = 2 = G 3 (s + 12 ) 2 + 43 Taking inverse Laplace transform we have

or

Option (D) is correct. Z (s) = R 1 sL = 2 sC s + We have been given Z (s) = 2 0.2s s + 0.1s + 2 Comparing with given we get 1 = 0.2 or C = 5 F C

Option (D) is correct. Writing in transform domain we have 1 Vc (s) = 1 s = 2 1 s 1 + + Vs (s) ^s h (s + s + 1) Since Vs (t) = d (t) " Vs (s) = 1 and Vc (s) = 2 1 (s + s + 1)

Vt = 2 e- sin c 3 t m 2 3 t 2

1 LC 1.41

Option (B) is correct. Let voltage across resistor be vR VR (s) = 1 1 = 2 s VS (s) ( s + s + 1) (s + s + 1) Since vs = d (t) " Vs (s) = 1 we get s VR (s) = 2 s = 1 2 (s + s + 1) (s + 2 ) + 43

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= or

(s + 12 ) (s + 12 ) 2 +

vR (t) = e- cos 1 2

(s +

1 2 3 1 2 2) + 4

Here we get also V0 = 0 Vi So frequency response of the circuit is as shown in fig and circuit is a Band pass filter.

3 t-1 2 e- sin 3 t 2 2# 3 2

t

= e- 2 =cos 1.42

3 4

-

Page 21

1 2

3 t - 1 sin 3 t 2 2 G 3

Option (C) is correct. From the problem statement we have z11 = v1 = 6 = 1. 5W i1 i = 0 4 z12 = v1 = 4.5 = 4.5W i2 i = 0 1 z21 = v2 = 6 = 1.5W i1 i = 0 4 z22 = v2 = 1.5 = 1.5W i2 i = 0 1 2

1

Option (D) is correct.

1.46

We know that bandwidth of series RLC circuit is R . Therefore L Bandwidth of filter 1 is B1 = R L1 Bandwidth of filter 2 is B2 = R = R = 4R L2 L1 L1 /4 Dividing above equation B1 = 1 B2 4

2

2

Thus z -parameter matrix is z11 z12 1.5 4.5 =z z G = =1.5 1.5 G 21 22 1.43

Option (A) is correct. From the problem statement we h12 = v1 = v2 i = 0 h22 = i2 = v2 i = 0 1

1

have 4. 5 = 3 1. 5 1 = 0.67 1.5

From z matrix, we have v1 = z11 i1 + z12 i2 v2 = z21 i1 + z22 i2 If v2 = 0 i2 = - z21 = - 1.5 =- 1 = h Then 21 i1 z22 1.5 or i2 =- i1 Putting in equation for v1, we get v1 i1

A I D

O N

.in

no w.

ww

Vth + Vth + Vth - 2i = 2 2 1 1 But from circuit i = Vth = Vth 1

co ia.

d

Therefore Vth + Vth + Vth - 2Vth = 2 2 1 1

v1 = (z11 - z12) i1 = h11 = z11 - z12 = 1.5 - 4.5 =- 3

or Vth = 4 volt From the figure shown below it may be easily seen that the short circuit current at terminal XY is isc = 2 A because i = 0 due to short circuit of 1 W resistor and all current will pass through short circuit.

v2 = 0

Hence h -parameter will be h11 h12 -3 3 =h h G = =- 1 0.67 G 21 22 1.44

Option (D) is correct. Here Vth is voltage across node also. Applying nodal analysis we get

1.47

Option (D) is correct. According to maximum Power Transform Theorem ZL = Zs* = (Rs - jXs)

1.45

Option (C) is correct. At w " 3 , capacitor acts as short circuited and circuit acts as shown in fig below

Therefore 1.48

Here we get V0 = 0 Vi

Rth = Vth = 4 = 2 W isc 2

Option (A) is correct. The voltage across capacitor is At t = 0+ , Vc (0+) = 0 At t = 3 , VC (3) = 5 V The equivalent resistance seen by capacitor as shown in fig is

At w " 0 , capacitor acts as open circuited and circuit look like as shown in fig below

Req = 20 20 = 10kW

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Page 22

Zins = 1 + R = 1 + sRC sC sC Thus pole is at origin and zero is at - 1 RC For parallel RC network input impedance is 1 R sC sC = Zin = 1 +R 1 + sRC sC

Time constant of the circuit is t = Req C = 10k # 4m = 0.04 s Using direct formula

or Now

-6

= 4 # 10 # (- 5 # 25e 1.49

Thus pole is at - 1 and zero is at infinity. RC

Vc (t) = VC (3) - [Vc (3) - Vc (0)] e-t/t = VC (3) (1 - e-t/t) + VC (0) e-t/t = 5 (1 - e-t/0.04) Vc (t) = 5 (1 - e-25t) dV (t) IC (t) = C C dt -25t

) = 0.5e

-25t

1.53

We know

= (5 - 3j) (5 + 3j) = =

mA

V (s) = sLI (s) - Li (0+) As per given in question - Li (0+) =- 1 mV Thus i (0+) = 1 mV = 0.5 A 2 mH

(5 - 3j) # (5 + 3j) 5 - 3j + 5 + 3j

(5) 2 - (3j) 2 = 25 + 9 = 3.4 10 10

1.54

VAB = Current # Impedance = 5+30c # 34 = 17+30c 1.50

Option (D) is correct. The network is shown in figure below.

Now and also From (1) and (2) Thus

V1 = AV2 - BI2 I1 = CV2 - DI2 V2 =- I2 RL we get V1 = AV2 - BI2 I1 CV2 - DI2

v = Ldi dt

Taking Laplace transform we get

Option (D) is correct. Impedance

Option (A) is correct.

Option (B) is correct. At initial all voltage are zero. So output is also zero. Thus v0 (0+) = 0 At steady state capacitor act as open circuit.

O N

A I D .in

co ia.

...(1) no d w. ...(2) w w ...(3)

Thus,

v0 (3) = 4 # vi = 4 # 10 = 8 5 5

The equivalent resistance and capacitance can be calculate after killing all source

Substituting value of V2 from (3) we get Input Impedance Zin = - A # I2 RL - BI2 - C # I2 RL - DI2 or Zin = ARL + B CRL + D 1.51

Option (B) is correct. The circuit is as shown below.

Req Ceq t v0 (t)

= 1 4 = 0.8 kW = 4 1 = 5 mF = Req Ceq = 0.8kW # 5mF = 4 ms

= v 0 (3) - [v 0 (3) - v 0 (0+)] e-t/t = 8 - (8 - 0) e-t/0.004 v0 (t) = 8 (1 - e-t/0.004) Volts At input port V1 = re I1 At output port V2 = r0 (I2 - bI1) =- r0 bI1 + r0 I2 Comparing standard equation V1 = z11 I1 + z12 I2 V2 = z21 I1 + z22 I2 z12 = 0 and z21 =- r0 b 1.52

Option (B) is correct. For series RC network input impedance is

1.55

Option (A) is correct. Here Z2 (s) = Rneg + Z1 (s) or Z2 (s) = Rneg + Re Z1 (s) + j Im Z1 (s) For Z2 (s) to be positive real, Re Z2 (s) $ 0 Thus Rneg + Re Z1 (s) $ 0 or Re Z1 (s) $- Rneg But Rneg is negative quantity and - Rneg is positive quantity. Therefore

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Re Z1 (s) $ Rneg Rneg # Re Z1 (jw)

or 1.56

Page 23

Here pole is s =- 1/RC and zero is at 3, therefore first critical frequency is a pole and last critical frequency is a zero.

For all w.

Option (C) is correct. Transfer function is

Option (A) is correct. Applying KCL we get

1.61

1 Y (s) 1 sC = = 2 1 U (s) s LC scR + 1 + R + sL + sC

i1 (t) + 5+0c = 10+60c

1 LC = s2 + R s + 1 L LC

or or

= j5 + j10 + j2 + j10 + j2 - j10 + j10 = j9

For no oscillations, x $ 1 R C $1 Thus 2 L or 1.57

L C

V1 = AV2 + BI2 I1 = CV2 + DI2 A B n 0 =C D G = = 0 1 G n x = 1 n

A I D

O N

Applying KCL at node we get Vab + Vab - 10 = 1 5 5

ia od

n

w

We have L = 1H and C = 1 # 10-6 400

Short circuit current from terminal ab is Isc = 1 + 10 = 3 A 5 Thus Rth = Vth = 7.5 = 2.5 W Isc 3

Resonant frequency 1 == 2p LC 2p

3 4 = 10 # 20 = 10 Hz 2p p 1.59

1 1 # 1 # 10 - 6 400

Option (C) is correct. Maximum power will be transferred when RL = Rs = 100W In this case voltage across RL is 5 V, therefore

Here current source being in series with dependent voltage source make it ineffective. 1.64

2 Pmax = V = 5 # 5 = 0.25 W R 100 1.60

or Vab = 7.5 = Vth n i Short o. circuit at terminal ab is shown below

.c

. ww

Option (B) is correct.

f0 =

Option (B) is correct. Open circuit at terminal ab is shown below

1.63

Option (B) is correct. For given transformer I2 = V1 = n I1 V2 1 or I1 = I2 and V1 = nV2 n Comparing with standard equation

Thus 1.58

R $2

Option (B) is correct. If L1 = j5W and L3 = j2W the mutual induction is subtractive because current enters from dotted terminal of j2W coil and exit from dotted terminal of j5W. If L2 = j2W and L3 = j2W the mutual induction is additive because current enters from dotted terminal of both coil. Thus Z = L1 - M13 + L2 + M23 + L3 - M31 + M32

1.62

Comparing with s2 + 2xwn s + wn2 = 0 we have Here 2xwn = R , L and wn = 1 LC Thus x = R LC = R C 2L 2 L

i1 (t) = 10+60c - 5+0c = 5 + 5 3j - 5 i1 (t) = 5 3 +90c = 10 3 +90c 2

Option (C) is correct. For stability poles and zero interlace on real axis. In RC series network the driving point impedance is Zins = R + 1 = 1 + sRC Cs sC Here pole is at origin and zero is at s =- 1/RC , therefore first critical frequency is a pole and last critical frequency is a zero. For RC parallel network the driving point impedance is R 1 R Cs = Zinp = 1 1 + sRC R+ Cs

Option (C) is correct. Here Va = 5 V because R1 = R2 and total voltage drop is 10 V. Now Vb = R3 # 10 = 1.1 # 10 = 5.238 V 2.1 R3 + R4 V = Va - Vb = 5 - 5.238 =- 0.238 V

1.65

Option (D) is correct. For h parameters we have to write V1 and I2 in terms of I1 and V2 . V1 = h11 I1 + h12 V2 I2 = h21 I1 + h22 V2 Applying KVL at input port V1 = 10I1 + V2 Applying KCL at output port V2 = I + I 1 2 20 or I2 =- I1 + V2 20

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Page 24

At t = 0 , i (t) = 0 At t = 12 , i (t) = 0.31 At t = 3 , i (t) = 0.5 Graph (C) satisfies all these conditions.

Thus from above equation we get h11 h12 10 1 =h h G = =- 1 0.05G 12 22 1.66

Option (B) is correct. RC = 0.1 # 10 - 6 # 103 = 10 - 4 sec Since time constant RC is very small, so steady state will be reached in 2 sec. At t = 2 sec the circuit is as shown in fig. Time constant

1.72

Option (D) is correct. We know that

where

V1 = z11 I1 + z12 I2 V2 = z11 I1 + z22 I2 z11 = V1 I1 I = 0 z21 = V2 I1 I = 0 2

1

Consider the given lattice network, when I2 = 0 . There is two similar path in the circuit for the current I1. So I = 1 I1 2

Vc = 3 V V2 =- Vc =- 3 V 1.67

1.68

1.69

1.70

Option (B) is correct. For a tree there must not be any loop. So a, c, and d don’t have any loop. Only b has loop. Option (D) is correct. The sign of M is as per sign of L If current enters or exit the dotted terminals of both coil. The sign of M is opposite of L If current enters in dotted terminal of a coil and exit from the dotted terminal of other coil. Thus Leq = L1 + L2 - 2M Option (A) is correct. Here w = 2 and V = 1+0c Y = 1 + jw C + 1 R jw L = 3 + j2 # 3 + 1 1 = 3 + j4 j2 # 4 -1 4 = 5+ tan = 5+53.11c 3

For z11 applying KVL at input port we get V1 = I (Za + Zb) Thus V1 = 1 I1 (Za + Zb) 2 z11 = 1 (Za + Zb) 2

A I D

O N

For Z21 applying KVL at output port we get n o.i V2 = Za I1 - Zb I1 c . 2 2 a di o .n Thus V2 = 1 I1 (Za - Zb) w 2 ww z21 = 1 (Za - Zb) 2 I = V * Y = (1+0c)( 5+53.1c) = 5+53.1c For this circuit z11 = z22 and z12 = z21. Thus Thus i (t) = 5 sin (2t + 53.1c) V R S Za + Zb Za - Zb W Option (A) is correct. z11 z12 2 2 =z z G = SS Za - Zb Za + Zb WW vi (t) = 2 sin 103 t 21 22 S 2 2 W Here w = 103 rad and Vi = 2 +0c X T 1 Here Za = 2j and Zb = 2W jwC 1 z11 z12 1+j j-1 Now V0 = .Vt = V + wCR i 1 1 j Thus =z z G = = j - 1 1 + j G R+ 21 22 jw C 1 1 + j # 103 # 10 - 3 = 1 - 45c

=

2 + 0c

1.73

v (t) = Ri (t) +

v0 (t) = sin (103 t - 45c) 1.71

Option (C) is correct. Input voltage

or

V (s)

Vi (s) = 1 s

v (t)

Z (s) = s + 2 V (s) 1 I (s) = i = s + 2 s (s + 2) I (s) = 1 ; 1 - 1 E 2 s s+2

Taking inverse Laplace transform i (t) = 1 (1 - e-2t) u (t) 2

Ldi (t) 1 + dt C

#0

3

i (t) dt

Taking L.T. on both sides,

vi (t) = u (t)

Taking Laplace transform Impedance

Option (B) is correct. Applying KVL,

Hence

1 s 2 +1 s

or 1.74

I (s)

I (s) vc (0+) = RI (s) + LsI (s) - Li (0 ) + + sC sC = u (t) thus V (s) = 1 s I (s) 1 = I (s) + sI (s) - 1 + s s I (s) 2 = 6s + s + 1@ s = 2s+2 s +s+1 +

Option (B) is correct.

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Page 25

Characteristics equation is 2

Differentiating with respect to t , we get 2di (t) 2d2 i (t) cos t = + i (t) + dt dt2

6

s + 20s + 10 = 0 Comparing with s2 + 2xwn s + wn2 = 0 we have

Thus Now 1.75

wn = 106 = 103 2xw = 20 2x = 203 = 0.02 10 Q = 1 = 1 = 50 2x 0.02

Option (A) is correct. For current i there is 3 similar path. So current will be divide in three path

1.81

Option (D) is correct. H (s) =

V0 (s) Vi (s)

1 1 sC = = 2 1 s LC + sCR + 1 R + sL + sC 1 = 2 -2 -4 s (10 # 10 ) + s (10-4 # 10 4) + 1 106 = -6 2 1 = 2 10 s + s + 1 s + 106 s + 106 1.76

Option (D) is correct. Impedance of series RLC circuit at resonant frequency is minimum, not zero. Actually imaginary part is zero. Z = R + j ` wL - 1 j wC At resonance wL - 1 = 0 and Z = R that is purely resistive. wC Thus S1 is false Now quality factor Q =R C L Since G = 1 , Q = 1 C G L R

Vab - b i # 1l - b i # 1l - b 1 # 1l = 0 3 6 3 Vab = R = 1 + 1 + 1 = 5 W eq i 6 3 6 3

Option (B) is correct. Number of loops = b - n + 1 = minimum number of equation Number of branches = b = 8 Number of nodes = n = 5 Minimum number of equation

Option ( ) is correct. Data are missing in question as L1 &L2 are not given.

1.82

Option (A) is correct. At t = 0 - circuit is in steady state. So inductor act as short circuit and capacitor act as open circuit.

1.83

A I D

O N

If G - then Q . provided C and L are constant. Thus S2 is also false. 1.77

so, we get

.in

no w.

co ia.

d

ww

At t = 0 - ,

i1 (0 -) = i2 (0 -) = 0

vc (0 -) = V At t = 0+ the circuit is as shown in fig. The voltage across capacitor and current in inductor can’t be changed instantaneously. Thus

= 8-5+1 = 4 1.78

Option (C) is correct. For maximum power transfer Thus

1.79

ZL = ZS* = Rs - jXs ZL = 1 - 1j

Option (B) is correct. Q = 1 R

L C

At t = 0+ ,

When R, L and C are doubled, 2L = 1 Q' = 1 2R 2C 2R Thus Q' = 100 = 50 2 1.80

1.84

L =Q C 2

Option (C) is correct. Applying KVL we get, di (t) 1 + i (t) dt dt C di (t) + i (t) dt sin t = 2i (t) + 2 dt

#

sin t = Ri (t) + L or

i1 = i2 =- V 2R

Option (C) is correct. When switch is in position 2, as shown in fig in question, applying KVL in loop (1), RI1 (s) + V + 1 I1 (s) + sL [I1 (s) - I2 (s)] = 0 s sC or I1 (s) 8R + 1 + sL B - I2 (s) sL = - V s sc z11 I1 + z12 I2 = V1

Applying KVL in loop 2, sL [I2 (s) - I1 (s)] + RI2 (s) + 1 I2 (s) = 0 sC

#

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1.85

1.86

Z12 I1 + Z22 I2 = V2 or - sLI1 (s) + 8R + sL + 1 BI2 (s) = 0 sc Now comparing with Z11 Z12 I1 V1 =Z Z G=I G = =V G 21 22 2 2 we get V R - sL W I1 (s) SR + sL + 1 -V sC W S sH = G = > R + sL + 1 WW I2 (s) - sL SS 0 sC X T Option (B) is correct. Zeros =- 3 Pole1 =- 1 + j Pole 2 =- 1 - j K (s + 3) Z (s) = (s + 1 + j)( s + 1 - j) K (s + 3) K (s + 3) = = 2 2 (s + 1) - j (s + 1) 2 + 1 From problem statement Z (0) w = 0 = 3 Thus 3K = 3 and we get K = 2 2 2 (s + 3) Z (s) = 2 s + 2s + 2

Page 26

z11 = V1 I1

Now

I2 = 0

= 2 + 0.5 + 0.25 = 2.75

z12 = R3 = 0.25 1.88

Option (A) is correct. Applying KCL at for node 2,

V2 + V2 - V1 = V1 5 5 5 or V2 = V1 = 20 V Voltage across dependent current source is 20 thus power delivered by it is PV2 # V1 = 20 # 20 = 80 W 5 5 It deliver power because current flows from its +ive terminals. 1.89

Option (C) is correct. When switch was closed, in steady state, iL (0 -) = 2.5 A

Option (C) is correct. v (t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c) 1 4444 2 4444 3 1 4444 4 2 4444 43 v1

A I D

v2

Thus we get w1 = 1 and w2 = 2 Now Z1 = R + jw1 L = 1 + j1 Z2 = R + jw2 L = 1 + j2 v (t) v (t) i (t) = 1 + 2 Z1 Z2

O N

10 2 cos (t + 10c) 10 5 cos (2t + 10c) .no = + w 1+j 1 + j2 w

.int = 0+ , oAt c . ia

d

w

10 2 cos (t + 10c) 10 5 cos (2t + 10c) = + 12 + 22 + tan-1 1 12 + 22 tan-1 2

10 2 cos (t + 10c) 10 5 cos (2t + 10c) + 2 + tan-1 45c 5 tan-1 2 i (t) = 10 cos (t - 35c) + 10 cos (2t + 10c - tan-1 2)

1.90

iL (0+) = iL (0 -) = 2.5 A and all this current of will pass through 2 W resistor. Thus Vx =- 2.5 # 20 =- 50 V

Option (A) is correct. For maximum power delivered, RL must be equal to Rth across same terminal.

=

1.87

Option (A) is correct. Using 3- Y conversion Applying KCL at Node, we get 0.5I1 = Vth + I1 20 or 2 # 1 = 2 = 0. 5 2+1+1 4 R2 = 1 # 1 = 1 = 0.25 2+1+1 4 R3 = 2 # 1 = 0.5 2+1+1 R1 =

but Thus

Vth + 10I1 = 0 I1 = Vth - 50 40 Vth + Vth - 50 = 0 4

or Vth = 10 V For Isc the circuit is shown in figure below.

Now the circuit is as shown in figure below.

Isc = 0.5I1 - I1 =- 0.5I1

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but

Page 27

I1 =- 50 =- 1.25 A 40 Isc =- 0.5 # - 12.5 = 0.625 A Rth = Vth = 10 = 16 W Isc 0.625

1.91

Option (D) is correct. IP , VP " Phase current and Phase voltage IL, VL " Line current and line voltage Now VP = c VL m and IP = IL 3 So, Power = 3VP IL cos q 1500 = 3 c VL m (IL) cos q 3 also IL = c VL m 3 ZL

Now applying nodal analysis we have e0 - 80 + e0 + e0 - 16 = 0 10 + 2 12 6

Option (A) is correct.

1.96

jwC I2 = Em +01c = Em +0c 1 + jwCR2 R2 + jwC +90c + tan-1 wCR2 E m wC I2 = + (90c - tan-1 wCR2) 2 2 2 1 + w C R2 At w = 0 I2 = 0 and at w = 3, I2 = Em R2 Only fig. given in option (A) satisfies both conditions. +I2 =

1500 = 3 c VL mc VL m cos q 3 3 ZL ZL =

(400) 2 (.844) = 90 W 1500

As power factor is leading So, cos q = 0.844 " q = 32.44 As phase current leads phase voltage Option (C) is correct. Applying KCL, we get e0 - 12 + e0 + e0 = 0 4 4 2+2 or 1.93

e0 = 4 V

Option (A) is correct. The star delta circuit is shown as below

Option (A) is correct.

1.97

ZL = 90+ - q = 90+ - 32.44c 1.92

4e0 = 112 e0 = 112 = 28 V 4

or

Xs = wL = 10 W For maximum power transfer

A I D

O N

1.98

.i

n

. ww

w

Rs2 + Xs2 =

102 + 102 = 14.14 W

Option (C) is correct. Applying KVL in LHS loop n

co ia.

od

RL =

or

E1 = 2I1 + 4 (I1 + I2) - 10E1 E1 = 6I1 + 4I2 11 11

Thus z11 = 6 11 Applying KVL in RHS loop E2 = 4 (I1 + I2) - 10E1 = 4 (I1 + I2) - 10 c 6I1 + 4I2 m 11 11 =- 16I1 + 4I2 11 11

Here and

Now

1.94

Thus z21 =- 16 11

ZAB = ZBC = ZCA = 3 Z ZAB ZCA ZA = ZAB + ZBC + ZCA ZAB ZBC ZB = ZAB + ZBC + ZCA ZBC ZCA ZC = ZAB + ZBC + ZCA ZA = ZB = ZC =

1.99

Option (D) is correct. At w = 0 , circuit act as shown in figure below.

3Z 3Z = Z 3Z+ 3Z+ 3Z 3

Option (C) is correct. y11 y12 y1 + y3 - y3 =y y G = = - y y + y G 21 22 3 2 3

V0 = RL Vs RL + Rs

(finite value)

At w = 3 , circuit act as shown in figure below:

y12 =- y3 y12 =- 1 =- 0.05 mho 20 1.95

Option (D) is correct. We apply source conversion the circuit as shown in fig below.

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Page 28 1.104

Option (A) is correct. For a 2-port network the parameter h21 is defined as h21 = I2 I1 V = 0 (short circuit) 2

V0 = RL Vs RL + Rs At resonant frequency w = V0 = 0 .

(finite value)

1 circuit acts as shown in fig and LC

Applying node equation at node a we get Va - V1 + Va - 0 + Va - 0 = 0 R R R 3Va = V1 & Va = V1 3 V1 - V1 V 3 = 2V1 1 - Va Now I1 = = R R 3R 0 - V1 - V 0 V 3 = a 1 and I2 = = R R 3R - V1 /3R - 1 I2 Thus = h21 = = 2 I1 V = 0 2V1 /3R

Thus it is a band reject filter. 1.100

Option (D) is correct. Applying KCL we get Now

1.101

iL = eat + ebt V (t) = vL = L diL = L d [eat + ebt] = aeat + bebt dt dt

2

1.105

Option (A) is correct. Going from 10 V to 0 V

Option (A) is correct. Applying node equation at node A Vth - 100 (1 + j0) Vth - 0 =0 + 3 4j

A I D

O N

or or

no w.

.in oBy c simplifying . ia

4jVth - 4j100 + 3Vth = 0 Vth (3 + 4j) = 4j100 4j100 Vth = 3 + 4j

d

Vth =

ww

or 1.102

1.103

10 + 5 + E + 1 = 0 E =- 16 V

Option (C) is correct. This is a reciprocal and linear network. So we can apply reciprocity theorem which states “Two loops A & B of a network N and if an ideal voltage source E in loop A produces a current I in loop B , then interchanging positions an identical source in loop B produces the same current in loop A. Since network is linear, principle of homogeneity may be applied and when volt source is doubled, current also doubles. Now applying reciprocity theorem i = 2 A for 10 V V = 10 V, i = 2 A V =- 20 V, i =- 4 A

4j100 3 - 4j 3 + 4j # 3 - 4j

Vth = 16j (3 - j4) 1.106

Option (C) is correct. For maximum power transfer RL should be equal to RTh at same terminal. so, equivalent Resistor of the circuit is

Req = 5W 20W + 4W Req = 5.20 + 4 = 4 + 4 = 8 W 5 + 20 1.107

Option (C) is correct. Tree is the set of those branch which does not make any loop and connects all the nodes. abfg is not a tree because it contains a loop l node (4) is not connected

1.108

1.109

Option (D) is correct. Delta to star conversion Rab Rac = 5 # 30 = 150 = 3 W R1 = 50 Rab + Rac + Rbc 5 + 30 + 15 Rab Rbc = 5 # 15 = 1.5 W R2 = Rab + Rac + Rbc 5 + 30 + 15 Rac Rbc = 15 # 30 = 9 W R3 = Rab + Rac + Rbc 5 + 30 + 15 Option (C) is correct. No. of branches = n + l - 1 = 7 + 5 - 1 = 11 Option (B) is correct.

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Page 29

In nodal method we sum up all the currents coming & going at the node So it is based on KCL. Furthermore we use ohms law to determine current in individual branch. Thus it is also based on ohms law. 1.110

1.111 1.112

1.113

Option (A) is correct. Superposition theorem is applicable to only linear circuits. Option (B) is correct.

so equivalent circuit is

1.120

Option (C) is correct. Taking b as reference node and applying KCL at a we get Vab - 1 + Vab = 3 2 2

-t

16e 4 = 4

-t

#0 3 4e 4

= 16 J

Zeq = (4W 8W) = 4 # 8 = 8 3 4+8

Vab - 1 + Vab = 6 Vab = 6 + 1 = 3.5 V 2

Option (B) is correct. Current in A2 , Inductor current can be Current in A 3 , Total current

1.121

Option (A) is correct. Option (B) is correct. The given figure is shown below.

I = i 0 + i1 = 7 + 5 = 12 A Applying KCL at node f so

ww

.no

w

I =- i 4 i 4 =- 12 amp

di

1.123

I3 = 4 I1 = I 2 + I 3 I1 = 4 - 3j I =

A I D

O N

Applying KCL at node a we have

I2 = 3 amp defined as I2 =- 3j

(4) 2 + (3) 2 = 5 amp

Option (C) is correct. For a tree we have (n - 1) branches. Links are the branches which from a loop, when connect two nodes of tree. so if total no. of branches = b in No. of links = b - (n - 1) = b - n + 1 . o a.cTotal no. of links in equal to total no. of independent loops.

1.122

1.116

#0

3

Option (B) is correct. For reciprocal network y12 = y21 but here y12 =- 12 ! y21 = 12 . Thus circuit is non reciprocal. Furthermore only reciprocal circuit are passive circuit.

or

1.115

#0

V R2 (t) = R

Option (B) is correct. It is a balanced whetstone bridge R1 R 3 b R2 = R 4 l

or

1.114

E

3

Option (B) is correct. In the steady state condition all capacitors behaves as open circuit & Inductors behaves as short circuits as shown below :

Option (A) is correct.

Thus voltage across capacitor C1 is VC = 100 # 40 = 80 V 10 + 40 1

so 1.117

1.118

1.119

Now the circuit faced by capacitor C2 and C 3 can be drawn as below :

V = 3 - 0 = 3 volt

Option (D) is correct. Can not determined V without knowing the elements in box. Option (A) is correct. The voltage V is the voltage across voltage source and that is 10 V. Option (B) is correct. Voltage across capacitor -t

VC (t) = VC (3) + (VC (0) - VC (3)) e RC Here VC (3) = 10 V and (VC (0) = 6 V. Thus -t

Now

-t

2

-t

VC (t) = 10 + (6 - 10) e RC = 10 - 4e RC = 10 - 4e 8 VR (t) = 10 - VC (t) -t

Voltage across capacitor C2 and C 3 are VC = 80 C 3 = 80 # 3 = 48 volt 5 C2 + C3 VC = 80 C2 = 80 # 2 = 32 volt 5 C2 + C3 3

-t

= 10 - 10 + 4e RC = 4e RC Energy absorbed by resistor

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UNIT 2

Page 30 2.6

In the CMOS circuit shown, electron and hole mobilities are equal, and M1 and M2 are equally sized. The device M1 is in the linear region if

ELECTRONICS DEVICES

2013 2.1

2.2

2.3

2.4

ONE MARK

In a forward biased pn junction diode, the sequence of events that best describes the mechanism of current flow is (A) injection, and subsequent diffusion and recombination of minority carriers (B) injection, and subsequent drift and generation of minority carriers (C) extraction, and subsequent diffusion and generation of minority carriers (D) extraction, and subsequent drift and recombination of minority carriers

(A) Vin < 1.875 V (C) Vin > 3.125 V

(B) 1.875 V < Vin < 3.125 V (D) 0 < Vin < 5 V

Common Data For Q. 2 and 3 : In the three dimensional view of a silicon n -channel MOS transistor shown below, d = 20 nm . The transistor is of width 1 mm . The depletion width formed at every p-n junction is 10 nm. The relative permittivity of Si and SiO 2 , respectively, are 11.7 and 3.9, and e0 = 8.9 # 10-12 F/m .

In IC technology, dry oxidation (using dry oxygen) as compared to wet oxidation (using steam or water vapor) produces (A) superior quality oxide with a higher growth rate (B) inferior quality oxide with a higher growth rate (C) inferior quality oxide with a lower growth rate (D) superior quality oxide with a lower growth rate

A I D

In a MOSFET operating in the saturation region, the channel length modulation effect causes n (A) an increase in the gate-source capacitance o.i c . (B) a decrease in the transconductance dia o .n (C) a decrease in the unity-gain cutoff frequency w ww (D) a decrease in the output resistance 2.7 The gate source overlap capacitance is approximately (A) 0.7 fF (B) 0.7 pF 2013 TWO MARKS (C) 0.35 fF (D) 0.24 pF The small-signal resistance (i.e., dVB /dID ) in kW offered by the The source-body junction capacitance is approximately n-channel MOSFET M shown in the figure below, at a bias point of 2.8 (A) 2 fF (B) 7 fF VB = 2 V is (device data for M: device transconductance parameter ' 2 kN = mn C 0x ^W/L h = 40 mA/V , threshold voltage VTN = 1 V , and (C) 2 pF (D) 7 pF neglect body effect and channel length modulation effects)

O N

2011 2.9

(A) 12.5 (C) 50

(B) 25 (D) 100

2012 2.5

2.10

TWO MARKS

The source of a silicon (ni = 1010 per cm3) n -channel MOS transistor has an area of 1 sq mm and a depth of 1 mm . If the dopant density in the source is 1019 /cm3 , the number of holes in the source region with the above volume is approximately (A) 107 (B) 100 (C) 10

(D) 0

2.11

ONE MARK

Drift current in the semiconductors depends upon (A) only the electric field (B) only the carrier concentration gradient (C) both the electric field and the carrier concentration (D) both the electric field and the carrier concentration gradient A Zener diode, when used in voltage stabilization circuits, is biased in (A) reverse bias region below the breakdown voltage (B) reverse breakdown region (C) forward bias region (D) forward bias constant current mode A silicon PN junction is forward biased with a constant current at room temperature. When the temperature is increased by 10ºC, the forward bias voltage across the PN junction

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Page 31

(A) increases by 60 mV (B) decreases by 60 mV (C) increases by 25 mV (D) decreases by 25 mV 2011

is lower (C) Reverse breakdown voltage is lower and depletion capacitance is higher (D) Reverse breakdown voltage is higher and depletion capacitance is higher TWO MARKS

Statements for Linked Answer Question : 3.10 & 3.11 : The silicon sample with unit cross-sectional area shown below is in thermal equilibrium. The following information is given: T = 300 K electronic charge = 1.6 # 10-19 C , thermal voltage = 26 mV and electron mobility = 1350 cm2 / V-s

Common Data For Q. 3.12 & 3.13 : The channel resistance of an N-channel JFET shown in the figure below is 600 W when the full channel thickness (tch ) of 10 μm is available for conduction. The built-in voltage of the gate P+ N junction (Vbi ) is - 1 V . When the gate to source voltage (VGS ) is 0 V, the channel is depleted by 1 μm on each side due to the built in voltage and hence the thickness available for conduction is only 8 μm

The magnitude of the electric field at x = 0.5 mm is (A) 1 kV/cm (B) 5 kV/cm (C) 10 kV/cm (D) 26 kV/cm

2.18

The magnitude of the electron of the electron drift current density at x = 0.5 mm is (A) 2.16 # 10 4 A/cm2 (B) 1.08 # 10 4 A/m2 (C) 4.32 # 103 A/cm2 (D) 6.48 # 102 A/cm2

2.19

2.12

2.13

The channel resistance when VGS =- 3 V is (A) 360 W (B) 917 W (C) 1000 W (D) 3000 W The channel resistance when VGS = 0 V is (A) 480 W (B) 600 W (C) 750 W (D) 1000 W 2010

2.14

2.15

2.17

w

ww

ONE MARK

Thin gate oxide in a CMOS process in preferably grown using (A) wet oxidation (B) dry oxidation (C) epitaxial oxidation (D) ion implantation TWO MARKS

In a uniformly doped BJT, assume that NE , NB and NC are the emitter, base and collector doping in atoms/cm3 , respectively. If the emitter injection efficiency of the BJT is close unity, which one of the following condition is TRUE (A) NE = NB = NC (B) NE >> NB and NB > NC (C) NE = NB and NB < NC (D) NE < NB < NC Compared to a p-n junction with NA = ND = 1014 /cm3 , which one of the following statements is TRUE for a p-n junction with NA = ND = 1020 /cm3 ? (A) Reverse breakdown voltage is lower and depletion capacitance is lower (B) Reverse breakdown voltage is higher and depletion capacitance

2009 n o.i c . ia In an

d

2.20

.no

2.21

At room temperature, a possible value for the mobility of electrons in the inversion layer of a silicon n -channel MOSFET is (A) 450 cm2 / V-s (B) 1350 cm2 / V-s (C) 1800 cm2 / V-s (D) 3600 cm2 / V-s

2010 2.16

O N

A I D

ONE MARK

n-type silicon crystal at room temperature, which of the following can have a concentration of 4 # 1019 cm - 3 ? (A) Silicon atoms (B) Holes (C) Dopant atoms (D) Valence electrons The ratio of the mobility to the diffusion coefficient in a semiconductor has the units (A) V - 1 (B) cm.V1 (C) V.cm - 1

(D) V.s

2009 2.22

TWO MARKS

Consider the following two statements about the internal conditions in a n -channel MOSFET operating in the active region. S1 : The inversion charge decreases from source to drain S2 : The channel potential increases from source to drain. Which of the following is correct? (A) Only S2 is true (B) Both S1 and S2 are false (C) Both S1 and S2 are true, but S2 is not a reason for S1 (D) Both S1 and S2 are true, and S2 is a reason for S1

Common Data For Q. 3.13 and 3.14 Consider a silicon p - n junction at room temperature having the following parameters: Doping on the n -side = 1 # 1017 cm - 3 Depletion width on the n -side = 0.1mm Depletion width on the p -side = 1.0mm

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Page 32

Intrinsic carrier concentration = 1.4 # 1010 cm - 3 Thermal voltage = 26 mV Permittivity of free space = 8.85 # 10 - 14 F.cm - 1 Dielectric constant of silicon = 12 2.23

2.24

The built-in potential of the junction (A) is 0.70 V (B) is 0.76 V (C) is 0.82 V (D) Cannot be estimated from the data given The peak electric field in the device is (A) 0.15 MV . cm - 1, directed from p -region to n -region (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 2008

2.25

2.26

2.28

Which of the following is NOT associated with a p - n junction ? (A) Junction Capacitance (B) Charge Storage Capacitance (C) Depletion Capacitance (D) Channel Length Modulations

O N

The drain current of MOSFET in saturation is given by ID = K (VGS - VT ) 2 where K is a constant. The magnitude of the transconductance gm is K (VGS - VT ) 2 VDS

(B) 2K (VGS - VT )

(C)

Id VGS - VDS

(D)

The cross section of a JFET is shown in the following figure. Let Vc be - 2 V and let VP be the initial pinch -off voltage. If the width W is doubled (with other geometrical parameters and doping levels remaining the same), then the ratio between the mutual trans conductances of the initial and the modified JFET is

n o.i c . ia (A) 4

A silicon wafer has 100 nm of oxide on it and is furnace at a d temperature above 1000c C for further oxidation in dry oxygen. The.no w oxidation rate ww (A) is independent of current oxide thickness and temperature (B) is independent of current oxide thickness but depends on tem- 2.32 perature (C) slows down as the oxide grows (D) is zero as the existing oxide prevents further oxidation

(A)

Silicon is doped with boron to a concentration of 4 # 1017 atoms cm3 . Assume the intrinsic carrier concentration of silicon to be 1.5 # 1010 / cm 3 and the value of kT/q to be 25 mV at 300 K. Compared to undopped silicon, the fermi level of doped silicon (A) goes down by 0.31 eV (B) goes up by 0.13 eV (C) goes down by 0.427 eV (D) goes up by 0.427 eV

A I D

Which of the following is true? (A) A silicon wafer heavily doped with boron is a p+ substrate (B) A silicon wafer lightly doped with boron is a p+ substrate

2008 2.29

2.31

ONE MARK

(C) A silicon wafer heavily doped with arsenic is a p+ substrate (D) A silicon wafer lightly doped with arsenic is a p+ substrate 2.27

2.30

K (VGS - VT ) 2 VGS

(C) e

1 - 2/Vp o 1 - 1/2Vp

1 - 2/Vp (B) 1 e 2 1 - 1/2Vp o 1 - (2 - Vp ) (D) 1 - [1 (2 Vp )]

Consider the following assertions. S1 : For Zener effect to occur, a very abrupt junction is required. S2 : For quantum tunneling to occur, a very narrow energy barrier is required. Which of the following is correct ? (A) Only S2 is true (B) S1 and S2 are both true but S2 is not a reason for S1 (C) S1 and S2 and are both true but S2 is not a reason for S1 (D) Both S1 and S2 are false 2007

TWO MARKS

2.33

The measured trans conductance gm of an NMOS transistor operating in the linear region is plotted against the gate voltage VG at a constant drain voltage VD . Which of the following figures represents the expected dependence of gm on VG ? 2.34

ONE MARK

The electron and hole concentrations in an intrinsic semiconductor are ni per cm3 at 300 K. Now, if acceptor impurities are introduced with a concentration of NA per cm3 (where NA >> ni , the electron concentration per cm3 at 300 K will be) (A) ni (B) ni + NA 2 (C) NA - ni (D) ni NA + In a p n junction diode under reverse biased the magnitude of electric field is maximum at (A) the edge of the depletion region on the p-side (B) the edge of the depletion region on the n -side (C) the p+ n junction

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Page 33

(D) the centre of the depletion region on the n -side

(C) 1 mm 2.41

2007 2.35

TWO MARKS

Group I lists four types of p - n junction diodes. Match each device in Group I with one of the option in Group II to indicate the bias condition of the device in its normal mode of operation. Group - I Group-II (P) Zener Diode (1) Forward bias (Q) Solar cell (2) Reverse bias (R) LASER diode (S) Avalanche Photodiode (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 (D) P - 2, Q - 1, R - 2, S - 2

2.36

2.37

2.38

Consider the following statements about the C - V characteristics plot : S1 : The MOS capacitor has as n -type substrate S2 : If positive charges are introduced in the oxide, the C - V polt will shift to the left. Then which of the following is true? (A) Both S1 and S2 are true (B) S1 is true and S2 is false (C) S1 is false and S2 is true (D) Both S1 and S2 are false 2006

2.42

Group I lists four different semiconductor devices. match each device in Group I with its charactecteristic property in Group II Group-I Group-II (P) BJT (1) Population iniversion (Q) MOS capacitor (2) Pinch-off voltage (R) LASER diode (3) Early effect (S) JFET (4) Flat-band voltage (A) P - 3, Q - 1, R - 4, S - 2 (B) P - 1, Q - 4, R - 3, S - 2 (C) P - 3, Q - 4, R - 1, S - 2 (D) P - 3, Q - 2, R - 1, S - 4

(D) 1.143 mm

2.43

ONE MARK

The values of voltage (VD) across a tunnel-diode corresponding to peak and valley currents are Vp, VD respectively. The range of tunneldiode voltage for VD which the slope of its I - VD characteristics is negative would be (A) VD < 0 (B) 0 # VD < Vp (C) Vp # VD < Vv (D) VD $ Vv The concentration of minority carriers in an extrinsic semiconductor under equilibrium is (A) Directly proportional to doping concentration (B) Inversely proportional to the doping concentration (C) Directly proportional to the intrinsic concentration (D) Inversely proportional to the intrinsic concentration

A I D

O N

2.44 Under level injection assumption, the injected minority carrier in low . A p+ n junction has a built-in potential of 0.8 V. The depletion layer o current for an extrinsic semiconductor is essentially the .c a width a reverse bias of 1.2 V is 2 mm. For a reverse bias of 7.2 V, the i d (A) Diffusion current (B) Drift current no depletion layer width will be . w (C) Recombination current (D) Induced current (A) 4 mm (B) 4.9 mm ww 2.45 (C) 8 mm (D) 12 mm The phenomenon known as “Early Effect” in a bipolar transistor refers to a reduction of the effective base-width caused by The DC current gain (b) of a BJT is 50. Assuming that the emitter (A) Electron - hole recombination at the base injection efficiency is 0.995, the base transport factor is (B) The reverse biasing of the base - collector junction (A) 0.980 (B) 0.985 (C) The forward biasing of emitter-base junction (C) 0.990 (D) 0.995 (D) The early removal of stored base charge during saturation-tocut off switching

Common Data For Q. 2.29, 2.30 and 2.31 : The figure shows the high-frequency capacitance - voltage characteristics of Metal/Sio 2 /silicon (MOS) capacitor having an area of 1 # 10 - 4 cm 2 . Assume that the permittivities (e0 er ) of silicon and Sio2 are 1 # 10 - 12 F/cm and 3.5 # 10 - 13 F/cm respectively.

2.39

2.40

The gate oxide thickness in the MOS capacitor is (A) 50 nm (B) 143 nm (C) 350 nm (D) 1 mm The maximum depletion layer width in silicon is (A) 0.143 mm (B) 0.857 mm

2006 2.46

In the circuit shown below, the switch was connected to position 1 at t < 0 and at t = 0 , it is changed to position 2. Assume that the diode has zero voltage drop and a storage time ts . For 0 < t # ts, vR is given by (all in Volts)

(A) vR =- 5 (C) 0 # vR < 5 2.47

TWO MARKS

(B) vR =+ 5 (D) - 5 # vR < 0

The majority carriers in an n-type semiconductor have an average drift velocity v in a direction perpendicular to a uniform magnetic field B . The electric field E induced due to Hall effect acts in the direction

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(A) v # B (C) along v 2.48

Page 34

(B) B # v (D) opposite to v

mode. The dominant charge in the channel is due to the presence of (A) holes (B) electrons (C) positively charged icons (D) negatively charged ions

Find the correct match between Group 1 and Group 2 Group 1 Group 2 E - Varactor diode 1. Voltage reference F - PIN diode 2. High frequency switch G - Zener diode 3. Tuned circuits H - Schottky diode 4. Current controlled attenuator

2.56

(A) E - 4, F - 2, G - 1, H - 3 (B) E - 3, F - 4, G - 1, H - 3 (C) E - 2, F - 4, G - 1, H - 2 (D) E - 1, F - 3, G - 2, H - 4 2.49

2.50

2.51

2.52

2.54

2.55

2004 2.57

ONE MARK

The impurity commonly used for realizing the base region of a silicon n - p - n transistor is (A) Gallium (B) Indium (C) Boron (D) Phosphorus

A I D

ONE MARK

The bandgap of Silicon at room temperature is (A) 1.3 eV (B) 0.7 eV (C) 1.1 eV (D) 1.4 eV

2.58

O N

If for a silicon npn transistor, the base-to-emitter voltage (VBE ) is 0.7 V and the collector-to-base voltage (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

n

A Silicon PN junction at a temperature of 20c C has a reverse o.i c . 2.59 saturation current of 10 pico - Ameres (pA). The reserve saturation dia Consider the following statements S1 and S2. o S1 : The b of a bipolar transistor reduces if the base width is in.n current at 40cC for the same bias is approximately w creased. (A) 30 pA (B) 40 pA ww S2 : The b of a bipolar transistor increases if the dopoing concen(C) 50 pA (D) 60 pA tration in the base is increased. The primary reason for the widespread use of Silicon in semiconductor Which remarks of the following is correct ? device technology is (A) S1 is FALSE and S2 is TRUE (A) abundance of Silicon on the surface of the Earth. (B) Both S1 and S2 are TRUE (B) larger bandgap of Silicon in comparison to Germanium. (C) Both S1 and S2 are FALSE (C) favorable properties of Silicon - dioxide (SiO2) (D) S1 is TRUE and S2 is FALSE (D) lower melting point 2.60 Given figure is the voltage transfer characteristic of 2005

2.53

(A) 1 V and the device is in active region (B) - 1 V and the device is in saturation region (C) 1 V and the device is in saturation region (D) - 1 V and the device is an active region

A heavily doped n - type semiconductor has the following data: Hole-electron ratio : 0.4 Doping concentration : 4.2 # 108 atoms/m3 Intrinsic concentration : 1.5 # 10 4 atoms/m 3 The ratio of conductance of the n -type semiconductor to that of the intrinsic semiconductor of same material and ate same temperature is given by (A) 0.00005 (B) 2000 (C) 10000 (D) 20000 2005

For an n -channel MOSFET and its transfer curve shown in the figure, the threshold voltage is

TWO MARKS

A Silicon sample A is doped with 1018 atoms/cm 3 of boron. Another sample B of identical dimension is doped with 1018 atoms/cm 3 phosphorus. The ratio of electron to hole mobility is 3. The ratio of conductivity of the sample A to B is (A) 3 (B) 1 3 (C) 2 (D) 3 3 2 A Silicon PN junction diode under reverse bias has depletion region of width 10 mm. The relative permittivity of Silicon, er = 11.7 and the permittivity of free space e0 = 8.85 # 10 - 12 F/m. The depletion capacitance of the diode per square meter is (A) 100 mF (B) 10 mF (C) 1 mF (D) 20 mF

(A) an NOMS inverter with enhancement mode transistor as load (B) an NMOS inverter with depletion mode transistor as load (C) a CMOS inverter (D) a BJT inverter 2.61

Assuming VCEsat = 0.2 V and b = 50 , the minimum base current (IB) required to drive the transistor in the figure to saturation is

A MOS capacitor made using p type substrate is in the accumulation

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Page 35

Coulomb) (A) 800 A/cm 2 (C) 200 A/cm 2

(B) 8 A/cm 2 (D) 2 A/cm 2

2003

n -type silicon is obtained by doping silicon with (A) Germanium (B) Aluminium (C) Boron (D) Phosphorus

2.69

(A) 56 mA (C) 60 mA

(B) 140 mA (D) 3 mA

The Bandgap of silicon at 300 K is (A) 1.36 eV (B) 1.10 eV (C) 0.80 eV (D) 0.67 eV

2.70

2004 2.62

2.63

2.64

2.65

2.66

2.67

2.68

TWO MARKS

In an abrupt p - n junction, the doping concentrations on the p side and n -side are NA = 9 # 1016 /cm 3 respectively. The p - n junction is 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

Consider an abrupt p - n junction. Let Vbi be the built-in potential of this junction and VR be the applied reverse bias. If the junction capacitance (Cj ) is 1 pF for Vbi + VR = 1 V, then for Vbi + VR = 4 V, Cj will be (A) 4 pF (B) 2 pF (C) 0.25 pF (D) 0.5 pF

(C) 5.00 # 1020 /m 3

A I D 2.73

co ia.

d

2.74

2003

2.75

2.76

TWO MARKS

An n -type silicon bar 0.1 cm long and 100 mm2 i cross-sectional area has a majority carrier concentration of 5 # 1020 /m 2 and the carrier mobility is 0.13 m2 /V-s at 300 K. If the charge of an electron is 1.5 # 10 - 19 coulomb, then the resistance of the bar is (A) 106 Ohm (B) 10 4 Ohm (C) 10 - 1 Ohm

The drain of an n-channel MOSFET is shorted to the gate so that VGS = VDS . The threshold voltage (VT ) of the MOSFET is 1 V. If the drain current (ID) is 1 mA for VGS = 2 V, then for VGS = 3 V, ID is (A) 2 mA (B) 3 mA (C) 9 mA (D) 4 mA

The neutral base width of a bipolar transistor, biased in the active region, is 0.5 mm. The maximum electron concentration and the diffusion constant in the base are 1014 / cm 3 and Dn = 25 cm 2 / sec respectively. Assuming negligible recombination in the base, the collector current density is (the electron charge is 1.6 # 10 - 19

For an n - channel enhancement type MOSFET, if the source is connected at a higher potential than that of the bulk (i.e. VSB > 0 ), the threshold voltage VT of the MOSFET will (A) remain unchanged (B) decrease (C) nchange polarity (D) increase

.i

no w.

The longest wavelength that can be absorbed by silicon, which has the bandgap of 1.12 eV, is 1.1 mm. If the longest wavelength that can be absorbed by another material is 0.87 mm, then bandgap of this material is (A) 1.416 A/cm 2 (B) 0.886 eV (C) 0.854 eV (D) 0.706 eV

(D) 3.00 # 10 - 5 /m 3

Choose proper substitutes for X and Y to make the following statement correct Tunnel diode and Avalanche photo diode are operated in X bias ad Y bias respectively (A) X: reverse, Y: reverse (B) X: reverse, Y: forward (C) X: forward, Y: reverse (D) X: forward, Y: forward

2.72

O N

Consider the following statements Sq and S2. ww S1 : The threshold voltage (VT ) of MOS capacitor decreases with increase in gate oxide thickness. S2 : The threshold voltage (VT ) of a MOS capacitor decreases with increase in substrate doping concentration. Which Marks of the following is correct ? (A) S1 is FALSE and S2 is TRUE (B) Both S1 and S2 are TRUE (C) Both S1 and S2 are FALSE (D) S1 is TRUE and S2 is FALSE

The intrinsic carrier concentration of silicon sample at 300 K is 1.5 # 1016 /m 3 . If after doping, the number of majority carriers is 5 # 1020 /m 3 , the minority carrier density is (A) 4.50 # 1011/m 3 (B) 3.333 # 10 4 /m 3

2.71

The resistivity of a uniformly doped n -type silicon sample is 0.5W mc. If the electron mobility (mn) is 1250 cm 2 /V-sec and the charge of an electron is 1.6 # 10 - 19 Coulomb, the donor impurity concentration (ND) in the sample is (A) 2 # 1016 /cm 3 (B) 1 # 1016 /cm 3 (C) 2.5 # 1015 /cm 3 (D) 5 # 1015 /cm 3

ONE MARK

(D) 10 - 4 Ohm

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 = 2mm . 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 (A) zero (B) -112 A/cm 2 (C) +1120 A/cm 2 (D) -1120 A/cm 2 Match items in Group 1 with items in Group 2, most suitably. Group 1 Group 2 P. LED 1. Heavy doping Q. Avalanche photo diode 2. Coherent radiation R. Tunnel diode 3. Spontaneous emission S. LASER 4. Current gain (A) P - 1, Q - 2, R - 4, S - 3 (B) P - 2, Q - 3, R - 1, S - 4 (C) P - 3 Q - 4, R - 1, S - 2 (D) P - 2, Q - 1, R - 4, S - 3

2.77

At 300 K, for a diode current of 1 mA, a certain germanium diode

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requires a forward bias of 0.1435 V, whereas a certain silicon diode requires a forward bias of 0.718 V. Under the conditions state above, the closest approximation of the ratio of reverse saturation current in germanium diode to that in silicon diode is (A) 1 (B) 5 3 (C) 4 # 10 (D) 8 # 103 2.78

2.79

2.80

2.81

2001

MOSFET can be used as a (A) current controlled capacitor (C) current controlled inductor

2.84

When the gate-to-source voltage (VGs) of a MOSFET with threshold voltage of 400 mV, working in saturation is 900 mV, the drain current is observed to be 1 mA. Neglecting the channel width modulation effect and assuming that the MOSFET is operating at saturation, 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 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) P - Q - R - S (B) Q - S - R - P (C) R - P - S - Q (D) S - R - Q - P

O N

2002

1999

1999

An npn transistor (with C = 0.3 pF ) has a unity-gain cutoff frequency fT of 400 MHz at a dc bias current Ic = 1 mA . The value of its Cm (in pF) is approximately (VT = 26 mV) (A)n 15 (B) 30 i . o (D) 96 a.c (C) 50

2.88

di

1998 2.89

2.90

The fT of a BJT is related to its gm, C p and C m as follows Cp + Cm gm gm (C) fT = Cp + Cm

2.91

If the transistor in the figure is in saturation, then

2.92

(A) (B) (C) (D)

IC IC IC IC

is always equal to bdc IB is always equal to - bde IB is greater than or equal to bdc IB is less than or equal to bdc IB

2.93

ONE MARK

The electron and hole concentrations in a intrinsic semiconductor are ni and pi respectively. When doped with a p-type material, these change to n and p, respectively, Then (A) n + p = ni + pi (B) n + ni = p + pi (C) npi = ni p (D) np = ni pi

(A) fT =

2.83

TWO MARKS

An n -channel JEFT has IDSS = 2 mA and Vp =- 4 V . Its transconductance gm (in milliohm) for an applied gate-to-source voltage VGS of - 2 V is (A) 0.25 (B) 0.5 (C) 0.75 (D) 1.0

2.87

In the figure, silicon diode is carrying a constant current of 1 mA. When the temperature of the diode is 20cC, VD is found to be 700 mV. If the temperature rises to 40cC, VD becomes approximately equal to

(B) 660 mV (D) 700 mV

ONE MARK

The early effect in a bipolar junction transistor is caused by (A) fast turn-on (B) fast turn-off (C) large collector-base reverse bias (D) large emitter-base forward bias

2.86

.no

ONE MARK

(A) 740 mV (C) 680 mV

(B) voltage controlled capacitor (D) voltage controlled inductor

A I D

The action of JFET in its equivalent circuit can best be represented as a (A) Current controlled current source (B) Current controlled voltage source (C) Voltage controlled voltage source (D) Voltage controlled current source

ww

ONE MARK

The effective channel length of MOSFET in saturation decreases with increase in (A) gate voltage (B) drain voltage (C) source voltage (D) body voltage

2.85

A particular green LED emits light of wavelength 5490 Ac. The energy bandgap of the semiconductor material used there is (Plank’s constant = 6.626 # 10 - 34 J - s ) (A) 2.26 eV (B) 1.98 eV (C) 1.17 eV (D) 0.74 eV

w

2.82

Page 36

2p (C p + C m) gm gm (D) fT = 2p (C p + C m) (B) fT =

The static characteristic of an adequately forward biased p-n junction is a straight line, if the plot is of (A) log I vs log V (B) log I vs V (C) I vs log V (D) I vs V A long specimen of p-type semiconductor material (A) is positively charged (B) is electrically neutral (C) has an electric field directed along its length (D) acts as a dipole Two identical FETs, each characterized by the parameters gm and rd are connected in parallel. The composite FET is then characterized by the parameters g g (B) m and rd (A) m and 2rd 2 2 2

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(C) 2gm and rd 2 2.94

2.95

2.96

2.98

2.99

2.100

(B) V (D) J/K ONE MARK 2.103

(B) halves (D) decreases by about 20 mV

An npn transistor has a beta cut-off frequency fb of 1 MHz and common emitter short circuit low-frequency current gain bo of 200 it unity gain frequency fT and the alpha cut-off frequency fa respectively are (A) 200 MHz, 201 MHz (B) 200 MHz, 199 MHz (C) 199 MHz, 200 MHz (D) 201 MHz, 200 MHz A silicon n MOSFET has a threshold voltage of 1 V and oxide thickness of Ao. [er (SiO 2) = 3.9, e0 = 8.854 # 10-14 F/cm, q = 1.6 # 10-19 C] The region under the gate is ion implanted for threshold voltage tailoring. The dose and type of the implant (assumed to be a sheet charge at the interface) required to shift the threshold voltage to - 1 V are (A) 1.08 # 1012 /cm2 , p-type (B) 1.08 # 1012 /cm2 , n-type (C) 5.4 # 1011 /cm2 , p-type (D) 5.4 # 1011 /cm2 , n-type

The intrinsic carrier density at 300 K is 1.5 # 1010 /cm3 , in silicon. For n -type silicon doped to 2.25 # 1015 atoms/cm3 , the equilibrium electron and hole densities are (A) n = 1.5 # 1015 /cm3, p = 1.5 # 1010 /cm3 (B) n = 1.5 # 1010 /cm3, p = 2.25 # 1015 /cm3 (C) n = 2.25 # 1015 /cm3, p = 1.0 # 1015 /cm3 (D) n = 1.5 # 1010 /cm3, p = 1.5 # 1010 /cm3

A I D

ONE MARK

The p-type substrate in a conventional pn -junction isolated integrated circuit should be connected to n o.i c (A) nowhere, i.e. left floating . ia d o (B) a DC ground potential .n w (C) the most positive potential available in the circuit ww (D) the most negative potential available in the circuit

O N

If a transistor is operating with both of its junctions forward biased, but with the collector base forward bias greater than the emitter base forward bias, then it is operating in the (A) forward active mode (B) reverse saturation mode (C) reverse active mode (D) forward saturation mode The common-emitter short-circuit current gain b of a transistor (A) is a monotonically increasing function of the collector current IC (B) is a monotonically decreasing function of IC (C) increase with IC , for low IC , reaches a maximum and then decreases with further increase in IC (D) is not a function of IC A n -channel silicon (Eg = 1.1 eV) MOSFET was fabricated using n +poly-silicon gate and the threshold voltage was found to be 1 V. Now, if the gate is changed to p+ poly-silicon, other things remaining the same, the new threshold voltage should be (A) - 0.1 V (B) 0 V (C) 1.0 V (D) 2.1 V 1996

2.101

2.102

-1

For a MOS capacitor fabricated on a p-type semiconductor, strong inversion occurs when (A) surface potential is equal to Fermi potential (B) surface potential is zero (C) surface potential is negative and equal to Fermi potential in magnitude (D) surface potential is positive and equal to twice the Fermi potential

1996 2.97

(A) doubles (C) increases by about 20 mV

(D) 2gm and 2rd

q The units of are kT (A) V (C) J 1997

Page 37

TWO MARKS

In a bipolar transistor at room temperature, if the emitter current is doubled the voltage across its base-emitter junction

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Page 38

SOLUTIONS 2.1

2.2

2.3

2.4

Option (A) is correct. The potential barrier of the pn junction is lowered when a forward bias voltage is applied, allowing electrons and holes to flow across the space charge region (Injection) when holes flow from the p region across the space charge region into the n region, they become excess minority carrier holes and are subject to diffuse, drift and recombination processes. Option (D) is correct. In IC technology, dry oxidation as compared to wet oxidation produces superior quality oxide with a lower growth rate Option (D) is correct. In a MOSFET operating in the saturation region, the channel length modulation effect causes a decrease in output resistance. Option (A) is correct. Given, VB = 2V VTN = 1V So, we have Drain voltage

Since all the parameters of PMOS and NMOS are equal. So, mn = mp COX bW l = COX bW l = COX bW l L M1 L M2 L Given that M1 is in linear region. So, we assume that M2 is either in cutoff or saturation. Case 1 : M2 is in cut off So, I 2 = I1 = 0 Where I1 is drain current in M1 and I2 is drain current in M2 . m C 2 Since, I1 = p OX bW l82VSD ^VSG - VTp h - V SD B 2 L m C 2 0 = p OX bW l [2VSD ^VSG - VTp h - V SD & ] 2 L Solving it we get,

A I D &

2 ^VSG - VTp h = VSD 2 ^5 - Vin - 1h = 5 - VD Vin = VD + 3 2

VD = 2 volt & VG = 2 volt VS = 0 (Ground) For I1 = 0 , VD = 5 V Therefore, VGS = 2 > VTN So, Vin = 5 + 3 = 4 V 2 and VDS = 2 > VGS - VTN infor the NMOS . So o So, the MOSFET is in the saturation region. Therefore, drain cura.c i d VGS = Vin - 0 = 4 - 0 = 4 V and VGS > VTn rent is no . 2 So it can’t be in cutoff region. ID = kN ^VGS - VTN h ww w Case 2 : M2 must be in saturation region. 2 or, ID = kN ^VB - 1h So, I1 = I 2 Differentiating both side with respect to ID mp COX W mn COX W 1 = kN 2 ^VB - 1hdVB 2 2 dID 2 (VSG - VTp) VSD - V SD @ = 2 L (VGS - VTn) 2 L6 2 Since, VBQ = 2 volt (at D.C. Voltage) & 2 (VSG - VTp) VSD - V SD = (VGS - VTn) 2 Hence, we obtain & 2 (5 - Vin - 1) (5 - VD) - (5 - VD) 2 = (Vin - 0 - 1) 2 dVB = 1 & 2 (4 - Vin) (5 - VD) - (5 - VD) 2 = (Vin - 1) 2 dID 2kN ^VB - 1h 1 Substituting VD = VDS = VGS - VTn and for N -MOS & VD = Vin - 1 = 2 # 40 # 10-6 # ^2 - 1h & 2 (4 - Vin) (6 - Vin) - (6 - Vin) 2 = (Vin - 1) 2 3 = 12.5 # 10 W & 48 - 36 - 8Vin =- 2Vin + 1 = 12.5 kW & 6Vin = 11 Option (D) is correct. & Vin = 11 = 1.833 V For the semiconductor, n 0 p 0 = n i2 6 2 20 n So for M2 to be in saturation Vin < 1.833 V or Vin < 1.875 V p 0 = i = 1019 = 10 per cm3 n 0 10 2.7 Option (B) is correct. Volume of given device, V = Area # depth Gate source overlap capacitance. = 1 mm 2 # 1 mm Co = dWeox e0 (medium Sio 2 ) tox = 10-8 cm2 # 10-4 cm -9 -6 -12 -12 3 = 10 cm = 20 # 10 # 1 # 10 #-93.9 # 8.9 # 10 1 # 10 So total no. of holes is, = 0.69 # 10-15 F p = p 0 # V = 10 # 10-12 = 10-11 2.8 Option (B) is correct. Which is approximately equal to zero. Source body junction capacitance. Option (A) is correct. Given the circuit as below : Cs = Aer e0 d

O N

2.5

2.6

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A = (0.2 mm + 0.2 mm + 0.2 mm) # 1 mm + 2 (0.2 mm # 0.2 mm)

Page 39

= 7 # 10-15 F 2.9

2.10

2.11

2.12

= 0.68 mm2 d = 10 nm (depletion width of all junction) -12 8.9 # 10-12 Cs = 0.68 # 10 # 11.7 -# 9 10 # 10

Option (C) is correct. Drift current Id = qnmn E It depends upon Electric field E and carrier concentration n

Option (C) is correct. Reverse bias breakdown or Zener effect occurs in highly doped PN junction through tunneling mechanism. In a highly doped PN junction, the conduction and valence bands on opposite sides of the junction are sufficiently close during reverse bias that electron may tunnel directly from the valence band on the p-side into the conduction band on n -side. Breakdown voltage VB \ 1 NA ND So, breakdown voltage decreases as concentration increases Depletion capacitance 1/2 ees NA ND C =' 1 2 (Vbi + VR) (NA + ND)

2.17

Option (B) is correct. Zener diode operates in reverse breakdown region.

Thus C \ NA ND Depletion capacitance increases as concentration increases

Option (D) is correct. For every 1c C increase in temperature, forward bias voltage across diode decreases by 2.5 mV. Thus for 10c C increase, there us 25 mV decreases. Option (B) is correct. Full channel resistance is r L r # = 600 W W#a If VGS is applied, Channel resistance is r L where b = a c1 rl = # W#b Pinch off voltage, qN Vp = D a2 2e If depletion on each side is d = 1 μm at VGS = 0 . qN Vj = D d2 2e qN qND or 1 = D (1 # 10-6) 2 & = 1012 2e 2e

A I D 2.19

...(1)

ww

...(2)

2.21

Thus

E =

Jd = ND mn eE = 1016 # 1350 # 1.6 # 10-19 # 10 # 1013 4 2 # 10 A/cm

Option (C) is correct. Only dopant atoms can have concentration of 4 # 1019 cm - 3 in n type silicon at room temperature. Option (A) is correct. 2 Unit of mobility mn is = cm V. sec 2 Unit of diffusion current Dn is = cm sec 2 2 m Thus unit of n is = cm / cm = 1 = V-1 V $ sec sec V Dn

2.22

or Vp =- 25 V At VGS =- 3 V ; - 3 mm = 3.26 mm b = 5 b1 - 25 l rL rL a 5 = rl = # b = 600 # 3.26 = 917 W Wa W#b

1 = 10 kV/cm 1 mm Option (A) is correct. Electron drift current density

n o.i c . ia = 2.16

VGS od Vp mw.n2.20

Vp = 1012 # (5 # 10-6) 2

Option (C) is correct. At VGS = 0 V ,

Option (C) is correct. Sample is in thermal equilibrium so, electric field

2.18

Now from equation (2), we have

2.14

Option (B) is correct. Emitter injection efficiency is given as 1 g = 1 + NB NE To achieve g = 1, NE >> NB

2.16

O N

2.13

Option (B) is correct. Dry oxidation is used to achieve high quality oxide growth.

2.15

2.23

Option (D) is correct. Both S1 and S2 are true and S2 is a reason for S1. Option (B) is correct. We know that or

NA WP = ND WN 17 -6 NA = ND WN = 1 # 10 # 0.1-6# 10 = 1 # 1016 WP 1 # 10

The built-in potential is D Vbi = VT 1nc NA N n i2 m 17 16 # 10 = 0.760 = 26 # 10-3 ln e 1 # 10 # 1 10 o 2 (1.4 # 10 )

since 2b = 8 mm b = 4 mm rL a = 600 5 = 750 W rl = #4 Wa # b

Option (A) is correct. At room temperature mobility of electrons for Si sample is given mn = 1350 cm2 /Vs . For an n -channel MOSFET to create an inversion layer of electrons, a large positive gate voltage is to be applied. Therefore, induced electric field increases and mobility decreases. So, Mobility mn < 1350 cm2 /Vs for n -channel MOSFET

2.24

Option (B) is correct. The peak electric field in device is directed from p to n and is from p to n E =- eND xn es from n to p = eND xn es

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-19

MV/cm 2.25

2.26

2.27

2.28

2.29

2.30

17

-5

# 1 # 10 = 1.6 # 10 # 1 #-10 14 8.85 # 10 # 12

Page 40

Thus = 0.15

or

Option (D) is correct. Channel length modulation is not associated with a p - n junction. It is being associated with MOSFET in which effective channel length decreases, producing the phenomenon called channel length modulation.

2.34

2.35

Option (A) is correct. Trivalent impurities are used for making p - type semiconductors. So, Silicon wafer heavily doped with boron is a p+ substrate. Option (D) is correct. Oxidation rate is zero because the existing oxide prevent the further oxidation.

nNA = ni2 2 n = ni NA

Option (C) is correct. The electric field has the maximum value at the junction of p+ n . Option (B) is correct. Zener diode and Avalanche diode works in the reverse bias and laser diode works in forward bias. In solar cell diode works in forward bias but photo current is in reverse direction. Thus Zener diode : Reverse Bias Solar Cell : Forward Bias Laser Diode : Forward Bias Avalanche Photo diode : Reverse Bias

Option (C) is correct. In BJT as the B-C reverse bias voltage increases, the B-C space charge region width increases which xB (i.e. neutral base width) > A change in neutral base width will change the collector current. A Option (C) is correct. reduction in base width will causes the gradient in minority carrier As VD = constant concentration to increase, which in turn causes an increased in the Thus Which is straight line. gm \ (VGS - VT ) diffusion current. This effect si known as base modulation as early effect. Option (C) is correct. In JFET the gate to source voltage that must be applied to achieve E2 - E1 = kT ln NA ni pinch off voltage is described as pinch off voltage and is also called 17 as turn voltage or threshold voltage. NA = 4 # 10 In LASER population inversion occurs on the condition when ni = 1.5 # 1010 concentration of electrons in one energy state is greater than that in 17 E2 - E1 = 25 # 10-3 e ln 4 # 10 10 = 0.427 eV lower energy state, i.e. a non equilibrium condition. 1.5 # 10 n oIn.i MOS capacitor, flat band voltage is the gate voltage that must be c Hence fermi level goes down by 0.427 eV as silicon is doped with . ia applied to create flat ban condition in which there is no space charge d o boron. .n region in semiconductor under oxide. w w Option (C) is correct. w Therefore 2 BJT : Early effect Pinch off voltage VP = eW ND es MOS capacitor : Flat-band voltage Let VP = VP1 LASER diode : Population inversion 2 2 W V W P1 1 Now = 2= JFET : Pinch-off voltage VP2 W2 (2W) 2 2.37 Option (A) is correct. or 4VP1 = VP2 W = K V + VR Initial transconductance Now 2m = K 0.8 + 1.2 gm = Kn ;1 - Vbi - VGS E Vp From above two equation we get For first condition gm1 W = 0.8 + 7.2 = 8 = 2 2m 0.8 + 1.2 2 0 - (- 2) 2 = Kn =1 = Kn ;1 G E VP1 VP1 or W2 = 4 m m For second condition 2.38 Option (B) is correct. 0 - (- 2) b 2 gm2 = Kn =1 = K2 ;1 = 50 = 50 a= VP2 G 4VP1 E b + 1 50 + 1 51 1 - 2/VP1 Current Gain = Base Transport Factor # Emitter injection Effigm1 Dividing =f p gm2 1 - 1/ (2VP1) ciency Hence VP = VP1 a = b1 # b2 2.36

Option (B) is correct. gm = 2ID = 2 K (VGS - VT ) 2 = 2K (VGS - VT ) 2VGS 2VGS

A I D

2.31

2.32 2.33

O N

Option (A) is correct.

or

Option (D) is correct. As per mass action law

2.39

np = ni2 If acceptor impurities are introduces p = NA

50 b1 = a = = 0.985 51 # 0.995 b2

Option (A) is correct. At low voltage when there is no depletion region and capacitance is decide by SiO2 thickness only, C = e0 er1 A D

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or 2.40

Page 41

-13 10-4 = 50 nm D = e0 er1 A = 3.5 # 10 -# C 7 # 10 12

reduction in base width will causes the gradient in minority carrier concentration to increases, which in turn causes an increases in the diffusion current. This effect si known as base modulation as early effect.

Option (B) is correct. The construction of given capacitor is shown in fig below 2.46

2.47

When applied voltage is 0 volts, there will be no depletion region and we get C1 = 7 pF When applied voltage is V , a depletion region will be formed as shown in fig an total capacitance is 1 pF. Thus CT = 1 pF or CT = C1 C2 = 1 pF C1 + C2 1 = 1 + 1 or CT C1 C2 Substituting values of CT and C1 we get C2 = 7 pF 6 - 12 -4 Now D2 = e0 er2 A = 1 # 710 #- 1210 = 6 # 10 - 4 cm C2 7 6 # 10 = 0.857 mm 2.41

2.42

Option (A) is correct. For t < 0 diode forward biased and VR = 5 . At t = 0 diode abruptly changes to reverse biased and current across resistor must be 0. But in storage time 0 < t < ts diode retain its resistance of forward biased. Thus for 0 < t < ts it will be ON and VR =- 5 V Option (B) is correct. According to Hall effect the direction of electric field is same as that of direction of force exerted. or

2.48

E =- v # B E = B#v

Option (B) is correct. The varacter diode is used in tuned circuit as it can provide frequently stability. PIN diode is used as a current controlled attenuator. Zener diode is used in regulated voltage supply or fixed voltage reference. Schottkey diode has metal-semiconductor function so it has fast switching action so it is used as high frequency switch Varactor diode : Tuned circuits PIN Diode : Current controlled attenuator Zener diode : Voltage reference Schottky diode : High frequency switch

A I D

O N

Option (C) is correct. n 2.49 Option o.i (D) is correct. c Depletion region will not be formed if the MOS capacitor has n type . mP ia = 0.4 substrate but from C-V characteristics, C reduces if V is increased. no d We have mn . Thus depletion region must be formed. Hence S1 is false ww Conductance of n type semiconductor w If positive charges is introduced in the oxide layer, then to equalize sn = nqmn the effect the applied voltage V must be reduced. Thus the C - V Conductance of intrinsic semiconductor plot moves to the left. Hence S2 is true. si = ni q (mn + mp) Option (C) is correct. nmn sn = n Ratio is = For the case of negative slope it is the negative resistance region s n ( m + m ) n 1 + ^ i i n p i 8 4 4 . 2 10 # = = 2 # 10 1.5 # 10 4 (1 + 0.4) 2.50

mp mn h

Option (C) is correct. For silicon at 0 K, Eg0 = 1.21 eV At any temperature

2.43

EgT = Eg0 - 3.6 # 10 - 4 T

Option (A) is correct. For n -type p is minority carrier concentration

At T = 300 K,

ni2

np = np = Constant p \ 1 n

Eg300 = 1.21 - 3.6 # 10 - 4 # 300 = 1.1 eV This is standard value, that must be remembered.

Since ni is constant 2.51

Thus p is inversely proportional to n . 2.44

2.45

Option (A) is correct. Diffusion current, since the drift current is negligible for minority carrier. Option (B) is correct. In BJT as the B-C reverse bias voltage increases, the B-C space charge region width increases which xB (i.e. neutral base width) > A change in neutral base width will change the collector current. A

Option (B) is correct. The reverse saturation current doubles for every 10cC rise in temperature as follows : I0 (T) = I 01 # 2(T - T )/10 Thus at 40c C, I0 = 40 pA 1

2.52

2.53

Option (A) is correct. Silicon is abundant on the surface of earth in the from of SiO2 . Option (B) is correct. sn = nqmn

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sp = pqmp sp m = p =1 sn mn 3 2.54

2.56

2.57

2.58

2.59

(n = p) 2.64

2.65

2.66

a 1-a

O N

d

2.67

or 2.68

3 m # 1016 = 0.3 m m Wp = Wn # ND = NA 9 # 1016 Option (B) is correct. Conductivity s = nqun or resistivity r = 1 = 1 s nqmn Thus n = 1 qrmn 16 1 = = 10 /cm 3 1.6 # 10 - 19 # 0.5 # 1250

2.63

Option (B) is correct. Concentration gradient 1014 dn = = 2 # 1018 -4 dx 0.5 # 10 q = 1.6 # 10 - 19 C Dn = 25 1014 dn = dx 0.5 # 10 - 4 JC = qDn dn = 1.6 # 10 - 19 # 25 # 2 # 1018 = 8 A/cm 2 dx

2.69

Wp NA = Wn ND or

Eg \ 1 l Eg2 = l1 = 1.1 Eg1 l2 0.87 Eg2 = 1.1 # 1.12 = 1.416 eV 0.87

Thus

VCC - IC RC - VCE = 0 IC = VCC - VCE = 3 - 0.2 = 2.8 mA RC 1k IB = IC = 2.8m = 56 mA b 50

Option (B) is correct. We know that

2 ID2 = (3 - 1) = 4 ID1 (2 - 1) 2 ID2 = 4IDI = 4 mA

or n o.i c . ia Option (A) is correct.

Option (A) is correct. Applying KVL we get

Now

Option (D) is correct. We know that

A I D

Option (C) is correct.

or

Option (C) is correct. Increase in gate oxide thickness makes difficult to induce charges in channel. Thus VT increases if we increases gate oxide thickness. Hence S1 is false. Increase in substrate doping concentration require more gate voltage because initially induce charges will get combine in substrate. Thus VT increases if we increase substrate doping concentration. Hence S2 is false.

ID = K (VGS - VT ) 2 2 IDS = (VGS2 - VT ) Thus IDI (VGS1 - VT ) 2 Substituting the values we have

.no

w a -" b ww a ." b . If the base width increases, recombination of carrier in base region increases and a decreases & hence b decreases. If doping in base region increases, recombination of carrier in base increases and a decreases thereby decreasing b . Thus S1 is true and S2 is false.

2.62

or

Option (A) is correct. Here emitter base junction is forward biased and base collector junction is reversed biased. Thus transistor is operating in normal active region.

Thus

2.61

Now

Option (C) is correct. Trivalent impurities are used for making p type semiconductor. Boron is trivalent.

b =

eeS NA ND Cj = ; 2 (Vbi + VR)( NA + ND) E 1 Cj \ (Vbi + VR) C j2 (Vbi + VR) 1 1 =1 = = (Vbi + VR) 2 4 2 C j1 C Cj2 = j1 = 1 = 0.5 pF 2 2

Thus

Option (C) is correct. From the graph it can be easily seen that Vth = 1 V Now VGS = 3 - 1 = 2 V and VDS = 5 - 1 = 4 V Since VDS > VGS $ VDS > VGS - Vth Thus MOSFET is in saturation region.

Option (D) is correct.

Option (D) is correct. We know that 1 2

C = e0 er A d C = e0 er = 8.85 # 10-12 # 11.7 = 10.35 m F d A 10 # 10-6

Option (B) is correct. In accumulation mode for NMOS having p -substrate, when positive voltage is applied at the gate, this will induce negative charge near p - type surface beneath the gate. When VGS is made sufficiently large, an inversion of electrons is formed and this in effect forms and n - channel.

We have

2.60

For n type semiconductor n = ND

Option (B) is correct.

or 2.55

Page 42

2.70

Option (D) is correct. Pentavalent make n -type semiconductor and phosphorous is pentavalent. Option (C) is correct. For silicon at 0 K Eg0 = 1.21 eV At any temperature EgT = Eg0 - 3.6 # 10 - 4 T At T = 300 K, Eg300 = 1.21 - 3.6 # 10 - 4 # 300 = 1.1 eV This is standard value, that must be remembered.

2.71

Option (A) is correct.

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By Mass action law

Page 43 2.80

np = p = 2.72

ni2 ni2

16 .5 # 1016 = 4.5 # 1011 = 1.5 # 10 # 120 n 5 # 10

Option (C) is correct. Tunnel diode shows the negative characteristics in forward bias. It is used in forward bias. Avalanche photo diode is used in reverse bias.

2.73

Option (D) is correct.

2.74

Option (A) is correct.

2.81

2.82

From above relation we have R = = 106 W 2.75

1 0.1 # 10 - 2 = 20 nqmn A 5 # 10 # 1.6 # 10 - 19 # 0.13 # 100 # 10 - 12

Option (D) is correct. 16

17

dn = 6 # 10 - 10 dx 2 # 10 - 4 - 0

=- 2 # 1020 2.83

Since no electric field is present, E = 0 and we get So, Jn = qDn dn dx

2.84

2.77

3 T = T2 - T1 = 40 - 20 = 20cC 3 VD =- 2.5 # 20 = 50 mV VD = 700 - 50 = 650 mV

Option (D) is correct. Condition for saturation is IC < bIB Option (B) is correct. The metal area of the gate in conjunction with the insulating dielectric oxide layer and semiconductor channel, form a parallel plate capacitor. It is voltage controlled capacitor because in active region the current voltage relationship is given by

A I D

= 1.6 # 10 - 19 # 35 # (- 2 # 1020) =- 1120 A/cm 2 2.76

Option (B) is correct. At constant current the rate of change of voltage with respect to temperature is dV =- 2.5 mV per degree centigrade dT Here Thus Therefore,

Jn = nqme E + Dn q dn dx

Now

Option (D) is correct. For a JFET in active region we have 2 IDS = IDSS c1 - VGS m VP

From above equation it is clear that the action of a JFET is voltage controlled current source.

rl R = , r = 1 and a = nqun A s

We that

Option (B) is correct. In n -well CMOS fabrication following are the steps : (A) n - well implant (B) Source drain diffusion (C) Metalization (D) Passivation

Option (C) is correct. LED works on the principal of spontaneous emission. In the avalanche photo diode due to the avalanche effect there is IDS = K (VGS - VT ) 2 n i o. large current gain. cOption 2.85 (D) is correct. . a i Tunnel diode has very large doping. d In MOSFET the body (substrate) is connected to power supply in .no LASER diode are used for coherent radiation. w such a way to maintain the body (substrate) to channel junction in w w Option (C) is correct. cutoff condition. The resulting reverse bias voltage between source V and body will have an effect on device function. The reverse bias We know that I = Io `e h V - 1j will widen the depletion region resulting the reduction in channel where h = 1 for germanium and h = 2 silicon. As per question length. V V Io `e e - 1j = Io `e hV - 1j 2.86 Option (C) is correct. V 0.718 At a given value of vBE , increasing the reverse-bias voltage on the Io # 10 - 1 or = eVhV - 1 = e 2 #026.1435 = 4 # 103 collector-base junction and thus increases the width of the depletion Io e 26 # 10 - 1 e hV - 1 region of this junction. This in turn results in a decrease in the Option (A) is correct. effective base width W . Since IS is inversely proportional to W , IS -34 8 increases and that iC increases proportionally. This is early effect. Eg = hc = 6.626 # 10 # -310# 10 = 3.62 J l 54900 # 10 2.87 Option (B) is correct. -19 Eg (J) 3 . 62 10 # For an n -channel JEFT trans-conductance is In eV Eg (eV) = = = 2.26 eV e 1.6 # 10-19 - 2IDSS 1 - VGS = - 2 # 2 # 10-3 1 - (- 2) g = m = Alternatively -4 VP b VP l (- 4)G

O N

D1 T

si

Dsi hVT

DGe

n

T

Ge

Dsi

si

-3

T

DGe

si

2.78

-3

T

1.24 Eg = 1.24 eV = = 2.26 eV l (mm) 5490 # 10-4 mm 2.79

Option (D) is correct. We know that

= 10-3 # 1 = 0.5 mho 2 2.88

ID = K (VGS - VT ) 2 ID2 = (VGS2 - VT ) Thus ID1 (VGS1 - VT ) 2 Substituting the values we have

2

Option (A) is correct. We have Now or

2

or

ID2 = (1.4 - 0.4) = 4 ID1 (0.9 - 0.4) 2 ID2 = 4IDI = 4 mA

or or

gm = IC = 1 26 VT gm fT = 2p (C p + C m) 1/26 400 = 2p (0.3 # 10-12 + C m) 1 = 15.3 # 10-12 (0.3 # 10-12 + C m) = 2p # 26 # 400 C m 15.3 # 10-12 - 0.3 # 10-12 = 15 # 10-12 15 pF

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Page 44

for p-channel [p+ polysilicon used in gate] VTP = VSD (sat) - VGS so VTP =- VDS (sat) + VGS so threshold voltage will be same.

Option (D) is correct. For any semiconductor (Intrinsic or extrinsic) the product n p remains constant at a given temperature so here np = ni pi

2.90

Option (D) is correct. fT =

2.91

2.101

gm 2p (C p + C m)

or or

I = I 0 (eV/kT - 1)

or

I + 1 = eV/kT I0 kT log b I + 1l = V I0

Now

So if we plot log I vs V we get a straight line. 2.92

2.93 2.94

2.95 2.96

or

Option (B) is correct. A specimen of p - type or n - type is always electrical neutral.

2.102

Option (D) is correct. ni = 1.5 # 10 /cm Nd = 2.25 # 1015 atoms/cm3 For n type doping we have electron concentration n - Nd = 2.25 # 1015 atom/cm3 For a given temperature Hole concentration

2.97

2.98

10 2

Option (A) is correct. Unity gain frequency is given by

A I D

3

np = n i2

O N

fT = fB # b = 106 # 200 = 200 MHz a-cutoff frequency is given by f fb fa = b = = fb (b + 1) 1-a b 1b+1

.in

no w.

d

co ia.

2.103

= 106 # (200 + 1) = 201 MHz

Option (A) is correct.

(1.5 # 10 ) w3 p =n = /cm = 1.0 # 105w n 2.25 # 1015 2 i

Option (D) is correct. In p n -junction isolated circuit we should have high impedance, so that p n junction should be kept in reverse bias. (So connect p to negative potential in the circuit) Option (B) is correct.

If both junction are forward biased and collector base junction is more forward biased then IC will be flowing out wards (opposite direction to normal mode) the collector and it will be in reverse saturation mode. 2.99

Option (C) is correct. For normal active mode we have b = IC IB For small values of IC , if we increases IC , b also increases until we reach (IC ) saturation. Further increases in IC (since transistor is in saturation mode know) will increases IB and b decreases.

2.100

BE

/kT

>> 1

= (VBE ) 1 + 15 m volt Thus if emitter current is doubled the base emitter junction voltage is increased by 15 mV.

Option (B) is correct. The unit of q is e and unit of kT is eV. Thus unit of e/kT is e/eV = V-1 .

10

BE

Now if emitter current is double i.e. IE 2 = 2IE1 (VBE ) 2 = (VBE ) 1 + (25 # 0.60) m volt

Option (C) is correct.

Option (C) is correct. We have

IE = I 0 (eV /kT - 1) IE = I 0 eV /kT eV VBE = kT ln b IE l I0 (VBE ) 1 = kT ln b IE 1 l I0 (VBE ) 2 = kT ln b IE 2 l I0 (VBE ) 2 - (VBE ) 1 = kT ;ln b IE 2 lE = kT ln b 2IE 1 l IE 1 IE 1 BE

Option (B) is correct. For a Forward Bias p-n junction, current equation or

Option (C) is correct. Emitter current is given by

Option (C) is correct. For a n -channel mosfet thresholds voltage is given by VTN = VGS - VDS (sat)

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UNIT 3

Page 45

(A) 125 and 125 (C) 250 and 125

2013 3.1

The ac schematic of an NMOS common-source state is shown in the figure below, where part of the biasing circuits has been omitted for simplicity. For the n -channel MOSFET M, the transconductance gm = 1 mA/V , and body effect and channel length modulation effect are to be neglected. The lower cutoff frequency in HZ of the circuit is approximately at

3.4

ANALOG CIRCUITS

ONE MARK

In the circuit shown below what is the output voltage ^Vouth if a silicon transistor Q and an ideal op-amp are used?

(A) 8 (C) 50 (A) - 15 V (C) + 0.7 V 3.2

(B) 125 and 250 (D) 250 and 250

(B) - 0.7 V (D) + 15 V

In the circuit shown below the op-amps are ideal. Then, Vout in Volts is

3.5

In a voltage-voltage feedback as shown below, which one of the following statements is TRUE if the gain k is increased?

(B) 32 (D) 200

A I D

O N

.in

no w.

co ia.

d

ww

(A) The input es (B) The input increases (C) The input decreases (D) The input es 2013 3.3

impedance increases and output impedance decreas-

(A) 4 (C) 8

impedance increases and output impedance also 3.6

impedance decreases and output impedance also impedance decreases and output impedance increas-

(B) 6 (D) 10

In the circuit shown below, Q1 has negligible collector-to-emitter saturation voltage and the diode drops negligible voltage across it under forward bias. If Vcc is + 5 V , X and Y are digital signals with 0 V as logic 0 and Vcc as logic 1, then the Boolean expression for Z is

TWO MARKS

In the circuit shown below, the knee current of the ideal Zener dioide is 10 mA. To maintain 5 V across RL , the minimum value of RL in W and the minimum power rating of the Zener diode in mW , respectively, are (A) XY (C) XY 3.7

(B) XY (D) XY

A voltage 1000 sin wt Volts is applied across YZ . Assuming ideal diodes, the voltage measured across WX in Volts, is

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(A) cos (wt) - 1 (C) 1 - cos (wt) (A) sin wt (C) ^sin wt - sin wt h /2 3.8

(B) _sin wt + sin wt i /2 (D) 0 for all t

3.12

(B) sin (wt) (D) 1 - sin (wt)

The impedance looking into nodes 1 and 2 in the given circuit is

In the circuit shown below, the silicon npn transistor Q has a very high value of b . The required value of R2 in kW to produce IC = 1 mA is

(A) 50 W (C) 5 kW

(B) 100 W (D) 10.1 kW

2012 3.13

(A) 20 (C) 40

3.9

O N ONE MARK

n

The current ib through the base of a silicon npn transistor is o.i c . 1 + 0.1 cos (10000pt) mA At 300 K, the rp in the small signal model dia no of the transistor is (A) low pass filter with f3dB = w.

1 rad/s (R1 + R2) C (B) high pass filter with f3dB = 1 rad/s R1 C (C) low pass filter with f3dB = 1 rad/s R1 C 1 (D) high pass filter with f3dB = rad/s (R1 + R2) C

ww

(A) 250 W (C) 25 W 3.10

(B) 27.5 W (D) 22.5 W

3.14

The voltage gain Av of the circuit shown below is

The i -v characteristics of the diode in the circuit given below are v - 0.7 A, v $ 0.7 V i = * 500 0A v < 0. 7 V

The current in the circuit is (A) 10 mA (C) 6.67 mA 3.11

The circuit shown is a

A I D

(B) 30 (D) 50

2012

TWO MARKS

(A) Av . 200 (C) Av . 20 (B) 9.3 mA (D) 6.2 mA

The diodes and capacitors in the circuit shown are ideal. The voltage v (t) across the diode D1 is

(B) Av . 100 (D) Av . 10

2011 3.15

ONE MARK

In the circuit shown below, capacitors C1 and C2 are very large and are shorts at the input frequency. vi is a small signal input. The gain magnitude vo at 10 M rad/s is vi

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Page 47 3.19

For a BJT, the common base current gain a = 0.98 and the collector base junction reverse bias saturation current ICO = 0.6 mA . This BJT is connected in the common emitter mode and operated in the active region with a base drive current IB = 20 mA . The collector current IC for this mode of operation is (A) 0.98 mA (B) 0.99 mA (C) 1.0 mA (D) 1.01 mA

Statement for Linked Answer Questions: 4.6 & 4.7 (A) maximum (C) unity 3.16

In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage Vt = kT/q = 25 mV . The small signal input vi = Vp cos ^wt h where Vp = 100 mV.

(B) minimum (D) zero

The circuit below implements a filter between the input current ii and the output voltage vo . Assume that the op-amp is ideal. The filter implemented is a

3.20

(A) low pass filter (C) band stop filter

A I D

(B) band pass filter (D) high pass filter

3.21

O N

2011 3.17

The ac output voltage vac is (A) 0.25 cos ^wt h mV (C)in2 cos (wt) mV

.

(B) 1 cos (wt) mV (D) 22 cos (wt) mV

In the circuit shown below, for the MOS transistors, mn Cox = 100 mA/V 2 .co a i and the threshold voltage VT = 1 V . The voltage Vx at the source of no d 2010 ONE MARK w. the upper transistor is w w 3.22 The amplifier circuit shown below uses a silicon transistor. The capacitors CC and CE can be assumed to be short at signal frequency and effect of output resistance r0 can be ignored. If CE is disconnected from the circuit, which one of the following statements is true

(A) 1 V (C) 3 V 3.18

TWO MARKS

The bias current IDC through the diodes is (A) 1 mA (B) 1.28 mA (C) 1.5 mA (D) 2 mA

(B) 2 V (D) 3.67 V

For the BJT, in the circuit shown below, Q1 b = 3, VBEon = 0.7 V, VCEsat = 0.7 V . The switch is initially closed. At time t = 0 , the switch is opened. The time t at which Q1 leaves the active region is (A) The input resistance Ri increases and magnitude of voltage gainAV decreases (B) The input resistance Ri decreases and magnitude of voltage gain AV increases (C) Both input resistance Ri and magnitude of voltage gain AV decreases (D) Both input resistance Ri and the magnitude of voltage gain AV increases 3.23

(A) 10 ms (C) 50 ms

(B) 25 ms (D) 100 ms

In the silicon BJT circuit shown below, assume that the emitter area of transistor Q1 is half that of transistor Q2

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Page 48

The value of current Io is approximately (A) 0.5 mA (B) 2 mA (C) 9.3 mA (D) 15 mA 3.24

Assuming the OP-AMP to be ideal, the voltage gain of the amplifier shown below is

(A) - R2 R1 R || R 3 (C) - 2 R1

(B) - R 3 R1 (D) -b R2 + R 3 l R1

2010

A I D 2009

TWO MARKS

Common Data For Q. 4.11 & 4.12 :

3.28

O N

In the circuit below, the diode is ideal. The voltage V is given by

in

Consider the common emitter amplifier shown below with the folco. . a lowing circuit parameters: di o n b = 100, gm = 0.3861 A/V, r0 = 259 W, RS = 1 kW, RB = 93 kW, w. w w RC = 250 kW, RL = 1 kW, C1 = 3 and C2 = 4.7 mF (A) min (Vi, 1) (C) min (- Vi, 1) 3.29

3.25

3.26

3.27

TWO MARKS

(B) max (Vi, 1) (D) max (- Vi, 1)

In the following a stable multivibrator circuit, which properties of v0 (t) depend on R2 ?

The resistance seen by the source vS is (A) 258 W (B) 1258 W (C) 93 kW (D) 3 The lower cut-off frequency due to C2 is (A) 33.9 Hz (B) 27.1 Hz (C) 13.6 Hz (D) 16.9 Hz The transfer characteristic for the precision rectifier circuit shown below is (assume ideal OP-AMP and practical diodes)

(A) Only the frequency (B) Only the amplitude (C) Both the amplitude and the frequency (D) Neither the amplitude nor the frequency

Statement for Linked Answer Question 4.16 and 4.17 Consider for CMOS circuit shown, where the gate voltage v0 of the n-MOSFET is increased from zero, while the gate voltage of the p -MOSFET is kept constant at 3 V. Assume, that, for both transistors, the magnitude of the threshold voltage is 1 V and the product of the trans-conductance parameter is 1mA. V - 2

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Page 49

2008

In the following limiter circuit, an input voltage Vi = 10 sin 100pt is applied. Assume that the diode drop is 0.7 V when it is forward biased. When it is forward biased. The zener breakdown voltage is 6.8 V The maximum and minimum values of the output voltage respectively are

3.34

3.30

3.31

3.32

For small increase in VG beyond 1V, which of the following gives the correct description of the region of operation of each MOSFET (A) Both the MOSFETs are in saturation region (B) Both the MOSFETs are in triode region (C) n-MOSFETs is in triode and p -MOSFET is in saturation region (D) n- MOSFET is in saturation and p -MOSFET is in triode region

(A) 6.1 V, - 0.7 V (C) 7.5 V, - 0.7 V

Estimate the output voltage V0 for VG = 1.5 V. [Hints : Use the appropriate current-voltage equation for each MOSFET, based on the answer to Q.4.16] (A) 4 - 1 (B) 4 + 1 2 2 (C) 4 - 3 (D) 4 + 3 2 2

ONE MARK

(B) 0.7 V, - 7.5 V (D) 7.5 V, - 7.5 V

2008

For the circuit shown in the following figure, transistor M1 and M2 are identical NMOS transistors. Assume the M2 is in saturation and the output is unloaded.

3.35

In the circuit shown below, the op-amp is ideal, the transistor has VBE = 0.6 V and b = 150 . Decide whether the feedback in the circuit is positive or negative and determine the voltage V at the output of the op-amp.

TWO MARSK

O N

A I D .in

no w.

d

co ia.

ww

3.36

The current Ix is related to Ibias as (A) Ix = Ibias + Is (B) Ix = Ibias (C) Ix = Ibias - cVDD - Vout m (D) Ix = Ibias - Is RE Consider the following circuit using an ideal OPAMP. The I-V V characteristic of the diode is described by the relation I = I 0 _eV - 1i where VT = 25 mV, I0 = 1m A and V is the voltage across the diode (taken as positive for forward bias). For an input voltage Vi =- 1 V , the output voltage V0 is t

(A) Positive feedback, V = 10 V (B) Positive feedback, V = 0 V (C) Negative feedback, V = 5 V (D) Negative feedback, V = 2 V 3.33

A small signal source Vi (t) = A cos 20t + B sin 106 t is applied to a transistor amplifier as shown below. The transistor has b = 150 and hie = 3W . Which expression best approximate V0 (t)

(A) 0 V (C) 0.7 V 3.37

(B) 0.1 V (D) 1.1 V

The OPAMP circuit shown above represents a

(A) V0 (t) =- 1500 (A cos 20t + B sin 106 t) (B) V0 (t) = - 1500( A cos 20t + B sin 106 t) (C) V0 (t) =- 1500B sin 106 t (D) V0 (t) =- 150B sin 106 t

(A) high pass filter (C) band pass filter

(B) low pass filter (D) band reject filter

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3.39

Page 50

Two identical NMOS transistors M1 and M2 are connected as shown below. Vbias is chosen so that both transistors are in saturation. The equivalent gm of the pair is defied to be 2Iout at constant Vout 2Vi The equivalent gm of the pair is

(A) the sum of individual gm ' s of the transistors (B) the product of individual gm ’s of the transistors (C) nearly equal to the gm of M1 g (D) nearly equal to m of M2 g0 Consider the Schmidt trigger circuit shown below A triangular wave which goes from -12 to 12 V is applied to the inverting input of OPMAP. Assume that the output of the OPAMP swings from +15 V to -15 V. The voltage at the non-inverting input switches between

(A) - 12V to +12 V (C) -5 V to +5 V

In a transconductance amplifier, it is desirable to have (A) a large input resistance and a large output resistance (B) a large input resistance and a small output resistance (C) a small input resistance and a large output resistance (D) a small input resistance and a small output resistance

3.43

2007

For the Op-Amp circuit shown in the figure, V0 is

3.44

A I D

O N

(B) -7.5 V to 7.5 V (D) 0 V and 5 V

Statement for Linked Answer Question 3.26 and 3.27:

(A) -2 V .in -0.5 V o(C)

.c

ia od

n

. ww

3.45

w

TWO MARKS

(B) -1 V (D) 0.5 V

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

In the following transistor circuit, VBE = 0.7 V, r3 = 25 mV/IE , and b and all the capacitances are very large

(A) cut-off (C) normal active 3.46

3.40

3.41

The value of DC current IE is (A) 1 mA (C) 5 mA

(B) 2 mA (D) 10 mA

In the Op-Amp circuit shown, assume that the diode current follows the equation I = Is exp (V/VT ). For Vi = 2V, V0 = V01, and for Vi = 4V, V0 = V02 . The relationship between V01 and V02 is

The mid-band voltage gain of the amplifier is approximately (A) -180 (B) -120 (C) -90 (D) -60 2007

3.42

(B) saturation (D) reverse active

(A) V02 = 2 Vo1 (C) Vo2 = Vo1 1n2

ONE MARK

The correct full wave rectifier circuit is

3.47

(B) Vo2 = e2 Vo1 (D) Vo1 - Vo2 = VT 1n2

In the CMOS inverter circuit shown, if the trans conductance parameters of the NMOS and PMOS transistors are W kn = kp = mn Cox Wn = mCox p = 40mA/V2 Ln Lp and their threshold voltages ae VTHn = VTHp = 1 V the current I is

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initially uncharged. At t = 0 the switch S is closed. The Vc across the capacitor at t = 1 millisecond is In the figure shown above, the OP-AMP is supplied with ! 15V .

(A) 0 A (C) 45 mA 3.48

(B) 25 mA (D) 90 mA

For the Zener diode shown in the figure, the Zener voltage at knee is 7 V, the knee current is negligible and the Zener dynamic resistance is 10 W. If the input voltage (Vi) range is from 10 to 16 V, the output voltage (V0) ranges from

(A) 0 Volt (C) 9.45 Volts

For the circuit shown below, assume that the zener diode is ideal with a breakdown voltage of 6 volts. The waveform observed across R is

3.54

(A) 7.00 to 7.29 V (C) 7.14 to 7.43 V

(B) 6.3 Volt (D) 10 Volts

(B) 7.14 to 7.29 V (D) 7.29 to 7.43 V

Statement for Linked Answer Questions 4.35 & 4.36: Consider the Op-Amp circuit shown in the figure.

3.49

3.50

The transfer function V0 (s)/ Vi (s) is (A) 1 - sRC (B) 1 + sRC 1 + sRC 1 - sRC 1 1 (C) (D) 1 - sRC 1 + sRC

3.52

no w.

co ia.

d

ww

Common Data For Q. 4.41, 4.42 and 4.43 : In the transistor amplifier circuit shown in the figure below, the transistor has the following parameters: bDC = 60 , VBE = 0.7V, hie " 3 The capacitance CC can be assumed to be infinite. In the figure above, the ground has been shown by the symbol 4

ONE MARK

The input impedance (Zi) and the output impedance (Z0) of an ideal trans-conductance (voltage controlled current source) amplifier are (A) Zi = 0, Z0 = 0 (B) Zi = 0, Z0 = 3 (C) Zi = 3, Z0 = 0 (D) Zi = 3, Z0 = 3 An n-channel depletion MOSFET has following two points on its ID - VGs curve: (i) VGS = 0 at ID = 12 mA and (ii) VGS =- 6 Volts at ID = 0 mA Which of the following Q point will given the highest trans conductance gain for small signals? (A) VGS =- 6 Volts (B) VGS =- 3 Volts (C) VGS = 0 Volts (D) VGS = 3 Volts 2006

3.53

O N

.in

If Vi = V1 sin (wt) and V0 = V2 sin (wt + f), then the minimum and maximum values of f (in radians) are respectively (A) - p and p (B) 0 and p 2 2 2 (C) - p and 0 (D) - p and 0 2 2006

3.51

A I D

TWO MARKS

For the circuit shown in the following figure, the capacitor C is

3.55

3.56

Under the DC conditions, the collector-or-emitter voltage drop is (A) 4.8 Volts (B) 5.3 Volts (C) 6.0 Volts (D) 6.6 Volts If bDC is increased by 10%, the collector-to-emitter voltage drop (A) increases by less than or equal to 10% (B) decreases by less than or equal to 10% (C) increase by more than 10%

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(D) decreases by more than 10% 3.57

(B) increases the common mode gain only. (C) decreases the differential mode gain only. (D) decreases the common mode gain only.

The small-signal gain of the amplifier vc is vs (A) -10 (B) -5.3 (C) 5.3 (D) 10

3.64

Common Data For Q. 4.44 & 4.45:

For an npn transistor connected as shown in figure VBE = 0.7 volts. Given that reverse saturation current of the junction at room temperature 300 K is 10 - 13 A, the emitter current is

A regulated power supply, shown in figure below, has an unregulated input (UR) of 15 Volts and generates a regulated output Vout . Use the component values shown in the figure. (A) 30 mA (C) 49 mA 3.65

3.58

3.59

3.61

3.62

A I D

If the unregulated voltage increases by 20%, the power dissipation across the transistor Q1 (A) increases by 20% (B) increases by 50% (C) remains unchanged (D) decreases by 20%

O N ONE MARK

The input resistance Ri of the amplifier shown in the figure is

(A) 30 kW 4

(B) 10 kW

(C) 40 kW

(D) infinite

(A) Bias current of the inverting input only (B) Bias current of the inverting and non-inverting inputs only (C) Input offset current only (D) in Both the bias currents and the input offset current

co. . a di The Op-amp circuit shown in the figure is filter. The type of filter

3.66

.no

w ww

and its cut. Off frequency are respectively

(A) high pass, 1000 rad/sec. (C) high pass, 1000 rad/sec

The effect of current shunt feedback in an amplifier is to (A) increase the input resistance and decrease the output resistance (B) increases both input and output resistance (C) decrease both input and output resistance (D) decrease the input resistance and increase the output resistance

3.67

(B) Low pass, 1000 rad/sec (D) low pass, 10000 rad/sec

The circuit using a BJT with b = 50 and VBE = 0.7V is shown in the figure. The base current IB and collector voltage by VC and respectively

The cascade amplifier is a multistage configuration of (A) CC - CB (B) CE - CB (C) CB - CC (D) CE - CC 2005

3.63

The voltage e0 is indicated in the figure has been measured by an ideal voltmeter. Which of the following can be calculated ?

The power dissipation across the transistor Q1 shown in the figure is (A) 4.8 Watts (B) 5.0 Watts (C) 5.4 Watts (D) 6.0 Watts

2005 3.60

(B) 39 mA (D) 20 mA

(A) 43 mA and 11.4 Volts (C) 45 mA and 11 Volts

TWO MARKS

In an ideal differential amplifier shown in the figure, a large value of (RE ). (A) increase both the differential and common - mode gains.

3.68

(B) 40 mA and 16 Volts (D) 50 mA and 10 Volts

The Zener diode in the regulator circuit shown in the figure has a Zener voltage of 5.8 volts and a zener knee current of 0.5 mA. The maximum load current drawn from this current ensuring proper

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Page 53

functioning over the input voltage range between 20 and 30 volts, is

(A) 23.7 mA (C) 13.7 mA 3.69

(B) 14.2 mA (D) 24.2 mA

Both transistors T1 and T2 show in the figure, have a b = 100 , threshold voltage of 1 Volts. The device parameters K1 and K2 of T1 and T2 are, respectively, 36 mA/V2 and 9 mA/V 2 . The output voltage Vo i s

(A) 1 V (C) 3 V

(B) 2 V (D) 4 V

Common Data For Q. 4.58, 4.59 and 4.60 :

A I D 2004

Given, rd = 20kW , IDSS = 10 mA, Vp =- 8 V

ONE MARK

An ideal op-amp is an ideal (A) voltage controlled current source (B) nvoltage controlled voltage source o.i current controlled current source c (C) . dia (D) current controlled voltage source o .n 3.74

O N w

ww

3.70

3.71

3.72

3.73

Zi and Z0 of the circuit are respectively (A) 2 MW and 2 kW (B) 2 MW and 20 kW 11 (C) infinity and 2 MW (D) infinity and 20 kW 11 ID and VDS under DC conditions are respectively (A) 5.625 mA and 8.75 V (B) 1.875 mA and 5.00 V (C) 4.500 mA and 11.00 V (D) 6.250 mA and 7.50 V

3.75

3.76

Transconductance in milli-Siemens (mS) and voltage gain of the amplifier are respectively (A) 1.875 mS and 3.41 (B) 1.875 ms and -3.41 (C) 3.3 mS and -6 (D) 3.3 mS and 6 Given the ideal operational amplifier circuit shown in the figure indicate the correct transfer characteristics assuming ideal diodes with zero cut-in voltage.

Voltage series feedback (also called series-shunt feedback) results in (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 (D) decrease in input impedance and increase in output impedance The circuit in the figure is a

(A) low-pass filter (C) band-pass filter

(B) high-pass filter (D) band-reject filter

2004 3.77

TWO MARKS

A bipolar transistor is operating in the active region with a collector current of 1 mA. Assuming that the b of the transistor is 100 and the thermal voltage (VT ) is 25 mV, the transconductance (gm) and the input resistance (rp) of the transistor in the common emitter configuration, are (A) gm = 25 mA/V and rp = 15.625 kW (B) gm = 40 mA/V and rp = 4.0 kW (C) gm = 25 mA/V and rp = 2.5 k W

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(D) gm = 40 mA/V and rp = 2.5 kW 3.78

The value of C required for sinusoidal oscillations of frequency 1 kHz in the circuit of the figure is

(A) IC = 1 mA, VCE = 4.7 V (C) IC = 1 mA, VCE = 2.5 V

(A) 1 mF 2p 1 mF (C) 2p 6 3.79

2003

(B) 2p mF

3.83

(D) 2p 6 mF

In the op-amp circuit given in the figure, the load current iL is

3.80

3.81

3.82

(B) Vs R2 (D) Vs R1

O N

The circuit shown in the figure is best described as a

.in

co ia.

od

In the voltage regulator shown in the figure, the load current can.n ww vary from 100 mA to 500 mA. Assuming that the Zener diode iswideal (i.e., the Zener knee current is negligibly small and Zener resistance 3.85 is zero in the breakdown region), the value of R is

(A) 7 W (B) 70 W (C) 70 W (D) 14 W 3 In a full-wave rectifier using two ideal diodes, Vdc and Vm are the dc and peak values of the voltage respectively across a resistive load. If PIV is the peak inverse voltage of the diode, then the appropriate relationships for this rectifier are (A) Vdc = Vm , PIV = 2Vm (B) Idc = 2 Vm , PIV = 2Vm p p (C) Vdc = 2 Vm , PIV = Vm (D) Vdc Vm , PIV = Vm p p Assume that the b of transistor is extremely large and VBE = 0.7V, IC and VCE in the circuit shown in the figure

ONE MARK

Choose the correct match for input resistance of various amplifier configurations shown below : Configuration Input resistance CB : Common Base LO : Low CC : Common Collector MO : Moderate CE : Common Emitter HI : High (A) CB - LO, CC - MO, CE - HI (B) CB - LO, CC - HI, CE - MO (C) CB - MO, CC - HI, CE - LO (D) CB - HI, CC - LO, CE - MO

A I D 3.84

(A) - Vs R2 (C) - Vs RL

(B) IC = 0.5 mA, VCE = 3.75 V (D) IC = 0.5 mA, VCE = 3.9 V

(A) bridge rectifier (C) frequency discriminator

If the input to the ideal comparators shown in the figure is a sinusoidal signal of 8 V (peak to peak) without any DC component, then the output of the comparators has a duty cycle of

(A) 1/2 (C) 1/6 3.86

3.87

(B) ring modulator (D) voltage double

(B) 1/3 (D) 1/2

If the differential voltage gain and the common mode voltage gain of a differential amplifier are 48 dB and 2 dB respectively, then common mode rejection ratio is (A) 23 dB (B) 25 dB (C) 46 dB (D) 50 dB Generally, the gain of a transistor amplifier falls at high frequencies due to the (A) internal capacitances of the device (B) coupling capacitor at the input (C) skin effect (D) coupling capacitor at the output 2003

3.88

TWO MARKS

An amplifier without feedback has a voltage gain of 50, input resistance of 1 k W and output resistance of 2.5 kW. The input

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input resistance of 1 kW and output resistance of 250 W are cascaded. The opened circuit voltages gain of the combined amplifier is (A) 49 dB (B) 51 dB (C) 98 dB (D) 102 dB

resistance of the current-shunt negative feedback amplifier using the above amplifier with a feedback factor of 0.2, is (B) 1 kW (A) 1 kW 11 5 (C) 5 kW 3.89

(D) 11 kW 3.94

In the amplifier circuit shown in the figure, the values of R1 and R2 are such that the transistor is operating at VCE = 3 V and IC = 1.5 mA when its b is 150. For a transistor with b of 200, the operating point (VCE , IC ) is

An ideal sawtooth voltages waveform of frequency of 500 Hz and amplitude 3 V is generated by charging a capacitor of 2 mF in every cycle. The charging requires (A) Constant voltage source of 3 V for 1 ms (B) Constant voltage source of 3 V for 2 ms (C) Constant voltage source of 1 mA for 1 ms (D) Constant voltage source of 3 mA for 2 ms 2002

3.95

(A) (2 V, 2 mA) (C) (4 V, 2 mA) 3.90

(B) (3 V, 2 mA) (D) (4 V, 1 mA)

The oscillator circuit shown in the figure has an ideal inverting amplifier. Its frequency of oscillation (in Hz) is

ONE MARK

In a negative feedback amplifier using voltage-series (i.e. voltagesampling, series mixing) feedback. (A) Ri decreases and R0 decreases (B) Ri decreases and R0 increases (C) Ri increases and R0 decreases (D) Ri increases and R0 increases (Ri and R0 denote the input and output resistance respectively)

A 741-type opamp has a gain-bandwidth product of 1 MHz. A noninverting amplifier suing this opamp and having a voltage gain of 20 dB will exhibit a -3 dB bandwidth of (A) 50 kHz (B) 100 kHz (C) 1000 kHz (D) 1000 kHz 17 7.07 3.97 Three in identical RC-coupled transistor amplifiers are cascaded. If . o c of the amplifiers has a frequency response as shown in the ia. each d figure, the overall frequency response is as given in o 3.96

1 (2p 6 RC) 1 (C) ( 6 RC) (A)

3.91

(D)

6 (2pRC)

O N

n

. ww

w

(B) 6 V (D) 12 V

If the op-amp in the figure is ideal, the output voltage Vout will be equal to

(A) 1 V (C) 14 V 3.93

1 (2pRC)

The output voltage of the regulated power supply shown in the figure is

(A) 3 V (C) 9 V 3.92

(B)

A I D

(B) 6 V (D) 17 V

2002

Three identical amplifiers with each one having a voltage gain of 50,

3.98

TWO MARKS

The circuit in the figure employs positive feedback and is

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intended to generate sinusoidal oscillation. If at a frequency V (f) 1 f0, B (f) = 3 f = +0c, then to sustain oscillation at this frequency V0 (f) 6

(A) R2 = 5R1 (C) R2 = R1 6 3.99

3.100

3.104

3.105

3.106

A zener diode regulator in the figure is to be designed to meet the specifications: IL = 10 mA V0 = 10 V and Vin varies from 30 V to 50 V. The zener diode has Vz = 10 V and Izk (knee current) =1 mA. For satisfactory operation

O N

(B) 2000W # R # 2200W (D) R $ 4000W

3.102

(A) gm r0 (C) gm rp 3.103

The transistor shunt regulator shown in the figure has a regulated output voltage of 10 V, when the input varies from 20 V to 30 V. The relevant parameters for the zener diode and the transistor are : Vz = 9.5 , VBE = 0.3 V, b = 99 , Neglect the current through RB . Then the maximum power dissipated in the zener diode (Pz ) and the transistor (PT ) are

.in

no w.

d

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(B) -16 (D) -6

(A) (B) (C) (D)

Pz = 75 mW, PT = 7.9 W Pz = 85 mW, PT = 8.9 W Pz = 95 mW, PT = 9.9 W Pz = 115 mW, PT = 11.9 W

The oscillator circuit shown in the figure is

4 (A) Hartely oscillator with foscillation = 79.6 MHz (B) Colpitts oscillator with foscillation = 50.3 MHz (C) Hartley oscillator with foscillation = 159.2 MHz (D) Colpitts oscillator with foscillation = 159.3 MHz

ONE MARK

The current gain of a BJT is

An npn BJT has gm = 38 mA/V, C m = 10-14 F, C p = 4 # 10-13 F, and DC current gain b0 = 90 . For this transistor fT and fb are (A) fT = 1.64 # 108 Hz and fb = 1.47 # 1010 Hz (B) fT = 1.47 # 1010 Hz and fb = 1.64 # 108 Hz (C) fT = 1.33 # 1012 Hz and fb = 1.47 # 1010 Hz (D) fT = 1.47 # 1010 Hz and fb = 1.33 # 1012 Hz

co ia.

3.107

2001

TWO MARKS

A I D

The voltage gain Av = v0 of the JFET amplifier shown in the figure vt is IDSS = 10 mA Vp =- 5 V(Assume C1, C2 and Cs to be very large

(A) +16 (C) +8

Consider the following two statements : Statement 1 : A stable multi vibrator can be used for generating square wave. Statement 2: Bistable multi vibrator can be used for storing binary information. (A) Only statement 1 is correct (B) Only statement 2 is correct (C) Both the statements 1 and 2 are correct (D) Both the statements 1 and 2 are incorrect 2001

An amplifier using an opamp with a slew-rate SR = 1 V/m sec has a gain of 40 dB. If this amplifier has to faithfully amplify sinusoidal signals from dc to 20 kHz without introducing any slew-rate induced distortion, then the input signal level must not exceed. (A) 795 mV (B) 395 mV (C) 79.5 mV (D) 39.5 mV

(A) R # 1800W (C) 3700W # R # 4000W 3.101

(B) R2 = 6R1 (D) R2 = R1 5

Page 56

gm r g (D) m rp

(B)

3.108

The inverting OP-AMP shown in the figure has an open-loop gain of 100.

Thee ideal OP-AMP has the following characteristics. (A) Ri = 3, A = 3, R0 = 0 (B) Ri = 0, A = 3, R0 = 0 (C) Ri = 3, A = 3, R0 = 3 (D) Ri = 0, A = 3, R0 = 3

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(C) parasitic inductive elements (D) the Early effect If the op-amp in the figure, is ideal, then v0 is

3.114

The closed-loop gain V0 is Vs (A) - 8 (C) - 10 3.109

(B) - 9 (D) - 11 (A) zero (C) - (V1 + V2) sin wt

In the figure assume the OP-AMPs to be ideal. The output v0 of the circuit is

The configuration of the figure is a

3.115

(A) 10 cos (100t) (C) 10 - 4

(B) 10

t

#0 cos (100t) dt

t

#0 cos (100t) dt

(D) 10 - 4 d cos (100t) dt

2000 3.110

(B) (V1 - V2) sin wt (D) (V1 + V2) sin wt

A I D

ONE MARK

In the differential amplifier of the figure, if the source resistance of the current source IEE is infinite, then the common-mode gain is

O N

(A) precision integrator (B) Hartely oscillator (C) Butterworth high pass filter (D) Wien-bridge oscillator

Assume that the op-amp of the figure is ideal. If vi is a triangular wave, then v0 will be

3.116

.in

no w.

co ia.

d

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(A) square wave (C) parabolic wave 3.117

(A) zero

(B) infinite (D) Vin1 + Vin2 2VT

(C) indeterminate 3.111

In the circuit of the figure, V0 is

(B) triangular wave (D) sine wave

The most commonly used amplifier is sample and hold circuits is (A) a unity gain inverting amplifier (B) a unity gain non-inverting amplifier (C) an inverting amplifier with a gain of 10 (D) an inverting amplifier with a gain of 100 2000

3.118

(A) -1 V (C) +1 V 3.112

3.113

(B) 2 V (D) +15 V

TWO MARKS

In the circuit of figure, assume that the transistor is in the active region. It has a large b and its base-emitter voltage is 0.7 V. The value of Ic is

Introducing a resistor in the emitter of a common amplifier stabilizes the dc operating point against variations in (A) only the temperature (B) only the b of the transistor (C) both temperature and b (D) none of the above The current gain of a bipolar transistor drops at high frequencies because of (A) transistor capacitances (B) high current effects in the base

(A) Indeterminate since Rc is not given (B) 1 mA

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(C) 5 mA 3.119

Page 58

(D) 10 mA

type

If the op-amp in the figure has an input offset voltage of 5 mV and an open-loop voltage gain of 10000, then v0 will be

(A) current series (C) voltage series (A) 0 V (C) + 15 V or -15 V

(B) 5 mV (D) +50 V or -50 V

1999 3.120

3.121

3.122

3.127

ONE MARK

The first dominant pole encountered in the frequency response of a compensated op-amp is approximately at (A) 5 Hz (B) 10 kHz (C) 1 MHz (D) 100 MHz Negative feedback in an amplifier (A) reduces gain (B) increases frequency and phase distortions (C) reduces bandwidth (D) increases noise

3.128

3.129

(B) current shunt (D) voltage shunt

In a differential amplifier, CMRR can be improved by using an increased (A) emitter resistance (B) collector resistance (C) power supply voltages (D) source resistance From a measurement of the rise time of the output pulse of an amplifier whose is a small amplitude square wave, one can estimate the following parameter of the amplifier (A) gain-bandwidth product (B) slow rate (C) upper 3–dB frequency (D) lower 3–dB frequency The emitter coupled pair of BJT’s given a linear transfer relation between the differential output voltage and the differential output voltage and the differential input voltage Vid is less a times the thermal voltage, where a is (A) 4 (B) 3 (C) 2 (D) 1

A I D

In the cascade amplifier shown in the given figure, if the commonemitter stage (Q1) has a transconductance gm1 , and the common base stage (Q2) has a transconductance gm2 , then the overall transconductance g (= i 0 /vi) of the cascade amplifier is

In a shunt-shunt negative feedback amplifier, as compared to the basic amplifier (A) both, input and output impedances,decrease .in input impedance decreases but output impedance increases o(B) c . dia (C) input impedance increase but output o .n w (D) both input and output impedances increases. w 3.130

O N w

(A) gm1 g (C) m1 2 3.123

(B) gm2 g (D) m2 2

1998 3.131

Crossover distortion behavior is characteristic of (A) Class A output stage (B) Class B output stage (C) Class AB output stage (D) Common-base output stage

A multistage amplifier has a low-pass response with three real poles at s =- w1 - w2 and w3 . The approximate overall bandwidth B of the amplifier will be given by (B) 1 = 1 + 1 + 1 (A) B = w1 + w2 + w3 w1 w2 w3 B (C) B = (w1 + w2 + w3) 1/3

1999 3.124

3.125

An amplifier has an open-loop gain of 100, an input impedance of 1 kW,and an output impedance of 100 W. A feedback network with a feedback factor of 0.99 is connected to the amplifier in a voltage series feedback mode. The new input and output impedances, respectively, are (A) 10 W and 1W (B) 10 W and 10 kW (C) 100 kW and 1 W (D) 100 kW and 1 kW A dc power supply has a no-load voltage of 30 V, and a full-load voltage of 25 V at a full-load current of 1 A. Its output resistance and load regulation, respectively, are (A) 5 W and 20% (B) 25 W and 20% (C) 5 W and 16.7% (D) 25 W and 16.7% 1998

3.126

TWO MARK

ONE MARK

The circuit of the figure is an example of feedback of the following

3.132

3.133

3.134

TWO MARKS

One input terminal of high gain ground and a sinusoidal voltage output of comparator will be (A) a sinusoid (C) a half rectified sinusoid

(D) B =

w12 + w22 + w23

comparator circuit is connected to is applied to the other input. The (B) a full rectified sinusoid (D) a square wave

In a series regulated power supply circuit, the voltage gain Av of the ‘pass’ transistor satisfies the condition (A) Av " 3 (B) 1 1 Thus for Vi > 1 diode is off and V = 1V Option (B) and (C) doesn’t satisfy this condition. Let Vi < 1. In this case diode will be on and voltage across diode will be zero and V = Vi Thus V = min (Vi, 1)

Input resistance seen by source vs R in = vs = Rs + Rs || rs is 3.29

= (1000 W) + (93 kW || 259 W) = 1258 W 3.26

Option (B) is correct. Cut-off frequency due to C2 fo =

3.30

1 2p (RC + RL) C2

fo 1 = 271 Hz 2 # 3.14 # 1250 # 4.7 # 10-6 Lower cut-off frequency f fL . o = 271 = 27.1 Hz 10 10

3.31

=

3.27

3.32

Option (A) is correct. The R2 decide only the frequency. Option (D) is correct. For small increase in VG beyond 1 V the n - channel MOSFET goes into saturation as VGS "+ ive and p - MOSFET is always in active region or triode region. Option (C) is correct. Option (D) is correct. The circuit is shown in fig below

Option (B) is correct. The circuit is as shown below

A I D

O N

I = 20 - 0 + Vi - 0 = 5 + Vi 4R R R If I > 0, diode D2 conducts So, for 5 + VI > 0 & VI > - 5, D2 conducts 2 Equivalent circuit is shown below Current

no w.

The voltage at non inverting terminal is 5 V because OP AMP is ideal and inverting terminal is at 5 V. .in IC = 10 - 5 = 1 mA oThus c . 5k a i

d

ww

3.33

VE = IE RE = 1m # 1.4k = 1.4V IE = IC = 0.6 + 1.4 = 2V Thus the feedback is negative and output voltage is V = 2V . Option (D) is correct. The output voltage is V0 = Ar Vi .-

hfe RC Vi hie

Here RC = 3 W and hie = 3 kW Thus Output is Vo = 0 . If I < 0 , diode D2 will be off 5 + VI < 0 & V < - 5, D is off I 2 R The circuit is shown below

.- 150 (A cos 20t + B sin 106 t) Since coupling capacitor is large so low frequency signal will be filtered out, and best approximation is V0 .- 150B sin 106 t 3.34

0 - Vi + 0 - 20 + 0 - Vo = 0 R 4R R

3.28

or

Vo =- Vi - 5

At Vi =- 5 V, At Vi =- 10 V,

Vo = 0 Vo = 5 V

Option (A) is correct. Let diode be OFF. In this case 1 A current will flow in resistor and

V0 . - 150 # 3k Vi 3k

3.35

Option (C) is correct. For the positive half of Vi , the diode D1 is forward bias, D2 is reverse bias and the zener diode is in breakdown state because Vi > 6.8 . Thus output voltage is V0 = 0.7 + 6.8 = 7.5 V For the negative half of Vi, D2 is forward bias thus Then V0 =- 0.7 V Option (B) is correct. By Current mirror,

^ L h2 Ibias W ^ L h1 W

Ix = Since MOSFETs are identical,

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Thus Hence 3.36

Page 67

W W b L l =b L l 2 2

KCL at non inverting terminal side we have 15 - V1 + V0 - V1 = V1 - (- 15) 10 10 10 or V1 = V0 3

Ix = Ibias

Option (B) is correct. The circuit is using ideal OPAMP. The non inverting terminal of OPAMP is at ground, thus inverting terminal is also at virtual ground.

If V0 swings from -15 to +15 V then V1 swings between -5 V to +5 V. Option (A) is correct. For the given DC values the Thevenin equivalent circuit is as follows

3.40

Thus current will flow from -ive terminal (0 Volt) to -1 Volt source. Thus the current I is 0 - (- 1) I = = 1 100k 100k

The Thevenin resistance and voltage are VTH = 10 # 9 = 3 V 10 + 20 and total RTH = 10k # 20k = 6.67 kW 10k + 20k

The current through diode is I = I 0 _eV - 1i Now VT = 25 mV and I0 = 1 mA V

t

Thus or Now V 3.37

V I = 10-6 8e 25 # 10 - 1B = 1 5 10 V = 0.06 V V0 = I # 4k + V = 1 # 4k + 0.06 = 0.1 100k

Since b is very large, therefore IB is small and can be ignored Thus IE = VTH - VBE = 3 - 0.7 = 1 mA RE 2.3k

-3

Option (B) is correct. The circuit is using ideal OPAMP. The non inverting terminal of OPAMP is at ground, thus inverting terminal is also at virtual ground.

O N

A I D

Option (D) is correct. The small signal model is shown in fig below

3.41

.in

no w.

co ia.

d

ww

gm =

IC = 1m = 1 A/V VT 25m 25

Vo =- gm Vp # (3k 3k ) =- 1 Vin (1.5k) 25 or

Thus we can write vi = R1 + sL or

-v

3.42

R2 sR2 C2 + 1

IC . IE

Vp = Vin

=- 60Vin Am = Vo =- 60 Vin

Option (C) is correct. The circuit shown in (C) is correct full wave rectifier circuit.

v0 =R2 vi (R1 + sL)( sR2 C2 + 1)

and from this equation it may be easily seen that this is the standard form of T.F. of low pass filter K H (s) = (R1 + sL)( sR2 C2 + 1) and form this equation it may be easily seen that this is the standard form of T.F. of low pass filter H (s) = 2 K as + bs + b 3.38

3.39

Option ( ) is correct. The current in both transistor are equal. Thus gm is decide by M1. Hence (C) is correct option.

3.43

3.44

Option (A) is correct. In the transconductance amplifier it is desirable to have large input resistance and large output resistance. Option (C) is correct. We redraw the circuit as shown in fig.

Option (C) is correct. Let the voltage at non inverting terminal be V1, then after applying

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Page 68

We have VZ = 7 volt, VK = 0, RZ = 10W Circuit can be modeled as shown in fig below

Since Vi is lies between 10 to 16 V, the range of voltage across 200 kW V200 = Vi - VZ = 3 to 9 volt The range of current through 200 kW is 3 = 15 mA to 9 = 45 mA 200k 200k

Applying voltage division rule v+ = 0.5 V v+ = vv- = 0.5 V i = 1 - 0.5 = 0.5 mA 1k i = 0.5 - v0 = 0.5 mA 2k

We know that Thus Now and

v0 = 0.5 - 1 =- 0.5 V

or 3.45

The range of variation in output voltage 15m # RZ = 0.15 V to 45m # RZ = 0.45 Thus the range of output voltage is 7.15 Volt to 7.45 Volt

Option (B) is correct. If we assume b very large, then IB = 0 and IE = IC ; VBE = 0.7 V. We assume that BJT is in active, so applying KVL in Base-emitter loop IE = 2 - VBE = 2 - 0.7 = 1.3 mA 1k RE Since b is very large, we have IE = IC , thus IC = 1.3 mA Now applying KVL in collector-emitter loop 10 - 10IC - VCE - IC = 0 or VCE =- 4.3 V Now VBC = VBE - VCE = 0.7 - (- 4.3) = 5 V Since VBC > 0.7 V, thus transistor in saturation.

3.46

V+ =

or or

O N

di

or 3.47

3.51

3.48

3.52 3.53

Vo1 - Vo2 = VT 1n 4 - VT 1n 2 Is R Is R Vo1 - Vo2 = VT 1n 4 = VT 1n2 2

Option (C) is correct.

V0 = H (s) = 1 - sRC Vi 1 + sRC 1 - jwRC H (jw) = 1 + jwRC

fmin fmax

=- 2 tan - 2 wRC = - p (at w " 3) = 0( at w = 0)

Option (D) is correct. In the transconductance amplifier it is desirable to have large input impedance and large output impedance. Option (C) is correct. Option (D) is correct. The voltage at inverting terminal is V- = V+ = 10 V Here note that current through the capacitor is constant and that is I = V- = 10 = 10 mA 1k 1k

Option (D) is correct. We have Vthp = Vthp = 1 V WP W and = N = 40mA/V2 LP LN From figure it may be easily seen that Vas for each NMOS and PMOS is 2.5 V mA Thus ID = K (Vas - VT ) 2 = 40 2 (2.5 - 1) 2 = 90 m A V

1 V 1 + sCR i

Option (C) is correct.

Minimum value, Maximum value,

For the first condition

Subtracting above equation

1 V 1 + sCR i

+H (jw) = f =- tan - 1 wRC - tan - 1 wRC

For the first condition

VD = 0 - Vo1 = VT 1n 4 Is R

Vi =

Applying voltage division rule (V + Vi) V+ = R1 (V0 + Vi) = o R1 + R1 2 (Vo + Vi) 1 or V = 1 + sCR i 2 Vo =- 1 + 2 or Vi 1 + sRC V0 = 1 - sRC n o.i c Vi 1 + sRC . a

T

VD = 0 - Vo1 = VT 1n 2 Is R

R+

1 sC

A I D

Vi = I eV /V s R VD = VT 1n Vi Is R D

1 sC

V- = V+ =

Now

Option (D) is correct. .no3.50 w Here the inverting terminal is at virtual ground and the current ww in resistor and diode current is equal i.e. IR = ID

Option (A) is correct. The voltage at non-inverting terminal is

3.49

Thus the voltage across capacitor at t = 1 msec is 1m 1m VC = 1 Idt = 1 10mdt C 0 1m 0 Im = 10 4 dt = 10 V

#

#

#0

3.54

Option (A) is correct. In forward bias Zener diode works as normal diode.

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Page 69

Thus for negative cycle of input Zener diode is forward biased and it conducts giving VR = Vin . For positive cycle of input Zener diode is reversed biased when 0 < Vin < 6 , Diode is OFF and VR = 0 when Vin > 6 Diode conducts and voltage across diode is 6 V. Thus voltage across is resistor is VR = Vin - 6 Only option (B) satisfy this condition.

and IC remain same. There will be change in VCE Thus, VCE - 18 - 9 = 9 V IC = 0.9 A Power dissipation P = VCE IC = 9 # 0.9 = 8.1 W Thus % increase in power is 8.1 - 5.4 # 100 = 50% 5.4

3.55

Option (B) is correct. Since the inverting terminal is at virtual ground, the current flowing through the voltage source is Is = Vs 10k Vs = 10 kW = R or in Is

3.60

Option (C) is correct. The circuit under DC condition is shown in fig below

Option (D) is correct. The effect of current shunt feedback in an amplifier is to decrease the input resistance and increase the output resistance as : Rif = Ri 1 + Ab

3.61

Applying KVL we have

3.56

VCC - RC (IC + IB) - VCE = 0 and VCC - RB IB - VBE = 0 Substituting IC = bIB in (1) we have

...(1) ...(2)

VCC - RC (bIB + IB) - VCE = 0 Solving (2) and (3) we get VCE = VCC - VCC - VBE RB 1+ RC (1 + b) Now substituting values we get 12 - 0.7 VCE = 12 = 5.95 V 53 1+ 1 + (1 + 60)

...(3)

Option (B) is correct.

ww

b' = 110 # 60 = 66 100

We have

where 3.62

O N

3.57 3.58

.c

ia od

3.63

n w.

= 5.29 - 59.5 # 100 =- 4.3% 5.95

Option (B) is correct. The CE configuration has high voltage gain as well as high current gain. It performs basic function of amplifications. The CB configuration has lowest Ri and highest Ro . It is used as last step to match a very low impedance source and to drain a high impedance load n Thus o.i cascade amplifier is a multistage configuration of CE-CB

A I D

...(4)

Option (D) is correct. Common mode gain ACM =- RC 2RE And differential mode gain

Substituting b' = 66 with other values in (iv) in previous solutions 12 - 0.7 VCE = 12 = 5.29 V 53 1+ 1 + (1 + 66) Thus change is

Rof = R0 (1 + Ab) Ri " Input resistance without feedback Rif " Input resistance with feedback.

ADM =- gm RC Thus only common mode gain depends on RE and for large value of RE it decreases. 3.64

Option (C) is correct.

IE = Is `e nV - 1j VBE

Option (A) is correct.

T

= 10

Option (C) is correct. The Zener diode is in breakdown region, thus V+ = VZ = 6 V = Vin R We know that Vo = Vin c1 + f m R1 or Vout = Vo = 6`1 + 12k j = 9 V 24k

3.65

- 13

0.7 c e1 # 26 # 10 - 1m = 49 mA -3

Option (C) is correct. The circuit is as shown below

The current in 12 kW branch is negligible as comparison to 10 W. Thus Current IC . IE . = Vout = 9 = 0.9 A RL 10 Now VCE = 15 - 9 = 6 V The power dissipated in transistor is P = VCE IC = 6 # 0.9 = 5.4 W 3.59

Option (B) is correct. If the unregulated voltage increase by 20%, them the unregulated voltage is 18 V, but the VZ = Vin = 6 remain same and hence Vout

Writing equation for I- have e 0 - V- = I 1M or e0 = I- (1M) + VWriting equation for I+ we have

...(1)

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Page 70

0 - V+ = I+ 1M

From the figure we have

or V+ = - I+ (1M) Since for ideal OPAMP V+ = V- , from (1) and (2) we have

Zin = 2 MW

...(2)

e0 = I- (1M) - I + (1M) = (I- - I+) (1M) = IOS (1M) Thus if e0 has been measured, we can calculate input offset current IOS only. 3.66

3.67

Z0 = rd RD = 20k 2k = 20 kW 11

and

Option (A) is correct. The circuit in DC condition is shown below

3.71

Option (C) is correct. At low frequency capacitor is open circuit and voltage acr s noninverting terminal is zero. At high frequency capacitor act as short circuit and all input voltage appear at non-inverting terminal. Thus, this is high pass circuit. The frequency is given by 1 = 1000 w = 1 = 3 RC 1 # 10 # 1 # 10 - 6 rad/sec

Since the FET has high input resistance, gate current can be neglect and we get VGS =- 2 V Since VP < VGS < 0 , FET is operating in active region 2 (- 2) 2 Now ID = IDSS c1 - VGS m = 10 c1 (- 8) m VP = 5.625 mA

Option (B) is correct. The circuit under DC condition is shown in fig below

Now

VDS = VDD - ID RD = 20 - 5.625 m # 2 k

= 8.75 V Option (B) is correct. The transconductance is

A I D 3.72

O N

Applying KVL we have

VCC - RB IB - VBE - RE IE = 0 or VCC - RB IB - VBE - RE (b + 1) IB = 0 Since IE = IB + bIB or IB = VCC - VBE RB + (b + 1) RE 20 - 0.7 = = 40m A 430k + (50 + 1)1 k

Vmax = 30 V i.e. Vmax - VZ = I + I L Z 1k 30 - 5.8 = I = 0.5 m L 1k

3.70

3.73

So,

2 ID IDSS

= 2 5.625mA # 10mA = 1.875 mS 8 A =- gm (rd RD) = 1.875ms # 20 K =- 3.41 11

Option (B) is correct. Only one diode will be in ON conditions When lower diode is in ON condition, then Vu = 2k Vsat = 2 10 = 8 V 2.5k 2.5

3.74

3.75

Option (B) is correct. An ideal OPAMP is an ideal voltage controlled voltage source. Option (C) is correct. In voltage series feed back amplifier, input impedance increases by factor (1 + Ab) and output impedance decreases by the factor (1 + Ab). Rif = Ri (1 + Ab) Ro Rof = (1 + Ab)

Option (D) is correct. Option (B) is correct. The small signal model is as shown below

VP

when upper diode is in ON condition Vu = 2k Vsat = 2 (- 10) =- 5 V 2.5k 4

IL = 24.2 - 0.5 = 23.7 mA

or 3.69

w

Option (A) is correct. The maximum load current will be at maximum input voltage i.e.

or

n

. ww

n o.i

.c The gain is

ia od

IC = bIB = 50 # 40m = 2 mA VC = VCC - RC IC = 20 - 2m # 2k = 16 V

Now 3.68

or,

gm =

3.76

3.77

Option (A) is correct. This is a Low pass filter, because V0 = 0 At w = 3 Vin V0 = 1 and at w = 0 Vin Option (D) is correct. When IC >> ICO

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Page 71

IC = 1mA = 0.04 = 40 mA/V VT 25mV b rp = = 100 - 3 = 2.5 kW gm 40 # 10

gm =

3.78

3.79

Option (A) is correct. The given circuit is wein bridge oscillator. The frequency of oscillation is 2pf = 1 RC 1 or = 1 m C = 1 = 3 3 2pRf 2p 2p # 10 # 10

VT =

Since b is large is large, IC . IE , IB . 0 and IE = VT - VBE = 1 - 0.7 = 3 mA RE 300

Option (A) is correct. The circuit is as shown below

Option (B) is correct. For the different combinations the table is as follows

3.83

We know that for ideal OPAMP V- = V+ Applying KCL at inverting terminal V- - Vs + V- - V0 = 0 R1 R1

or 2V+ - Vo + IL R2 = 0 Since V- = V+ , from (1) and (2) we have Vs + IL R2 = 0

3.80

...(1)

O N ww

CE

CC

CB

Ai

High

High

Unity

Av

High

Unity

High

Ri

Medium

High

Low

Ro

Medium

Low

High

A I D

Option in (B) is correct. . o .cIf the input is sinusoidal signal of 8 V (peak to peak) then

3.85

ia

d .no

...(2)w

CE

Option (D) is correct. This circuit having two diode and capacitor pair in parallel, works as voltage doubler.

3.84

IL =- Vs R2

or

VCE = 5 - 2.2kIC - 300IE = 5 - 2.2k # 1m - 300 # 1m = 2.5 V

Now

or 2V- - Vo = Vs Applying KCL at non-inverting terminal V+ V - Vo =0 + IL + + R2 R2

R1 V = 1 #5 = 1 V R1 + R2 C 4+1

Vi = 4 sin wt The output of comparator will be high when input is higher than Vref = 2 V and will be low when input is lower than Vref = 2 V. Thus the waveform for input is shown below

Option (D) is correct. If IZ is negligible the load current is 12 - Vz = I L R as per given condition 100 mA # 12 - VZ # 500 mA R At IL = 100 mA 12 - 5 = 100 mA R

VZ = 5 V

R = 70W

or

At IL = 500 mA 12 - 5 = 500 mA R

From fig, first crossover is at wt1 and second crossover is at wt2 where

VZ = 5 V

4 sin wt1 = 2V

or R = 14 W Thus taking minimum we get

Thus

R = 14 W 3.81 3.82

Option (B) is correct. Option (C) is correct. The Thevenin equivalent is shown below 3.86

wt1 = sin - 1 1 = p 2 6 wt2 = p - p = 5p 6 6 5p p -6 Duty Cycle = 6 =1 2p 3

Thus the output of comparators has a duty cycle of 1 . 3 Option (C) is correct. CMMR = Ad Ac

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Page 72

V+ = V- = 8 V 3

20 log CMMR = 20 log Ad - 20 log Ac = 48 - 2 = 46 dB Where Ad "Differential Voltage Gain and AC " Common Mode Voltage Gain

or

3.87

Now applying KCL at inverting terminal we get V- - 2 + V- - Vo = 0 1 5

Option (B) is correct. The gain of amplifier is Ai =

- gm gb + jwC

3.93

Thus the gain of a transistor amplifier falls at high frequencies due to the internal capacitance that are diffusion capacitance and transition capacitance. 3.88

3.89

Option (C) is correct. The equivalent circuit of 3 cascade stage is as shown in fig.

Option (A) is correct. We have Ri = 1kW, b = 0.2, A = 50 Ri Thus, Rif = = 1 kW (1 + Ab) 11 Option (A) is correct. The DC equivalent circuit is shown as below. This is fixed bias circuit operating in active region. or or

A I D or

In first case

20 log AV = 20 log 8000 = 98 dB

#

#

Option (C) is correct. If we see th figure we find that the voltage at non-inverting terminal is 3 V by the zener diode and voltage at inverting terminal will be 3 V. Thus Vo can be get by applying voltage division rule, i.e. 20 V = 3 20 + 40 o or

3.92

V3 = 40 # 40V1 Vo = 50V3 = 50 # 40 # 40V1 AV = Vo = 50 # 40 # 40 = 8000 V1

3.94 Option (D) is correct. VCC - IC1 R2 - VCE1 = 0 If a constant current is made to flow in a capacitor, the output or 6 - 1.5mR2 - 3 = 0 voltage is integration of input current and that is sawtooth waveform n or R2 = 2kW oas.i below : c . t ia IB1 = IC1 = 1.5m = 0.01 mA d VC = 1 idt o b1 150 n C 0 . In second case IB2 will we equal to IB1 as there is no in R1. www The time period of wave form is Thus IC2 = b2 IB2 = 200 # 0.01 = 2 mA T = 1 = 1 = 2 m sec f 500 VCE2 = VCC - IC2 R2 = 6 - 2m # 2 kW = 2 V 20 # 10 1 Thus 3= idt Option (A) is correct. 2 # 106 0 The given circuit is a R - C phase shift oscillator and frequency of or i (2 # 10 - 3 - 0) = 6 # 10 - 6 its oscillation is or i = 3 mA 1 f = Thus the charging require 3 mA current source for 2 msec. 2p 6 RC

O N

3.91

1k 50V1 = 40V1 1k + 0.25k 1k 50V2 = 40V2 V3 = 1k + 0.25k V2 =

Similarly

3.90

Vo = 6V- - 10 = 6 # 8 - 10 = 6 V 3

or

V0 = 9 V

3.95

-3

Option (C) is correct. In voltage-amplifier or voltage-series amplifier, the Ri increase and Ro decrease because Rif = Ri (1 + Ab) Ro Rof = (1 + Ab)

3.96

Option (B) is correct. The circuit is as shown below

Option (B) is correct. Let x be the gain and it is 20 db, therefore 20 log x = 20 or x = 10 Since Gain band width product is 106 Hz, thus So, bandwidth is 6 6 BW = 10 = 10 = 105 Hz = 100 kHz 10 Gain

3.97

V+ =

8 (3) = 8 kW 1+8 3

Option (A) is correct. In multistage amplifier bandwidth decrease and overall gain increase. From bandwidth point of view only options (A) may be correct because lower cutoff frequency must be increases and higher must

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be decreases. From following calculation we have We have fL = 20 Hz and fH = 1 kHz For n stage amplifier the lower cutoff frequency is fL 20 f = = Ln

1 2n

Hz

-1

1 23

-1

Page 73 3.101

3.98

and Thus Now

1

2 2 - 1 = 0.5 kHz

Option (A) is correct. As per Barkhousen criterion for sustained oscillations Ab $ 1 and phase shift must be or 2pn . V (f) Now from circuit A= O = 1 + R2 Vf (f) R1 V (f) b (f) = 1 +0 = f 6 VO (f)

3.99

3.102

3.103

Slew Rate or

dVO = AV Vm w = AV Vm 2pf c dt m max Vm = SR AV V2pf

1 = -6 10 # 100 # 2p # 20 # 103 or VM = 79.5 mV 3.100

Option (A) is correct. The circuit is shown as below

I For satisfactory operations Vin - V0 R When Vin = 30 V, 30 - 10 R 20 or R or when Vin = 50 V

3.104

Ri " 3 R0 " 0 A"3

Option (C) is correct. Both statements are correct because (1) A stable multivibrator can be used for generating square wave, because of its characteristic (2) Bi-stable multivibrator can store binary information, and this multivibrator also give help in all digital kind of storing.

A I D

Option (B) is correct. If fT is the frequency at which the short circuit common emitter gain n attains o.i unity magnitude then c . gm 38 # 10 - 3 dia = fT = o 2p (Cm + Cp) 2p # (10 - 14 + 4 # 10 - 13) .n w or = 1.47 # 1010 Hz ww If fB is bandwidth then we have 10 f fB = T = 1.47 # 10 = 1.64 # 108 Hz b 90 3.105

O N

3.106

Option (C) is correct. If we neglect current through RB then it can be open circuit as shown in fig.

= IZ + IL > IZ + IL $ (10 + 1) mA

[IZ + IL = I]

Maximum power will dissipate in Zener diode when current through it is maximum and it will occur at Vin = 30 V I = Vin - Vo = 30 - 10 = 1 A 20 20

$ 11 mA

R # 1818 W or 50 - 10 $ (10 + 1) mA R 40 $ 11 # 10 - 3 R

or Thus R # 1818W

Option (A) is correct. The ideal op-amp has following characteristic :

and

Option (C) is correct. Let the gain of OPAMP be AV then we have 20 log AV = 40 dB or AV = 100 Let input be Vi = Vm sin wt then we have Now

Option (C) is correct. The current gain of a BJT is hfe = gm rp

R2 = 5R1

VO = VV Vi = Vm sin wt dVO = A V w cos wt V m dt

=- 2ms # 3k =- 6

So,

Thus from above equation for sustained oscillation 6 = 1 + R2 R1 or

= 10 mA and VP =- 5 V =0 = ID RS = 1 # 2.5W = 2.5 V = VG - VS = 0 - 2.5 =- 2.5 V gm = 2IDSS 81 - ` - 2.5 jB = 2 mS VP -5 AV = V0 =- gm RD Vi

IDSS VG VS VGS

Now

= 39.2 . 40

The higher cutoff frequency is fHn = fH

Option (D) is correct. We have

I IC + IZ = bIB + IZ = bIZ + IZ = (b + 1) IZ IZ = I = 1 = 0.01 A b+1 99 + 1

Since IC = bIB since IB = IZ

Power dissipated in zener diode is PZ = VZ IZ = 9.5 # 0.01 = 95 mW IC = bIZ = 99 # 0.1 = 0.99 A VCE = Vo = 10 V

R # 3636W

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Power dissipated in transistor is

Page 74 3.110

PT = VC IC = 10 # 0.99 = 9.9 W 3.107

3.108

Option (B) is correct. From the it may be easily seen that the tank circuit is having 2-capacitors and one-inductor, so it is colpits oscillator and frequency is 1 f = 2p LCeq Ceq = C1 C2 = 2 # 2 = 1 pF 4 C1 + C2 1 f = 2p 10 # 10 - 6 # 10 - 12 9 = 1 # 10 = 50.3 MHz 2p 10

Option (A) is correct. Common mode gain is AC = aRC REE Since source resistance of the current source is infinite REE = 3 , common mode gain AC = 0

3.111

3.112

Option (D) is correct. The circuit is as shown below 3.113

Option (D) is correct. In positive feed back it is working as OP-AMP in saturation region, and the input applied voltage is +ve. So, V0 =+ Vsat = 15 V Option (C) is correct. With the addition of RE the DC abis currents and voltages remain closer to the point where they were set by the circuit when the outside condition such as temperature and transistor parameter b change. Option (A) is correct. At high frequency gm + jw (C) 1 Ai \ Capacitance 1 Ai a frequency Ai =or,

Let V- be the voltage of inverting terminal, since non inverting terminal a at ground, the output voltage is ...(1) Vo = AOL VNow applying KCL at inverting terminal we have V- - Vs + V- - V0 = 0 ...(2) R1 R2

and

' gbc

Thus due to the transistor capacitance current gain of a bipolar transistor drops.

A I D

Option (C) is correct. As OP-AMP is ideal, the inverting terminal at virtual ground due to ground at non-inverting terminal. Applying KCL at inverting .in oterminal c . dia sC (v1 sin wt - 0) + sC (V2 sin wt - 0) + sC (Vo - 0) = 0 o n or Vo =- (V1 + V2) sin wt w.

O N

3.109

From (1) and (2) we have VO = A = - R2 CL Vs R - R2 + R1 ROL Substituting the values we have ww - 10k =- 1000 . - 11 ACL = 10 k 1 k 89 + 1k 100k Option (A) is correct. The first OPAMP stage is the differentiator and second OPAMP stage is integrator. Thus if input is cosine term, output will be also cosine term. Only option (A) is cosine term. Other are sine term. However we can calculate as follows. The circuit is shown in fig

3.114

3.115

3.116

3.117

3.118

Option (D) is correct. There is R - C , series connection in parallel with parallel R - C combination. So, it is a wein bridge oscillator because two resistors R1 and R2 is also in parallel with them. Option (A) is correct. The given circuit is a differentiator, so the output of triangular wave will be square wave. Option (B) is correct. In sampling and hold circuit the unity gain non-inverting amplifier is used. Option (D) is correct. The Thevenin equivalent is shown below

Applying KCL at inverting terminal of first OP AMP we have V1 = - wjL = - 100 # 10 # 10 - 3 = - 1 R 10 10 VS - jVS or V1 = = j cos 100t 10 Applying KCL at inverting terminal of second OP AMP we have VO = - 1/jwC V1 100 1 == j10 j100 # 10 # 10 - 6 # 100 or V0 = j10V2 = j10 (- j cos 100t) V0 = 10 cos 100t

R1 V = 5 # 15 = 5 V 10 + 5 R1 + R2 C Since b is large is large, IC . IE , IB . 0 and IE = VT - VBE RE 4. 3 = 5 - 0.7 = = 10 mA 0.430kW 0.430KW VT =

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3.120 3.121

3.122

Page 75

Output resistance = 25 = 25 W 1

Option (C) is correct. The output voltage will be input offset voltage multiplied by open by open loop gain. Thus So V0 = 5mV # 10, 000 = 50 V But V0 = ! 15 V in saturation condition So, it can never be exceeds ! 15 V So, V0 = ! Vset = ! 15V

Option (D) is correct. This is a voltage shunt feedback as the feedback samples a portion of output voltage and convert it to current (shunt).

3.126

Option (A) is correct. In a differential amplifier CMRR is given by (1 + b) IQ R 0 CMRR = 1 ;1 + E 2 VT b So where R 0 is the emitter resistance. So CMRR can be improved by increasing emitter resistance.

3.127

Option (A) is correct. Option (A) is correct. Negative feedback in amplifier reduces the gain of the system. Option (A) is correct. By drawing small signal equivalent circuit

Option (C) is correct. We know that rise time (tr ) is

3.128

tr = 0.35 fH where fH is upper 3 dB frequency. Thus we can obtain upper 3 dB frequency it rise time is known. Option (D) is correct. In a BJT differential amplifier for a linear response Vid < VT .

3.129

Option (D) is correct. In a shunt negative feedback amplifier. Input impedance Ri R in = (1 + bA)

3.130

by applying KCL at E2 gm1 Vp 1

Vp = gm2 Vp rp

A I D

2

2

2

where

i 0 =- gm2 Vp

at C2 from eq (1) and (2)

2

i 0 =- i 0 gm2 rp

gm1 Vp + 1

2

gm1 Vp =- i 0 :1 + 1 D gm2 rp 1

2

O N

.no

gm2 rp = b >> 1 ww w so gm1 Vp =- i 0 i 0 =- g m1 Vp i0 = g a Vp = Vi m1 Vi Option (B) is correct. Crossover behavior is characteristic of calss B output stage. Here 2 transistor are operated one for amplifying +ve going portion and other for -ve going portion. 2

1

3.123

3.124

Option (C) is correct. In Voltage series feedback mode input impedance is given by R in = Ri (1 + bv Av) where bv = feedback factor , Av = openloop gain and Ri = Input impedance So, R in = 1 # 103 (1 + 0.99 # 100) = 100 kW Similarly output impedance is given by R0 ROUT = R 0 = output impedance (1 + bv Av) 100 Thus ROUT = = 1W (1 + 0.99 # 100)

3.125

So, n i Rin < Ri . o a.cSimilarly

di

3.131

R0 (1 + bA)

3.132

3.133

3.134

3.135

Option (A) is correct. Option (D) is correct. Comparator will give an output either equal to + Vsupply or - Vsupply . So output is a square wave. Option (C) is correct. In series voltage regulator the pass transistor is in common collector configuration having voltage gain close to unity. Option (D) is correct. In bridge rectifier we do not need central tap transformer, so its less expensive and smaller in size and its PIV (Peak inverse voltage) is also greater than the two diode circuit, so it is also suitable for higher voltage application. Option (C) is correct. In the circuit we have V2 = IS # RD 2 and

Option (B) is correct. Regulation = Vno - load - Vfuel - load Vfull - load = 30 - 25 # 100 = 20% 25

ROUT =

ROUT < R 0 Thus input & output impedances decreases.

1

1

Ri = input impedance of basic amplifier b = feedback factor A = open loop gain

V1 = IS # RD V2 = 1 2 V1 V1 = 2V2

3.136

Option (C) is correct.

3.137

Option (C) is correct.

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Page 76

The equivalent circuit of given amplifier circuit (when CE is connected, RE is short-circuited)

Ib = 0 - 3 + 10 - 3 4 4 10 6 Ib = = 1 amp 4

so current

3.141

Option (D) is correct. By applying node equation at terminal (2) and (3) of OP -amp

Input impedance Ri = RB || r p Voltage gain AV = gm RC Now, if CE is disconnected, resistance RE appears in the circuit

Va - Q Va - V0 =0 + 5 10 2Va - 4 + Va - V0 = 0 V0 = 3Va - 4 Va - V0 + Va - 0 = 0 100 10

Input impedance R in = RB || [rp + (b + 1)] RE Input impedance increases gm RC Voltage gain Voltage gain decreases. AV = 1 + gm R E 3.138

3.139

Option (A) is correct. In common emitter stage input impedance is high, so in cascaded amplifier common emitter stage is followed by common base stage.

A I D So

Option (C) is correct. We know that collect-emitter break down voltage is less than compare to collector base breakdown voltage.

O N

Va - V0 + 10Va = 0 11Va = V0 Va = V0 11 V0 = 3V0 - 4 11 8V0 =- 4 11

V0 =- 5.5 Volts n BVCEO < BVCBO o.i c . both avalanche and zener break down. Voltage are higher than 3.142 dia Option (B) is correct. o n Circuit with diode forward resistance looks BVCEO .So BVCEO limits the power supply. w. 3.140

Option (C) is correct.

ww

So the DC current will If we assume consider the diode in reverse bias then Vn should be greater than VP . VP < Vn by calculating

IDC = 3.143

VP = 10 # 4 = 5 Volt 4+4

Vm p (R f + RL)

Option (D) is correct. For the positive half cycle of input diode D1 will conduct & D2 will be off. In negative half cycle of input D1 will be off & D2 conduct so output voltage wave from across resistor (10 kW) is –

Vn = 2 # 1 = 2 Volt here VP > Vn (so diode cannot be in reverse bias mode).

apply node equation at node a Va - 10 + Va + Va = 2 1 4 4 6Va - 10 = 8 Va = 3 Volt

Ammeter will read rms value of current so I rms = Vm (half wave rectifier) pR 4 = 0.4 mA = p (10 kW) p 3.144

Option (D) is correct. In given circuit positive feedback is applied in the op-amp., so it works as a Schmitt trigger.

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Page 77

Option (D) is correct. Gain with out feedback factor is given by V0 = kVi after connecting feedback impedance Z

given input impedance is very large, so after connecting Z we have Ii = Vi - V0 V0 = kVi Z Ii = Vi - kVi Z input impedance Zin = Vi = Z Ii (1 - k) 3.146 3.147

Option (A) is correct. Option (A) is correct. For the circuit, In balanced condition It will oscillated at a frequency w 1 = 1 = 105 rad/ sec -3 -6 LC 10 # 10 # .01 # 10 In this condition R1 = R 3 R2 R4 5 =R 100 1 =

4

R = 20 kW = 2 # 10 W 3.148

Option (C) is correct. V0 kept constant at so current in 50 W resistor

V0 = 6 volt I = 9-6 50 W

A I D

O N

no w.

.in

co ia.

d

ww

I = 60 m amp Maximum allowed power dissipation in zener PZ = 300 mW Maximum current allowed in zener PZ = VZ (IZ ) max = 300 # 10-3 & = 6 (IZ ) max = 300 # 10-3 & = (IZ ) max = 50 m amp Given knee current or minimum current in zener In given circuit

(IZ ) min I IL (IL) min (IL) max

= 5 m amp = IZ + I L = I - IZ = I - (IZ ) max = (60 - 50) m amp = 10 m amp = I - (IZ ) min = (60 - 5) = 55 m amp

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UNIT 4

Page 78

2012

ONE MARK

Consider the given circuit

4.4

DIGITAL CIRCUITS

2013 4.1

4.2

A bulb in a staircase has two switches, one switch being at the ground floor and the other one at the first floor. The bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch. The logic of switching of the bulb resembles (A) and AND gate (B) an OR gate (C) an XOR gate (D) a NAND gate For 8085 microprocessor, the following program is executed. MVI A, 05H; MVI B, 05H; PTR: ADD B; DCR B; JNZ PTR; ADI 03H; HLT; At the end of program, accumulator contains (A) 17H (B) 20H (C) 23H (D) 05H 2013

4.3

ONE MARK

In this circuit, the race around (A) does not occur (B) occur when CLK = 0 (C) occur when CLK = 1 and A = B = 1 (D) occur when CLK = 1 and A = B = 0

In the circuit shown

4.6

A I D

O N TWO MARKS

The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit input B . The number of combinations for which the output is logic 1, is (A) 4 (B) 6 (C) 8 (D) 10

4.5

.in

co ia.

There are four chips each of 1024 bytes connected to a 16 bit address o d .n w bus as shown in the figure below, RAMs 1, 2, 3 and 4 respectively ww are mappped to addresses

4.7

(A) Y = A B + C (C) Y = (A + B ) C

(B) Y = (A + B) C (D) Y = AB + C

/

In the sum of products function f (X, Y, Z) = (2, 3, 4, 5), the prime implicants are (A) XY, XY (B) XY, X Y Z , XY Z (C) XY Z , XYZ, XY (D) XY Z , XYZ, XY Z , XY Z 2012

4.8

(A) 0C00H-0FFFH, 1C00H-1FFFH, 2C00H-2FFFH, 3C00H3FFFH (B) 1800H-1FFFH, 2800H-2FFFH, 3800H-3FFFH, 4800H-4FFFH (C) 0500H-08FFH, 1500H-18FFH, 3500H-38FFH, 5500H-58FFH (D) 0800H-0BFFH, 1800H-1BFFH, 2800H-2BFFH, 3800H-3BFFH

In the CMOS circuit shown, electron and hole mobilities are equal, and M1 and M2 are equally sized. The device M1 is in the linear region if

(A) Vin < 1.875 V (C) Vin > 3.125 V 4.9

TWO MARKS

(B) 1.875 V < Vin < 3.125 V (D) 0 < Vin < 5 V

The state transition diagram for the logic circuit shown is

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Page 79

2011 4.13

2011 4.10

TWO MARKS

The output of a 3-stage Johnson (twisted ring) counter is fed to a digital-to analog (D/A) converter as shown in the figure below. Assume all states of the counter to be unset initially. The waveform which represents the D/A converter output Vo is

ONE MARK

The output Y in the circuit below is always ‘1’ when

A I D

Two D flip-flops are connected as a synchronous counter that goes through the following QB QA sequence 00 " 11 " 01 " 10 " 00 " .... (A) two or more of the inputs P, Q, R are ‘0’ The connections to the inputs DA and DB are (B) two or more of the inputs P, Q, R are ‘1’ (A) .inDA = QB, DB = QA o c . (C) any odd number of the inputs P, Q, R is ‘0’ ia (B) DA = Q A, DB = Q B d o (D) any odd number of the inputs P, Q, R is ‘1’ .n (C) DA = (QA Q B + Q A QB), DB = QA w w w (D) DA = (QA QB + Q A Q B), DB = Q B When the output Y in the circuit below is “1”, it implies that data has 4.15 An 8085 assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is 4.14

4.11

O N

(A) changed from “0” to “1” (C) changed in either direction 4.12

(B) changed from “1” to “0” (D) not changed (A) 8CH (C) 23H

The logic function implemented by the circuit below is (ground implies a logic “0”)

(B) 64H (D) 15H

2010 4.16

(A) F = AND ^P, Q h (C) F = XNOR ^P, Q h

ONE MARK

Match the logic gates in Column A with their equivalents in Column B

(B) F = OR ^P, Q h (D) F = XOR ^P, Q h

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Page 80

For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is

4.21

(A) P-2, Q-4, R-1, S-3 (C) P-2, Q-4, R-3, S-1 4.17

(B) P-4, Q-2, R-1, S-3 (D) P-4, Q-2, R-3, S-1

In the circuit shown, the device connected Y5 can have address in the range

(A) 00H (C) 67H

(B) 45H (D) E7H

2009

The full form of the abbreviations TTL and CMOS in reference to logic families are (A) Triple Transistor Logic and Chip Metal Oxide Semiconductor (B) Tristate Transistor Logic and Chip Metal Oxide Semiconductor (C) Transistor Transistor Logic and Complementary Metal Oxide Semiconductor (D) Tristate Transistor Logic and Complementary Metal Oxide Silicon

4.22

(A) 2000 - 20FF (C) 2E00 - 2EFF 4.18

(B) 2D00 - 2DFF (D) FD00 - FDFF

A I D

O N

(B) A = 1, B = 0, C = 0 (D) A = 0, B = 0, C = 1

2010 4.19

.c

ia od

n

. ww

2009

w

4.24

TWO MARKS

Assuming that the flip-flop are in reset condition initially, the count sequence observed at QA , in the circuit shown is

4.25

4.26

(A) 0010111... (C) 0101111... 4.20

In a microprocessor, the service routine for a certain interrupt starts from a fixed location of memory which cannot be externally set, but the interrupt can be delayed or rejected Such an interrupt is (A) non-maskable and non-vectored (B) maskable and non-vectored (C) non-maskable and vectored .in maskable and vectored o(D)

4.23

For the output F to be 1 in the logic circuit shown, the input combination should be

(A) A = 1, B = 1, C = 0 (C) A = 0, B = 1, C = 0

ONE MARK

TWO MARKS

If X = 1 in logic equation 6X + Z {Y + (Z + XY )}@ {X + X (X + Y)} = 1 , then (A) Y = Z (B) Y = Z (C) Z = 1 (D) Z = 0 What are the minimum number of 2- to -1 multiplexers required to generate a 2- input AND gate and a 2- input Ex-OR gate (A) 1 and 2 (B) 1 and 3 (C) 1 and 1 (D) 2 and 2 What are the counting states (Q1, Q2) for the counter shown in the figure below

(B) 0001011... (D) 0110100....

The Boolean function realized by the logic circuit shown is (A) 11, 10, 00, 11, 10,... (C) 00, 11, 01, 10, 00...

(B) 01, 10, 11, 00, 01... (D) 01, 10, 00, 01, 10...

Statement for Linked Answer Question 5.18 & 5.19 : (A) F = Sm (0, 1, 3, 5, 9, 10, 14) (C) F = Sm (1, 2, 4, 5, 11, 14, 15)

(B) F = Sm (2, 3, 5, 7, 8, 12, 13) (D) F = Sm (2, 3, 5, 7, 8, 9, 12)

Two products are sold from a vending machine, which has two push buttons P1 and P2 . When a buttons is pressed, the price of the corresponding product

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is displayed in a 7 - segment display. If no buttons are pressed, '0' is displayed signifying ‘Rs 0’. If only P1 is pressed, ‘2’ is displayed, signifying ‘Rs. 2’ If only P2 is pressed ‘5’ is displayed, signifying ‘Rs. 5’ If both P1 and P2 are pressed, 'E' is displayed, signifying ‘Error’ The names of the segments in the 7 - segment display, and the glow of the display for ‘0’, ‘2’, ‘5’ and ‘E’ are shown below.

Page 81

The two numbers represented in signed 2’s complement form are P + 11101101 and Q = 11100110 . If Q is subtracted from P , the value obtained in signed 2’s complement is (A) 1000001111 (B) 00000111 (C) 11111001 (D) 111111001

4.31

Which of the following Boolean Expressions correctly represents the relation between P, Q, R and M1

4.32

Consider (1) push buttons pressed/not pressed in equivalent to logic 1/0 respectively. (2) a segment glowing/not glowing in the display is equivalent to logic 1/0 respectively. 4.27

4.28

4.29

If segments a to g are considered as functions of P1 and P2 , then which of the following is correct (A) g = P 1 + P2, d = c + e (B) g = P1 + P2, d = c + e (C) g = P1 + P2, e = b + c (D) g = P1 + P2, e = b + c

(A) M1 = (P OR Q) XOR R (B) M1 = (P AND Q) X OR R (C) M1 = (P NOR Q) X OR R (D) M1 = (P XOR Q) XOR R

What are the minimum numbers of NOT gates and 2 - input OR gates required to design the logic of the driver for this 7 - Segment display (A) 3 NOT and 4 OR (B) 2 NOT and 4 OR (C) 1 NOT and 3 OR (D) 2 NOT and 3 OR

A I D

O N

.in

co ia.

Refer to the NAND and NOR latches shown in the figure. The no d . inputs (P1, P2) for both latches are first made (0, 1) and then, afterwaw w few seconds, made (1, 1). The corresponding stable outputs (Q1, Q2) are

4.34

(A) NAND: first (0, 1) then (0, 1) NOR: first (1, 0) then (0, 0) (B) NAND : first (1, 0) then (1, 0) NOR : first (1, 0) then (1, 0) (C) NAND : first (1, 0) then (1, 0) NOR : first (1, 0) then (0, 0) (D) NAND : first (1, 0) then (1, 1) NOR : first (0, 1) then (0, 1) 2008 4.30

For the circuit shown in the figure, D has a transition from 0 to 1 after CLK changes from 1 to 0. Assume gate delays to be negligible Which of the following statements is true

4.33

For each of the positive edge-triggered J - K flip flop used in the following figure, the propagation delay is 3 t .

TWO MARKS

The logic function implemented by the following circuit at the terminal OUT is

(A) P NOR Q (C) P OR Q

(A) Q goes to 1 at the CLK transition and stays at 1 (B) Q goes to 0 at the CLK transition and stays 0 (C) Q goes to 1 at the CLK tradition and goes to 0 when D goes to 1 (D) Q goes to 0 at the CLK transition and goes to 1 when D goes to 1

Which of the following wave forms correctly represents the output at Q1 ?

(B) P NAND Q (D) P AND Q

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Page 82

2710 LXI H, 30A0 H 2713 DAD H 2414 PCHL All address and constants are in Hex. Let PC be the contents of the program counter and HL be the contents of the HL register pair just after executing PCHL. Which of the following statements is correct ? PC = 2715H PC = 30A0H (B) (A) HL = 30A0H HL = 2715H PC = 6140H PC = 6140H (C) (D) HL = 6140H HL = 2715H 2007 4.39

Statement For Linked Answer Question 5.26 & 5.27 : In the following circuit, the comparators output is logic “1” if V1 > V2 and is logic "0" otherwise. The D/A conversion is done as 3

per the relation VDAC = 2n - 1 bn Volts, where b3 (MSB), b1, b2 and 0 The counter starts from the b0 (LSB) are the countern =outputs. clear state.

/

4.40

ONE MARK

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 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

A I D 2007

TWO MARKS

The Boolean expression Y = ABC D + ABCD + ABC D + ABC D can be minimized to (A) Y = ABC D + ABC + AC D (B)nY = ABC D + BCD + ABC D .i o c (C) Y = ABCD + BC D + ABC D . ia d (D) Y = ABCD + BC D + ABC D .no 4.41

O N w

ww

4.35

4.36

4.37

The stable reading of the LED displays is (A) 06 (B) 07 (C) 12 (D) 13

4.42

The magnitude of the error between VDAC and Vin at steady state in volts is (A) 0.2 (B) 0.3 (C) 0.5 (D) 1.0

(A) (B) (C) (D)

For the circuit shown in the following, I0 - I3 are inputs to the 4:1 multiplexers, R(MSB) and S are control bits. The output Z can be represented by 4.43

(A) (B) (C) (D) 4.38

In the following circuit, X is given by

X = ABC + ABC + ABC + ABC X = ABC + ABC + ABC + ABC X = AB + BC + AC X = AB + BC + AC

The circuit diagram of a standard TTL NOT gate is shown in the figure. Vi = 25 V, the modes of operation of the transistors will be

PQ + PQS + QRS PQ + PQR + PQS PQR + PQR + PARS + QRS PQR + PQRS + PQRS + QRS

An 8085 executes the following instructions

(A) Q1: revere active; Q2: normal active; Q3: saturation; Q4: cut-off

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Page 83

(B) Q1: revere active; Q2: saturation; Q3: saturation; Q4: cut-off (C) Q1: normal active; Q2: cut-off; Q3: cut-off; Q4: saturation (D) Q1: saturation; Q2: saturation; Q3: saturation; Q4: normal active 4.44

5: 6: 7: 8:

The following binary values were applied to the X and Y inputs of NAND latch shown in the figure in the sequence indicated below : X = 0,Y = 1; X = 0, Y = 0; X = 1; Y = 1 The corresponding stable P, Q output will be.

4.48

4.49

(A) (B) (C) (D) 4.45

P = 1, Q = 0; P = 1, Q = 0; P = 1, Q = 0; P = 1, Q = 0;

P = 1, Q = 0; P = 1, Q = 0 or P = 0, Q = 1 P = 0, Q = 1; or P = 0, Q = 1; P = 0, Q = 1 P = 1, Q = 1; P = 1, Q = 0 or P = 0, Q = 1 P = 1, Q = 1; P = 1, Q = 1

4.50

An 8255 chip is interfaced to an 8085 microprocessor system as an I/O mapped I/O as show in the figure. The address lines A0 and A1 of the 8085 are used by the 8255 chip to decode internally its thee ports and the Control register. The address lines A3 to A7 as well as the IO/M signal are used for address decoding. The range of addresses for which the 8255 chip would get selected is

ANI 9BH CPI 9FH STA 3010H HLT

The contents of the accumulator just execution of the ADD instruction in line 4 will be (A) C3H (B) EAH (C) DCH (D) 69H After execution of line 7 of the program, the status of the CY and Z flags will be (A) CY = 0, Z = 0 (B) CY = 0, Z = 1 (C) CY = 1, Z = 0 (D) CY = 1, Z = 1 For the circuit shown, the counter state (Q1 Q0) follows the sequence

(A) 00, 01, 10, 11, 00 (C) 00, 01, 11, 00, 01

(B) 00, 01, 10, 00, 01 (D) 00, 10, 11, 00, 10

A I D 2006

4.51

(A) F8H - FBH (C) F8H - FFH

(B) F8GH - FCH (D) F0H - F7H

O N

Statement for Linked Answer Question 5.37 and 5.38 :

ONE MARK

The number of product terms in the minimized sum-of-product expression obtained through the following K - map is (where, "d" denotes in don’t care states)

o.

no

. ww

.c dia

w

In the Digital-to-Analog converter circuit shown in the figure below, VR = 10V and R = 10kW

(A) 2 (C) 4

(B) 3 (D) 5

2006 4.52

4.46

4.47

The current is (A) 31.25mA (C) 125mA

(B) 62.5mA (D) 250mA

The voltage V0 is (A) - 0.781 V (C) - 3.125 V

(B) - 1.562 V (D) - 6.250 V

Statement for Linked Answer Questions 5.39 & 5.40 : An 8085 assembly language program is given below. Line 1: MVI A, B5H 2: MVI B, OEH 3: XRI 69H 4: ADD B

An I/O peripheral device shown in Fig. (b) below is to be interfaced to an 8085 microprocessor. To select the I/O device in the I/O address range D4 H - D7 H, its chip-select (CS ) should be connected to the output of the decoder shown in as below :

(A) output 7 (C) output 2 4.53

TWO MARKS

(B) output 5 (D) output 0

For the circuit shown in figures below, two 4 - bit parallel - in serial - out shift registers loaded with the data shown are used to feed the data to a full adder. Initially, all the flip - flops are in clear state. After applying two clock pulse, the output of the full-adder should

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be

(A) Q 1 and Q0 (C) Q1 Q0 and Q 1 Q0 4.58

(A) S = 0, C0 = 0 (C) S = 1, C0 = 0 4.54

4.55

(B) Q 0 and Q1 (D) Q 1 Q 0 and Q1 Q0

The point P in the following figure is stuck at 1. The output f will be

(B) S = 0, C0 = 1 (D) S = 1, C0 = 1

A new Binary Coded Pentary (BCP) number system is proposed in which every digit of a base-5 number is represented by its corresponding 3-bit binary code. For example, the base-5 number 24 will be represented by its BCP code 010100. In this numbering system, the BCP code 10001001101 corresponds of the following number is base-5 system (A) 423 (B) 1324 (C) 2201 (D) 4231

(A) ABC (C) ABC

(B) A (D) A

2005 4.59

A 4 - bit D/A converter is connected to a free - running 3 - big UP counter, as shown in the following figure. Which of the following waveforms will be observed at V0 ? 4.60

ONE MARK

Decimal 43 in Hexadecimal and BCD number system is respectively (A) B2, 0100 011 (B) 2B, 0100 0011 (C) 2B, 0011 0100 (D) B2, 0100 0100 The Boolean function f implemented in the figure using two input multiplexes is

A I D

O N

.in

In the figure shown above, the ground has been shown by the sym-.no w bol 4 ww

co ia.

d

(A) ABC + ABC (C) ABC + ABC

(B) ABC + ABC (D) ABC + ABC

2005 4.61

4.56

4.57

Following is the segment of a 8085 assembly language program LXI SP, EFFF H CALL 3000 H : : : 3000 H LXI H, 3CF4 PUSH PSW SPHL POP PSW RET On completion of RET execution, the contents of SP is (A) 3CF0 H (B) 3CF8 H (C) EFFD H (D) EFFF H Two D - flip - flops, as shown below, are to be connected as a synchronous counter that goes through the following sequence 00 " 01 " 11 " 10 " 00 " ... The inputs D0 and D1 respectively should be connected as,

The transistors used in a portion of the TTL gate show in the figure have b = 100 . The base emitter voltage of is 0.7 V for a transistor in active region and 0.75 V for a transistor in saturation. If the sink current I = 1 A and the output is at logic 0, then the current IR will be equal to

(A) 0.65 mA (C) 0.75 mA 4.62

TWO MARKS

(B) 0.70 mA (D) 1.00 mA

The Boolean expression for the truth table shown is

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0109 H is (A) 20 H (C) 00 H 4.67

(A) B (A + C)( A + C ) (C) B (A + C )( A + C) 4.63

4.64

(B) B (A + C )( A + C) (D) B (A + C)( A + C )

The present output Qn of an edge triggered JK flip-flop is logic 0. If J = 1, then Qn + 1 (A) Cannot be determined (B) Will be logic 0 (C) will be logic 1 (D) will rave around

4.68

4.69

4.65

ONE MARK

A master - slave flip flop has the characteristic that (A) change in the output immediately reflected in the output (B) change in the output occurs when the state of the master is affected (C) change in the output occurs when the state of the slave is affected (D) both the master and the slave states are affected at the same time The range of signed decimal numbers that can be represented by 6-bits 1’s complement number is (A) -31 to +31 (B) -63 to +63 (C) -64 to +63 (D) -32 to +31

A I D

(B) 111 (D) 101

A digital system is required to amplify a binary-encoded audio signal. The user should be able to control the gain of the amplifier from .in minimum to a maximum in 100 increments. The minimum o c . number of bits required to encode, in straight binary, is dia (A) 8 o (B) 6 .n w (C) 5 (D) 7 ww

What memory address range is NOT represents by chip # 1 and chip # 2 in the figure A0 to A15 in this figure are the address lines and CS means chip select.

O N

4.70

4.71

4.72

(A) 0100 - 02FF (C) F900 - FAFF

(B) 1500 - 16FF (D) F800 - F9FF

Choose the correct one from among the alternatives A, B, C, D after matching an item from Group 1 most appropriate item in Group 2. 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 (A) P - 3, Q - 2, R - 1 (B) P - 3, Q - 1, R - 2 (C) P - 2, Q - 1, R - 3 (D) P - 1, Q - 2, R - 2 The figure the internal schematic of a TTL AND-OR-OR-Invert (AOI) gate. For the inputs shown in the figure, the output Y is

(A) 0 (C) AB

Statement For Linked Answer Questions 5.57 & 5.58 : The following program starts at location 0100H. LXI SP, OOFF LXI H, 0701 MVI A, 20H SUB M The content of accumulator when the program counter reaches

(B) 1 (D) AB

2004

Consider an 8085 microprocessor system. 4.66

If in addition following code exists from 019H onwards, ORI 40 H ADD M What will be the result in the accumulator after the last instruction is executed ? (A) 40 H (B) 20 H (C) 60 H (D) 42 H 2004

The given figure shows a ripple counter using positive edge triggered flip-flops. If the present state of the counter is Q2 Q1 Q0 = 001 then is next state Q2 Q1 Q will be

(A) 010 (C) 100

(B) 02 H (D) FF H

4.73

4.74

TWO MARKS

11001, 1001, 111001 correspond to the 2’s complement representation of which one of the following sets of number (A) 25,9, and 57 respectively (B) -6, -6, and -6 respectively (C) -7, -7 and -7 respectively (D) -25, -9 and -57 respectively In the modulo-6 ripple counter shown in figure, the output of the 2- input gate is used to clear the J-K flip-flop

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Page 86

The 2-input gate is

the accumulator (C) Contents of location 8529 are complemented and stored in location 8529 (D) Contents of location 5892 are complemented and stored in location 5892 4.81

(A) a NAND gate (C) an OR gate 4.75

4.76

4.77

4.78

4.79

4.80

(B) a NOR gate (D) a AND gare

The minimum number of 2- to -1 multiplexers required to realize a 4- to -1 multiplexers is (A) 1 (B) 2 (C) 3 (D) 4 The Boolean expression AC + BC is equivalent to (A) AC + BC + AC (B) BC + AC + BC + ACB (C) AC + BC + BC + ABC (D) ABC + ABC + ABC + ABC A Boolean function f of two variables x and y is defined as follows : 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 2input OR gates (each having unit cost) would have a total cost of (A) 1 unit (B) 4 unit (C) 3 unit (D) 2 unit

It is desired to multiply the numbers 0AH by 0BH and store the result in the accumulator. The numbers are available in registers B and C respectively. A part of the 8085 program for this purpose is given below : MVI A, 00H LOOP --------------HLT END The sequence of instructions to complete the program would be (A) JNX LOOP, ADD B, DCR C (B) ADD B, JNZ LOOP, DCR C (C) DCR C, JNZ LOOP, ADD B (D) ADD B, DCR C, JNZ LOOP 2003

4.82

ONE MARK

The number of distinct Boolean expressions of 4 variables is (A) 16 (B) 256 (C) 1023 (D) 65536

A I D

The minimum number of comparators required to build an 8-bits flash ADC is The 8255 Programmable Peripheral Interface is used as described .in 8 (B) 63 o(A) below. c . a (D) 256 di (C) 255 (i) An A/D converter is interface to a microprocessor through an o n . 8255. w 4.84 The output of the 74 series of GATE of TTL gates is taken from a w w The conversion is initiated by a signal from the 8255 on Port C. A BJT in signal on Port C causes data to be stobed into Port A. (A) totem pole and common collector configuration (ii) Two computers exchange data using a pair of 8255s. Port A (B) either totem pole or open collector configuration works as a bidirectional data port supported by appropriate hand(C) common base configuration shaking signals. The appropriate modes of operation of the 8255 for (i) and (ii) (D) common collector configuration would be 4.85 Without any additional circuitry, an 8:1 MUX can be used to obtain (A) Mode 0 for (i) and Mode 1 for (ii) (A) some but not all Boolean functions of 3 variables (B) Mode 1 for (i) and Mode 2 for (ii) (B) all functions of 3 variables but non of 4 variables (C) Mode for (i) and Mode 0 for (ii) (C) all functions of 3 variables and some but not all of 4 variables (D) Mode 2 for (i) and Mode 1 for (ii) (D) all functions of 4 variables The number of memory cycles required to execute the following 4.86 A 0 to 6 counter consists of 3 flip flops and a combination circuit of 8085 instructions 2 input gate (s). The common circuit consists of (i) LDA 3000 H (A) one AND gate (ii) LXI D, FOF1H would be (B) one OR gate (A) 2 for (i) and 2 for (ii) (B) 4 for (i) and 3 for (ii) (C) one AND gate and one OR gate (C) 3 for (i) and 3 for (ii) (D) 3 for (i) and 4 for (ii) (D) two AND gates

O N

Consider the sequence of 8085 instructions given below LXI H, 9258 MOV A, M CMA MOV M, A Which one of the following is performed by this sequence ? (A) Contents of location 9258 are moved to the accumulator (B) Contents of location 9258 are compared with the contents of

4.83

2003 4.87

TWO MARKS

The circuit in the figure has 4 boxes each described by inputs P, Q, R and outputs Y, Z with Y = P 5 Q 5 R and Z = RQ + PR + QP The circuit acts as a

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Page 87

(A) 1111 (C) 1000 4.91

(A) 4 bit adder giving P + Q (B) 4 bit subtractor giving P - Q (C) 4 bit subtractor giving Q-P (D) 4 bit adder giving P + Q + R 4.88

4.89

4.90

The DTL, TTL, ECL and CMOS famil GATE 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

The correct column is (A) P (C) R

If the function W, X, Y and Z are as follows W = R + PQ + RS X = PQRS + PQRS + PQRS Y = RS + PR + PQ + 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 A 4 bit ripple counter and a bit synchronous counter are made using flip flops having a propagation delay of 10 ns each. If the worst case delay in the ripple 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

4.92

(B) 1011 (D) 0010

(B) Q (D) S

The circuit shown in figure converts

A I D

(A) BCD to binary code (C) Excess -3 to gray code

(B) Binary to excess - 3 code (D) Gray to Binary code

In the circuit shown in the figure, A is parallel-in, parallel-out 4 bit register, which loads at the rising edge of the clock C . The input 4.93 In an in 8085 microprocessor, the instruction CMP B has been executed . lines are connected to a 4 bit bus, W . Its output acts at input to a o cwhile the content of the accumulator is less than that of register B ia. . As a result 16 # 4 ROM whose output is floating when the input to a partial d no table of the contents of the ROM is as follows (A) Carry flag will be set but Zero flag will be reset w.

O N ww

Data

0011

1111

0100

1010

1011

1000

0010

1000

Address

0

2

4

6

8

10

11

14

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

(B) Carry flag will be rest but Zero flag will be set (C) Both Carry flag and Zero flag will be rest (D) Both Carry flag and Zero flag will be set 4.94

The circuit shown in the figure 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 nearest multiple of 5%) for the tolerance of the DAC is (A) ! 35% (B) ! 20% (C) ! 10% (D) ! 5% 2002 4.95

ONE MARK

4 - bit 2’s complement representation of a decimal number is 1000. The number is (A) +8 (B) 0 (C) -7

(D) -8

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 4.96

If the input to the digital circuit (in the figure) consisting of a cascade of 20 XOR - gates is X , then the output Y is equal to

(A) 0 (C) X 4.97

OUT PORT1 HLT NEXT : XRA B JP START OUT PORT2 HTL The execution of above program in an 8085 microprocessor will result in (A) an output of 87H at PORT1 (B) an output of 87H at PORT2 (C) infinite looping of the program execution with accumulator data remaining at 00H (D) infinite looping of the program execution with accumulator data alternating between 00H and 87H

(B) 1 (D) X

The number of comparators required in a 3-bit comparators type ADC (A) 2 (B) 3 (C) 7 (D) 8 2002

4.98

Page 88

TWO MARKS

The circuit in the figure has two CMOS NOR gates. This circuit functions as a:

2001

The 2’s complement representation of -17 is (A) 101110 (B) 101111 (C) 111110 (D) 110001

4.102

For the ring oscillator shown in the figure, the propagation delay of each inverter is 100 pico sec. What is the fundamental frequency of the oscillator output

4.103

(A) flip-flop (C) Monostable multivibrator 4.99

(B) Schmitt trigger (D) astable multivibrator

A I D

The gates G1 and G2 in the figure have propagation delays of 10 ns and 20 ns respectively. If the input V1, makes an output change from logic 0 to 1 at time t = t0 , then the output waveform V0 is

O N

no w.

ww

4.100

If the input X3, X2, X1, X0 to the ROM in the figure are 8 4 2 1 BCD numbers, then the outputs Y3, Y2, Y1, Y0 are

(A) 10 MHz (C) .in 1 GHz

co

. dia

4.104

4.101

(B) 2 4 2 1 BCD numbers (D) none of the above

(B) 100 MHz (D) 2 GHz

Ab 8085 microprocessor based system uses a 4K # 8 bit RAM whose starting address is AA00H. The address of the last byte in this RAM is (A) OFFFH (B) 1000H (C) B9FFH (D) BA00H 2001

4.105

4.106

TWO MARKS

In the TTL circuit in the figure, S2 and S0 are select lines and X7 and X0 are input lines. S0 and X0 are LSBs. The output Y is

(A) indeterminate (C) A 5 B (A) gray code numbers (C) excess - 3 code numbers

ONE MARKS

(B) A 5 B (D) C (A 5 B ) + C (A 5 B)

In the figure, the LED

Consider the following assembly language program MVI B, 87H MOV A, B START : JMP NEXT MVI B, 00H XRA B

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(A) emits light when both S1 and S2 are closed (B) emits light when both S1 and S2 are open (C) emits light when only of S1 and S2 is closed (D) does not emit light, irrespective of the switch positions.

Page 89

4.107

(A) 1,0,1 (C) 1,1,1

The number of hardware interrupts (which require an external signal to interrupt) present in an 8085 microprocessor are (A) 1 (B) 4 (C) 5 (D) 13

4.112

The digital block in the figure is realized using two positive edge triggered D-flip-flop. Assume that for t < t0, Q1 = Q2 = 0 . The circuit in the digital block is given by

(B) 0,0,1 (D) 0,1,1

In the microprocessor, the RST6 instruction transfer the program execution to the following location : (A)30 H (B) 24 H (C) 48 H (D) 60 H

4.113

2000

The contents of register (B) and accumulator (A) of 8085 microprocessor are 49J are 3AH respectively. The contents of A and status of carry (CY) and sign (S) after execution SUB B instructions are (A) A = F1, CY = 1, S = 1 (B) A = 0F, CY = 1, S = 1 (C) A = F0, CY = 0, S = 0 (D) A = 1F, CY = 1, S = 1

4.114

For the logic circuit shown in the figure, the simplified Boolean expression for the output Y is

4.115

4.108

A I D

O N w

(A) 5 V; 3 V; 7 V (C) 5 V; 5 V; 5 V

4.110

4.111

(B) A (D) C

For the 4 bit DAC shown in the figure, the output voltage V0 is

(B) 4 V; 3 V; 4 V (D) 4 V; 4 V; 4 V

2000 4.109

(A) nA + B + C o.i c . ia (C) B

d

In the DRAM cell in the figure, the Vt of the NMOSFET is 1 V. For .no the following three combinations of WL and BL voltages. ww 4.116

TWO MARKS

(A) 10 V (C) 4 V

ONE MARKS

An 8 bit successive approximation analog to digital communication has full scale reading of 2.55 V and its conversion time for an analog input of 1 V is 20 ms. The conversion time for a 2 V input will be (A) 10 ms (B) 20 ms (C) 40 ms (D) 50 ms

4.117

(B) 5 V (D) 8 V

A sequential circuit using D flip-flop and logic gates is shown in the figure, where X and Y are the inputs and Z is the inputs. The circuit is

The number of comparator in a 4-bit flash ADC is (A) 4 (B) 5 (C) 15 (D) 16 For the logic circuit shown in the figure, the required input condition (A, B, C) to make the output (X) = 1 is (A) S - R Flip-Flop with inputs X = R and Y = S (B) S - R Flip-Flop with inputs X = S and Y = R (C) J - K Flip-Flop with inputs X = J and Y = K

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Page 90

(D) J - K Flip-Flop with input X = K and Y = J 4.118

In the figure, the J and K inputs of all the four Flip-Flips are made high. The frequency of the signal at output Y is

(A) 0.833 kHz (C) 0.91 kHz

4.120

4.121

4.122

4.127

4.125

4.128

A Darlington emitter follower circuit is sometimes used in the output stage of a TTL gate in order to (A) increase its IOL (B) reduce its IOH (C) increase its speed of operation (D) reduce power dissipation

If CS = A15 A14 A13 is used as the chip select logic of a 4 K RAM in an 8085 system, then its memory range will be (A) 3000 H - 3 FFF H (B) 7000 H - 7 FFF H (C) 5000 H - 5 FFF H and 6000 H - 6 FFF H (D) 6000 H - 6 FFF H and 7000 H - 7 FFF H

4.129

ONE MARK

The minimum number of 2-input NAND gates required to implement of Boolean function Z = ABC , assuming that A, B and C are available, is (A) two (B) three (C) five (D) six The noise margin of a TTL gate is about (A) 0.2 V (B) 0.4 V (C) 0.6 V (D) 0.8 V In the figure is A = 1 and B = 1, the input B is now replaced by a sequence 101010....., the output x and y will be

A I D

Commercially available ECL gears use two ground lines and one negative supply in order to (A) reduce power dissipation (B) increase fan-out .in fixed at 0 and 1, respectively o(A) (C) reduce loading effect c . ia (B) x = 1010.....while y = 0101...... (D) eliminate the effect of power line glitches or the biasing circuit no d w. (C) x = 1010.....and y = 1010...... w w The resolution of a 4-bit counting ADC is 0.5 volts. For an analog (D) fixed at 1 and 0, respectively input of 6.6 volts, the digital output of the ADC will be

1999

4.124

ONE MARK

The logical expression y = A + AB is equivalent to (A) y = AB (B) y = AB (C) y = A + B (D) y = A + B

(A) 1011 (C) 1100

4.123

4.126

(B) mod-5 up counter (D) mod-5 down counter

1998

(B) 1.0 kHz (D) 0.77 kHz

1999 4.119

(A) mod-3 up counter (C) mod-3 down counter

(B) 1101 (D) 1110

O N

4.130

TWO MARKS

The minimized form of the logical expression (ABC + ABC + ABC + ABC ) is (A) AC + BC + AB (B) AC + BC + AB (C) AC + BC + AB (D) AC + BC + AB

4.131

For a binary half-subtractor having two inputs A and B, the correct set of logical expressions for the outputs D (= A minus B) and X (= borrow) are (A) D = AB + AB, X = AB (B) D = AB + AB + AB , X = AB (C) D = AB + AB , X = AB (D) D = AB + AB , X = AB

An equivalent 2’s complement representation of the 2’s complement number 1101 is (A) 110100 (B) 01101 (C) 110111 (D) 111101 The threshold voltage for each transistor in the figure is 2 V. For this circuit to work as an inverter, Vi must take the values

(A) - 5 V and 0 V (C) - 0 V and 3 V

The ripple counter shown in the given figure is works as a 4.132

4.133

(B) - 5 V and 5 V (D) 3 V and 5 V

An I/O processor control the flow of information between (A) cache memory and I/O devices (B) main memory and I/O devices (C) two I/O devices (D) cache and main memories Two 2’s complement number having sign bits x and y are added and the sign bit of the result is z . Then, the occurrence of overflow

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Page 91

1997

is indicated by the Boolean function (A) xyz (B) x y z (C) x yz + xyz (D) xy + yz + zx 4.134

4.135

4.136

4.137

Each cell of a static Random Access Memory contains (A) 6 MOS transistors (B) 4 MOS transistors and 2 capacitors (C) 2 MOS transistors and 4 capacitors (D) 1 MOS transistors and 1 capacitors

4.140

The advantage of using a dual slope ADC in a digital voltmeter is that (A) its conversion time is small (B) its accuracy is high (C) it gives output in BCD format (D) it does not require a

A 2 bit binary multiplier can be implemented using (A) 2 inputs ANSs only (B) 2 input XORs and 4 input AND gates only (C) Two 2 inputs NORs and one XNO gate (D) XOR gates and shift registers

4.141

For the identity AB + AC + BC = AB + AC , the dual form is (A) (A + B) (A + C) (B + C) = (A + B) (A + C) (B) (A + B ) (A + C ) (B + C ) = (A + B ) (A + C ) (C) (A + B) (A + C) (B + C) = (A + B ) (A + C ) (D) AB + AC + BC = AB + AC

In standard TTL, the ‘totem pole’ stage refers to (A) the multi-emitter input stage (B) the phase splitter (C) the output buffer (D) open collector output stage

4.142

An instruction used to set the carry Flag in a computer can be classified as (A) data transfer (B) arithmetic (C) logical (D) program control

The inverter 74 ALSO4 has the following specifications IOH max =- 0.4 A, IOL max = 8 mA, IIH max = 20 mA, IIL max =- 0.1 mA The fan out based on the above will be (A) 10 (B) 20 (C) 60 (D) 100

4.143

The figure is shows a mod-K counter, here K is equal to

A I D

The output of the logic gate in the figure is

4.144

(A) 1 (C) 3 4.138

(B) 2 (D) 4

O N

no w.

The current I through resistance r in the circuit shown in the figure ww is

4.139

n (A) o.i 0 c . ia

d

4.145

4.146

(A) - V 12R (C) V 6R

ONE MARK

(B) V 12R (D) V 3T

(B) 1 (D) F

(C) A

In an 8085 mP system, the RST instruction will cause an interrupt (A) only if an interrupt service routine is not being executed (B) only if a bit in the interrupt mask is made 0 (C) only if interrupts have been enabled by an EI instruction (D) None of the above The decoding circuit shown in the figure is has been used to generate the active low chip select signal for a microprocessor peripheral. (The address lines are designated as AO to A7 for I/O address)

The K -map for a Boolean function is shown in the figure is the number of essential prime implicates for this function is

The peripheral will correspond to I/O address in the range (A) 60 H to 63 H (B) A4 to A 7H (C) 30 H to 33 H (D) 70 H to 73 H 4.147

(A) 4 (C) 6

The following instructions have been executed by an 8085 mP

(B) 5 (D) 8

ADDRESS (HEX)

INSTRUCTION

6010

LXI H, 8 A 79 H

6013

MOV A, L

6015

ADDH

6016

DAA

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6017

MOV H, A

6018

PCHL

Page 92

The boolean function A + BC is a reduced form of (A) AB + BC (B) (A + B) : (A + C) (C) AB + ABC (D) (A + C) : B

4.152

From which address will the next instruction be fetched ? (A) 6019 (B) 6379 (C) 6979 (D) None of the above 4.148

A signed integer has been stored in a byte using the 2’s complement format. We wish to store the same integer in a 16 bit word. We should (A) copy the original byte to the less significant byte of the word and fill the more significant with zeros (B) copy the original byte to the more significant byte of the word and fill the less significant byte with zeros (C) copy the original byte to the less significant byte of the word and make each fit of the more significant byte equal to the most significant bit of the original byte (D) copy the original byte to the less significant byte as well as the more significant byte of the word 1997

4.149

1996

(A) ABCDE (C) A : (B + C) + D : E

The total number of memory accesses involved (inclusive of the opcode fetch) when an 8085 processor executes the instruction LDA 2003 is (A) 1 (B) 2 (C) 3 (D) 4

4.156

1996 n

(B) (AB + C ) : (D + E ) (D) (A + B ) : C + D : E

In a J–K flip-flop we have J = Q and K = 1. Assuming the flip flop was initially cleared and then clocked for 6 pulses, the sequence at the Q output will be

4.158

(B) 011001 (D) 010101

The gate delay of an NMOS inverter is dominated by charge time rather than discharge time because (A) the driver transistor has larger threshold voltage than the load transistor (B) the driver transistor has larger leakage currents compared to the load transistor (C) the load transistor has a smaller W/L ratio compared to the driver transistor (D) none of the above

.i .co

ia

4.157

4.159

4.151

A 12-bit ADC is operating with a 1 m sec clock period and the total conversion time is seen to be 14 m sec . The ADC must be of the (A) flash type (B) counting type (C) intergrating type (D) successive approximation type

4.155

d .no

w

(A) 010000 (C) 010010

A pulse train can be delayed by a finite number of clock periods using (A) a serial-in serial-out shift register (B) a serial-in parallel-out shift register (C) a parallel-in serial-out shift register (D) a parallel-in parallel-out shift register

4.154

A I D

O N ww

4.150

Schottky clamping is resorted in TTl gates (A) to reduce propagation delay (B) to increase noise margins (C) to increase packing density (D) to increase fan-out

4.153

TWO MARKS

For the NMOS logic gate shown in the figure is the logic function implemented is

4.160

ONE MARK

TWO MARKS

A dynamic RAM cell which hold 5 V has to be refreshed every 20 m sec, so that the stored voltage does not fall by more than 0.5 V . If the cell has a constant discharge current of 1 pA, the storage capacitance of the cell is (A) 4 # 10-6 F (B) 4 # 10-9 F (C) 4 # 10-12 F (D) 4 # 10-15 F A 10-bit ADC with a full scale output voltage of 10.24 V is designed to have a ! LSB/2 accuracy. If the ADC is calibrated at 25c C and the operating temperature ranges from 0c C to 25c C , then the maximum net temperature coefficient of the ADC should not exceed (A) ! 200 mV/cC (B) ! 400 mV/cC (C) ! 600 mV/cC (D) ! 800 mV/cC

A memory system of size 26 K bytes is required to be designed using memory chips which have 12 address lines and 4 data lines each. The number of such chips required to design the memory system is (A) 2 (B) 4 (C) 8 (D) 13 The following sequence of instructions are executed by an 8085 microprocessor: 1000 LXI SP, 27 FF 1003 CALL 1006 1006 POP H The contents of the stack pointer (SP) and the HL, register pair on completion of execution of these instruction are (A) SP = 27 FF, HL = 1003 (B) SP = 27 FD, HL = 1003 (C) SP = 27 FF, HL = 1006 (D) SP = 27 FD, HL = 1006

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Page 93

SOLUTIONS 4.1

Therefore, we have the address range as A15 A14 A13 A12 A11 A10 A 9

B

Y(Bulb)

up(1)

up(1)

OFF(0)

Down(0)

Down(0)

OFF(0)

up(1)

Down(0)

ON(1)

05

05 + 05

04

05 + 05 + 04

03

05 + 05 + 04 + 03

02

05 + 05 + 04 + 03 + 02

01

05 + 05 + 04 + 03 + 02 + 01

00

System is out of loop

A1

A0

0

0

0

0

1

0

0

0

0

0

0

0

0

0

0

0

to

0

0

0

0

1

0

1

1

1

1

1

1

1

1

1

1

A15

A14

A13

A12

A11

A10

0

0

0

1

1

0

A8

A7

A6

A5

A4

A3

A2

A1

A0

From

0

0

0

1

1

0

0

0

0

0

0

0

0

0

0

0

to

0

0

0

1

1

0

1

1

1

1

1

1

1

1

1

1

.in

no w.

ww

A = A + 03 H = 14 + 03 = 17 H Option (D) is correct. For chip-1, we have the following conclusions: it is enable when (i) S1 S 0 = 0 0 and (ii) Input = 1 For S1 S 0 = 0 0 We have A13 = A12 = 0 and for I/p = 1we obtain

A2

A I D

i.e., A = 05 + 05 + 04 + 03 + 02 + 01 = 144 At this stage, the 8085 microprocessor exits from the loop and reads the next instruction. i.e., the accumulator is being added to 03 H. Hence, we obtain 4.3

A3

Option (A) is correct. The given circuit is

4.4

O N

Output of ADD B (Stored value at A)

A4

In hexadecimal it is from 1800 H to 1BFFH . There is no need to obtain rest of address ranged as only (D) is matching to two results.

i.e., the XOR gate

Content in B

A5

From

A15 A14 A13 A12 A11 A10 A 9

Y = A5B Option (A) is correct. The program is being executed as follows MVI A, 0.5H; A = 05H MVI B, 0.5H; B = 05H At the next instruction, a loop is being introduced in which for the instruction “DCR B” if the result is zero then it exits from loop so, the loop is executed five times as follows :

A6

Therefore, the address range is

Down(0) up(1) ON(1) When the switches A and B are both up or both down, output will be zero (i.e. Bulb will be OFF). Any of the switch changes its position leads to the ON state of bulb. Hence, from the truth table, we get

4.2

A7

In Hexadecimal & 0800 H to 0BFFH Similarly, for chip 2, we obtain the range as follows E = 1 for S1 S 0 = 0 1 so, A13 = 0 and A12 = 1 and also the I/P = 1 for A10 = 0 , A11 = 1, A14 = 0 , A15 = 0 so, the fixed I/ps are

Option (C) is correct. Let A denotes the position of switch at ground floor and B denotes the position of switch at upper floor. The switch can be either in up position or down position. Following are the truth table given for different combinations of A and B A

A8

4.5

co ia.

d

Condition for the race-around It occurs when the output of the circuit (Y1, Y2) oscillates between ‘0’ and ‘1’ checking it from the options. 1. Option (A): When CLK = 0 Output of the NAND gate will be A1 = B1 = 0 = 1. Due to these input to the next NAND gate, Y2 = Y1 : 1 = Y1 and Y1 = Y2 : 1 = Y2 . If Y1 = 0 , Y2 = Y1 = 1 and it will remain the same and doesn’t oscillate. If Y2 = 0 , Y1 = Y2 = 1 and it will also remain the same for the clock period. So, it won’t oscillate for CLK = 0 . So, here race around doesn’t occur for the condition CLK = 0 . 2. Option (C): When CLK = 1, A = B = 1 A1 = B1 = 0 and so Y1 = Y2 = 1 And it will remain same for the clock period. So race around doesn’t occur for the condition. 3. Option (D): When CLK = 1, A = B = 0 So, A1 = B1 = 1 And again as described for Option (B) race around doesn’t occur for the condition. Option ( ) is correct.

A10 = 1 or A10 = 0 A11 = 1 A14 = 1 or A14 = 0 A15 = 1 or A15 = 0 Since, A 0 - A 9 can have any value 0 or 1

Y = 1, when A > B A = a1 a 0, B = b1 b 0

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a1

a0

b1

b0

Y

0

1

0

0

1

1

0

0

0

1

1

0

0

1

1

1

1

0

0

1

1

1

0

1

1

1

1

1

0

1

Page 94

So it can’t be in cutoff region. Case 2 : M2 must be in saturation region. So, I1 = I 2 mp COX W mn COX W 2 2 2 (VSG - VTp) VSD - V SD @ = 2 L (VGS - VTn) 2 L6 2 & 2 (VSG - VTp) VSD - V SD = (VGS - VTn) 2 & 2 (5 - Vin - 1) (5 - VD) - (5 - VD) 2 = (Vin - 0 - 1) 2 & 2 (4 - Vin) (5 - VD) - (5 - VD) 2 = (Vin - 1) 2 Substituting VD = VDS = VGS - VTn and for N -MOS & VD = Vin - 1

Total combination = 6 4.6

4.7

2 (4 - Vin) (6 - Vin) - (6 - Vin) 2 = (Vin - 1) 2 48 - 36 - 8Vin =- 2Vin + 1 & 6Vin = 11 & Vin = 11 = 1.833 V 6 So for M2 to be in saturation Vin < 1.833 V or Vin < 1.875 V & &

Option (A) is correct. Parallel connection of MOS & OR operation Series connection of MOS & AND operation The pull-up network acts as an inverter. From pull down network we write Y = (A + B) C = (A + B) + C = A B + C Option (A) is correct. Prime implicants are the terms that we get by solving K-map

F = XY + XY 1prime 44 2 44 3 implicants 4.8

Option (D) is correct. Let Qn + 1 is next state and Qn is the present state. From the given below figure.

4.9

If A = 0, If A = 1, So state diagram is

A I D

Option (A) is correct. Given the circuit as below :

O N

Since all the parameters of PMOS and NMOS are equal. So, mn = mp COX bW l = COX bW l = COX bW l L M1 L M2 L Given that M1 is in linear region. So, we assume that M2 is either in cutoff or saturation. Case 1 : M2 is in cut off So, I 2 = I1 = 0 Where I1 is drain current in M1 and I2 is drain current in M2 . m C 2 Since, I1 = p OX bW l82VSD ^VSG - VTp h - V SD B 2 L m C 2 0 = p OX bW l [2VSD ^VSG - VTp h - V SD & ] 2 L Solving it we get,

co ia.

d

4.10

& For So, So for the NMOS

I1 = 0 , VD = 5 V Vin = 5 + 3 = 4 V 2

Option (B) is correct. The given circuit is shown below:

(PQ QR ) PR = (PQ + QR PR ) = PQ + QR + PR = PQ + QR + PR If any two or more inputs are ‘1’ then output y will be 1. 4.11

2 ^VSG - VTp h = VSD 2 ^5 - Vin - 1h = 5 - VD Vin = VD + 3 2

&

Qn + 1 = Qn

.in

no w.

ww

D = Y = AX 0 + AX1 Qn + 1 = D = AX 0 + AX1 Qn + 1 = A Qn + AQn X 0 = Q , X1 = Q Qn + 1 = Qn (toggle of previous state)

4.12

Option (A) is correct. For the output to be high, both inputs to AND gate should be high. The D-Flip Flop output is the same, after a delay. Let initial input be 0; (Consider Option A) st then Q = 1 (For 1 D-Flip Flop). This is given as input to 2nd FF. Let the second input be 1. Now, considering after 1 time interval; The output of 1st Flip Flop is 1 and 2nd FF is also 1. Thus Output = 1. Option (D) is correct. F = S1 S 0 I 0 + S1 S 0 I1 + S1 S 0 I 2 + S1 S 0 I 3 I0 = I3 = 0 ( S1 = P, S 0 = Q ) F = PQ + PQ = XOR (P, Q)

VGS = Vin - 0 = 4 - 0 = 4 V and VGS > VTn 4.13

Option (A) is correct.

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Page 95

All the states of the counter are initially unset.

4.17

State Initially are shown below in table :

4.14

Q2

Q1

Q0

0

0

0

0

1

0

0

4

1

1

0

6

1

1

1

7

0

1

1

3

0

0

1

1

0

0

0

0

Option (B) is correct. Since G2 is active low input, output of NAND gate must be 0 G2 = A15 : A14 A13 A12 A11 = 0 So, A15 A14 A13 A12 A11 = 00101 To select Y5 Decoder input ABC = A 8 A 9 A10 = 101 Address range A15 A14 A13 A12 A11 A10 A 9 A 8 ...............A 0 0011101........A 0 S S 2 D ^2D00 - 2DFF h

4.18

Option (D) is correct. The sequence is QB QA 00 " 11 " 01 " 10 " 00 " ... QB

QA

QB (t + 1)

QA (t + 1)

0

0

1

1

1

1

0

1

0

1

1

0

1 QB ^t + 1h

0

0

0

4.19

Option (A) (B) (C) are correct. In the circuit F = (A 5 B) 9 (A 9 B) 9 C For two variables A5B = A9B So, (A 5 B) 9 (A 9 B) = 0 (always) F = 09C = 0$C+1$C = C So, F = 1 when C = 1 or C = 0 Option (D) is correct. Let QA (n), QB (n), QC (n) are present states and QA (n + 1), QB (n + 1), QC (n + 1) are next states of flop-flops. In the circuit

A I D

QA (n + 1) = QB (n) 9 QC (n) QB (n + 1) QA (n) n o.i c . QC (n + 1) QB (n) dia o Initially all flip-flops are reset .n w w 1st clock pulse w

O N

QB ^t + 1h = Q A

QA = 0 9 0 = 1 QB = 0 QC = 0 2 nd clock pulse QA = 0 9 0 = 1 QB = 1 QC = 0

DA = Q A Q B + QA QB 4.15

4.16

Option (C) is correct. Initially Carry Flag, C = 0 MVI A, 07 H ; A = 0000 0111 RLC ; Rotate left without carry. A = 0000 1110 MVO B, A ; B = A = 0000 1110 RLC ; A = 0001 1100 RLC ; A = 0011 1000 ADD B ; A = 0011 1000 + 0000 1110 ; 0100 0110 ; RRC ; Rotate Right with out carry, A = 0010 0011 Thus A = 23 H Option ( ) is correct.

3 rd clock pulse QA = 1 9 0 = 0 QB = 1 QC = 1 4 th clock pulse

So, sequence 4.20

QA = 1 9 1 = 1 QB = 0 QC = 1 QA = 01101.......

Option (D) is correct. Output of the MUX can be written as F = I 0 S 0 S1 + I1 S 0 S1 + I 2 S 0 S1 + I 3 S 0 S1 Here, I 0 = C, I1 = D, I2 = C , I 3 = CD and S 0 = A, S1 = B So, F = C A B + D A B + C A B + C DA B Writing all SOP terms

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Page 96

F = A B C D + A B C D + A BCD + A B C D 1 44 2 44 3 1 44 2 44 3 S 1 44 2 4 43 m m m m 3

7

2

5

+A B C D + A B C D + ABC D 1 44 2 4 4 3 1 44 2 44 3 S m m m 9

F = / m (2, 3, 5, 7, 8, 9, 12) 4.21

12

8

Option (C) is correct. By executing instruction one by one MVI A, 45 H & MOV 45 H into accumulator, A = 45 H STC & Set carry, C = 1 CMC & Complement carry flag, C = 0 RAR & Rotate accumulator right through carry

The truth table is as shown below. Sequence is 00, 11, 10, 00 ... CLK

J1

K1

Q1

J2

K2

Q2

1

1

1

0

1

1

0

2

1

1

1

1

1

1

3

0

0

1

0

1

0

4

1

1

0

1

1

0

Option (B) is correct. The given situation is as follows

4.27

A = 00100010 XRA B & XOR A and B A = A 5 B = 00100010 5 01000101 = 01100111 = 674 4.22

4.23

4.24

4.25

Option (C) is correct. TTL " Transistor - Transistor logic CMOS " Complementary Metal Oxide Semi-conductor

The truth table is as shown below

Option (D) is correct. Vectored interrupts : Vectored interrupts are those interrupts in which program control transferred to a fixed memory location. Maskable interrupts : Maskable interrupts are those interrupts which can be rejected or delayed by microprocessor if it is performing some critical task.

IA

D O

P2

a

b

c

d

e

f

g

0

0

1

1

1

1

1

1

0

0

1

1

0

1

1

0

1

1

1

0

1

1

0

1

1

0

1

1

0

0

1

1

1

1

n o.i 1

1 c Option (D) is correct. . a di From truth table o We have 6X + Z {Y + (Z + XY )}@ [X + Z (X + Y)] = 1 n w. Substituting X = 1 and X = 0 we get w w [1 + Z {Y + (Z + 1Y )}][ 0 + Z (1 + Y)] = 1 or [1][ Z (1)] = 1 1 + A = 1 and 0 + A = A or Z =1)Z=0

N

Option (A) is correct. The AND gate implementation by 2:1 mux is as follows

and

a b c d

=1 = P 1 P 2 + P1 P 2 = P 2 = P1 P2 + P1 P2 = P1 = 1 = c+e

c = P1 P2 = P1 + P2

1 NOT Gate 1 NOT Gate 1 OR GATE

g = P1 + P2 LED d is 1 all condition and also it depends on d = c+e

The EX - OR gate implementation by 2:1 mux is as follows

4.28

4.29

Y = BI0 + BI1 = AB + BA

we can write

1 OR GATE f = P1 P2 = P1 + P2 1 OR GATE g = P1 P2 = P1 + P2 Thus we have g = P1 + P2 and d = 1 = c + e . It may be observed easily from figure that Led g does not glow only when both P1 and P2 are 0. Thus

Y = AI 0 + AI1 = AB

4.26

P1

Option (D) is correct. As shown in previous solution 2 NOT gates and 3-OR gates are required. Option (C) is correct. For the NAND latche the stable states are as follows

Option (A) is correct. The given circuit is as follows.

For the NOR latche the stable states are as follows

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Page 97

The output Q0 of first FF occurs after time 3 T and it is as shown below

4.30

Option (D) is correct. From the figure shown below it may be easily seen upper MOSFET are shorted and connected to Vdd thus OUT is 1 only when the node S is 0,

The output Q1 of second FF occurs after time 3 T when it gets input (i.e. after 3 T from t1) and it is as shown below

Option (D) is correct.

4.35

VDAC =

We have

n=0

Since the lower MOSFETs are shorted to ground, node S is 0 only when input P and Q are 1. This is the function of AND gate. 4.31

4.32

So and

Clock

b3 b3 b2 b0

VDAC

1

0001

0

2

0010

0.5

3

0011

1

4

0100

1.5

5

0101

2

6

0110

2.5

7

0111

3

8

1000

3.5

9

1001

4

10

1010

4.5

11

1011

5

12

1100

5.5

13

1101

6

14

1110

6.5

IA

D O

Option (D) is correct. The circuit is as shown below

N

4.33

or VDAC = 0.5b0 + b1 + 2b2 + 4b3 The counter outputs will increase by 1 from 0000 till Vth > VDAC . The output of counter and VDAC is as shown below

Option (B) is correct. MSB of both number are 1, thus both are negative number. Now we get 11101101 = (- 19) 10 and 11100110 = (- 26) 10 P - Q = (- 19) - (- 26) = 7 Thus 7 signed two’s complements form is (7) 10 = 00000111

3

/ 2n - 1bn = 2- 1b0 + 20 b1 + 21b2 + 22 b3

no w.

n o.i

c

. dia

ww

X = PQ Y = (P + Q) Z = PQ (P + Q) = (P + Q )( P + Q) = PQ + PQ = P 5 Q M1 = Z 5 R = (P 5 Q) 5 R

and when VADC = 6.5 V (at 1101), the output of AND is zero and the counter stops. The stable output of LED display is 13.

Option (A) is correct. The circuit is as shown below

4.36

4.37

Option (B) is correct. The VADC - Vin at steady state is = 6.5 - 6.2 = 0.3V Option (A) is correct. Z = I0 RS + I1 RS + I2 RS + I3 RS = (P + Q ) RS + PRS + PQRS + PRS = PRS + QRS + PRS + PQRS + PRS The k - Map is as shown below

The truth table is shown below. When CLK make transition Q goes to 1 and when D goes to 1, Q goes to 0 4.34

Option (B) is correct. Since the input to both JK flip-flop is 11, the output will change every time with clock pulse. The input to clock is Z = PQ + PQS + QRS 4.38

Option (C) is correct.

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Page 98

2710H LXI H, 30A0H ; Load 16 bit data 30A0 in HL pair 2713H DAD H ; 6140H " HL 2714H PCHL ; Copy the contents 6140H of HL in PC Thus after execution above instruction contests of PC and HL are same and that is 6140H 4.39

4.40

Option (C) is correct. MSB of Y is 1, thus it is negative number and X is positive number Now we have X = 01110 = (14) 10 and Y = 11001 = (- 7) 10 X + Y = (14) + (- 7) = 7 In signed two’s complements from 7 is (7) 10 = 000111

The current from voltage source is I = VR = 10 = 1 mA R 10k

Option (B) is correct. Y = AB + CD = AB .CD This is SOP form and we require only 3 NAND gate

4.41

This current will be divide as shown below

Option (A) is correct. The circuit is as shown below

A I D Now

Y = AB + AB and

4.42

O N

Y = ABCD + ABCD + ABC D + ABC D = ABCD + ABC D + ABC D + ABC D = ABCD + ABC D + BC D (A + A) = ABCD + ABC D + BC D 4.43

4.44

4.45

4.46

Option (C) is correct. The net current in inverting terminal of OP - amp is I - = 1 + 1 = 5I n 4 16 16 i o. c . V0 =- R # 5I =- 3.125 dia So that 16 o n 4.47

X = YC + YC = (AB + AB ) C + (AB + AB ) C = (AB + AB) C + (AB + AB ) C = ABC + ABC + ABC + ABC

Option (D) is correct.

.

w ww

4.48

A+A = 1

Option (B) is correct. In given TTL NOT gate when Vi = 2.5 (HIGH), then Q1 " Reverse active Q2 " Saturation Q3 " Saturation Q4 " cut - off region Option (C) is correct. For X = 0, Y = 1 For X = 0, Y = 0 For X = 1, Y = 1

-3 i = I = 1 # 10 = 62.5 m A 16 16

Option (B) is correct. Since the inverting terminal is at virtual ground the resistor network can be reduced as follows

; Move B5H to A ; Move 0EH to B ; [A] XOR 69H and store in A ; Contents of A is CDH ; Add the contents of A to contents of B and ; store in A, contents of A is EAH ; [a] AND 9BH, and store in A, ; Contents of A is 8 AH ; Compare 9FH with the contents of A ; Since 8 AH < 9BH, CY = 1 ; Store the contents of A to location 3010

4 : ADDB 5 : ANI 9BH 6 : CPI 9FH

P = 1, Q = 0 P = 1, Q = 1 P = 1, Q = 0 or P = 0, Q = 1

Option (C) is correct. Chip 8255 will be selected if bits A3 to A7 are 1. Bit A0 to A2 can be 0 or. 1. Thus address range is 11111000 F8H 11111111 FFH

Option (B) is correct. Line 1 : MVI A, B5H 2 : MVI B, 0EH 3 : XRI 69H

7 : STA 3010 H H

8 : HLT ; Stop Thus the contents of accumulator after execution of ADD instruction is EAH. 4.49

4.50

Option (C) is correct. The CY = 1 and Z = 0 Option (A) is correct. For this circuit the counter state (Q1, Q0) follows the sequence 00, 01, 10, 00 ... as shown below Clock 1st

D1 D0 01

Q1 Q0

Q1 NOR Q0

00

1

10

0

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2nd

10

01

0

3rd

00

00

0

Page 99

After execution of POP PSW, SP + 2 " SP After execution of RET, SP + 2 " SP Thus the contents of SP will be 3CF4H + 4 = 3CF8H Option (A) is correct. The inputs D0 and D1 respectively should be connected as Q1 and Q0 where Q0 " D1 and Q1 " D0

4.57

Option (D) is correct. If the point P is stuck at 1, then output f is equal to A

4.58

4.51

4.52

4.53

4.54

Option (A) is correct. As shown below there are 2 terms in the minimized sum of product expression. 0

0

1

0

d

0

0

0

0

d

1

1

0

0

1

Option (B) is correct. Dividing 43 by 16 we get 2 16 43 32 11 11 in decimal is equivalent is B in hexamal. Thus 4310 * 2B16 Now 410 * 01002 310 * 00112 Thus 4310 * 01000011BCD

4.59

g

Option (B) is correct. The output is taken from the 5th line. Option (D) is correct. After applying two clock poles, the outputs of the full adder is S = 1 , C0 = 1 A B Ci S Co 1st 1 0 0 0 1 2nd 1 1 1 1 1 Option (D) is correct. 100010011001 S SSS 4

4.55

1

2

3

1

A I D

Option (A) is correct. The diagram is as shown in fig

4.60

O N

.in

co ia.

Option (B) is correct. d In this the diode D2 is connected to the ground. The following table .no w shows the state of counter and D/A converter ww Q2 Q1 Q0

D3 = Q2

D2 = 0

D1 = Q1

D0 = Q0

Vo

000

0

0

0

0

0

001

0

0

0

1

1

010

0

0

1

0

2

011

0

0

1

1

3

100

1

0

0

0

8

101

1

0

0

1

9

110

1

0

1

0

10

111

1

0

1

1

11

000

0

0

0

0

0

001

0

0

0

1

1

f' = BC + BC f = f'A + f'0 = f'A = ABC + ABC 4.61

Option (C) is correct. The circuit is as shown below

Thus option (B) is correct 4.56

Option (B) is correct. LXI, EFFF H ; Load SP with data EFFH CALL 3000 H ; Jump to location 3000 H : : : 3000H LXI H, 3CF4 ; Load HL with data 3CF4H PUSH PSW ; Store contnets of PSW to Stack POP PSW ; Restore contents of PSW from stack PRE ; stop Before instruction SPHL the contents of SP is 3CF4H.

If output is at logic 0, the we have V0 = 0 which signifies BJT Q3 is in saturation and applying KVL we have or or 4.62

4.63

VBE3 = IR # 1k 0.75 = IR # 1k IR = 0.75 mA

Option (A) is correct. We have f = ABC + ABC = B (AC + AC ) = B (A + C)( A + C ) Option (C) is correct.

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Page 100

Characteristic equation for a jk flip-flop is written as Qn + 1 Qn Qn + 1 Qn + 1 Qn + 1

Where So, 4.64

4.65 4.66

Counter " Frequency division Decoder " Addressing in memory chips.

= JQ n + K Qn is the present output is next output = 10 + K : 0 =1

4.72

Qn = 0 4.73

Option (C) is correct. Since T2 T1 T0 is at 111, at every clock Q2 Q1 Q0 will be changes. Ir present state is 011, the next state will be 100. Option (D) is correct. Option (C) is correct. 0100H LXI SP, 00FF 0103H LXI H, 0701 0106H MVI A, 20H 0108 H SUB M

; Load SP with 00FFG ; Load HL with 0107H ; Move A with 20 H ; Subtract the contents of memory ; location whose address is stored in HL ; from the A and store in A 0109H ORI 40H ; 40H OR [A] and store in A 010BH ADD M ; Add the contents of memeory location ; whose address is stored in HL to A ; and store in A HL contains 0107H and contents of 0107H is 20H Thus after execution of SUB the data of A is 20H - 20H = 00 4.67

4.74

4.75

Option (A) is correct. For the TTL family if terminal is floating, then it is at logic 1. Thus Y = (AB + 1) = AB .0 = 0 Option (C) is correct. 11001 1001 111001 00110 0110 000110 +1 +1 +1 00111 0111 000111 7 7 7 Thus 2’s complement of 11001, 1001 and 111001 is 7. So the number given in the question are 2’s complement correspond to -7. Option (C) is correct. In the modulo - 6 ripple counter at the end of sixth pulse (i.e. after 101 or at 110) all states must be cleared. Thus when CB is 11 the all states must be cleared. The input to 2-input gate is C and B and the desired output should be low since the CLEAR is active low Thus when C and B are 0, 0, then output must be 0. In all other case the output must be 1. OR gate can implement this functions. Option (C) is correct. Number of MUX is 4 = 2 and 2 = 1. Thus the total number 3 3 2 multiplexers is required.

A I D

4.76 Option (D) is correct. Option (C) is correct. Before ORI instruction the contents of A is 00H. On execution the AC + BC = AC1 + BC 1 ORI 40H the contents of A will be 40H = AC (B + B ) + BC (A + A) 00H = 00000000 = ACB + ACB + BC A + BC A 40H = 01000000 .in (D) is correct. 4.77 oOption c . ORI 01000000 f (x, y) = xy + xy + xy = x (y + y) + xy dia We have o n . After ADD instruction the contents of memory location whose adw = x + xy w w dress is stored in HL will be added to and will be stored in A or f (x, y) = x + y 40H + 20 H = 60 H Here compliments are not available, so to get x we use NOR gate. Option (C) is correct. Thus desired circuit require 1 unit OR and 1 unit NOR gate giving A master slave D-flip flop is shown in the figure. total cost 2 unit.

O N

4.68

4.78

In the circuit we can see that output of flip-flop call be triggered only by transition of clock from 1 to 0 or when state of slave latch is affected. 4.69

4.70

Option (D) is correct. The minimum number of bit require to encode 100 increment is or

4.71

Form (ii) the mode is 1. Mode 2 : Bi-directional data transfer This mode is used to transfer data between two computer. In this mode port A can be configured as bidirectional port. Port A uses five signal from port C as hand shake signal. For (1), mode is 2

Option (A) is correct. The range of signed decimal numbers that can be represented by n - bits 1’s complement number is - (2n - 1 - 1) to + (2n - 1 - 1). Thus for n = 6 we have Range =- (26 - 1 - 1) to + (26 - 1 - 1) =- 31 to + 31

2n $ 100 n $7

Option (B) is correct. Shift Register " Serial to parallel data conversion

Option (D) is correct. For 8255, various modes are described as following. Mode 1 : Input or output with hand shake In this mode following actions are executed 1. Two port (A & B) function as 8 - bit input output ports. 2. Each port uses three lines from C as a hand shake signal 3. Input & output data are latched.

4.79

Option (B) is correct. LDA 16 bit & Load accumulator directly this instruction copies data byte from memory location (specified within the instruction) the accumulator. It takes 4 memory cycle-as following. 1. in instruction fetch 2. in reading 16 bit address 1. in copying data from memory to accumulator

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Page 101

LXI D, (F0F1) 4 & It copies 16 bit data into register pair D and E. It takes 3 memory cycles. 4.80

4.81

4.82

Option (A) is correct. LXI H, 9258H MOV A, M CMa MOV M, A This program complement

= R + S + (P + Q )( P + Q + R)( P + Q + S) = R + S + PQ + PQ + PQS + PR + PQR + PRS + PQ + PQS + PQR + QRS

; 9258H " HL ; (9258H) " A ; A"A ; A"M the data of memory location 9258H.

= R + S + PQ + PQS + PR + PQR + PRS + PQS + PQR + QRS

Option (D) is correct. MVI A, 00H ; Clear accumulator LOOP ADD B ; Add the contents of B to A DCR C ; Decrement C JNZ LOOP ; If C is not zero jump to loop HLT END This instruction set add the contents of B to accumulator to contents of C times.

= R + S + PQ (1 + S) + PR (1 + P ) + PRS + PQS + PQR + QRS = R + S + PQ + PR + PRS + PQS + PQR + QRS

Option (D) is correct. The number of distinct boolean expression of n variable is 22n . Thus

= R + S + PQ + PR (1 + Q ) + PQS + QRS = R + S + PQ + PR + PQS + QRS Thus W = Z and X = Z 4.89

22 = 216 = 65536 4

4.83

Option (C) is correct. In the flash analog to digital converter, the no. of comparators is equal to 2n - 1, where n is no. of bit.s So,

4.84

4.85

4.86

A I D

2n - 1 = 28 - 1 = 255

Option (B) is correct. When output of the 74 series gate of TTL gates is taken from BJT then the configuration is either totem pole or open collector configuration .

O N

4.88

S = tpd = 10 ns

Option in (C) is correct. . o c Option (D) is correct. ia. After t = t1, at first rising edge of clock, the output of shift register d is 0110, which in input to address line of ROM. At 0110 is applied o A 2n: 1 MUX can implement all logic functions of (n + 1) variablew.n to register. So at this time data stroed in ROM at 1010 (10), 1000 without andy additional circuitry. Here n = 3 . Thus a 8 : 1 MUX ww will be on bus. can implement all logic functions of 4 variable. When W has the data 0110 and it is 6 in decimal, and it’s data Option (D) is correct. value at that add is 1010 Counter must be reset when it count 111. This can be implemented then 1010 i.e. 10 is acting as odd, at time t2 and data at that by following circuitry movement is 1000. 4.90

4.91

4.87

Option (B) is correct. Propagation delay of flip flop is tpd = 10 nsec Propagation delay of 4 bit ripple counter R = 4tpd = 40 ns and in synchronous counter all flip-flop are given clock simultaneously, so

Option (B) is correct. We have Y = P5Q5R Z = RQ + PR + QP Here every block is a full subtractor giving P - Q - R where R is borrow. Thus circuit acts as a 4 bit subtractor giving P - Q .

4.92

4.93

Option (A) is correct. W = R + PQ + RS X = PQRS + PQRS + PQRS

4.94

Y = RS + PR + PQ + PQ = RS + PR $ PQ $ PQ = RS + (P + R )( P + Q)( P + Q) = RS + (P + PQ + PR + QR )( P + Q) = RS + PQ + QR (P + P ) + QR = RS + PQ + QR Z = R + S + PQ + PQR + PQS

Option (B) is correct. The DTL has minimum fan out and CMOS has minimum power consumption. Propagation delay is minimum in ECL. Option (D) is correct. Let input be 1010; output will be 1101 Let input be 0110; output will be 0100 Thus it convert gray to Binary code. Option (A) is correct. CMP B & Compare the accumulator content with context of Register B CY is set and zero flag will be reset. If A < R Option (A) is correct. Vo =- V1 :R bo + R b1 + R b2 + R b 3D R 2R 4R 4R Exact value when V1 = 5 , for maximum output VoExact =- 5 :1 + 1 + 1 + 1 D =- 9.375 2 4 8 Maximum Vout due to tolerance Vo max =- 5.5 :110 + 110 + 110 + 110 D 90 2 # 90 4 # 90 8 # 90

= R + S + PQ $ PQR $ PQS

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Page 102

=- 12.604 Tolerance 4.95

4.96

4.97

4.98

17 = 010001 Its 1’s complement is 101110 So 2’s compliment is 101110 + 1

= 34.44% = 35%

Option (D) is correct. If the 4- bit 2’s complement representation of a decimal number is 1000, then the number is -8 Option (B) is correct. Output of 1 st XOR = = X $ 1 + X $ 1 = X Output of 2 nd XOR = X X + XX = 1 So after 4,6,8,...20 XOR output will be 1.

101111 4.103

Option (C) is correct. In the comparator type ADC, the no. of comparators is equal to 2n - 1 , where n is no. of bit.s So, 23 - 1 = 7

4.104

Option (C) is correct. The circuit is as shown below

AA00H + 1000H - 0001H = B9FFH 4.105

The circuit shown is monostable multivibrator as it requires an external triggering and it has one stable and one quasistable state. 4.99

Option (C) is correct. The propagation delay of each inverter is tpd then The fundamental frequency of oscillator output is 1 = 1 GHz f = 1 = 2ntpd 2 # 5 # 100 # 10 - 12 Option (C) is correct. 4K # 8 bit means 102410 location of byte are present Now 102410 * 1000H It starting address is AA00H then address of last byte is

Option (B) is correct. They have prorogation delay as respectively, G1 " 10 nsec G2 " 20 nsec For abrupt change in Vi from 0 to 1 at time t = t0 we have to assume the output of NOR then we can say that option (B) is correct waveform.

Option (D) is correct.

or 4.106

Y = I0 + I3 + I5 + I6 = C BA + C AB + CBA + CBA Y = C (A 5 B ) + C (A 5 B)

Option (D) is correct. For the LED to glow it must be forward biased. Thus output of NAND must be LOW for LED to emit light. So both input to NAND must be HIGH. If any one or both switch are closed, output of AND will be LOW. If both switch are open, output of XOR will be LOW. So there can’t be both input HIGH to NAND. So LED doesn’t emit light.

A I D

O N

.in (C) is correct. oOption c . ia

4.107

no w.

d

The output of options (C) satisfy the given conditions

ww

4.100

4.101

Option (B) is correct. Let X3 X2 X1 X0 be 1001 then Y3 Y2 Y1 Y0 will be 1111. Let X3 X2 X1 X0 be 1000 then Y3 Y2 Y1 Y0 will be 1110 Let X3 X2 X1 X0 be 0110 then Y3 Y2 Y1 Y0 will be 1100 So this converts 2-4-2-1 BCD numbers. Option (B) is correct. MVI B, 87H MOV A, B START : JMP NEXT XRA B JP START

NEXT :

JMP NEXT XRA

JP START OUT PORT2 4.102

; B = 87 ; A = B = 87 ; Jump to next ; A 5 B " A, ; A = 00, B = 87 ; Since A = 00 is positive ; so jump to START ;Jump to NEXT ; unconditionally ; B ; A 5 B " A, A = 87 , ; B = 87 H ; will not jump as D7 , of A is 1 ; A = 87 " PORT2

Option (B) is correct. The two’s compliment representation of 17 is

4.108 4.109

4.110

4.111

Option (B) is correct. Option (B) is correct. Conversion time of successive approximate analog to digital converters is independent of input voltage. It depends upon the number of bits only. Thus it remains unchanged. Option (C) is correct. In the flash analog to digital converter, the no. of comparators is equal to 2n - 1, where n is no. of bits. So, 2 4 - 1 = 15 Option (D) is correct. As the output of AND is X = 1, the all input of this AND must be 1. Thus ...(1) AB + AB = 1 ...(2) BC + BC = 1 ...(3) C =1 From (2) and (3), if C = 1, then B = 1

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Page 103

Vo = 8 # 5 = 5V 8

If B = 1, then from (1) A = 0 . Thus A = 0, B = 1 and C = 1 4.112

4.113

4.114

Option (C) is correct. Interrupt is a process of data transfer by which an external device can inform the processor that it is ready for communication. 8085 microprocessor have five interrupts namely TRAP, INTR, RST 7.5, RST 6.5 and RST 5.5

Z = XQ + YQ Comparing from the truth table of J - K FF Y = J,

Option (A) is correct. For any RST instruction, location of program transfer is obtained in following way. RST x & (x ) 8) 10 " convert in hexadecimal So for RST 6 & (6 ) 8) 10 = (48) 10 = (30) H

X =K

Option (A) is correct. Accumulator contains A = 49 H Register B = 3 AH SUB B = A minus B

Carry so here outputA Carry CY Sign flag S

=1 =0F =1 =1

Option (C) is correct. The circuit is as shown below :

Y

Z

0

0

Q

0

1

0

1

0

1

1

1

Q1

Option (D) is correct. We have y = A + AB we know from Distributive property

4.119

A I D Thus

O N

no w.

d

4.121 4.122

Y = B + (B + C ) = B (B + C ) = B

x + yz = (x + y) (x + z) y = (A + A) (A + B) = A + B

Option (C) is correct. Darligton emitter follower provides a low output impedance in both logical state (1 or 0). Due to this low output impedance, any stray capacitance is rapidly charged and discharged, so the output state .in o c . ia changes quickly. It improves speed of operation.

4.120

ww

4.116

X

Option (B) is correct. In the figure the given counter is mod-10 counter, so frequency of output is 10k = 1k 10

4.118

A = 49 H = 01001001 B = 3 AH = 00111010 2’s complement of (- B) = 11000110 A - B = A + (- B) 010 010 01 & +1 1 0 0 0 1 1 0 0 0 0 0 1111

4.115

Option (D) is correct. The truth table is shown below

4.117

Option (D) is correct. Option (B) is correct. For ADC we can write Analog input = (decimal eq of digital output) # resol 6.6 = (decimal eq. of digital output) # 0.5 6.6 = decimal eq of digital. output 0.5 13.2 = decimal equivalent of digital output so output of ADC is = 1101.

Option (B) is correct. The circuit is as shown below

4.123

Option (A) is correct. We use the K -map as below.

So given expression equal to The voltage at non-inverting terminal is V+ = 1 + 1 = 5 8 2 8 V- = V+ = 5 8 Now applying voltage divider rule V- = 1k V% = 1 Vo 1k + 7k 8

= AC + BC + AB 4.124

Option (C) is correct. For a binary half-subtractor truth table si given below.

...(1)

...(2)

From (1) and (2) we have

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= 4+1 = 5

from truth table we can find expressions of D & X D = A 5 B = AB + AB X = AB 4.125

Page 104

Option (B) is correct. For TTL worst cases low voltages are

4.128

VOL (max) = 0.4 V VIL (max) = 0.8 V Worst case high voltages are

Option (D) is correct. From the given figure we can write the output

VOH (min) = 2.4 V VIH (min) = 2 V The difference between maximum input low voltage and maximum output low voltage is called noise margin. It is 0.4 V in case of TTL. Option (D) is correct. From the figure we can see If A =1 B=0 then y =1 x=0 If A =1 B=1 then also y =1 x=0 so for sequence B = 101010....output x and y will be fixed at 0 and 1 respectively.

4.129

For the state 010 all preset = 1 and output QA QB QC = 111 so here total no. of states = 5 (down counter) 4.126

Option (B) is correct. We have 4 K RAM (12 address lines)

Option (D) is correct. Given 2’s complement no. 1101; the no. is 0011 for 6 digit output we can write the no. is – 000011 2’s complement representation of above no. is 111101

4.130

A I D

Option (A) is correct.

4.131

Option (B) is correct. An I/O Microprocessor controls data flow between main memory and the I/O device which wants to communicate.

4.132

O N w

ww

so here chip select logic CS = A15 A14 A13 address range (111) A15 A14 A13 A12 A11 A10 A 9 A 8 A7 A6 A5 A 4 A 3 A2 A1 A 0 initial 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 address &7000H final 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 address &7FFFH so address range is (7 0 0 0 H – 7 F F F H) 4.127

ia

4.134

d .no

4.135

4.136

4.137

Z = ABC Z = ABC = ACB = AC + B

Thus Z = AC + B we have Z = X + Y (1 NOR gate) where X = AC (1 NAND gate) To implement a NOR gate we required 4 NAND gates as shown below in figure.

Option (B) is correct. Dual slope ADC is more accurate. Option (A) is correct. Dual form of any identity can be find by replacing all AND function to OR and vice-versa. so here dual form will be (A + B) (A + C) (B + C) = (A + B) (A + C)

Option (C) is correct. Given boolean function is Now

Option n (D) is correct.

.i .co

4.133

Option (B) is correct. Carry flag will be affected by arithmetic instructions only. Option (C) is correct. This is a synchronous counter. we can find output as QA QB 0 0 1 0 0 1 0 0 h So It counts only three states. It is a mod-3 counter. K =3

4.138 4.139

Option (B) is correct. Option (A) is correct. Essential prime implicates for a function is no. of terms that we get by solving K -map. Here we get 4 terms when solve the K -map.

here total no. of NAND gates required

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Page 105

Carry = 1 DAA (Carry Flag is set, so DAA adds 6 to high order four bits) 0 1111 0 0 1 DAA add 1 0 0 0 1 0 1 0 A = 0 0 0 0 0 0 1 1 = 63 H MOV H, A (copy contain of A to H) H = 63 H PCHL (Load program counter by HL pair) PC = 6379 H

4.140 4.141

Option (C) is correct.

4.148

y = B D + A C D + C AB + CA B so no of prime implicates is 4

Option (C) is correct. NMOS In parallel makes OR Gate & in series makes AND so here we can have

4.149

Option (A) is correct. Option (B) is correct. For a 2 bit multiplier

C3

B1 # A1 A 0 B1 # A1 B1 A1 B 0 C2 C1

F = A (B + C) + DE we took complement because there is another NMOS given above (works as an inverter)

B0 A0 A0 B0

Option (D) is correct. For a J -K flip flop we have characteristic equation as Q (t + 1) = JQ (t) + KQ (t) Q (t) & Q (t + 1) are present & next states. In given figure J = Q (t), K = 1 so

4.150

C0

This multiplication is identical to AND operation and then addition. 4.142

4.143

Option (C) is correct. In totem pole stage output resistance will be small so it acts like a output buffer. Option (B) is correct. Consider high output state fan out = IOH max = 400 mA = 20 IIH max 20 mA Consider low output state fan out = IOL max = 8 mA = 80 IIL max 0.1 mA Thus fan out is 20

4.144

Option (A) is correct. The given gate is ex-OR so output

4.145

A I D

O N

4.151

d

4.147

4.153

Option (C) is correct.

Option (A) is correct. Here only for the range 60 to 63 H chipselect will be 0, so peripheral will correspond in this range only chipselect = 1 for rest of the given address ranges. Option (B) is correct. By executing instructions one by one LXI H, 8A79 H (Load HL pair by value 8A79) H = 8AH L = 79 H MOV A, L (copy contain of L to accumulator) A = 79 H ADDH (add contain of H to accumulator)

By distributive property in boolean algebra we have

ww

EI = Enabled Interput flag ,RST will cause an Interrupt only it we enable EI . 4.146

Option (C) is correct.

n o.i (B) is correct. Option c . ia

4.152

no w.

F = AB + AB B = 0 so, F = A1 + A0 = A

Here input

Q (t + 1) = Q (t) Q (t) + 0Q (t) Q (t + 1) = Q (t)[complement of previous state] we have initial input Q (t) = 0 so for 6 clock pulses sequence at output Q will be 010101

4.154 4.155

Option (A) is correct. The current in a p n junction diode is controlled by diffusion of majority carriers while current in schottky diode dominated by the flow of majority carrier over the potential barrier at metallurgical junction. So there is no minority carrier storage in schottky diode, so switching time from forward bias to reverse bias is very short compared to p n junction diode. Hence the propagation delay will reduces. Option (B) is correct. Option (D) is correct. The total conversion time for different type of ADC are given as– t is clock period For flash type & 1t Counter type & (2n - t) = 4095 m sec n = no.of bits Integrating type conver time > 4095 m sec successive approximation type nt = 12 m sec here n = 12 so

A = 79 H = 0 1111 0 0 1 H = 8AH = add 1 0 0 0 1 0 1 0 =A=

(A + BC) = (A + B) (A + C) (A + B) (A + C) = AA + AC + AB + BC = A (1 + C) + AB + BC = A + AB + BC = A (1 + B) + BC = A + BC

nt = 12

0 0 0 0 0 0 11

12t = 12

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Page 106

so this is succ. app. type ADC. 4.156

4.157

Option (D) is correct. LDA 2003 (Load accumulator by a value 2003 H) so here total no. of memory access will be 4. 1 = Fetching instruction 2 = Read the value from memory 1 = write value to accumulator Option (D) is correct. Storage capacitance -12 C = i = 1 # 10 5 - 0.5 dv b dt l b 20 10-3 l # -12

= 1 # 10 4.158

= 4.4 # 10-15 F

Option (A) is correct.

or or = 200 mV/cC 4.159

-3

# 20 # 10 4.5

Accuracy ! 1 LSB = Tcoff # DT 2 1 10.24 = T coff # DT 2 # 210 10.24 Tcoff = 2 # 1024 # (50 - 25) cC

Option (D) is correct.

A I D

210 # 8 = 13 No. of chips = 26 # 12 2 #4 4.160

Option (C) is correct. Given instruction set 1000 LXI SP 27FF 1003 CALL 1006 1006 POP H First Instruction will initialize the SP by a value 27FF SP ! 27FF CALL 1006 will “Push PC” and Load PC by value 1006 PUSH PC will store value of PC in stack

O N

no w.

.in

co ia.

d

ww

PC = 1006

now POP H will be executed which load HL pair by stack values HL SP SP SP

= 1006 and = SPl + 2 = SPl + 2 = SP - 2 + 2 = SP = 27FF

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

UNIT 5

Page 107

2013 5.8

SIGNALS & SYSTEMS 5.9

2013 5.1

5.2

ONE MARK

Two systems with impulse responses h1 ^ t h and h2 ^ t h are connected in cascade. Then the overall impulse response of the cascaded system is given by (A) product of h1 ^ t h and h2 ^ t h (B) sum of h1 ^ t h and h2 ^ t h (C) convolution of h1 ^ t h and h2 ^ t h (D) subtraction of h2 ^ t h from h1 ^ t h

^t - 1h2 (C) u ^t - 1h 2 5.3

5.4

5.5

5.6

2 (D) t - 1 u ^t - 1h 2

For a periodic signal v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p/4h, the fundamental frequency in rad/s (A) 100 (B) 300 (C) 500 (D) 1500

The impulse response of a continuous time system is given by h ^ t h = d ^t - 1h + d ^t - 3h. The value of the step response at t = 2 is (A) 0 (B) 1 (C) 2 (D) 3 A system described by the differential equation 2 dy dy + 5 + 6y ^ t h = x ^ t h. Let x ^ t h be a rectangular pulse given by dt dt2 1 0 0 , we need to (A) change the initial condition to - y ^0 h and the forcing function to 2x ^ t h (B) change the initial condition to 2y ^0 h and the forcing function to - x ^ t h (C) change the initial condition to j 2 y ^0 h and the forcing function to j 2 x ^ t h (D) change the initial condition to - 2y ^0 h and the forcing function into - 2x ^ t h

A I D

O N

A band-limited signal with a maximum frequency of 5 kHz is to co. . a be sampled. According to the sampling theorem, the sampling 5.11 di The DFT of a vector 8a b c dB is the vector 8a b g dB. Consider o n the product frequency which is not valid is w. w w (A) 5 kHz (B) 12 kHz R V Sa b c d W (C) 15 kHz (D) 20 kHz Sd a b c W 8p q r sB = 8a b c dBSc d a b W Which one of the following statements is NOT TRUE for a continuous S W b c d a S W time causal and stable LTI system? T X The DFT of the vector 8p q r sB is a scaled version of (A) All the poles of the system must lie on the left side of the jw axis (B) 9 a b (A) 9a2 b2 g2 d2C g dC (B) Zeros of the system can lie anywhere in the s-plane (C) 8a + b b + d d + g g + aB (D) 8a b g dB (C) All the poles must lie within s = 1 (D) All the roots of the characteristic equation must be located on 2012 ONE MARK the left side of the jw axis. 5.12 The unilateral Laplace transform of f (t) is 2 1 . The unilateral s +s+1 Assuming zero initial condition, the response y ^ t h of the system Laplace transform of tf (t) is given below to a unit step input u ^ t h is (B) - 2 2s + 1 2 (A) - 2 s 2 (s + s + 1) (s + s + 1) (C) 2 s (D) 2 2s + 1 2 (s + s + 1) 2 (s + s + 1) (A) u ^ t h 2 (C) t u ^ t h 2

5.7

5.10

The impulse response of a system is h ^ t h = tu ^ t h. For an input u ^t - 1h, the output is 2 t ^t - 1h (B) (A) t u ^ t h u ^t - 1h 2 2

TWO MARKS

5.13

(B) tu ^ t h (D) e-t u ^ t h

Let g ^ t h = e- pt , and h ^ t h is a filter matched to g ^ t h. If g ^ t h is applied as input to h ^ t h, then the Fourier transform of the output is (B) e- pf /2 (A) e- pf 2

2

(C) e

-p f

2

(D) e

-2pf2

If x [n] = (1/3) n - (1/2) n u [n], then the region of convergence (ROC) of its z -transform in the z -plane will be (B) 1 < z < 1 (A) 1 < z < 3 3 3 2 (C) 1 < z < 3 (D) 1 < z 2 3 2012

5.14

TWO MARKS

The input x (t) and output y (t) of a system are related as y (t) =

t

# x (t) cos (3t) dt . The system is -3

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(D) 0.5 [1 - exp (- 2t)] u (t) + u (t - 6)

(A) time-invariant and stable (B) stable and not time-invariant (C) time-invariant and not stable (D) not time-invariant and not stable 5.15

5.16

Page 108

5.22

The Fourier transform of a signal h (t) is H (jw) = (2 cos w) (sin 2w) /w . The value of h (0) is (A) 1/4 (B) 1/2 (C) 1 (D) 2 Let y [n] denote the convolution of h [n] and g [n], where h [n] = (1/2) n u [n] and g [n] is a causal sequence. If y [0] = 1 and y [1] = 1/2, then g [1] equals (A) 0 (B) 1/2 (C) 1 (D) 3/2 2011

1 - 0.6z-1 z-1 (1 - 0.4z-1) z-1 (1 - 0.4z-1) (C) (1 - 0.6z-1) (A)

5.23

ONE MARK 2

5.17

The differential equation 100 ddty - 20 dy dt + y = x (t) describes a system with an input x (t) and an output y (t). The system, which is initially relaxed, is excited by a unit step input. The output y ^ t h can be represented by the waveform

Two systems H1 (Z ) and H2 (Z ) are connected in cascade as shown below. The overall output y (n) is the same as the input x (n) with a one unit delay. The transfer function of the second system H2 (Z ) is

(B)

z-1 (1 - 0.6z-1) (1 - 0.4z-1)

(D)

1 - 0.4 z-1 z-1 (1 - 0.6z-1)

The first six points of the 8-point DFT of a real valued sequence are 5, 1 - j 3, 0, 3 - j 4, 0 and 3 + j 4 . The last two points of the DFT are respectively (A) 0, 1 - j 3 (B) 0, 1 + j 3 (C) 1 + j3, 5 (D) 1 - j 3, 5

2

2010 5.24

ONE MARK

The trigonometric Fourier series for the waveform f (t) shown below contains

A I D

O N

.in

no w.

co ia.

d

ww

5.18

5.19

5.20

The trigonometric Fourier series of an even function does not have the (A) dc term (B) cosine terms (C) sine terms (D) odd harmonic terms A system is defined by its impulse response h (n) = 2n u (n - 2). The system is (A) stable and causal (B) causal but not stable (C) stable but not causal (D) unstable and non-causal If the unit step response of a network is (1 - e- at), then its unit impulse response is (A) ae- at (B) a-1 e- at (C) (1 - a-1) e- at (D) (1 - a) e- at 2011

5.21

TWO MARKS

An input x (t) = exp (- 2t) u (t) + d (t - 6) is applied to an LTI system with impulse response h (t) = u (t) . The output is (A) [1 - exp (- 2t)] u (t) + u (t + 6) (B) [1 - exp (- 2t)] u (t) + u (t - 6) (C) 0.5 [1 - exp (- 2t)] u (t) + u (t + 6)

(A) only (B) only (C) only (D) only 5.25

5.26

5.27

cosine terms and zero values for the dc components cosine terms and a positive value for the dc components cosine terms and a negative value for the dc components sine terms and a negative value for the dc components

Consider the z -transform x (z) = 5z2 + 4z-1 + 3; 0 < z < 3. The inverse z - transform x [n] is (A) 5d [n + 2] + 3d [n] + 4d [n - 1] (B) 5d [n - 2] + 3d [n] + 4d [n + 1] (C) 5u [n + 2] + 3u [n] + 4u [n - 1] (D) 5u [n - 2] + 3u [n] + 4u [n + 1] Two discrete time system with impulse response h1 [n] = d [n - 1] and h2 [n] = d [n - 2] are connected in cascade. The overall impulse response of the cascaded system is (B) d [n - 4] (A) d [n - 1] + d [n - 2] (C) d [n - 3] (D) d [n - 1] d [n - 2] For a N -point FET algorithm N = 2m which one of the following statements is TRUE ? (A) It is not possible to construct a signal flow graph with both input and output in normal order (B) The number of butterflies in the m th stage in N/m (C) In-place computation requires storage of only 2N data (D) Computation of a butterfly requires only one complex multiplication.

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2010 5.28

5.29

5.30

TWO MARKS

5.32

5.33

TWO MARKS t

#0 f (t) dt is

(A) sF (s) - f (0)

(B) 1 F (s) s

5.38

-3

Fourier Transform (DFT) of a discrete time

(A) [0, - 2 + 2j , 2, - 2 - 2j ] (C) [6, 1 - 3j , 2, 1 + 3j ]

(B) [2, 2 + 2j , 6, 2 - 2j ] (D) [6, - 1 + 3j , 0, - 1 - 3j ]

An LTI system having transfer function s +s 2+s 1+ 1 and input x (t) = sin (t + 1) is in steady state. The output is sampled at a rate ws rad/s to obtain the final output {x (k)}. Which of the following is true ? (A) y (.) is zero for all sampling frequencies ws (B) y (.) is nonzero for all sampling frequencies ws (C) y (.) is nonzero for ws > 2 , but zero for ws < 2 (D) y (.) is zero for ws > 2 , but nonzero for w2 < 2 2

2

2008 5.39

5.40

Given that F (s) is the one-side Laplace transform of f (t), the Laplace transform of

d

ww

The ROC of z -transform of the discrete time sequence n n x (n) = b 1 l u (n) - b 1 l u (- n - 1) is 3 2 (A) 1 < z < 1 (B) z > 1 3 2 2 (C) z < 1 (D) 2 < z < 3 3

3

n The o.i 4-point Discrete c . ia sequence {1,0,2,3} is

5.37

no w.

A function is given by f (t) = sin2 t + cos 2t . Which of the following is true ? (A) f has frequency components at 0 and 1 Hz 2p (B) f has frequency components at 0 and 1 Hz p (C) f has frequency components at 1 and 1 Hz 2p p (D) f has frequency components at 0.1 and 1 Hz 2p p

# x (t - t) g (2t) dt where h (t) is shown in the graph.

Which of the following four properties are possessed by the system ? BIBO : Bounded input gives a bounded output. Causal : The system is causal, LP : The system is low pass. LTI : The system is linear and time-invariant. (A) Causal, LP (B) BIBO, LTI (C) BIBO, Causal, LTI (D) LP, LTI

A I D

O N

(D) 1 [F (s) - f (0)] s

F (t) dt

equation y (t) =

ONE MARK

The Fourier series of a real periodic function has only (P) cosine terms if it is even (Q) sine terms if it is even (R) cosine terms if it is odd (S) sine terms if it is odd Which of the above statements are correct ? (A) P and S (B) P and R (C) Q and S (D) Q and R

s

Consider a system whose input x and output y are related by the

5.36

The transfer function of a discrete time LTI system is given by 2 - 3 z-1 4 H (z) = 3 -1 1 - z + 1 z-2 8 4 Consider the following statements: S1: The system is stable and causal for ROC: z > 1/2 S2: The system is stable but not causal for ROC: z < 1/4 S3: The system is neither stable nor causal for ROC: 1/4 < z < 1/2 Which one of the following statements is valid ? (A) Both S1 and S2 are true (B) Both S2 and S3 are true (C) Both S1 and S3 are true (D) S1, S2 and S3 are all true

#0

A system with transfer function H (z) has impulse response h (.) defined as h (2) = 1, h (3) =- 1 and h (k) = 0 otherwise. Consider the following statements. S1 : H (z) is a low-pass filter. S2 : H (z) is an FIR filter. Which of the following is correct? (A) Only S2 is true (B) Both S1 and S2 are false (C) Both S1 and S2 are true, and S2 is a reason for S1 (D) Both S1 and S2 are true, but S2 is not a reason for S1

5.35

A continuous time LTI system is described by dx (t) d 2 y (t) dy (t) + 4x (t) +4 + 3y (t) = 2 2 dt dt dt Assuming zero initial conditions, the response y (t) of the above system for the input x (t) = e-2t u (t) is given by (A) (et - e3t) u (t) (B) (e-t - e-3t) u (t) (C) (e-t + e-3t) u (t) (D) (et + e3t) u (t)

2009 5.34

(C)

3s + 1 Given f (t) = L-1 ; 3 . If lim f (t) = 1, then the value t"3 s + 4s2 + (k - 3) s E of k is (A) 1 (B) 2 (C) 3 (D) 4

2009 5.31

Page 109

ONE MARK

The input and output of a continuous time system are respectively denoted by x (t) and y (t). Which of the following descriptions corresponds to a causal system ? (A) y (t) = x (t - 2) + x (t + 4) (B) y (t) = (t - 4) x (t + 1) (C) y (t) = (t + 4) x (t - 1) (D) y (t) = (t + 5) x (t + 5) The impulse response h (t) of a linear time invariant continuous time system is described by h (t) = exp (at) u (t) + exp (bt) u (- t) where u (- t) denotes the unit step function, and a and b are real constants. This system is stable if (A) a is positive and b is positive (B) a is negative and b is negative (C) a is negative and b is negative (D) a is negative and b is positive

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2008 5.41

5.42

5.43

Page 110

TWO MARKS

(C) 5 (1 - e-5n)

A linear, time - invariant, causal continuous time system has a rational transfer function with simple poles at s =- 2 and s =- 4 and one simple zero at s =- 1. A unit step u (t) is applied at the input of the system. At steady state, the output has constant value of 1. The impulse response of this system is (A) [exp (- 2t) + exp (- 4t)] u (t) (B) [- 4 exp (- 2t) - 12 exp (- 4t) - exp (- t)] u (t) (C) [- 4 exp (- 2t) + 12 exp (- 4t)] u (t) (D) [- 0.5 exp (- 2t) + 1.5 exp (- 4t)] u (t) The signal x (t) is described by 1 for - 1 # t # + 1 x (t) = ) 0 otherwise Two of the angular frequencies at which its Fourier transform becomes zero are (A) p, 2p (B) 0.5p, 1.5p (C) 0, p (D) 2p, 2.5p

5.47

The impulse response h (t) of linear time - invariant continuous time system is given by h (t) = exp (- 2t) u (t), where u (t) denotes the unit step function.

5.49

The output of this system, to the sinusoidal input x (t) = 2 cos 2t for all time t , is (A) 0 (B) 2-0.25 cos (2t - 0.125p) (C) 2-0.5 cos (2t - 0.125p) (D) 2-0.5 cos (2t - 0.25p)

A I D

ONE MARK

1 If the Laplace transform of a signal Y (s) = , then its final s (s - 1) value is (A) - 1 (B) 0 (C) (D) Unbounded .in 1

Let x (t) be the input and y (t) be the output of a continuous time system. Match the system properties P1, P2 and P3 with system relations R1, R2, R3, R4 Properties Relations .co a i P1 : Linear but NOT time - invariant R1 : y (t) = t2 x (t) d no . P2 : Time - invariant but NOT linear R2 : y (t) = t x (t) 2007 TWO MARKS w w P3 : Linear and time - invariant R3 : y (t) = x (t) w 5.51 The 3-dB bandwidth of the low-pass signal e-t u (t), where u (t) is the R4 : y (t) = x (t - 5) unit step function, is given by (A) (P1, R1), (P2, R3), (P3, R4) (A) 1 Hz (B) 1 2 - 1 Hz (B) (P1, R2), (P2, R3), (P3, R4) 2p 2p (C) (P1, R3), (P2, R1), (P3, R2) (C) 3 (D) 1 Hz (D) (P1, R1), (P2, R2), (P3, R3) 5.52 A 5-point sequence x [n] is given as x [- 3] = 1, x [- 2] = 1, x [- 1] = 0, {x (n)} is a real - valued periodic sequence with a period N . x (n) x [0] = 5 and x [1] = 1. Let X (eiw) denoted the discrete-time Fourier and X (k) form N-point Discrete Fourier Transform (DFT) pairs. p N-1 transform of x [n]. The value of X (e jw) dw is 1 The DFT Y (k) of the sequence y (n) = x (r) x (n + r) is -p / N r=0 (A) 5 (B) 10p N-1 (A) X (k) 2 (B) 1 / X (r) X (k + r) (C) 16p (D) 5 + j10p N r=0

O N

5.45

The frequency response H (w) of this system in terms of angular frequency w, is given by H (w) 1 (A) (B) sin w 1 + j2w w jw 1 (C) (D) 2 + jw 2 + jw

2007

5.50

5.44

The expression and the region of convergence of the z -transform of the sampled signal are 5z , z < e-0.05 (B) (A) 5z 5 , z < e-5 z - e-0.05 z-e 5z , z > e-0.05 (C) (D) 5z -5 , z > e-5 z-e z - e-0.05

Statement for Linked Answer Question 6.33 & 6.34:

5.48

A discrete time linear shift - invariant system has an impulse response h [n] with h [0] = 1, h [1] =- 1, h [2] = 2, and zero otherwise The system is given an input sequence x [n] with x [0] = x [2] = 1, and zero otherwise. The number of nonzero samples in the output sequence y [n], and the value of y [2] are respectively (A) 5, 2 (B) 6, 2 (C) 6, 1 (D) 5, 3

(D) 5e-5n

#

N-1 (C) 1 / X (r) X (k + r) N r=0

(D) 0

5.53

Statement for Linked Answer Question 6.31 and 6.32: In the following network, the switch is closed at t = 0- and the sampling starts from t = 0 . The sampling frequency is 10 Hz.

5.54

5.55

5.46

The samples x (n), n = (0, 1, 2, ...) are given by (B) 5e-0.05n (A) 5 (1 - e-0.05n)

The z -transform X (z) of a sequence x [n] is given by X [z] = 1 -0.25z . It is given that the region of convergence of X (z) includes the unit circle. The value of x [0] is (A) - 0.5 (B) 0 (C) 0.25 (D) 05 -1

A Hilbert transformer is a (A) non-linear system (C) time-varying system

(B) non-causal system (D) low-pass system

The frequency response of a linear, time-invariant system is given by H (f) = 1 + j510pf . The step response of the system is t (A) 5 (1 - e-5t) u (t) (B) 5 61 - e- 5@ u (t) t (C) 1 (1 - e-5t) u (t) (D) 1 ^1 - e- 5 h u (t) 2 5

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

2006 5.56

5.57

Page 111

ONE MARK

2005

Let x (t) * X (jw) be Fourier Transform pair. The Fourier Transform of the signal x (5t - 3) in terms of X (jw) is given as j3w j3w jw jw (A) 1 e- 5 X b l (B) 1 e 5 X b l 5 5 5 5 jw jw (C) 1 e-j3w X b l (D) 1 e j3w X b l 5 5 5 5 The Dirac delta function d (t) is defined as 1 t=0 (A) d (t) = ) 0 otherwise 3 t=0 (B) d (t) = ) 0 otherwise 3 1 t=0 (C) d (t) = ) and d (t) dt = 1 -3 0 otherwise 3 3 t=0 (D) d (t) = ) and d (t) dt = 1 -3 0 otherwise If the region of convergence of x1 [n] + x2 [n] is region of convergence of x1 [n] - x2 [n] includes (A) 1 < z < 3 (B) 2 < z 3 3 (C) 3 < z < 3 (D) 1 < z 2 3

5.64

5.65

ONE MARK

Choose the function f (t); - 3 < t < 3 for which a Fourier series cannot be defined. (A) 3 sin (25t) (B) 4 cos (20t + 3) + 2 sin (710t) (C) exp (- t ) sin (25t) (D) 1 The function x (t) is shown in the figure. Even and odd parts of a unit step function u (t) are respectively,

#

#

5.58

5.59

2006

5.61

5.62

5.63

0 s + w0 The final value of f (t) would be (A) 0 (B) 1 (C) - 1 # f (3) # 1 (D) 3

A system with input x [n] and output y [n] is given as y [n] = (sin pn) x [n] . The system is (A) linear, stable and invertible (B) non-linear, stable and non-invertible (C) linear, stable and non-invertible (D) linear, unstable and invertible 5 6

The unit step response of a system starting from rest is given by c (t) = 1 - e-2t for t $ 0 . The transfer function of the system is 1 (A) (B) 2 1 + 2s 2+s (C) 1 (D) 2s 2+s 1 + 2s The unit impulse response of a system is f (t) = e-t, t $ 0 . For this system the steady-state value of the output for unit step input is equal to (A) - 1 (B) 0 (C) 1 (D) 3

5.68

Let x (n) = ( 12 ) n u (n), y (n) = x2 (n) and Y (e jw) be the Fourier transform of y (n) then Y (e j0) (A) 1 4 (B) 2 (C) 4 (D) 4 3

5.69

The power in the signal s (t) = 8 cos (20p - p2 ) + 4 sin (15pt) is (A) 40 (B) 41 (C) 42 (D) 82 2005

5.70

TWO MARKS

The output y (t) of a linear time invariant system is related to its input x (t) by the following equations

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 112

y (t)= 0.5x (t - td + T) + x (t - td ) + 0.5x (t - td + T) The filter transfer function H (w) of such a system is given by (A) (1 + cos wT) e-jwt (B) (1 + 0.5 cos wT) e-jwt (C) (1 - cos wT) e-jwt (D) (1 - 0.5 cos wT) e-jwt d

d

d

d

5.71

Match the following and choose the correct combination. Group 1 E. Continuous and aperiodic signal F. Continuous and periodic signal G. Discrete and aperiodic signal H. Discrete and periodic signal Group 2 1. Fourier representation is continuous and aperiodic 2. Fourier representation is discrete and aperiodic 3. Fourier representation is continuous and periodic 4. Fourier representation is discrete and periodic (A) (B) (C) (D)

5.72

E - 3, E - 1, E - 1, E - 2,

F - 2, F - 3, F - 2, F - 1,

G - 4, G - 2, G - 3, G - 4,

H-1 H-4 H-4 H-3

A signal x (n) = sin (w0 n + f) is the input to a linear time- invariant system having a frequency response H (e jw). If the output of the system Ax (n - n0) then the most general form of +H (e jw) will be (A) - n0 w0 + b for any arbitrary real (B) - n0 w0 + 2pk for any arbitrary integer k (C) n0 w0 + 2pk for any arbitrary integer k (D) - n0 w0 f

O N

A I D

Statement of linked answer question 6.59 and 6.60 :

Forn a signal x (t) the Fourier transform is .i o Fourier transform of X (3f + 2) is given by .c

5.75

ia

d .no

w

ww

The Fourier transform of y (2n) will be (A) e-j2w [cos 4w + 2 cos 2w + 2] (B) cos 2w + 2 cos w + 2 -jw (C) e [cos 2w + 2 cos w + 2] (D) e-j2w [cos 2w + 2 cos + 2]

5.74

(A) 1 x` t j e j3pt 2 2

(C) 3x (3t) e-j4pt

X (f). Then the inverse

j4pt (B) 1 x` t j e - 3 3 3

(D) x (3t + 2)

A sequence x (n) has non-zero values as shown in the figure.

2004 5.76

x ( n2 - 1),

5.73

The sequence y (n) = * 0,

For n even will be

5.77

For n odd

5.78

ONE MARK

The impulse response h [n] of a linear time-invariant system is given by h [n] = u [n + 3] + u [n - 2) - 2n [n - 7] where u [n] is the unit step sequence. The above system is (A) stable but not causal (B) stable and causal (C) causal but unstable (D) unstable and not causal The z -transform of a system is H (z) = z -z0.2 . If the ROC is z < 0.2 , then the impulse response of the system is (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] The Fourier transform of a conjugate symmetric function is always (A) imaginary (B) conjugate anti-symmetric (C) real (D) conjugate symmetric 2004

5.79

5.80

TWO MARKS

Consider the sequence x [n] = [- 4 - j51 + j25]. The conjugate antisymmetric part of the sequence is (A) [- 4 - j2.5, j2, 4 - j2.5] (B) [- j2.5, 1, j2.5] (C) [- j2.5, j2, 0] (D) [- 4, 1, 4] A causal LTI system is described by the difference equation 2y [n] = ay [n - 2] - 2x [n] + bx [n - 1]

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

The system is stable only if (A) a = 2 , b < 2 (C) a < 2 , any value of b 5.81

Page 113

x (n) n$1 y (n) = *0, n=0 x (n + 1) n # - 1

(B) a > 2, b > 2 (D) b < 2 , any value of a

where x (n) is the input and y (n) is the output. The above system has the properties (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

The impulse response h [n] of a linear time invariant system is given as -2 2

n = 1, - 1

h [ n] = * 4 2 n = 2, - 2 0 otherwise If the input to the above system is the sequence e jpn/4 , then the output is (A) 4 2 e jpn/4 (B) 4 2 e-jpn/4 (C) 4e jpn/4 (D) - 4e jpn/4 5.82

Common Data For Q. 6.73 & 6.74 :

Let x (t) and y (t) with Fourier transforms F (f) and Y (f) respectively be related as shown in Fig. Then Y (f) is

The system under consideration is an RC low-pass filter (RC-LPF) with R = 1 kW and C = 1.0 mF. Let H (f) denote the frequency response of the RC-LPF. Let f1 be H (f1) the highest frequency such that 0 # f # f1 $ 0.95 . Then f1 H (0) (in Hz) is (A) 324.8 (B) 163.9 (C) 52.2 (D) 104.4

5.88

Let tg (f) be the group delay function of the given RC-LPF and f2 = 100 Hz. Then tg (f2) in ms, is (A) 0.717 (B) 7.17 (C) 71.7 (D) 4.505

5.89

(A) - 1 X (f/2) e-jpf 2

(B) - 1 X (f/2) e j2pf 2

(C) - X (f/2) e j2pf

(D) - X (f/2) e-j2pf

5.83

ONE MARK

N

5.86

ww

The Fourier series expansion of a real periodic signal with fundamental frequency f0 is given by gp (t) = cn e j2pf t . It is given n =- 3 that c3 = 3 + j5 . Then c-3 is (A) 5 + j3 (B) - 3 - j5 (C) - 5 + j3 (D) 3 - j5

/

5.85

no w.

5.91

5.92

5.93

A sequence x (n) with the z -transform X (z) = z 4 + z2 - 2z + 2 - 3z-4 is applied as an input to a linear, time-invariant system with the impulse response h (n) = 2d (n - 3) where 1, n = 0 d (n) = ) 0, otherwise The output at n = 4 is (A) - 6 (B) zero (C) 2 (D) - 4

5.87

TWO MARKS

Let P be linearity, Q be time-invariance, R be causality and S be stability. A discrete time system has the input-output relationship,

Which of the following cannot be the Fourier series expansion of a periodic signal? (A) x (t) = 2 cos t + 3 cos 3t (B) x (t) = 2 cos pt + 7 cos t (C) x (t) = cos t + 0.5 (D) x (t) = 2 cos 1.5pt + sin 3.5pt 1 The Fourier transform F {e-1 u (t)} is equal to . Therefore, 1 + j 2p f 1 is F' 1 + j2pt 1 (A) e f u (f) (B) e-f u (f) (C) e f u (- f) (D) e-f u (- f) A linear phase channel with phase delay Tp and group delay Tg must have (A) Tp = Tg = constant (B) Tp \ f and Tg \ f (C) Tp = constant and Tg \ f ( f denote frequency) (D) Tp \ f and Tp = constant 2002

5.94

2003

ONE MARK

d

0

Let x (t) be the input to a linear, time-invariant system. The required output is 4p (t - 2). The transfer function of the system should be (A) 4e j4pf (B) 2e-j8pf (C) 4e-j4pf (D) 2e j8pf

2002

Convolution of x (t + 5) with impulse function d (t - 7) is equal to (A) (B) x (t + 12) nx (t - 12) o.i c . (D) x (t + 2) ia (C) x (t - 2)

5.90

The Laplace transform of i (t) is given by 2 I (s) = s (1 + s) At t " 3, The value of i (t) tends to (A) 0 (B) 1 (C) 2 (D) 3

5.84

IA

D O

2003

5.95

TWO MARKS

The Laplace transform of continuous - time signal x (t) is X (s) = . If the Fourier transform of this signal exists, the x (t) is (A) e2t u (t) - 2e-t u (t) (B) - e2t u (- t) + 2e-t u (t) (C) - e2t u (- t) - 2e-t u (t) (D) e2t u (- t) - 2e-t u (t)

5-s s2 - s - 2

If the impulse response of discrete - time system is

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h [n] =- 5n u [- n - 1], then the system function H (z) is equal to (A) - z and the system is stable z-5 (B) z and the system is stable z-5 (C) - z and the system is unstable z-5 (D) z and the system is unstable z-5

Page 114

5.97

5.98

A linear time invariant system has an impulse response e2t, t > 0 . If the initial conditions are zero and the input is e3t , the output for t > 0 is (A) e3t - e2t (B) e5t (C) e3t + e2t (D) None of these

5.104

2001 5.96

A system with an input x (t) and output y (t) is described by the relations : y (t) = tx (t). This system is (A) linear and time - invariant (B) linear and time varying (C) non - linear and time - invariant (D) non - linear and time - varying

5.103

ONE MARK

The transfer function of a system is given by H (s) = 2 1 . The s ( s 2 ) impulse response of the system is (A) (t2 * e-2t) u (t) (B) (t * e2t) u (t) (C) (te-2 t) u (t) (D) (te-2t) u (t)

2000

The region of convergence of the z - transform of a unit step function is (A) z > 1 (B) z < 1 (C) (Real part of z ) > 0 (D) (Real part of z ) < 0

TWO MARKS

One period (0, T) each of two periodic waveforms W1 and W2 are shown in the figure. The magnitudes of the nth Fourier series coefficients of W1 and W2 , for n $ 1, n odd, are respectively proportional to

5.105

Let d (t) denote the delta function. The value of the integral 3 d (t) cos b 3t l dt is 2 -3

#

5.99

A I D

(B) - 1 (D) p2

(A) 1 (C) 0

If a signal f (t) has energy E , the energy of the signal f (2t) is equal to (A) 1 (B) E/2 (C) 2E (D) 4E

O N w

n-3 n-2 n-1 n-4

and and and and

n-2 n-3 n-2 n-2

Let u (t) be the step function. Which of the waveforms in the figure corresponds to the convolution of u (t) - u (t - 1) with u (t) - u (t - 2) ?

The impulse response functions of four linear systems S1, S2, S3, S4 are given respectively by h1 (t) = 1, h2 (t) = u (t), u (t) and h3 (t) = t+1 h 4 (t) = e-3t u (t) where u (t) is the unit step function. Which of these systems is time invariant, causal, and stable? (A) S1 (B) S2 (C) S3 (D) S4 ONE MARK

L [f (t)] = s2+ 2 , s +1 t h (t) = f (t) g (t - t) dt . 0 L [h (t)] is

Given

that

L [g (t)] =

#

2 (A) s + 1 s+3

(C) 5.102

5.106

TWO MARKS

2000 5.101

ia od

n

. ww

2001 5.100

.

(A) (B) (C) .in co(D)

and 5.107

(B)

s2 + 1 + s2+ 2 (s + 3)( s + 2) s + 1

s2 + 1 (s + 3) (s + 2)

1 s+3

0

(D) None of the above

o

#

o

o

2

The Fourier Transform of the signal x (t) = e-3t is of the following form, where A and B are constants : (B) Ae-Bf (A) Ae-B f (C) A + B f 2 (D) Ae-Bf

A system has a phase response given by f (w), where w is the angular frequency. The phase delay and group delay at w = w0 are respectively given by f (w0) df (w) d2 f (w0) (B) f (wo), (A) , w0 dw w = w dw2 w = w w df (w) (C) wo , (D) wo f (wo), f (l) f (wo) d (w) w = w -3 1999

2

5.108

The z -transform F (z) of the function f (nT) = anT is (A) z T (B) z T z-a z+a

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ONE MARK

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

5.109

5.110

z z (C) (D) -T z-a z + a-T If [f (t)] = F (s), then [f (t - T)] is equal to (A) esT F (s) (B) e-sT F (s) F (s) F (s) (C) (D) 1 - esT 1 - e-sT

2

5.117

(A) 1 f (w) 2p (C) 2pf (- w)

A I D 5.121

O N ww

3

# g (t) e-jwt dt o is -3

(B) 1 f (- w) 2p (D) None of the above

The Laplace Transform of eat cos (at) is equal to (s - a) (s + a) (A) (B) 2 2 (s - a) + a (s - a) 2 + a2 1 (C) (D) None of the above 2 ( s a ) n

.i

no w.

The trigonometric Fourier series of a even time function can have only (A) cosine terms (B) sine terms (C) cosine and sine terms (D) d.c and cosine terms

d

co ia.

5.122

5.123

1996

ONE MARK

The trigonometric Fourier series of an even function of time does not have the (A) dc term (B) cosine terms (C) sine terms (D) odd harmonic terms The Fourier transform of a real valued time signal has (A) odd symmetry (B) even symmetry (C) conjugate symmetry (D) no symmetry

A periodic signal x (t) of period T0 is given by 1, t < T1 x (t) = * 0, T1 < t < T0 2 The dc component of x (t) is (A) T1 T0 (C) 2T1 T0

5.116

ff (t) g (t) e =

ONE MARK

w , then the value of Limf (t) t"3 s + w2 (A) cannot be determined (B) is zero (C) is unity (D) is infinite If F (s) =

ONE MARK

The function f (t) has the Fourier Transform g (w). The Fourier Transform

5.120

3 jpnt (B) 1 / exp T0 n =- 3 T0 3 j2pnt (D) 1 / exp T0 n =- 3 T0

The z -transform of a signal is given by 1z-1 (1 - z-4) C (z) = 4 (1 - z-1) 2 Its final value is (A) 1/4 (B) zero (C) 1.0 (D) infinity

2 1

1997

3

1998

5.115

5.119

/ d (t - nT0) is given by

3 j2pnt (A) 1 / exp T0 n =- 3 T0 3 jpnt (C) 1 / exp T0 n =- 3 T0

5.114

2 2 A 22 + A 32 + ..... (D) c A 2 + A 3 + ..... m 2 2 A1 A + A 2 + A 3 + .... The Fourier transform of a function x (t) is X (f). The Fourier dX (t) transform of will be df dX (f) (A) (B) j2pfX (f) df X (f) (C) jfX (f) (D) jf

(C)

The Fourier series representation of an impulse train denoted by n =- 3

5.113

5.118

TWO MARKS

s (t) =

5.112

(z - 1) 2 z (D) z (z - 1) 2 A distorted sinusoid has the amplitudes A1, A2, A 3, .... of the fundamental, second harmonic, third harmonic,..... respectively. The total harmonic distortion is 2 2 (B) A 2 + A 3 + ..... (A) A2 + A 3 + .... A1 A1 (C)

A signal x (t) has a Fourier transform X (w). If x (t) is a real and odd function of t , then X (w) is (A) a real and even function of w (B) a imaginary and odd function of w (C) an imaginary and even function of w (D) a real and odd function of w 1999

5.111

Page 115

(B) T1 2T0 (D) T0 T1

The unit impulse response of a linear time invariant system is the unit step function u (t). For t > 0 , the response of the system to an excitation e-at u (t), a > 0 will be (A) ae-at (B) (1/a) (1 - e-at) (C) a (1 - e-at) (D) 1 - e-at The z-transform of the time function (A) z - 1 z

3

/ d (n - k) is k=0

(B)

z z-1

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 116

SOLUTIONS 5.1

Given, the maximum frequency of the band-limited signal fm = 5 kHz According to the Nyquist sampling theorem, the sampling frequency must be greater than the Nyquist frequency which is given as fN = 2fm = 2 # 5 = 10 kHz So, the sampling frequency fs must satisfy

Option (C) is correct. If the two systems with impulse response h1 ^ t h and h2 ^ t h are connected in cascaded configuration as shown in figure, then the overall response of the system is the convolution of the individual impulse responses.

fs $ fN fs $ 10 kHz only the option (A) doesn’t satisfy the condition therefore, 5 kHz is not a valid sampling frequency. 5.5

5.2

Option (C) is correct. Given, the input

x ^ t h = u ^t - 1h It’s Laplace transform is -s X ^s h = e s The impulse response of system is given

h^t h = t u^t h Its Laplace transform is H ^s h = 12 s Hence, the overall response at the output is Y ^s h = X ^s h H ^s h -s =e3 s its inverse Laplace transform is ^t - 1h2 y^t h = u ^t - 1h 2 5.3

Option (A) is correct. Given, the signal

Option (B) is correct. The Laplace transform of unit step fun n is U ^s h = 1 s So, the O/P of the system is given as Y ^s h = b 1 lb 1 l s s = 12 s in . o c For zero initial condition, we check ia. d dy ^ t h u^t h = no . dt w ww & U ^s h = SY ^s h - y ^0 h & U ^s h = s c 12 m - y ^0 h s 1 or, U ^s h = s ^y ^0 h = 0h 5.6

A I D

O N

Hence, the O/P is correct which is Y ^s h = 12 s its inverse Laplace transform is given by y ^ t h = tu ^ t h

v ^ t h = 30 sin 100t + 10 cos 300t + 6 sin ^500t + p4 h

So we have w1 = 100 rad/s w2 = 300 rad/s w3 = 500 rad/s Therefore, the respective time periods are T1 = 2p = 2p sec w1 100 T2 = 2p = 2p sec 300 w2 T3 = 2p sec 500 So, the fundamental time period of the signal is LCM ^2p, 2p, 2ph L.C.M. ^T1, T2 T3h = HCF ^100, 300, 500h or, T0 = 2p 100 Hence, the fundamental frequency in rad/sec is w0 = 2p = 100 rad/s 10 5.4

Option (A) is correct.

Option (C) is correct. For a system to be casual, the R.O.C of system transfer function H ^s h which is rational should be in the right half plane and to the right of the right most pole. For the stability of LTI system. All poles of the system should lie in the left half of S -plane and no repeated pole should be on imaginary axis. Hence, options (A), (B), (D) satisfies an LTI system stability and causality both. But, Option (C) is not true for the stable system as, S = 1 have one pole in right hand plane also.

5.7

No Option is correct. The matched filter is characterized by a frequency response that is given as H ^ f h = G * ^ f h exp ^- j2pfT h f where g^t h G^f h Now, consider a filter matched to a known signal g ^ t h. The fourier transform of the resulting matched filter output g 0 ^ t h will be G0 ^ f h = H^ f hG^ f h = G * ^ f h G ^ f h exp ^- j2pfT h = G ^ f h 2 exp ^- j2pfT h T is duration of g ^ t h Assume exp ^- j2pfT h = 1 So, G0 ^ f h = G_ f i 2 Since, the given Gaussian function is g ^ t h = e- pt Fourier transform of this signal will be f g ^ t h = e- pt e- pf = G ^ f h 2

2

2

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Page 117

change the initial condition to - 2y ^0 h and the forcing equation to - 2x ^ t h

Therefore, output of the matched filter is 2 G 0 ^ f h = e- pf 2

5.8

Option (B) is correct. Given, the impulse response of continuous time system h ^ t h = d ^t - 1h + d ^t - 3h From the convolution property, we know x ^ t h * d ^t - t 0h = x ^t - t 0h So, for the input x ^ t h = u ^ t h (Unit step fun n ) The output of the system is obtained as

5.11

y^t h = u^t h * h^t h

= u ^ t h * 6d ^t - 1h + d ^t - 3h@ = u ^t - 1h + u ^t - 3h at t = 2

y ^2 h = u ^2 - 1h + u ^2 - 3h =1

5.9

Option (B) is correct. Given, the differential equation d2y dy + 5 + 6y ^ t h = x ^ t h dt dt2 Taking its Laplace transform with zero initial conditions, we have s2 Y ^s h + 5sY ^s h + 6Y ^s h = X ^s h

A I D

....(1) Now, the input signal is

1 0 1/2

A I D

1 n - 1 = a u [n], z > a 1-z Thus system is causal. Since ROC of H (z ) includes unit circle, so it is stable also. Hence S1 is True n o.i ROC : z < 14 For c . dia o n n .n h [n] =-b 1 l u [- n - 1] + b 1 l u (n), z > 1 , z < 1 w 2 2 4 4 w

n n h [n] = b 1 l u [n] + b 1 l u [n], n > 0 2 4

O N w

Option (D) is correct. For an N-point FET algorithm butterfly operates on one pair of samples and involves two complex addition and one complex multiplication.

System is not causal. ROC of H (z ) does not include unity circle, so it is not stable and S 3 is True 5.31

5.32

Option (D) is correct.

t"3

By final value theorem lim f (t) = lim sF (s) = 1

t"3

or 5.29

Option (B) is correct. Given function is f (t) = sin2 t + cos 2t = 1 - cos 2t + cos 2t = 1 + 1 cos 2t 2 2 2

3s + 1 f (t) = L ; 3 s + 4s 2 + (k - 3) s E lim f (t) = 1

and

or

Option (A) is correct. The Fourier series of a real periodic function has only cosine terms if it is even and sine terms if it is odd.

-1

We have

or

Option (C) is correct. We have H (z) =

Inverse Z - transform

We know that Given that Inverse z-transform

H (z ) = H1 (z ) : H2 (z )

5.28

5.30

Option (A) is correct.

Response of cascaded system

5.27

y (t) = (e-t - e-3t) u (t)

s"0

s. (3s + 1) =1 s3 + 4s2 + (k - 3) s s (3s + 1) =1 lim 2 s " 0 s [s + 4s + (k - 3)] 1 =1 k-3 lim s"0

5.33

The function has a DC term and a cosine function. The frequency of cosine terms is w = 2 = 2pf " f = 1 Hz p The given function has frequency component at 0 and 1 Hz. p Option (A) is correct. n n x [n] = b 1 l u (n) - b 1 l u (- n - 1) 3 2 Taking z transform we have X (z) =

k =4

Option (B) is correct. System is described as dx (t) d 2 y (t) dt (t) + 4x (t) +4 + 3y (t) = 2 2 dt dt dt Taking Laplace transform on both side of given equation

=

n=3

n =- 1

n

/ b 13 l z-n - /

1 n -n b2l z

/ b 13 z-1l

1 -1 n b2z l

n=0 n=3

n

n=0

First term gives Second term gives

1 z-1 < 1 " 3 1 z-1 > 1 " 2

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-

n =- 3 n =- 1

/

n =- 3

1< z 3 1> z 2

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Page 120

In option (B) y (0) depends on x (1). In option (C) y (0) depends on x (- 1). In option (D) y (0) depends on x (5). Thus only in option (C) the value of y (t) at t = 0 depends on x (- 1) past value. In all other option present value depends on future value.

Thus its ROC is the common ROC of both terms. that is 1< z 0 that implies a < 0 and bt > 0 for t > 0 that implies b > 0 . Thus, a is negative and b is positive.

5.40

-

Here function is defined for 0 < t < t , Thus t F (s) L f (t) s 0

#

5.35

Option (A) is correct. We have h (2) = 1, h (3) =- 1 otherwise h (k) = 0 . The diagram of response is as follows :

Option (C) is correct.

5.41

K (s + 1) , and R (s) = 1 s (s + 2)( s + 4) K (s + 1) C (s) = G (s) R (s) = s (s + 2)( s + 4) K = K + - 3K 8s 4 (s + 2) 8 (s + 4) c (t) = K :1 + 1 e-2t - 3 e-4tD u (t) 8 4 8

G (s) =

Thus

At steady-state , c (3) = 1 K = 1 or K = 8 Thus 8 8 (s + 1) Then, = 12 - 4 G (s) = (s + 4) (s + 2) (s + 2)( s + 4)

It has the finite magnitude values. So it is a finite impulse response filter. Thus S2 is true but it is not a low pass filter. So S1 is false. 5.36

5.37

Option (B) is correct. Here h (t) ! 0 for t < 0 . Thus system is non causal. Again any bounded input x (t) gives bounded output y (t). Thus it is BIBO stable. Here we can conclude that option (B) is correct. Option (D) is correct. We have x [n] = {1, 0, 2, 3) and N = 4 X [k ] = For N = 4 ,

X [k ] =

N-1

/ x [ n] e

n=0 3

/ x [ n] e

-j2pnk/N

-j2pnk/4

n=0

X [ 0] =

Now

5.42

N

D O

k = 0, 1...N - 1

n

. ww

w

k = 0, 1,... 3

3

/ x [ n]

n=0

3

-jpn/2

5.43

n=0

= x [0] + x [1] e-jp/2 + x [2] e-jp + x [3] e-jp3/2 = 1 + 0 - 2 + j3 =- 1 + j3 X [ 2] =

3

/ x [ n] e

-jpn

n=0

X [ 3] =

/ x [ n] e

5.39

Option (A) is correct. Option (C) is correct. The output of causal system depends only on present and past states only. In option (A) y (0) depends on x (- 2) and x (4).

1

#-1 e-jwt 1dt

= 1 [e-jwt]-11 - jw = 1 (e-jw - e jw) = 1 (- 2j sin w) - jw - jw = 2 sin w w Option (D) is correct. Given h (n) = [1, - 1, 2] x (n) = [1, 0, 1] y (n) = x (n)* h (n) The length of y [n] is = L1 + L2 - 1 = 3 + 3 - 1 = 5

y (2) =

3

/ x (k) h (n - k)

3

/ x (k) h (2 - k) k =- 3

= x (0) h (2 - 0) + x (1) h (2 - 1) + x (2) h (2 - 2) = h (2) + 0 + h (0) = 1 + 2 = 3 There are 5 non zero sample in output sequence and the value of y [2] is 3.

n=0

5.38

=

k =- 3

-j3pn/2

= x [0] + x [1] e-j3p/2 + x [2] e-j3p + x [3] e-j9p/2 = 1 + 0 - 2 - j3 =- 1 - j3 Thus [6, - 1 + j3, 0, - 1 - j3]

#- 33e-jwt x (t) dt

for - 1 # t # + 1 otherwise

y (n) = x (n) * h (n) =

= x [0] + x [1] e-jp + x [2] e-j2p + x [3] e-jp3 = 1+0+2-3 = 0 3

1 x (t) = ) 0 n i transform is . oFourier We have

This is zero at w = p and w = 2p

= 1+0+2+3 = 6

/ x [ n] e

Option (A) is correct.

.c

ia od

= x [0] + x [1] + x [2] + x [3] x [1] =

IA

h (t) = L-1 G (s) = (- 4e-2t + 12e-4t) u (t)

5.44

Option (B) is correct. Mode function are not linear. Thus y (t) = x (t) is not linear but this functions is time invariant. Option (A) and (B) may be correct. The y (t) = t x (t) is not linear, thus option (B) is wrong and (a) is correct. We can see that R1: y (t) = t2 x (t) Linear and time variant.

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Page 121

R2: y (t) = t x (t) Non linear and time variant. R3: y (t) = x (t) Non linear and time invariant R4: y (t) = x (t - 5) Linear and time invariant 5.45

Magnitude at 3dB frequency is

Given :

y (n) = 1 N

N-1

or

r=0

or

/x (r) x (n + r)

It is Auto correlation. Hence 5.46

DFT

y (n) = rxx (n)

X (k) 2

Option (B) is correct. For discrete time Fourier transform (DTFT) when N " 3 p x [n] = 1 X (e jw) e jwn dw 2p - p

5.52

Option (B) is correct. Current through resistor (i.e. capacitor) is Here,

#

I = I (0+) e-t/RC I (0+) = V = 5 = 25mA R 200k

Putting n = 0 we get = 1 2p

RC = 200k # 10m = 2 sec I = 25e- m A t 2

= VR # R = 5e V Here the voltages across the resistor is input to sampler at frequency of 10 Hz. Thus x (n) = 5e 5.47

5.48

= 5e

-0.05n

H (jw) = = 5.49

#- 3 h (t) e 3

#0

-jwt

3 -2t -jwt

e

e

dt

dt =

#0

3 - (2 + jw) t

e

Option (D) is correct. H (jw) =

1 (2 + jw)

5.54

no w.

1 dt = w (2 + jw)w

-p

X (e jw) dw = 2px [0] = 2p # 5 = 10p

d

n o.i (A) is correct. c Option . ia

5.55

A Hilbert transformer is a non-linear system. Option (B) is correct. 5 1 + j10pf 5 = 5 = = 1 1 + 5s 5^s + 15 h s + =1 a 1 s ^s + 5 h =1 11 =5- 51 s ^s + 5 h s s+ 5 -t/5 = 5 (1 - e ) u (t)

H (f) = H (s)

5.56

1 s (s - 1) Final value theorem is applicable only when all poles of system lies in left half of S -plane. Here s = 1 is right s -plane pole. Thus it is unbounded. x (t) = e-t u (t) Taking Fourier transform X (jw) = 1 1 + jw X (jw) = 1 2 1+w

p

x (n) =- (0.5) (2) -n u (- n - 1) x (0) = 0 If we apply initial value theorem x (0) = lim X (z) = lim 0.5 -1 = 0.5 z"3 z " 31 - 2z That is wrong because here initial value theorem is not applicable because signal x (n) is defined for n < 0 .

Option (D) is correct. Y (s) =

Option (A) is correct.

#

X (e jw) e jw0 dw

A I D

O N

Magnitude response at w = 2 rad/sec is 1 H (jw) = = 1 2 2 2 +w 2 2 Input is x (t) = 2 cos (2t) Output is = 1 # 2 cos (2t - 0.25p) 2 2 = 1 cos [2t - 0.25p] 2

5.51

X (e jw) dw

-p

0. 5 1 - 2z-1 Since ROC includes unit circle, it is left handed system

The phase response at w = 2 rad/sec is +H (jw) =- tan-1 w =- tan-1 2 =- p =- 0.25p 2 2 4

5.50

-p

p

X (z) =

Option (C) is correct. Since x (n) = 5e-0.05n u (n) is a causal signal Its z transform is 5z 1 X (z) = 5 : -0.05 -1 D = z - e-0.05 1-e z Its ROC is e-0.05 z-1 > 1 " z > e-0.05 h (t) = e-2t u (t)

p

#

Option (B) is correct.

5.53

For t > 0

Option (C) is correct.

#

x [0] = 1 2p

or

- 2t

-n 2 # 10

1 = 1 2 1 + w2 w = 1 rad f = 1 Hz 2p

Thus

Option (A) is correct.

1 2

5.57

5.58

5.59

Step response

Y (s)

or

Y (s)

or

y (t)

1 5

Option (A) is correct. F x (t) X (jw) Using scaling we have F 1 X jw x (5t) 5 c 5 m Using shifting property we get F 1 X jw e- j35w x ;5 bt - 3 lE 5 5 b5l Option (D) is correct. Dirac delta function d (t) is defined at t = 0 and it has infinite value a t = 0 . The area of dirac delta function is unity. Option (D) is correct. The ROC of addition or subtraction of two functions x1 (n) and x2 (n) is R1 + R2 . We have been given ROC of addition of two function and has been asked ROC of subtraction of two function. It will be same. Option (A) is correct. As we have x (t) = sin t ,

thus w = 1

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Now or or Thus 5.60

1 s+1 H (jw) = 1 = 1 jw + 1 j+1 H (jw) = 1 + - 45c 2 1 y (t) = sin (t - p4 ) 2

Page 122

Thus ROC of x1 (n) + x2 (n) is R1 + R2 which is 5 < z < 6 6 5

H (s) =

5.67

5.68

x (n) = b 1 l u (n) 2 n

y (n) = x2 (n) = b 1 l u2 (n) 2 2n

w0 s + w2 L-1 F (s) = sin wo t f (t) = sin wo t Thus the final value is - 1 # f (3) # 1 2

Y (e jw) =

A I D

Option (C) is correct. H (s) =

5.69

O N

-t

y (t) = u (t) - e In steady state i.e. t " 3, y (3) = 1

5.65

co ia.

d

ww

5.70

=4 3

= 8 sin 20pt + 4 sin 15pt Here A1 = 8 and A2 = 4 . Thus power is 2 2 2 2 P = A1 + A2 = 8 + 4 = 40 2 2 2 2 Option (A) is correct.

Y (w) = 0.5e

-jw (- td + T)

or

g (t) + g (- t) 2 g (t) - g (- t) odd{g (t)} = 2

5.66

1 3 4 = 1 +b1l +b1l+b1l +b1l 4 4 4 4

s (t) = 8 cos ` p - 20pt j + 4 sin 15pt 2

X (w) + e-jwt X (w) + 0.5e-jw (- t - T) X (w) Y (w) = e-jwt [0.5e jwT + 1 + 0.5e-jwT ] X (w) d

d

d

= e-jwt [0.5 (e jwT + e-jwT ) + 1]

Ev{g (t)} =

Thus

n

y (t) = 0.5x (t - td + T) + x (t - td ) + 0.5x (t - td - T) Taking Fourier transform we have

Option (A) is correct.

g (t) = u (t) u (t) + u (- t) ue (t) = = 2 u (t) - u (- t) uo (t) = = 2

1 4

/ b 14 l e-jwn

Option (A) is correct.

.in

no w.

Option (C) is correct. Fourier series is defined for periodic function and constant. 3 sin (25t) is a periodic function. 4 cos (20t + 3) + 2 sin (710t) is sum of two periodic function and also a periodic function. e- t sin (25t) is not a periodic function, so FS can’t be defined for it. 1 is constant

Here

/ ` 14 j

1 1-

n=3 n=0

n

n=3

=

...(1)

Alternative : Taking z transform of (1) we get 1 Y (z) = 1 - 14 z-1 Substituting z = e jw we have 1 Y (e jw) = 1 - 14 e-jw Y (e j0) = 1 1 = 4 3 1- 4

BIBO stable

1 s+1 L x (t) = u (t) X (s) = 1 s Y (s) = H (s) X (s) = 1 # 1 = 1 - 1 s+1 s s s+1

5.64

Y (e j0) =

or

Option (B) is correct.

L

/ y (n) e-jwn

n=0

x (n) = d (n) y (n) = sin 0 = 0 (bounded)

h (t) = e-t

Y (e j0) =

or

c (t) = 1 - e-2t Taking Laplace transform C (s) 2 = C (s) = #s = 2 s (s + 2) s+2 U (s) 5.63

n=3

n =- 3

Option (C) is correct.

Let Now

2 n n y (n) = ;b 1 l E u (n) = b 1 l u (n) 2 4

or

y (n) = b sin 5 pn l x (n) 6

5.62

Option (D) is correct.

Option (C) is correct. F (s) =

5.61

Option (D) is correct. For causal system h (t) = 0 for t # 0 . Only (D) satisfy this condition.

d

= e-jwt [cos wT + 1] d

or 1 2 x (t) 2

Option (C) is correct. Here x1 (n) = ` 5 jn u (n) 6 1 X1 (z) = 1 - ^ 65 z-1h x2 (n) =-` 6 jn u (- n - 1) 5 1 X1 (z) = 1 1 - ^ 65 z-1h

5.71

ROC : R1 " z > 5 6 5.72

ROC : R2 " z < 6 5

H (w) =

Y (w) = e-jwt (cos wT + 1) X (w) d

Option (C) is correct. For continuous and aperiodic signal Fourier representation is continuous and aperiodic. For continuous and periodic signal Fourier representation is discrete and aperiodic. For discrete and aperiodic signal Fourier representation is continuous and periodic. For discrete and periodic signal Fourier representation is discrete and periodic. Option (B) is correct. y (n) = Ax (n - no) Taking Fourier transform

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Page 123

Y (e jw) = Ae-jw n X (e jw) Y (e jw) -jw n or H (e jw) = jw = Ae X (e ) jw Thus +H (e ) =- wo no For LTI discrete time system phase and frequency of H (e jw) are periodic with period 2p. So in general form o

o

o

o

q (w) =- no wo + 2pk 5.73

5.74

Option (A) is correct. From x (n) = [ 12 , 1, 2, 1, 1, 12 ] y (n) = x ^ n2 - 1h, n even = 0 , for n odd n =- 2 , y (- 2) = x ( -22 - 1) = x (- 2) = 12 n =- 1, y (- 1) = 0 n = 0, y (0) = x ( 20 - 1) = x (- 1) = 1 n = 1, y (1) = 0 n=2 y (2) = x ( 22 - 1) = x (0) = 2 n = 3, y (3) = 0 n=4 y (4) = x ( 24 - 1) = x (1) = 1 n = 5, y (5) = 0 n=6 y (6) = x ( 26 - 1) = x (2) = 12 Hence y (n) = 1 d (n + 2) + d (n) + 2d (n - 2) + d (n - 4) + 1 d (n - 6) 2 2 Option (C) is correct. Here y (n) is scaled and shifted version of x (n) and again y (2n) is scaled version of y (n) giving

N

Taking Fourier transform. Z (e jw) = 1 e jw + 1 + 2e-jw + e-2jw + 1 e-3jw 2 2 = e-jw b 1 e2jw + e jw + 2 + e-jw + 1 e-2jw l 2 2 = e-jw b e or 5.75

2jw

x (at)

F

xb 1 f l 3

F

We know that

Option (A) is correct. We have x (n) = [- 4 - j5, 1 + 2j,

IA

5.79

no w.

n o.i

c ia.

d

5.80

+ e-2jw + e jw + 2 + e-jw l 2

X (f) 1 X f a ca m 3X (3f)

x *( n) = [- 4 + j5, 1 - 2j, 4] -

x *( - n) = [4,

1 - 2j, - 4 + j5] -

xcas (n) =

x (n) - x* (- n) 2

= [- 4 - j 25 ,

4 - j 25 ]

2j -

Option (C) is correct. We have 2y (n) = ay (n - 2) - 2x (n) + bx (n - 1) Taking z transform we get

or a 0 , the response Y (s) = X (s) H (s) 1 (s + a) Y (s) = 1 1 = 1 :1 - 1 D a s s+a (s + a) s X (s) = L [x (t)] = so

Taking inverse Laplace, the response will be y (t) = 1 61 - e-at@ a 5.117

Option (B) is correct. We have

x [n] =

3

/ d (n - k) k=0

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X (z) =

3

/ x [n] z-n

=

k=0

3

Page 127

3

/ ; / d (n - k) z-nE n =- 3 k = 0

then

Option (B) is correct.

5.119

Option (B) is correct.

3

# x (t) e jwt dt = X (- jw) -3

Since d (n - k) defined only for n = k so 3 1 X (z) = / z-k = = z (z - 1) (1 - 1/z) k=0 5.118

X)(jw) =

F

X (f) x (t) by differentiation property; dx (t) = jwX (w) F; dt E dx (t) or = j2pfX (f) F; dt E 5.120

Option (C) is correct. F

We have g (w) f (t) by duality property of fourier transform we can write g (t)

F

# g (t) e-jwt dt = 2pf (- w)

F [g (t)] =

so

2pf (- w) 3

-3 5.121

Option (B) is correct. Given function x (t) = eat cos (at) L s cos (at) 2 s + a2

Now

x (t)

L

X (s)

es t x (t)

L

e cos (at)

L

X (s - s 0) (s - a) (s - a) 2 + a2

If then so 5.122

0

at

Option (C) is correct. For a function x (t), trigonometric fourier series is : x (t) = A 0 +

A I D

shifting in s-domain

3

O N

/ [An cos nwt + Bn sin nwt] n=1

no w.

.in

co ia.

d

ww

where A 0 = 1 # x (t) dt T0 =Fundamental period T0 T An = 2 # x (t) cos nwtdt T0 T Bn = 2 # x (t) sin nwtdt T0 T For an even function x (t), coefficient Bn = 0 for an odd function x (t), A0 = 0 0

0

0

An = 0 so if x (t) is even function its fourier series will not contain sine terms. 5.123

Option (C) is correct. The conjugation property allows us to show if x (t) is real, then X (jw) has conjugate symmetry, that is X (- jw) = X)(jw)

[ x (t) real]

Proof : 3

# x (t) e-jwt dt

X (jw) = replace w by - w then

-3 3

# x (t) e jwt dt

X (- jw) =

-3 3

X)(jw) = = if x (t) real x)(t) = x (t)

)

# x (t) e-jwt dtG

-3

=

3

# x)(t) e jwt dt -3

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Page 128

UNIT 6

The state diagram of a system is shown below. A system is o = AX + Bu ; described by the state-variable equations X y = CX + Du

CONTROL SYSTEMS

2013 6.1

ONE MARK

6.4

The Bode plot of a transfer function G ^s h is shown in the figure below.

6.5

The state-variable equations of the system shown in the figure above are o = >- 1 0 H X + >- 1H u o = >- 1 0 H X + >- 1H u X X (B) (A) -1 -1 1 -1 1 1 y = 61 - 1@ X + u y = 6- 1 - 1@ X + u o = >- 1 0 H X + >- 1H u o = >- 1 - 1H X + >- 1H u X X (C) (D) -1 -1 1 0 -1 1 y = 6- 1 - 1@ X - u y = 61 - 1@ X - u The state transition matrix eAt of the system shown in the figure above is e-t 0 e-t 0 (A) > -t -tH (B) > H -t te e - te e-t e-t 0 (C) > -t -tH e e

The gain _20 log G ^s h i is 32 dB 10 rad/s respectively. The phase (A) 39.8 s (C) 32 s

and - 8 dB at 1 rad/s and is negative for all w. Then G ^s h is (B) 392.8 s (D) 322 s

6.2

TWO MARKS

(B) 5 (D) 100

s+1 5s2 + 6s + 2 (C) 2 s + 1 s + 4s + 2

2012 6.7

6.8

s+1 s2 + 6s + 2 (D) 2 1 5s + 6s + 2

TWO MARKS

The feedback system shown below oscillates at 2 rad/s when

(A) K = 2 and a = 0.75 (C) K = 4 and a = 0.5

(B)

(B) K = 3 and a = 0.75 (D) K = 2 and a = 0.5

The state variable description of an LTI system is given by Jxo1N J 0 a1 0NJx1N J0N K O K OK O K O Kxo2O = K 0 0 a2OKx2O + K0O u Kxo O Ka 0 0OKx 3O K 1O 3 3 L P L PL P L P Jx1N K O y = _1 0 0iKx2O Kx 3O L P where y is the output and u is the input. The system is controllable for (A) a1 ! 0, a2 = 0, a 3 ! 0 (B) a1 = 0, a2 ! 0, a 3 ! 0 (C) a1 = 0, a 3 ! 0, a 3 = 0 (D) a1 ! 0, a2 ! 0, a 3 = 0 2011

6.9

Statement for Linked Answer Questions 4 and 5:

A system with transfer function G (s) =

.n

The signal flow graph for a system is given below. The transfer Y ^s h function for this system is U ^s h

(A)

ONE MARK

(s2 + 9) (s + 2) (s + 1) (s + 3) (s + 4) is excited by sin (wt). The steady-state output of the system is zero at .in w = 1 rad/s (B) w = 2 rad/s o(A) c . a i (D) w = 4 rad/s od (C) w = 3 rad/s 6.6

The open-loop transfer function of a dc motor is givenww as w w ^s h = 10 . When connected in feedback as shown below, the Va ^s h 1 + 10s approximate value of Ka that will reduce the time constant of the closed loop system by one hundred times as compared to that of the open-loop system is

(A) 1 (C) 10 6.3

A I D 2012

O N

2013

e-t - te-t (D) > H 0 e-t

ONE MARK

The root locus plot for a system is given below. The open loop transfer function corresponding to this plot is given by

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Page 129

100 . s (s + 10) 2 The plant is placed in a unity negative feedback configuration as shown in the figure below. The input-output transfer function of a plant H (s) =

s (s + 1) (s + 2) (s + 3) (s + 1) (B) G ^s h H ^s h = k s (s + 2) (s + 3) 2 1 (C) G ^s h H ^s h = k s (s - 1) (s + 2) (s + 3) (s + 1) (D) G ^s h H ^s h = k s (s + 2) (s + 3) 6.10

The gain margin of the system under closed loop unity negative feedback is (A) 0 dB (B) 20 dB (C) 26 dB (D) 46 dB

6.12

(A) G ^s h H ^s h = k

The signal flow graph that DOES NOT model the plant transfer function H (s) is

6.13

For the transfer function G (jw) = 5 + jw , the corresponding Nyquist plot for positive frequency has the form

A I D

O N

2010

. The transfer function Y (s) /R (s) of the system shown is

ia od

6.14

ONE MARK

in co.

n

. ww

w

2011 6.11

TWO MARKS

The block diagram of a system with one input u and two outputs y1 and y2 is given below.

6.15

A state space model of the above system in terms of the state vector x and the output vector y = [y1 y2]T is (A) xo = [2] x + [1] u ; y = [1 2] x 1 (B) xo = [- 2] x + [1] u; y = > H x 2 -2 0 1 (C) xo = > x + > H u ; y = 81 2B x H 0 -2 1 2 0 1 1 (D) xo = > H x + > H u ; y = > H x 0 2 1 2

6.16

1 s+1 (C) 2 (D) 2 s+1 s+3 Y (s) A system with transfer function has an output = s s+p X ( s ) p y (t) = cos a2t - k 3 for the input signal x (t) = p cos a2t - p k. Then, the system param2 eter p is (B) 2/ 3 (A) 3 (A) 0

(B)

(C) 1

(D)

3 /2

For the asymptotic Bode magnitude plot shown below, the system transfer function can be

Common Data For Q. 7.4 & 7.5

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(A) 10s + 1 0.1s + 1 (C) 100s 10s + 1

Page 130

(B) 100s + 1 0.1s + 1 (D) 0.1s + 1 10s + 1

(C) The system is uncontrollable for all values of p and q (D) We cannot conclude about controllability from the given data 2009

2010 6.17

TWO MARKS

6.22

TWO MARKS

The feedback configuration and the pole-zero locations of 2 G (s) = s2 - 2s + 2 s + 2s + 2 are shown below. The root locus for negative values of k , i.e. for - 3 < k < 0 , has breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) equal to

A unity negative feedback closed loop system has a plant with the and a controller Gc (s) in the transfer function G (s) = 2 1 s + 2s + 2 feed forward path. For a unit set input, the transfer function of the controller that gives minimum steady state error is (B) Gc (s) = s + 2 (A) Gc (s) = s + 1 s+2 s+1 (s + 1) (s + 4) (C) Gc (s) = (D) Gc (s) = 1 + 2 + 3s s (s + 2) (s + 3)

Common Data For Q. 7.10 & 7.11 : The signal flow graph of a system is shown below:

6.23

6.18

6.19

The state variable representation of the system can be -1 1 0 1 1 0 o= > xo = > x +> Hu H x x + u H > H (A) (B) -1 0 2 -1 0 2 yo = 80 0.5B x yo = [0 0.5] x 1 1 0 xo = > x +> Hu H (C) (D) -1 0 2 yo = 80.5 0.5B x The transfer function of the system is (B) (A) s2+ 1 s +1 (C) 2 s + 1 (D) s +s+1 2009

6.20

-1 xo = > -1 yo = 80.5

(C) ! 3 and 0c

(D) ! 3 and 45c

The unit step response of an under-damped second order system has steady state value of -2. Which one of the following transfer functions has theses properties ? (A) 2 - 2.24 (B) 2 - 3.82 s + 2.59s + 1.12 s + 1.91s + 1.91 - 382 (C) 2 - 2.24 (D) 2 s - 2.59s + 1.12 s - 1.91s + 1.91

O N

Common Data For Q. 7.16 and 7.17 :

no w.

The .in Nyquist plot of a stable transfer function G (s) is shown in the ofigure c . are interested in the stability of the closed loop system in ia

d

the feedback configuration shown.

ww

ONE MARK

The magnitude plot of a rational transfer function G (s) with real coefficients is shown below. Which of the following compensators has such a magnitude plot ?

6.24

6.25

(A) Lead compensator (C) PID compensator 6.21

(B) ! 2 and 45c

A I D

1 0 x +> Hu H 0 2 0.5B x

s-1 s2+1 s-1 s2+s+1

(A) ! 2 and 0c

(B) Lag compensator (D) Lead-lag compensator

Consider the system dx = Ax + Bu with A = =1 0G and B = = p G q 0 1 dt where p and q are arbitrary real numbers. Which of the following statements about the controllability of the system is true ? (A) The system is completely state controllable for any nonzero values of p and q (B) Only p = 0 and q = 0 result in controllability

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 (D) G (s) is marginally stable The gain and phase margins of G (s) for closed loop stability are (A) 6 dB and 180c (B) 3 dB and 180c (C) 6 dB and 90c (D) 3 dB and 90c 2008

6.26

ONE MARKS

Step responses of a set of three second-order underdamped systems all have the same percentage overshoot. Which of the following diagrams represents the poles of the three systems ?

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Page 131

6.30

Jx1N R b - g 0 VJx1N R0 0 V W u1 WK O S K O S (A) d K x2O = S g a 0 WK x2O+ S0 1 We o u2 dt K O S x3 S- a b 0 WWK x3O SS1 0 WW L P L P Jx1N RT0 a g XVJx1N TR1 0 XV W u1 WK O S K O S (B) d K x2O = S0 - a - g WK x2O+ S0 1 We o u2 dt K O S x3 S0 b - b WWK x3O SS0 0 WW L P TR L P Jx1N - a b 0 VXJx1N RT1 0 VX W u1 W S K O K O S (C) d K x2O = S- b - g 0 WK x2O+ S0 1 We o u2 dt K O S x3 S a g 0 WWK x3O SS0 0 WW L P TR L P Jx1N - a 0 b XVJx1N TR1 0XV W u1 WK O S K O S (D) d K x2O = S g 0 a WK x2O+ S0 1 We o u2 dt K O S x3 S- b 0 - a WWK x3O SS0 0 WW L P T X T XL P function A certain system has transfer G (s) = 2 s + 8 s + as - 4 where a is a parameter. Consider the standard negative unity feedback configuration as shown below

6.27

The pole-zero given below correspond to a

(A) Law pass filter (C) Band filter 2008 6.28

Which of the following statements is true? (A) The closed loop systems is never stable for any value of a (B) For some positive value of a, the closed loop system is stable, but not for all positive values. (C) For all positive values of a, the closed loop system is stable. (D) The closed loop system stable for all values of a, both positive and negative.

(B) High pass filter (D) Notch filter

A I D

TWO MARKS

Group I lists a set of four transfer functions. Group II gives a list of possible step response y (t). Match the step responses with the corresponding transfer functions.

O N

no w.

d

ww

6.32

6.33

(A) P - 3, Q - 1, R - 4, S - 2 (C) P - 2, Q - 1, R - 4, S - 2 6.29

The .innumber of open right half plane of 10 is G (s) = 5 4 3 s + 2s + 3s + 6s2 + 5s + 3 (A) 0 (B) 1 (C) 2 (D) 3

co ia.

6.31

The magnitude of frequency responses of an underdamped second order system is 5 at 0 rad/sec and peaks to 10 at 5 2 rad/sec. 3 The transfer function of the system is 500 (A) 2 (B) 2 375 s + 10s + 100 s + 5s + 75 720 (C) 2 (D) 2 1125 s + 12s + 144 s + 25s + 225 Group I gives two possible choices for the impedance Z in the diagram. The circuit elements in Z satisfy the conditions R2 C2 > R1 C1. The transfer functions V0 represents a kind of controller. Vi

(B) P - 3, Q - 2, R - 4, S - 1 (D) P - 3, Q - 4, R - 1, S - 2 Match the impedances in Group I with the type of controllers in Group II

A signal flow graph of a system is given below

The set of equalities that corresponds to this signal flow graph is

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(A) Q - 1, R - 2 (C) Q - 2, R - 3

(B) Q - 1, R - 3 (D) Q - 3, R - 2

Page 132 6.40

The state space representation of a separately excited DC servo motor dynamics is given as dw dt

> di H = =- 1 - 10G=ia G + =10Gu dt -1

1 w

0

o

2007 6.34

ONE MARK

2007 6.35

6.37

6.38

10 s2 + 11s + 11 (C) 2 10s + 10 s + 11s + 11

(A)

Statement for linked Answer Question 8.33 & 8.34 :

A control system with PD controller is shown in the figure. If the velocity error constant KV = 1000 and the damping ratio z = 0.5 , then the value of KP and KD are

Consider a linear system whose state space representation is 1 x (t) = Ax (t). If the initial state vector of the system is x (0) = = G, -2 e-2x then the system response is x (t) = > H. If the itial state vector - 2e-2t 1 of the system changes to x (0) = = G, then the system response -2 e-t becomes x (t) = > -tH -e

(B) KP = 100, KD = 0.9 (D) KP = 10, KD = 0.9

The transfer function of a plant is 6.41 The eigenvalue and eigenvector pairs (li vi) for the system are 5 1 1 1 1 T (s) = (A) e- 1 = Go and e- 2 = Go (B) e- 1, = Go and e2, = Go (s + 5)( s2 + s + 1) -1 -2 -1 -2 The second-order approximation of T (s) using dominant pole con1 1 1 1 cept is (C) e- 1, = Go and e- 2, = Go (D) e- 2 = Go and e1, = Go -1 -2 -1 -2 1 5 (A) (B) 6.42 The system matrix A is (s + 5)( s + 1) (s + 5)( s + 1) 1 1 .in 0 1 o(A) c (C) 2 5 (D) 2 1 (B) = =- 1 1G . a - 1 - 2G i s +s+1 s +s+1 d o 2 1 0 1 The open-loop transfer function of a plant is given as G (s) = s 1-w ..n 1 (C) = (D) = G w -1 -1 - 2 - 3G If the plant is operated in a unity feedback configuration, then w the lead compensator that an stabilize this control system is 10 (s - 1) 10 (s + 4) 2006 ONE MARK (B) (A) s+2 s+2 6.43 The open-loop function of a unity-gain feedback control system is 10 (s + 2) 2 (s + 2) (C) (D) given by s + 10 s + 10 K G (s) = (s + 1)( s + 2) A unity feedback control system has an open-loop transfer function K The gain margin of the system in dB is given by G (s) = s (s2 + 7s + 12) (A) 0 (B) 1

A I D

O N 2

The gain K for which s = 1 + j1 will lie on the root locus of this system is (A) 4 (B) 5.5 (C) 6.5 (D) 10 6.39

1 s2 + 11s + 11 (D) 2 1 s + s + 11 (B)

TWO MARKS

(A) KP = 100, KD = 0.09 (C) KP = 10, KD = 0.09 6.36

where w is the speed of the motor, ia is the armature current and w (s) of the motor u is the armature voltage. The transfer function U (s) is

If the closed-loop transfer function of a control system is given as s-5 , then It is T (s) (s + 2)( s + 3) (A) an unstable system (B) an uncontrollable system (C) a minimum phase system (D) a non-minimum phase system

The asymptotic Bode plot of a transfer function is as shown in the figure. The transfer function G (s) corresponding to this Bode plot is

(C) 20 2006 6.44

6.45

1 (s + 1)( s + 20) 100 (C) s (s + 1)( s + 20)

1 s (s + 1)( s + 20) 100 (D) s (s + 1)( 1 + 0.05s) (B)

6.46

TWO MARKS

Consider two transfer functions G1 (s) = 2 1 and s + as + b . G2 (s) = 2 s s + as + b The 3-dB bandwidths of their frequency responses are, respectively (B) a2 + 4b , a2 - 4b (A) a2 - 4b , a2 + 4b (C)

(A)

(D) 3

a2 - 4b , a2 - 4b

(D)

a 2 + 4b , a 2 + 4b

The Nyquist plot of G (jw) H (jw)for a closed loop control system, passes through (- 1, j0) point in the GH plane. The gain margin of the system in dB is equal to (A) infinite (B) greater than zero (C) less than zero (D) zero The positive values of K and a so that the system shown in the figures below oscillates at a frequency of 2 rad/sec respectively are

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(A) 1, 0.75 (C) 1, 1 6.47

6.48

Page 133

(B) 2, 0.75 (D) 2, 2

The transfer function of a phase lead compensator is given by Gc (s) = 1 + 3Ts where T > 0 The maximum phase shift provide by 1 + Ts such a compensator is (A) p (B) p 2 3 (C) p (D) p 4 6 A linear system is described by the following state equation 0 1 Xo (t) = AX (t) + BU (t), A = = - 1 0G The state transition matrix of the system is cos t sin t - cos t sin t (A) = (B) = G - sin t cos t - sin t - cos t G - cos t - sin t cos t - sin t (C) = (D) = G - sin t cos t cos t sin t G

2005

Consider a unity - gain feedback control system whose open - loop 1 transfer function is : G (s) = as + s2 The value of a so that the system has a phase - margin equal to p 4 is approximately equal to (A) 2.40 (B) 1.40 (C) 0.84 (D) 0.74

O N

A I D .in

no w.

6.50

6.51

ONE MARK

6.55

A linear system is equivalently represented by two sets of state equations : Xo = AX + BU and Wo = CW + DU The eigenvalues of the representations are also computed as [l] and [m]. Which one of the following statements is true ? (A) [l] = [m] and X = W (B) [l] = [m] and X ! W (C) [l] ! [m] and X = W (D) [l] = [m] and X ! W

6.52

Which one of the following polar diagrams corresponds to a lag network ?

co ia.

d

w With the value of a set for a phase - margin of p , the value ofw unit 4 - impulse response of the open - loop system at t = 1 second is equal to (A) 3.40 (B) 2.40 (C) 1.84 (D) 1.74 2005

TWO MARKS

The polar diagram of a conditionally stable system for open loop gain K = 1 is shown in the figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for

6.54

Statement for Linked Answer Questions 7.41 & 7.42 :

6.49

Despite the presence of negative feedback, control systems still have problems of instability because the (A) Components used have non- linearities (B) Dynamic equations of the subsystem are not known exactly. (C) Mathematical analysis involves approximations. (D) System has large negative phase angle at high frequencies.

6.53

6.56

6.57

(A) K < 5 and 1 < K < 1 (B) K < 1 and 1 < K < 5 2 8 8 2 (C) K < 1 and 5 < K (D) K > 1 and 5 > K 8 8 In the derivation of expression for peak percent overshoot - px Mp = exp e o # 100% 1 - x2 Which one of the following conditions is NOT required ? (A) System is linear and time invariant (B) The system transfer function has a pair of complex conjugate poles and no zeroes. (C) There is no transportation delay in the system. (D) The system has zero initial conditions. A ramp input applied to an unity feedback system results in 5% steady state error. The type number and zero frequency gain of the system are respectively (A) 1 and 20 (B) 0 and 20 (C) 0 and 1 (D) 1 and 1 20 20 2 A double integrator plant G (s) = K/s , H (s) = 1 is to be compensated to achieve the damping ratio z = 0.5 and an undamped natural frequency, wn = 5 rad/sec which one of the following compensator Ge (s) will be suitable ? (B) s + 99 (A) s + 3 s + 99 s+3

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(C) 6.58

s-6 s + 8.33

Page 134

(D) s - 6 s

excited with 10u (t). The time at which the output reaches 99% of its steady state value is (A) 2.7 sec (B) 2.5 sec (C) 2.3 sec (D) 2.1 sec

K (1 - s) An unity feedback system is given as G (s) = . s (s + 3) Indicate the correct root locus diagram. 6.65

6.66

A system has poles at 0.1 Hz, 1 Hz and 80 Hz; zeros at 5 Hz, 100 Hz and 200 Hz. The approximate phase of the system response at 20 Hz is (A) - 90c (B) 0c (C) 90c (D) - 180c Consider the signal flow graph shown in Fig. The gain x5 is x1

(A)

Statement for Linked Answer Question 40 and 41 :

6.59

6.60

The open loop transfer function of a unity feedback system is given by -2s G (s) = 3e s (s + 2) The gain and phase crossover frequencies in rad/sec are, respectively (A) 0.632 and 1.26 (B) 0.632 and 0.485 (C) 0.485 and 0.632 (D) 1.26 and 0.632

6.61

6.62

6.67

.in open-loop transfer function of a unity feedback system is oThe c . ia

6.68

no w.

d

The gain margin for the system with open-loop transfer function ww 2 (1 + s) , is G (s) H (s) = s2 (A) 3 (B) 0 (C) 1 (D) - 3

K Given G (s) H (s) = .The point of intersection of the s (s + 1)( s + 3) asymptotes of the root loci with the real axis is (A) - 4 (B) 1.33 (C) - 1.33 (D) 4

6.69

6.70

2004 6.63

TWO MARKS

Consider the Bode magnitude plot shown in the fig. The transfer function H (s) is

6.71

(A)

(s + 10) (s + 1)( s + 100)

(B)

10 (s + 1) (s + 10)( s + 100)

102 (s + 1) 103 (s + 100) (D) (s + 10)( s + 100) (s + 1)( s + 10) A causal system having the transfer function H (s) = 1/ (s + 2) is

K s (s2 + s + 2)( s + 3) The range of K for which the system is stable is (A) 21 > K > 0 (B) 13 > K > 0 4 (C) 21 < K < 3 (D) - 6 < K < 3 4 For the polynomial P (s) = s2 + s 4 + 2s3 + 2s2 + 3s + 15 the number of roots which lie in the right half of the s -plane is (A) 4 (B) 2 (C) 3 (D) 1 G (s) =

The state variable equations of a system are : xo1 =- 3x1 - x2 = u, xo2 = 2x1 and y = x1 + u . The system is (A) controllable but not observable (B) observable but not controllable (C) neither controllable nor observable (D) controllable and observable Given A = =

1 0 , the state transition matrix eAt is given by 0 1G

0 e-t (A) > -t H e 0

et 0 (B) = G 0 et

e-t 0 (C) > H 0 e-t

0 et (D) = t G e 0

2003

(C) 6.64

bedg 1 - (be + cf + dg) 1 - (be + cf + dg) + bedg (D) abcd

(B)

sin (- 4t) + 2 sin (- t) - 2 sin (- 4t) + 2 sin (- t) (A) 1 = G 3 - sin (- 4t) + sin (- t) 2 sin (- 4t) + sin (- t) sin (- 2t) sin (2t) (B) = sin (t) sin (- 3t)G sin (4t) + 2 sin (t) 2 sin (- 4t) - 2 sin (- t) (C) 1 = 2 sin (4t) + sin (t) G 3 - sin (- 4t) + sin (t) cos (- t) + 2 cos (t) 2 cos (- 4t) + 2 cos (- t) (D) 1 = G 3 - cos (- 4t) + cos (- t) - 2 cos (- 4t) + cos (t)

A I D

O N ONE MARK

abcd 1 - (be + cf + dg) + bedg -2 2 If A = = , then sin At is 1 - 3G (C)

Based on the above results, the gain and phase margins of the system will be (A) -7.09 dB and 87.5c (B) 7.09 dB and 87.5c (C) 7.09 dB and - 87.5c (D) - 7.09 and - 87.5c 2004

1 - (be + cf + dg) abcd

6.72

ONE MARK

Fig. shows the Nyquist plot of the open-loop transfer function G (s) H (s) of a system. If G (s) H (s) has one right-hand pole, the

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Page 135

closed-loop system is

(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 6.73

A PD controller is used to compensate a system. Compared to the uncompensated system, the compensated system has (A) a higher type number (B) reduced damping (C) higher noise amplification (D) larger transient overshoot 6.78

2003 6.74

TWO MARKS

The signal flow graph of a system is shown in Fig. below. The transfer function C (s)/ R (s) of the system is 6.79

(A)

6.76

(B)

s (s + 2) s + 29s + 6

(D)

A I D

6s 2 s + 29s + 6

tet (A) = G t et (C) = t G te

2

2

O N

6.82

(s + 0.1) 3 (s + 0.1) 3 7 (B) 10 (s + 10)( s + 100) (s + 10) 2 (s + 100) 2 (s + 0.1) (s + 0.1) 3 (C) (D) (s + 10) 2 (s + 100) (s + 10)( s + 100) 2 A second-order system has the transfer function C (s) = 2 4 R (s) s + 4s + 4 With r (t) as the unit-step function, the response c (t) of the system is represented by (A) 108

6.77

The zero-input response of a system given by the state-space equation 1 0 x1 x1 (0) 1 xo1 =xo G = =1 1G=x G and =x (0)G = = 0 G is 2 2 2 et (B) = G t t (D) = t G te

s (s + 27) s + 29s + 6 K .in The root locus of system G (s) H (s) = has the breako c 2002 ONE MARK . s (s + 2)( s + 3) ia away point located at d o .n6.80 (A) (- 0.5, 0) (B) (- 2.548, 0) Consider a system with transfer function G (s) = 2s + 6 . Its w ks + s + 6 ww damping ratio will be 0.5 when the value of k is (C) (- 4, 0) (D) (- 0.784, 0) (A) 2 (B) 3 6 The approximate Bode magnitude plot of a minimum phase system is shown in Fig. below. The transfer function of the system is (C) 1 (D) 6 6 6.81 Which of the following points is NOT on the root locus of a system k with the open-loop transfer function G (s) H (s) = s (s + 1)( s + 3) (A) s =- j 3 (B) s =- 1.5 (C) s =- 3 (D) s =- 3 (C)

6.75

6 2 s + 29s + 6

The gain margin and the phase margin of feedback system with 8 are G (s) H (s) = (s + 100) 3 (A) dB, 0c (B) 3, 3 (C) 3, 0c (D) 88.5 dB, 3

6.83

The phase margin of a system with the open - loop transfer function (1 - s) G (s) H (s) = (1 + s)( 2 + s) (A) 0c (B) 63.4c (C) 90c (D) 3 The transfer function Y (s)/ U (s) of system described by the state equation xo (t) =- 2x (t) + 2u (t) and y (t) = 0.5x (t) is 1 (A) 0.5 (B) (s - 2) (s - 2) 1 (C) 0.5 (D) (s + 2) (s + 2) 2002

6.84

TWO MARKS

The system shown in the figure remains stable when (A) k < - 1 (B) - 1 < k < 3 (C) 1 < k < 3 (D) k > 3

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6.85

100 The transfer function of a system is G (s) = . For a (s + 1)( s + 100) unit - step input to the system the approximate settling time for 2% criterion is

(A)100 sec (C) 1 sec 6.86

Page 136

(B) 4 sec (D) 0.01 sec

The characteristic polynomial of a system is 5

4

3

(A) only if 0 # k # 1 (C) only if k > 5

2

q (s) = 2s + s + 4s + 2s + 2s + 1 The system is (A) stable (C) unstable

(B) marginally stable (D) oscillatory

2001 6.92

6.87

The system with the open loop transfer 1 has a gain margin of G (s) H (s) = 2 (A) - 6 db s (s + s + 1) (B) 0 db (C) 35 db (D) 6 db 2001

6.88

6.89

function

ONE MARK

The Nyquist plot for the open-loop transfer function G (s) of a unity negative feedback system is shown in the figure, if G (s) has no pole in the right-half of s -plane, the number of roots of the system characteristic equation in the right-half of s -plane is (A) 0 (B) 1 (C) 2 (D) 3

A I D

Z3 (s) - Z3 (s) , in Z1 (s) + Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) . The equivalent of the block diagram in the figure is given is o - Z (s) - Z (s) c ia. (B) Z2 (s) - Z3 (3s) + Z4 (s) , Z1 (s) +3 Z3 (s) d .no w Z3 (s) Z3 (s) (C) , ww Z2 (s) + Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) - Z3 (s) Z3 (s) (D) , Z2 (s) - Z3 (s) + Z4 (s) Z1 (s) + Z3 (s) 6.93 The open-loop DC gain of a unity negative feedback system with closed-loop transfer function 2 s + 4 is s + 7s + 13 (A) 4 (B) 4 13 9

O N

(A)

6.94

6.91

TWO MARK

An electrical system and its signal-flow graph representations are shown the figure (A) and (B) respectively. The values of G2 and H , respectively are

(C) 4

6.90

(B) only if 1 < k < 5 (D) if 0 # k < 1 or k > 5

If the characteristic equation of a closed - loop system is s2 + 2s + 2 = 0 , then the system is (A) overdamped (B) critically damped (C) underdamped (D) undamped

The feedback control system in the figure is stable

(A) for all K $ 0 (C) only if 0 # K < 1

The root-locus diagram for a closed-loop feedback system is shown in the figure. The system is overdamped.

(D) 13

(B) only if K $ 0 (D) only if 0 # K # 1

2000 6.95

ONE MARK

An amplifier with resistive negative feedback has tow left half plane poles in its open-loop transfer function. The amplifier (A) will always be unstable at high frequency (B) will be stable for all frequency (C) may be unstable, depending on the feedback factor (D) will oscillate at low frequency.

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2000 6.96

TWO MARKS

6.98

ONE MARK

The gain margin (in dB) of a system a having the loop transfer function

(A) 0 (C) 6

2 is s (s + 1) (B) 3 (D) 3

1998

(A) controllable and observable (B) controllable, but not observable (C) observable, but not controllable (D) neither controllable nor observable

A I D

O N

6.102

6.103

TWO MARKS

An amplifier is assumed to have a single-pole high-frequency transfer function. The rise time of its output response to a step function input is 35 n sec . The upper 3 dB frequency (in MHz) for the amplifier to as sinusoidal input is approximately at (A) 4.55 (B) 10 (C) 20 (D) 28.6 If the closed - loop transfer function T (s) of a unity negative feedback system is given by an - 1 s + an T (s) = n n-1 s + a1 s + .... + an - 1 s + an then the steady state error for a unit ramp input is (A) an (B) an an - 1 an - 2 (C) an - 2 (D) zero an - 2 Consider the points s1 =- 3 + j4 and s2 =- 3 - j2 in the s-plane.

.

ia od

6.107

6.108

6.101

The transfer function of a tachometer is of the form (B) K (A) Ks s K (C) (D) n(s K i . + 1) s (s + 1) co

6.106

The phase margin (in degrees) of a system having the loop transferw.n ww function G (s) H (s) = 2 3 is s (s + 1) (A) 45c (B) - 30c (C) 60c (D) 30c 1999

ONE MARK

The number of roots of s3 + 5s2 + 7s + 3 = 0 in the left half of the s -plane is (A) zero (B) one (C) two (D) three

6.105

The system modeled described by the state equations is 0 1 0 x + > Hu X => H 2 -3 1 Y = 81 1B x

6.100

For the system described by the state equation R0V R 0 1 0V W S W S xo = S 0 0 1W x + S0W u SS1WW SS0.5 1 2WW X T X T If the control signal u is given by u = [- 0.5 - 3 - 5] x + v , then the eigen values of the closed-loop system will be (A) 0, - 1, - 2 (B) 0, - 1, - 3 (C) - 1, - 1, - 2 (D) 0, - 1, - 1

6.104

For a second order system with the closed-loop transfer function T (s) = 2 9 s + 4s + 9 the settling time for 2-percent band, in seconds, is (A) 1.5 (B) 2.0 (C) 3.0 (D) 4.0

G (s) H (s) =

6.99

Then, for a system with the open-loop transfer function G (s) H (s) = K 4 (s + 1) (A) s1 is on the root locus, but not s2 (B) s2 is on the root locus, but not s1 (C) both s1 and s2 are on the root locus (D) neither s1 nor s2 is on the root locus

1 A system described by the transfer function H (s) = 3 2 s + as + ks + 3 is stable. The constraints on a and k are. (A) a > 0, ak < 3 (B) a > 0, ak > 3 (C) a < 0, ak > 3 (D) a > 0, ak < 3 1999

6.97

Page 137

6.109

6.110

Consider a unity feedback control system with open-loop transfer K . function G (s) = s (s + 1) The steady state error of the system due to unit step input is (A) zero (B) K (C) 1/K (D) infinite

The transfer function of a zero-order-hold system is (A) (1/s) (1 + e-sT ) (B) (1/s) (1 - e-sT ) (C) 1 - (1/s) e-sT (D) 1 + (1/s) e-sT In the Bode-plot of a unity feedback control system, the value of phase of G (jw) at the gain cross over frequency is - 125c. The phase margin of the system is (A) - 125c (B) - 55c (C) 55c (D) 125c Consider a feedback control system with loop transfer function K (1 + 0.5s) G (s) H (s) = s (1 + s) (1 + 2s) The type of the closed loop system is (A) zero (B) one

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Page 138

(C) two (D) three 6.111

6.112

6.113

6.114

The transfer function of a phase lead controller is 1 + 3Ts . The 1 + Ts maximum value of phase provided by this controller is (A) 90c (B) 60c (C) 45c (D) 30c The Nyquist plot of a phase transfer function g (jw) H (jw) of a system encloses the (–1, 0) point. The gain margin of the system is (A) less than zero (B) zero (C) greater than zero (D) infinity 2s2 + 6s + 5 (s + 1) 2 (s + 2) The characteristic equation of the system is (A) 2s2 + 6s + 5 = 0 (B) (s + 1) 2 (s + 2) = 0 (C) 2s2 + 6s + 5 + (s + 1) 2 (s + 2) = 0 (D) 2s2 + 6s + 5 - (s + 1) 2 (s + 2) = 0 The transfer function of a system is

A I D

In a synchro error detector, the output voltage is proportional to [w (t)] n, where w (t) is the rotor velocity and n equals (A) –2 (B) –1 (C) 1 (D) 2

O N

no w.

d

w

w ONE MARK

1997 6.115

.in

co ia.

In the signal flow graph of the figure is y/x equals

(A) 3 (B) 5 2 (C) 2 (D) None of the above 6.116

A certain linear time invariant system has the state and the output equations given below 1 - 1 X1 0 Xo1 > o H = >0 1 H>X H + >1H u 2 X2 y = 81 1B: X1 D X2 If X1 (0) = 1, X2 (0) =- 1, u (0) = 0, then

dy dt

is t=0

(A) 1 (B) –1 (C) 0 (D) None of the above ***********

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Page 139

SOLUTIONS 6.1

Pk2 = ^1 h^s-1h^1 h^1 h = s-1 since, all the loops are touching to the paths Pk1 and Pk2 so, Dk 1 = Dk 2 = 1 Now, we have D = 1 - (sum of individual loops) + (sum of product of nontouching loops) Here, the loops are

Option (B) is correct. From the given plot, we obtain the slope as 20 log G2 - 20 log G1 Slope = log w2 - log w1

L1 = ^- 4h^1 h =- 4 L2 = ^- 4h^s-1h = 4s-1 L 3 = ^- 2h^s-1h^s-1h =- 2s-2 L 4 = ^- 2h^s-1h^1 h =- 2s-1 As all the loop L1, L2, L 3 and L 4 are touching to each other so, D = 1 - ^L1 + L2 + L 3 + L 4h

From the figure

and

20 log G2 20 log G1 w1 w2

=- 8 dB = 32 dB = 1 rad/s = 10 rad/s

So, the slope is

Slope = - 8 - 32 log 10 - log 1 =- 40 dB/decade Therefore, the transfer function can be given as G ^s h = k2 S at w = 1 G ^ jwh = k 2 = k w In decibel, 20 log G ^ jwh = 20 log k = 32 32

6.2

or, k = 10 = 39.8 Hence, the Transfer function is G ^s h = k2 = 392.8 s s Option (C) is correct. Given, open loop transfer function G ^s h = 10Ka = Ka 1 1 + 10s s + 10 By taking inverse Laplace transform, we have g ^ t h = e- t

= 1 - ^- 4 - 4s-1 - 2s-2 - 2s-1h = 5 + 6s1 + 2s2 From Mason’s gain formulae Y ^s h = SPk Dk D U ^s h s-2 + s-1 = 5 + 6s-1 + 2s-2 = 2s+1 5s + 6s + 2

A I D

20

1 10

O N

.in

no w.

ww

6.3

Option (A) is correct. For the given SFG, we have two forward paths

d

and

y = ^- 1h^1 h x2 + ^- 1h^1 h^- 1h x1 + ^1 h^- 1h^1 h^- 1h^1 h u = x1 - x 2 + u Hence, in matrix form we can write the state variable equations - 1 0 x1 -1 xo1 > o H = > 1 - 1H >x H + > 1 H u x2 2 x1 and y = 81 - 1B > H + u x2

1 10

Now, given that ka reduces open loop time constant by a factor of 100. i.e., tcl = tol 100 1 or, = 10 100 ka Hence, ka = 10

co ia.

So, from the state diagram, we obtain xo1 =- x1 - u xo2 =- x2 + ^1 h^- 1h^1 h^- 1h u + ^- 1h^1 h^- 1h x1 xo2 =- x2 + x1 + u

Comparing with standard form of transfer function, Ae-t/t , we get the open loop time constant, tol = 10 Now, we obtain the closed loop transfer function for the given system as G ^s h 10Ka H ^s h = = 1 + G ^s h 1 + 10s + 10Ka Ka = s + ^Ka + 101 h By taking inverse Laplace transform, we get h ^ t h = ka .e-^k + ht So, the time constant of closed loop system is obtained as tcl = 1 1 ka + 10 or, tcl = 1 ka (approximately) a

Option (A) is correct. For the shown state diagram we can denote the states x1 , x2 as below

6.4

which can be written in more general form as -1 0 -1 X +> H Xo = > 1 - 1H 1 y = 81 - 1B X + u 6.5

Option (A) is correct. From the obtained state-variable equations We have -1 0 A => 1 - 1H So,

S+1 0 SI - A = > - 1 S + 1H

Pk1 = ^1 h^s-1h^s-1h^1 h = s-2

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1 >S + 1 0 H ^S + 1h2 1 S + 1 R 1 V S 0 W S+1 W =S 1 1 W S S^S + 1h2 S + 1W T X Hence, the state transition matrix is obtained as eAt = L-1 ^SI - Ah-1 V_ ZR 1 ]]S 0 Wbb S+1 W = L-1 [S 1 1 W` S ]S^S + 1h2 S + 1Wb \T Xa e-1 0 = > -t -tH te e Option (C) is correct. (s2 + 9) (s + 2) G (s) = (s + 1) (s + 3) (s + 4) (- w2 + 9) (jw + 2) = (jw + 1) (jw + 3) (jw + 4) The steady state output will be zero if

^SI - Ah-1 =

and

6.6

G (jw) = 0 -w 2 + 9 = 0 6.7

Page 140

&

R 0 a 0VR0V R 0V 1 WS W S W S AB = S 0 0 a2WS0W = Sa2W SSa 0 0WWSS1WW SS 0WW 3 RT 0 0XT aX1 a2VWTRS0XVW RSa1 a2VW S A2 B = Sa2 a 3 0 0WS0W = S 0W SS 0 a a 0WWSS1WW SS 0WW 3 1 T XT X T X For controllability it is necessary that following matrix has a tank of n = 3 . R0 0 a a V 1 2W S 2 U = 6B : AB : A B@ = S0 a2 0W SS1 0 0WW So, a2 ! 0 X T a1 a 2 ! 0 & a1 ! 0 a 3 may be zero or not. 6.9

w = 3 rad/s

6.10

Option (A) is correct.

Option (B) is correct. For given plot root locus exists from - 3 to 3, So there must be odd number of poles and zeros. There is a double pole at s =- 3 Now poles = 0, - 2, - 3, - 3 zeros =- 1 k (s + 1) Thus transfer function G (s) H (s) = s (s + 2) (s + 3) 2 Option (A) is correct. We have G (jw) = 5 + jw Here s = 5 . Thus G (jw) is a straight line parallel to jw axis.

K (s + 1) [R (s) - Y (s)] s3 + as2 + 2s + 1 6.11 Option (B) is correct. K (s + 1) K (s + 1) R (s) Y (s) ;1 + 3 = Here x s + as2 + 2s + 1E s3 + as2 + 2s + 1 3 2 Y (s) [s + as + s (2 + k) + (1 + k)] = K (s + 1) R (s) y Y (s) Transfer Function, H (s) = R (s) .in oNow K (s + 1) c . y1 = 3 ia d s + as2 + s (2 + k) + (1 + k) .no y1 (s + 2) Routh Table : w ww yo1 + 2y1 xo + 2x xo xo Y (s) =

A I D

O N

For oscillation,

Auxiliary equation

=u =u =u =- 2x + u = [- 2] x + [1] u Drawing SFG as shown below

a (2 + K) - (1 + K) =0 a a = K+1 K+2 A (s) = as2 + (k + 1) = 0 s2 =- k + 1 = - k + 1 (k + 2) =- (k + 2) a (k + 1)

xo1 = [- 2] x1 + [1] u y1 = x1 ; y2 = 2x1

Thus

s = j k+2

y1 1 y = > H = > H x1 y2 2

jw = j k + 2

and 6.8

w = k + 2 = 2 (Oscillation frequency) k =2 a = 2 + 1 = 3 = 0.75 2+2 4

Option (D) is correct. General form of state equations are given as xo = Ax + Bu yo = Cx + Du For the given problem R 0 a 0V R0V 1 S W S W A = S 0 0 a2W, B = S0W SSa SS1WW 0 0WW 3 T X X GATE Electronics &T Communication

dy = y1 and xo = 1 dx 1 y1 x = > H = > H = > Hx 2 y2 2x = 1 u s+2

x1 = x

Here 6.12

Option (C) is correct. 100 s (s + 10) 2 100 Now G (jw) H (jw) = jw (jw + 10) 2 If wp is phase cross over frequency +G (jw) H (jw) = 180c We have

G (s) H (s) =

Thus

- 180c = 100 tan-1 0 - tan-1 3 - 2 tan-1 a

or or

- 180c =- 90 - 2 tan-1 (0.1wp) 45c = tan-1 (0.1wp)

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wp 10 k

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

or or Now

Page 141

tan 45c 0.1wp = 1 wp = 10 rad/se 100 G (jw) H (jw) = w (w2 + 100)

At w = wp

So,

100 G (jw) H (jw) = = 1 10 (100 + 100) 20 Gain Margin =- 20 log 10 G (jw) H (jw) =- 20 log 10 b 1 l 20 = 26 dB 6.13

6.14

Option (D) is correct. From option (D) TF = H (s) 100 100 = ! s (s2 + 100) s (s + 10) 2

Option (A) is correct. Initial slope is zero, so K = 1 At corner frequency w 1 = 0.5 rad/ sec , slope increases by + 20 dB/ decade, so there is a zero in the transfer function at w 1 At corner frequency w 2 = 10 rad/ sec , slope decreases by - 20 dB/ decade and becomes zero, so there is a pole in transfer function at w2 K a1 + s k w1 Transfer function H (s) = s a1 + w 2 k 1 a1 + s k (1 + 10s) 0.1 = = s (1 + 0.1s) 1 + a 0. 1 k 6.17 Option (D) is correct. Steady state error is given as sR (s) eSS = lim n s " 0 i 1 + G (s) GC (s) co. . a (unit step unit) R (s) = 1 di o s n . ...(1)w 1 ww eSS = lim s " 0 1 + G (s) GC (s) ...(2) 1 = lim s"0 GC (s) 1+ 2 s + 2s + 2 eSS will be minimum if lim GC (s) is maximum s"0 In option (D) lim GC (s) = lim 1 + 2 + 3s = 3 s s"0 s"0 So, eSS = lim 1 = 0 (minimum) s"0 3

A I D

E (s) = R (s) - H (s) E (s) (s + 1)

1 = R (s) - Y (s) s + 1D sE (s) = R (s) - Y (s) (s + 1) E (s) Y (s) = s+1

E (s) :1 -

O N

sY (s) = R (s) - Y (s) (s + 1) Y (s) = R (s)

From (1) and (2) Transfer function

Y (s) = 1 R (s) s + 1 6.15

p = 2/ 3

6.16

1 s+1

= R (s) - Y (s) +

qh = 9- p - a- p kC = p 3 2 6 p = p - tan-1 w apk 2 6 tan-1 a w k = p - p = p p 2 6 3 w = tan p = 3 a3k p 2 = 3 , (w = 2 rad/ sec) p

or

Option (B) is correct. From the given block diagram

H (s) = Y (s) - E (s) $

p = 2/ 3

or Alternative :

Option (B) is correct. Transfer function is given as Y (s) = s X (s) s + p jw H (jw) = jw + p

6.18

H (s) =

Option (D) is correct. Assign output of each integrator by a state variable

Amplitude Response H (jw) = Phase Response Input Output

or

w w2+ p 2

qh (w) = 90c - tan-1 a w k p x (t) = p cos a2t - p k 2 y (t) = H (jw) x (t - qh) = cos a2t - p k 3 w H (jw) = p = 2 w +p2 1 = 2 , (w = 2 rad/ sec) p 4+p2 4p 2 = 4 + p 2 & 3p 2 = 4

xo1 =- x1 + x2 xo2 =- x1 + 2u y = 0.5x1 + 0.5x2 State variable representation -1 1 0 x + > Hu xo = > H -1 0 2 yo = [0.5 0.5] x 6.19

Option (C) is correct.

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Page 142

By masson’s gain formula

Transfer function Y (s) H (s) = = U (s) Forward path given

- qd - qp1 + qz1 + qz2 = 180c - qd = 180c - (- qp1 + qz1 + q2) = 180c - (90c + 180 - 45c) =- 45c

/ PK DK D

6.23

P1 (abcdef ) = 2 # 1 # 1 # 0.5 = 12 s s s P2 (abcdef ) = 2 # 1 # 1 # 0.5 3 Loop gain L1 (cdc) =- 1 s L2 (bcdb) = 1 # 1 # - 1 = -21 s s s

kwn2 where x < 1 s + 2xwn s + wn2 Thus (A) or (B) may be correct For option (A) wn = 1.12 and 2xwn = 2.59 " x = 1.12 For option (B) wn = 1.91 and 2xwn = 1.51 " x = 0.69 T (s) =

6.24

D = 1 - [L1 + L2] = 1 - :- 1 - 12 D = 1 + 1 + 12 s s s s So,

6.20

6.21

D1 = 1, D2 = 2 Y (s) H (s) = = P1 D 1 + P2 D 2 D U (s) 1 :1+1:1 2 (1 + s) s =s = 2 1 1 (s + s + 1) 1+ + 2 s s

6.25

2

Option (B) is correct. The plot has one encirclement of origin in clockwise direction. Thus G (s) has a zero is in RHP. Option (C) is correct. The Nyzuist plot intersect the real axis ate - 0.5. Thus G. M. =- 20 log x =- 20 log 0.5 = 6.020 dB And its phase margin is 90c.

A I D

Option (C) is correct. Option (C) is correct. Transfer function for the given pole zero plot is: (s + Z1)( s + Z2) This compensator is roughly equivalent to combining lead and lad (s + P1)( s + P2) compensators in the same design and it is referred also as PID compensator. .in the plot Re (P1 and P2 )>(Z1 and Z2 ) oFrom c . Option (C) is correct. dia So, these are two lead compensator. o Hence both high pass filters and the system is high pass filter. .n p 1 0 w Here and B = = G A == ww 6.27 Option (C) is correct. q 0 1G 1 0 p p == G G G = 0 1 q q p q S = 8B AB B = = q pG

AB = =

6.26

O N

Percent overshoot depends only on damping ratio, x . 2

Mp = e- xp 1 - x If Mp is same then x is also same and we get x = cos q Thus q = constant The option (C) only have same angle.

S = pq - pq = 0 Since S is singular, system is completely uncontrollable for all values of p and q . 6.22

Option (B) is correct. For under-damped second order response

Option (B) is correct. The characteristic equation is 1 + G (s) H (s) = 0 K (s2 - 2s + 2) or =0 1+ s2 + 2s + 2 or s2 + 2s + 2 + K (s2 - 2s + 2) = 0 2 or K =- s2 + 2s + 2 s - 2s + 2 For break away & break in point differentiating above w.r.t. s we have 2 2 dK =- (s - 2s + 2)( 2s + 2) - (s + 2s + 2)( 2s - 2) = 0 ds (s2 - 2s + 2) 2

Thus

6.28

6.29

Option (D) is correct. P = 2 25 2xwn = 0, x = 0 " Undamped s + 25

Graph 3

Q=

62 s + 20s + 62

2xwn = 20, x > 1 " Overdamped

Graph 4

R=

62 s2 + 12s + 62

2xwn = 12, x = 1 " Critically

Graph 1

S=

72 s2 + 7s + 72

2xwn = 7, x < 1 " underdamped

Graph 2

2

Option (C) is correct. We labeled the given SFG as below :

(s2 - 2s + 2)( 2s + 2) - (s2 + 2s + 2)( 2s - 2) = 0

or s =! 2 Let qd be the angle of departure at pole P , then From this SFG we have

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6.30

xo1 =- gx1 + bx3 + m1 xo2 = gx1 + ax3 xo3 =- bx1 - ax3 + u2 R V R VR V R V Sx1 W S- g 0 b WSx1 W S0 1 W u1 Sx2 W = S g 0 a WSx2 W+ S0 0 We o Thus SSx WW SS- b 0 - a WWSSx WW SS1 0 WW u2 3 3 X T X T XT X T Option (C) is correct. The characteristic equation of closed lop transfer function is

Page 143

For R,

Since R2 C2 > R1 C1, it is lag compensator. Option (D) is correct. In a minimum phase system, all the poles as well as zeros are on the left half of the s -plane. In given system as there is right half zero (s = 5), the system is a non-minimum phase system.

6.34

1 + G (s) H (s) = 0 =0 1+ 2 s+8 s + as - 4 or s 2 + as - 4 + s + 8 = 0 or s2 + (a + 1) s + 4 = 0 This will be stable if (a + 1) > 0 " a > - 1. Thus system is stable for all positive value of a. 6.31

Option (B) is correct. We have Kv = lim sG (s) H (s)

6.35

s"0

1 + G (s) H (s) = 0 1000 = lims " 0 s

or

3

6

2

z4

5

3

1

3

21 5

7 5

z2

4 3

3

z1

- 74

z0

1

N

w

.no

ww

d

1

For Q ,

Z =

(sC2 R2 + 1) sC2

We have

5 (s + 5)( s2 + s + 1) 5 = = 2 1 2 s 5`1 + j (s + s + 1) s +s+1 5

T (s) =

Option (A) is correct. 1 = 1 ( s 1 )( s - 1) + s -1 The lead compensator C (s) should first stabilize the plant i.e. 1 remove term. From only options (A), C (s) can remove this (s - 1) term

5 2x 1 - x2

1

.in (D) is correct. Option

G (s) =

wr = wn 1 - 2x2

Option (B) is correct. The given circuit is a inverting amplifier and transfer function is Vo = - Z = - Z (sC1 R1 + 1) R Vi R1 sC R + 1 1

or

2xwn = 10 + 100KD KD = 0.9

. We can see easily that pole at s =- 0.5 ! j 23 is dominant then pole at s =- 5 . Thus we have approximated it. 6.37

We have wr = 5 2 , and mr = 10 . Only options (A) satisfy these 3 values. wn = 10, x = 1 2 where wr = 10 1 - 2` 1 j = 5 2 4 and Hence satisfied mr = 1 5 1 = 10 22 1- 4 3

Kp = 100

In given transfer function denominator is (s + 5)[( s + 0.5) 2 + 43 ]

Option (C) is correct. For underdamped second order system the transfer function is Kwn2 s2 + 2xwn s + wn2 It peaks at resonant frequency. Therefore

1 + G (s) H (s) = 0 (100 + KD s) 100 =0 1+ s (s + 10)

or s2 + (10 + 100KD) s + 10 4 = 0 Comparing with s2 + 2xwn + wn2 = 0 we get

co ia.

6.36

T (s) =

6.33

IA

D O

z5

mr =

(Kp + KD s) 100 = Kp s (s + 100)

Now characteristics equation is

3z5 + 5z 4 + 6z3 + 3z2 + 2z + 1 = 0 The routh table is shown below. As there are tow sign change in first column, there are two RHS poles.

Resonant frequency and peak at this frequency

s"0

(Kp + KD s) 100 = Kp s (s + 100)

Now characteristics equations is

1 + G (s) = 0 5 4 or s + 2s + 3s3 + 6s2 + 5s + 3 = 0 Substituting s = z1 we have

6.32

1000 = lim s

or

Option (C) is correct. The characteristic equation is

z

Vo =- (sC2 R2 + 1) (sC1 R1 + 1) PID Controller # Vi sC2 R1 R2 Z = (sC2 R2 + 1) (sC1 R1 + 1) Vo =R2 # Vi (sC2 R2 + 1) R1

Thus

satisfies. 6.38

2

10 (s - 1) 1 # (s + 1)( s - 1) (s + 2) 10 Only option (A) = (s + 1)( s + 2)

G (s) C (s) =

Option (D) is correct. For ufb system the characteristics equation is 1 + G (s) = 0 K or 1+ =0 2 s (s + 7s + 12) or s (s2 + 7s + 12) + K = 0 Point s =- 1 + j lie on root locus if it satisfy above equation i.e (- 1 + j)[( - 1 + j) 2 + 7 (- 1 + j) + 12) + K] = 0 or K =+ 10

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Page 144

Solving (1) and (2) set of equations we get p q 0 1 =r s G = =- 2 - 3G

Option (D) is correct. At every corner frequency there is change of -20 db/decade in slope which indicate pole at every corner frequency. Thus K G (s) = s (1 + s)`1 + s j 20

The characteristic equation lI - A = 0

Bode plot is in (1 + sT) form 20 log K = 60 dB = 1000 w w = 0. 1

6.40

or l (l + 3) + 2 = 0 or l =- 1, - 2 Thus Eigen values are - 1 and - 2 Eigen vectors for l1 =- 1

K =5

Thus Hence

l -1 =0 2 l+3

G (s) =

100 s (s + 1)( 1 + .05s)

Option (A) is correct. We have or and

dw dt dia dt

> H = =- 1 - 10G=in G + =10Gu -1

1 w

0

(l1 I - A) X1 = 0

dw =- w + i n dt dia =- w - 10i + 10u a dt

...(2)

or - x11 - x21 = 0 or x11 + x21 = 0 We have only one independent equation x11 =- x21. Let x11 = K , then x21 =- K , the Eigen vector will be x11 K 1 =x G = =- K G = K =- 1G 21

Taking Laplace transform (i) we get sw (s) =- w (s) = Ia (s) or (s + 1) w (s) = Ia (s) Taking Laplace transform (ii) we get or or or or 6.41

...(3)

sIa (s) =- w (s) - 10Ia (s) + 10U (s) w (s) = (- 10 - s) Ia (s) + 10U (s) = (- 10 - s)( s + 1) w (s) + 10U (s) w (s) =- [s2 + 11s + 10] w (s) + 10U (s) (s2 + 11s + 11) w (s) = 10U (s) w (s) = 2 10 U (s) (s + 11s + 11)

Thus

e-2t > d (- 2e-2t)H dt

t=0

or

p - 2q =- 2 and r - 2s = 4 1 For initial state vector x (0) = = G the system response is -1 e-t x (t) = > -tH -e

Thus

6.42

p q 1 == r s G=- 2G

We get

e-t > d (- e-t)H dt

t=0

- e- (0)

==

p q

6.43

6.44

...(i)

6.45

or - x11 - x21 = 0 or x11 + x21 = 0 We have only one independent equation x11 =- x21. Let x11 = K, then x21 =- K , the Eigen vector will be x12 K 1 =x G = =- 2K G = K =- 2G 22 Option (D) is correct. As shown in previous solution the system matrix is 0 1 A == - 2 - 3G Option (D) is correct. Given system is 2nd order and for 2nd order system G.M. is infinite. Option (D) is correct. Option (D) is correct. If the Nyquist polt of G (jw) H (jw) for a closed loop system pass through (- 1, j0) point, the gain margin is 1 and in dB GM =- 20 log 1 = 0 dB

p q 1 r s G=- 1G

6.46

Option (B) is correct. The characteristics equation is 1 + G (s) H (s) = 0

1

> e- (0) H = =r s G=- 1G

1+

-1 p-q = 1G = = r - s G

We get

Now Eigen vector for l2 =- 2 (l2 I - A) X2 = 0 l2 - 1 x12 or = 2 l + 3G=x G = 0 2 22 n i - 2 - 1 x11 coor. =0 . = a 2 1 G=x21G di

w

p q 1 - 2e-2 (0) > 4e-2 (0) H = =r s G=- 2G -2 p - 2q = 4 G = = r - 2s G

d dt

no

. ww

1 For initial state vector x (0) = = G the system response is -2 e-2t x (t) = > H - 2e-2t d dt

A I D

From (3)

O N

Option (A) is correct. We have xo (t) = Ax (t) p q Let A == r sG

l1 - 1 x11 = 2 l + 3G=x G = 0 1 21 - 1 - 1 x11 = 2 2 G=x G = 0 21

or

...(1)

p - q =- 1 and r - s = 1

...(2)

K (s + 1) =0 s + as2 + 2s + 1 3

s3 + as2 + (2 + K) s + K + 1 = 0

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The Routh Table is shown below. For system to be oscillatory stable a (2 + K) - (K + 1) =0 a or ...(1) a = K+1 K+2

Page 145

Option (C) is correct. For a = 0.84 we have

6.50

G (s) = 0.84s2 + 1 s Due to ufb system H (s) = 1 and due to unit impulse response R (s) = 1, thus C (s) = G (s) R (s) = G (s) = 0.84s2 + 1 = 12 + 0.84 s s s Taking inverse Laplace transform

Then we have as2 + K + 1 = 0 At 2 rad/sec we have s = jw " s2 =- w2 =- 4 , Thus - 4a + K + 1 = 0 Solving (i) and (ii) we get K = 2 and a = 0.75 .

6.47

...(2) At t = 1,

s3

1

2+K

s2

a

1+K

s1

(1 + K) a - (1 + K) a

s0

1+K

6.48

N

Option (A) is correct. s 0 0 1 s -1 (sI - A) = = -= == G G 0 s -1 0 1 sG (sI - A)

s s -1 s +1 1 = 2 G = > -1 = s +1 1 s s +1 2

2

1 s2 + 1 s s2 + 1

or

6.53

6.54 6.55

Thus or or

1+a wg2

6.56

wg = (2)

Option (A) is correct. Despite the presence of negative feedback, control systems still have problems of instability because components used have nonlinearity. There are always some variation as compared to ideal characteristics. Option (B) is correct. Option (C) is correct. The peak percent overshoot is determined for LTI second order closed loop system with zero initial condition. It’s transfer function is 2

Option (A) is correct.

s"0

R (s) 1 = lim 1 + G (s) s " 0 s + sG (s) ess = lim 1 = 5% = 1 s " 0 sG (s) 20 kv = 1 = lim sG (s) = 20 s"0 ess = lim s s"0

or But

Finite

kv is finite for type 1 system having ramp input.

(as awg = 1) 1 4

+T (jw) = 0

For ramp input we have R (s) = 12 s Now ess = lim sE (s)

=1

1 + 1 = wg2

At w = 3 ,

+T (jw) = tan-1 (wT) - tan-1 (wbT) T (jw) = 1 +T (jw) =- tan-1 0 = 0 T (jw) = 1 b

wn2 s + 2xwn s + wn2 Transfer function has a pair of complex conjugate poles and zeroes.

or awg = 1 At gain crossover frequency G (jwg) = 1 wg2

At w = 3 ,

b > 1; T > 0

1 + w2 T2 1 + w2 b2 T2

T (jw) =

T (s) =

p = p + tan-1 (w a) - p g 4 p = tan-1 (w a) g 4

2

and At w in = 0 , . o a.cAt w = 0 ,

di

H

Option (C) is correct. 1 G (s) = as + We have s2 +G (jw) = tan-1 (wa) - p Since PM is p i.e. 45c, thus 4 p = p + +G (jw ) w " Gain cross over Frequeng g 4 cy or

.no

w ww

cos t sin t f (t) = eAt = L-1 [(sI - A)] -1 = = - sin t cos t G 6.49

IA

D O

fmax = tan-1 a - 1 2 a -1 3 - 1 = tan = tan-1 1 2 3 3 fmax = p 6

-1

Option (D) is correct. The transfer function of a lag network is T (s) = 1 + sT 1 + sbT

6.52

T>0

The maximum phase sift is

or

Option (C) is correct. We have where l is set of Eigen values Xo = AX + BU o and where m is set of Eigen values W = CW + DU If a liner system is equivalently represented by two sets of state equations, then for both sets, states will be same but their sets of Eigne values will not be same i.e. X = W but l ! m

6.51

Option (D) is correct. The transfer function of given compensator is Gc (s) = 1 + 3Ts 1 + Ts Comparing with Gc (s) = 1 + aTs we get a = 3 1 + Ts

c (t) = (t + 0.84) u (t) c (1 sec) = 1 + 0.84 = 1.84

6.57

Option (A) is correct.

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 6.58

Option (C) is correct. Any point on real axis of s - is part of root locus if number of OL poles and zeros to right of that point is even. Thus (B) and (C) are possible option. The characteristics equation is

Page 146

or 6.61

1 + G (s) H (s) = 0 or or

1+

K (1 - s) =0 s (s + 3) 2 K = s + 3s 1-s

For break away & break in point dK = (1 - s)( 2s + 3) + s2 + 3s = 0 ds or - s2 + 2s + 3 = 0 which gives s = 3 , - 1 Here - 1 must be the break away point and 3 must be the break in point. 6.59

Option (D) is correct. -2s G (s) = 3e s (s + 2) -2jw or G (jw) = 3e jw (jw + 2) 3 G (jw) = w w2 + 4 Let at frequency wg the gain is 1. Thus 3 =1 wg (wg2 + 4)

or or or Now

wg4 + 4wg2 - 9 = 0 wg2 = 1.606 wg = 1.26 rad/sec +G (jw) =- 2w - p - tan-1 w 2 2

Let at frequency wf we have +GH =- 180c w - p =- 2wf - p - tan-1 f 2 2 w or 2wf + tan-1 f = p 2 2 w w 3 or 2wf + c f - 1 ` f j m = p 2 2 3 2 or

or 6.60

6.62

Option (D) is correct. The open loop transfer function is 2 (1 + s) G (s) H (s) = s2 Substituting s = jw we have 2 (1 + jw) ...(1) G (jw) H (jw) = - w2 +G (jw) H (jw) =- 180c + tan-1 w The frequency at which phase becomes - 180c, is called phase crossover frequency. Thus - 180 =- 180c + tan-1 wf or tan-1 wf = 0 or wf = 0 The gain at wf = 0 is w2 = 3 G (jw) H (jw) = 2 1 + 2 w 1 Thus gain margin is = = 0 and in dB this is - 3 . 3 Option (C) is correct. Centroid is the point where all asymptotes intersects. SReal of Open Loop Pole - SReal Part of Open Loop Pole s = SNo.of Open Loop Pole - SNo.of Open Loop zero = - 1 - 3 =- 1.33 3

A I D 6.63

O N

Option (C) is correct. The given bode plot is shown below

.in

no w.

co ia.

d

ww

At w = 1 change in slope is +20 dB " 1 zero at w = 1 At w = 10 change in slope is - 20 dB " 1 poles at w = 10 At w = 100 change in slope is - 20 dB " 1 poles at w = 100 K (s + 1) Thus T (s) = s s + 1) ( 10 + 1)( 100 Now 20 log10 K =- 20 " K = 0.1 0.1 (s + 1) 100 (s + 1) Thus = T (s) = s s ( 10 + 1)( 100 + 1) (s + 10)( s + 100)

5wf wf3 =p 2 2 24 5wf .p 2 2

6.64

wf = 0.63 rad

Option (D) is correct. The gain at phase crossover frequency wf is 3 3 G (jwg) = = 2 wf (wf + 4) 0.63 (0.632 + 4) or G (jwg) = 2.27 G.M. =- 20 log G (jwg) - 20 log 2.26 =- 7.08 dB Since G.M. is negative system is unstable. The phase at gain cross over frequency is w +G (jwg) =- 2wg - p - tan-1 g 2 2 =- 2 # 1.26 - p - tan-1 1.26 2 2

=- 4.65 rad or - 266.5c PM = 180c + +G (jwg) = 180c - 266.5c =- 86.5c

1 2

Option (C) is correct. We have r (t) = 10u (t) or R (s) = 10 s Now H (s) = 1 s+2 C (s) = H (s) $ R (s) = or

1 $ 10 10 s + 2 s s (s + 2)

C (s) = 5 - 5 s s+2

c (t) = 5 [1 - e-2t] The steady state value of c (t) is 5. It will reach 99% of steady state value reaches at t , where 5 [1 - e-2t] = 0.99 # 5 or

1 - e-2t = 0.99

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or or 6.65

6.66

Page 147

e-2t = 0.1 - 2t = ln 0.1 t = 2.3 sec

We have only one independent equation x12 = 2x22 Let x22 = K , then x12 = 2K . Thus Eigen vector will be 2 x12 2K =x G = = K G = K = 1 G 22

Option (A) is correct. Approximate (comparable to 90c) phase shift are Due to pole at 0.01 Hz " - 90c Due to pole at 80 Hz " - 90c Due to pole at 80 Hz " 0 Due to zero at 5 Hz " 90c Due to zero at 100 Hz " 0 Due to zero at 200 Hz " 0 Thus approximate total - 90c phase shift is provided.

Digonalizing matrix -1 2 x11 x12 M == = x21 x22 G = 1 1G 1 -2 M-1 = ` - 1 j= G 3 -1 -1

Now

Now Diagonal matrix of sin At is D where sin (l1 t) 0 sin (- 4t) 0 D == == G 0 sin (l2 t) 0 sin (l2 t)G

Option (C) is correct. Mason Gain Formula T (s) =

1 - 2 x12 =- 1 2 G=x G = 0 22

or

Spk 3 k 3

Now matrix

In given SFG there is only one forward path and 3 possible loop.

- 1 2 sin (- 4t) 0 1 -2 =-` 1 j= G G = = 1 1 0 sin (- t) - 1 - 1G 3

p1 = abcd 31 = 1 3= 1 - (sum of indivudual loops) - (Sum of two non touching loops)

- sin (- 4t) - 2 sin (- t) 2 sin (- 4t) - 2 sin (- t) =-` 1 j= sin (- 4t) + 2 sin (t) - 2 sin (- 4t) - sin (- t)G 3

= 1 - (L1 + L2 + L3) + (L1 L3) Non touching loop are L1 and L3 where L1 L2 = bedg Thus

6.67

C (s) p1 3 1 = 1 - (be + cf + dg) + bedg R (s) abcd = 1 - (be + cf + dg) + bedg

Option (A) is correct. We have

-2 2 A == 1 - 3G

Characteristic equation is [lI - A] = 0 or

- sin (- 4t) - 2 sin (- t) 2 sin (- 4t) - 2 sin (- t) =-` 1 j= sin (- 4t) - sin (- t) - 2 sin (- 4t) + 2 sin (- t)G 3

A I D

O N

sin (- 4t) + 2 sin (- t) - 2 sin (- 4t) + 2 sin (- t) = ` 1 j= Gs 3 - sin (- 4t + sin (- t) 2 sin (- 4t) + sin (- t)

no w.

d

ww

or - 2x11 - 2x21 = 0 or x11 + x21 = 0 We have only one independent equation x11 =- x21. Let x21 = K , then x11 =- K , the Eigen vector will be x11 -K -1 =x G = = K G = K = 1 G 21

1 + G (s) = 0 1 + G (s)

K =0 s (s2 + 2s + 2)( s + 3) s 4 + 4s3 + 5s2 + 6s + K = 0 The routh table is shown below. For system to be stable, (21 - 4K) 0 < K and 0 < 2/7 This gives 0 < K < 21 4 1+

l + 2 -2 =0 -1 l + 3

(l1 I - A) X1 = 0 l1 + 2 - 2 x11 = 1 l + 3G=x G = 0 1 21 - 2 - 2 x11 =- 1 - 1G=x G = 0 21

n Option o.i (A) is correct. c . ia For ufb system the characteristic equation is

6.68

or (l + 2)( l + 3) - 2 = 0 or l2 + 5l + 4 = 0 Thus l1 =- 4 and l2 =- 1 Eigen values are - 4 and - 1. Eigen vectors for l1 =- 4 or

B = sin At = MDM-1

6.69

s4

1

5

K

s3

4

6

0

s2

7 2

K

s1

21 - 4K 7/2

0

s0

K

Option (B) is correct. We have P (s) = s5 + s 4 + 2s3 + 3s + 15 The routh table is shown below. If e " 0+ then 2e +e 12 is positive and -15e2-e +2412e - 144 is negative. Thus there are two sign change in first column. Hence system has 2 root on RHS of plane. 2

Now Eigen vector for l2 =- 1 (l2 I - A) X2 = 0 l2 + 2 - 2 x12 or = - 1 l + 3G=x G = 0 2 22

s5

1

2

3

s4

1

2

15

s3

e

- 12

0

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s2

2e + 12 e

s1

-15e2 - 24e - 144 2e + 12

s0

15

Page 148

0

= =

0 6.75

6.70

Option (D) is correct. - 3 - 1 x1 x1 1 =x G = = 2 0 G=x G + = 0 Gu 2 2 x1 1 and Y = [1 0]= G + = Gu x2 2 -3 -1 1 Here , B = = G and C = [1 0] A == G 2 0 0 The controllability matrix is 1 -3 QC = [B AB ] = = 0 2G

s (s + 27) s + 29s + 6

or

which gives

Thus observable

6.72

6.73

6.74

6.76

e

s 0 1 0 s-1 0 -= == G G 0 s 0 1 0 s - 1G

1 0 (s - 1) s-1 1 = = = 0 (s - 1)G > 0 (s - 1) 2 -1

= L [(sI - A)]

K =- s (s2 + 5s2 + 6s) dK =- (3s2 + 10s + 6) = 0 ds s = - 10 ! 100 - 72 =- 0.784, - 2.548 6

The location of poles on s - plane is

Option (B) is correct.

At

1 + G (s) H (s) = 0 K 1+ =0 s (s + 2)( s + 3)

Since breakpoint must lie on root locus so s =- 0.748 is possible.

det Q0 ! 0

(sI - A)

h

Option (D) is correct. We have or

Q0 = [CT AT CT ] 1 -3 == !0 0 - 1G

-1

s + 27 s

Thus controllable

det QC ! 0 The observability matrix is

(sI - A) = =

^

1 + 29s + s62

2

We have

6.71

^ s +s27 h = 1 - ^ -s3 - 24s - s2 h + -s2 . -s3

-1

et 0 == G 0 et

Option (A) is correct.

0 1 s-1

Option (A) is correct. The given bode plot is shown below

A I D

H

O N

.in

co ia.

d Z = P-N no . w N " Net encirclement of (- 1 + j0) by Nyquist plot, ww P " Number of open loop poles in right hand side of s - plane Z " Number of closed loop poles in right hand side of s - plane Here N = 1 and P = 1 Thus Z =0 Hence there are no roots on RH of s -plane and system is always stable.

At w = 0.1 change in slope is + 60 dB " 3 zeroes at w = 0.1 At w = 10 change in slope is - 40 dB " 2 poles at w = 10 At w = 100 change in slope is - 20 dB " 1 poles at w = 100 K ( 0s.1 + 1) 3 Thus T (s) = s s ( 10 + 1) 2 ( 100 + 1) Now 20 log10 K = 20 or K = 10 10 ( 0s.1 + 1) 3 108 (s + 0.1) 3 Thus T (s) = s = s ( 10 + 1) 2 ( 100 + 1) (s + 10) 2 (s + 100)

Option (C) is correct. PD Controller may accentuate noise at higher frequency. It does not effect the type of system and it increases the damping. It also reduce the maximum overshoot.

Option (B) is correct. The characteristics equation is

6.77

s2 + 4s + 4 = 0 Comparing with

Option (D) is correct. Mason Gain Formula T (s) =

s2 + 2xwn + wn2 = 0 we get 2xwn = 4 and wn2 = 4 Thus x =1 ts = 4 = 4 = 2 1#2 xwn

Spk 3 k 3

In given SFG there is only forward path and 3 possible loop. p1 = 1 31 = 1 + 3 + 24 = s + 27 s s s L1 = - 2 , L2 = - 24 and L3 = - 3 s s s where L1 and L3 are non-touching C (s) This R (s) p1 3 1 = 1 - (loop gain) + pair of non - touching loops

6.78 6.79

Critically damped

Option (B) is correct. Option (C) is correct. We have 1 xo1 =xo G = =1 2 1 A == 1 s (sI - A) = = 0

0 x1 x1 (0) 1 and = == G G G G = 1 x2 x2 (0) 0 0 1G 0 1 0 s-1 0 -= G = = G s 1 1 - 1 s - 1G

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 1 (s - 1) 0 s-1 1 (sI - A) = > H = > +1 (s - 1) 2 + 1 (s - 1) (s - 1) t e 0 L-1 [(sI - A) -1] = eAt = = t t G te e et 0 1 et x (t) = eAt # [x (t0)] = = t t G= G = = t G te e 0 te

0

-1

6.80

6.83

s5

2

4

2

s4

1

2

1

s3

0

0

0

s2 s1 s0 Given x = 0.5

6.87

A I D

O N

or or

6.85

Option (B) is correct. Routh table is shown below. Here all element in 3rd row are zero, so system is marginal stable.

Option (B) is correct. The open loop transfer function is 1 G (s) H (s) = 2 s (s + s + 1) Option (B) is correct. Substituting s = jw we have Any point on real axis lies on the root locus if total number of poles 1 G (jw) H (jw) = and zeros to the right of that point is odd. Here s =- 1.5 does not jw (- w2 + jw + 1) lie on real axis because there are total two poles and zeros (0 and +G (jw) H (jw) =- p - tan-1 w 2 - 1) to the right of s =- 1.5 . 2 (1 - w ) Option (D) is correct. The frequency at which phase becomes - 180c, is called phase From the expression of OLTF it may be easily see that the maximum crossover frequency. magnitude is 0.5 and does not become 1 at any frequency. Thus gain wf Thus - 180 =- 90 - tan-1 cross over frequency does not exist. When gain cross over frequency 1 - wf2 does not exist, the phase margin is infinite. wf or - 90 =- tan-1 Option (D) is correct. 1 - wf2 n i . .coro 1 - w2f = 0 We have ...(i) xo (t) =- 2x (t) + 2u (t) a i d wf = 1 rad/sec Taking Laplace transform we get .no w The gain margin at this frequency wf = 1 is sX (s) =- 2X (s) + 2U (s) ww GM =- 20 log10 G (jwf) H (jwf) or (s + 2) X (s) = 2U (s) 2U (s) = 20 log10 (wf (1 - w2f) 2 + w2f or X (s) = (s + 2) =- 20 log 1 = 0 Now

6.84

6.86

ks2 + s + 6 = 0 s2 + 1 s + 6 = 0 K K

Comparing with s2 + 2xwn s + wn2 = 0 we have we get 2xwn = 1 and wn2 = 6 K K or 2 # 0.5 # 6 Kw = 1 K 6 = 1 & K =1 or K 6 K2

6.82

pole consideration is at s =- 1. Thus 1 =1 and Ts = 4 = 4 sec T T

H

Option (C) is correct. The characteristics equation is or

6.81

2

1 s-1

Page 149

y (t) = 0.5x (t) Y (s) = 0.5X (s) 0.5 # 2U (s) Y (s) = s+2 Y (s) 1 = (s + 2) U (s)

6.88

Z = P-N N " Net encirclement of (- 1 + j0) by Nyquist plot, P " Number of open loop poles in right had side of s - plane Z " Number of closed loop poles in right hand side of s - plane Here N = 0 (1 encirclement in CW direction and other in CCW) and P = 0 Thus Z = 0 Hence there are no roots on RH of s - plane.

Option (D) is correct. From Mason gain formula we can write transfer function as K Y (s) K s = = 3 R (s) 1 - ( s + -sK ) s - 3 (3 - K) For system to be stable (3 - K) < 0 i.e. K > 3 Option (B) is correct. The characteristics equation is (s + 1)( s + 100) = 0 s2 + 101s + 100 = 0 Comparing with s2 + 2xwn + wn2 = 0 we get 2xwn = 101 and wn2 = 100 Thus x = 101 20

6.89

Overdamped

For overdamped system settling time can be determined by the dominant pole of the closed loop system. In given system dominant

Option (A) is correct.

6.90

Option (D) is correct. Take off point is moved after G2 as shown below

Option (C) is correct. The characteristics equation is s2 + 2s + 2 = 0

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Page 150

Comparing with s2 + 2xwn + wn2 = 0 we get 2xwn = 2 and wn2 = 2

The Characteristic equation is 1 + G (s) H (s) = 0 K (s - 2) 1+ (s - 2) = 0 (s + 2) 2 or (s + 2) 2 + K (s - 2) 2 = 0 or (1 + K) s2 + 4 (1 - K) s + 4K + 4 = 0 Routh Table is shown below. For System to be stable 1 + k > 0 , and 4 + 4k > 0 and 4 - 4k > 0 . This gives - 1 < K < 1 As per question for 0 # K < 1

wn =

2 and x = 1 2 Since x < 1 thus system is under damped 6.91

6.92

Option (D) is correct. If roots of characteristics equation lie on negative axis at different positions (i.e. unequal), then system response is over damped. From the root locus diagram we see that for 0 < K < 1, the roots are on imaginary axis and for 1 < K < 5 roots are on complex plain. For K > 5 roots are again on imaginary axis. Thus system is over damped for 0 # K < 1 and K > 5 . Option (C) is correct. From SFG we have I1 (s) I2 (s) V0 (s) Now applying

= G1 Vi (s) + HI2 (s) = G2 I1 (s) = G3 I2 (s) KVL in given block diagram we have

...(1) ...(2) ...(3)

or

I1 (s) Z3 (S) = I2 (s)[ Z2 (s) + Z3 (s) + Z4 (s)] I1 (s) Z3 (s) Is (s) = Z3 (s) + Z2 (s) + Z4 (s)

Comparing (2) and (7) we have Z3 (s) G2 = Z3 (s) + Z2 (s) + Z4 (s) Comparing (1) and (6) we have Z3 (s) H = Z1 (s) + Z3 (s) 6.93

...(4) ...(5)

or

1 = s2 + 7s + 13 - 1 = s2 + 6s + 9 G (s) s+4 s+4 G (s) = 2 s + 4 s + 6s + 9

For DC gain s = 0 , thus Thus

G (0) = 4 9

Option (C) is correct. From the Block diagram transfer function is G (s) T (s) = 1 + G (s) H (s) K (s - 2) Where G (s) = (s + 2) and

H (s) = (s - 2)

s1

4 - 4k

0

s0

4 + 4k

IA

Thus a > 0 and aK > 3

...(6)

...(7)

in co.

.

ia od

n

. ww

w

6.97

Option (B) is correct. For unity negative feedback system the closed loop transfer function is G (s) , CLTF = G (s) " OL Gain = 2 s+4 1 + G (s) s + 7s + 13 2 1 + G (s) or = s + 7s + 13 G (s) s+4 or

6.94

N

4 + 4k

Option (B) is correct. The characteristics equation is s2 + as2 + ks + 3 = 0 The Routh Table is shown below For system to be stable a > 0 and aK - 3 > 0 a

D O

From (5) we have

1+k

Option (B) is correct. It is stable at all frequencies because for resistive network feedback factor is always less than unity. Hence overall gain decreases.

6.95

6.96

Vi (s) = I1 (s) Z1 (s) + [I1 (s) - I2 (s)] Z3 (s) 0 = [I2 (s) - I1 (s)] Z3 (s) + I2 (s) Z2 (s) + I2 (s) Z4 (s) From (4) we have or Vi (s) = I1 (s)[ Z1 (s) + Z3 (S)] - I2 (s) Z3 (S) Z3 (s) 1 or I1 (s) = Vi + I2 Z1 (s) + Z3 (s) Z1 (s) + Z3 (s)

s2

s3

1

K

s2

a

3

s1

aK - 3 a

0

s0

3

Option (B) is correct. Closed loop transfer function is given as T (s) = 2 9 s + 4s + 9 by comparing with standard form we get natural freq. wA2 = 9 wn = 3 2xwn = 4 4 = 2/3 2#3 for second order system the setting time for 2-percent band is given by 4 =4 =2 ts = 4 = xwn 3 # 2/3 2 x =

damping factor

6.98

Option (D) is correct. Given loop transfer function is G (s) H (s) =

2 s (s + 1)

2 jw (jw + 1) Phase cross over frequency can be calculated as G (jw) H (jw) =

So here

f (w) at w = w =- 180c f (w) =- 90c - tan-1 (w) p

- 90c - tan-1 (wp) =- 180c tan-1 (wp) = 90c

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Page 151

wp = 3

for unity feed back H (s) = 1

Gain margin 20 log 10 =

R (s) = 12 (unit Ramp) s 1 so E (s) = lim 12 s " 0 s 1 + G (s) n a1 sn - 1 + .... + an - 2 s2 = lim 12 s + n s"0 s s + a1 sn - 1 + .... + an = an - 2 an Here input

1 at w = wp G (jw) H (jw) G G.M. = 20 log 10 e

1 G (jw) H (jwp) o

2 =0 wp w2p + 1 G.M. = 20 log 10 b 1 l = 3 0

G (jwp) H (jwp) = so 6.99

6.103 6.104

Option (A) is correct. A ==

Here

Option (B) is correct.

0 1 0 , B = = G and C = [1 1] G 2 -3 1

6.105 6.106

The controllability matrix is QC = [B AB ] = =

Option (A) is correct. Option (A) is correct. By applying Routh’s criteria

0 1 1 - 3G

s3 + 5s2 + 7s + 3 = 0 Thus controllable

det QC ! 0 The observability matrix is 1 2 !0 Q0 = [CT AT CT ] = = 1 - 2G

Thus observable

det Q0 ! 0 6.100

G (s) H (s) = 2 3 s (s + 1) G (jw) H (jw) =

or

2 3 jw (jw + 1)

G (jw) H (jw) at w = w = 1 g

2 3 =1 w w2 + 1 12 = w2 (w2 + 1) w4 + w2 - 12 = 0 (w2 + 4) (w2 - 3) = 0 w2 = 3 and w2 =- 4 wg =

3

s1

7#5-3 5

s0

3

=

0

32 5

O N

6.108

H (s) = 1 (unity feedback)

R (s) = 1 s

3 f (w) at w = w =- 90 - tan-1 (wg)

6.109

g

Option (B) is correct. Option (C) is correct. Closed-loop transfer function is given by an - 1 s + an T (s) = n n-1 s + a1 s + ... + an - 1 s + an an - 1 s + an n n-1 + + ...an - 2 s2 s a 1s = an - 1 s + an 1+ n s + a1 sn - 1 + ...an - 2 s2 an - 1 s + an Thus G (s) H (s) = n s + a1 sn - 1 + ....an - 2 s2 For unity feed back H (s) = 1 an - 1 s + an Thus G (s) = n s + a1 sn - 1 + ....an - 2 s2 Steady state error is given by 1 E (s) = lim R (s) s"0 1 + G (s) H (s)

R (s) = input

Where

g

6.102

5

Option (A) is correct. Open loop transfer function is n K o.i c G (s) = . a i s ( s + 1) d o n Steady state error w. w sR (s) w E (s) = lim s " 0 1 + G (s) H (s)

=- 90 - tan-1 3 =- 90 - 60 =- 150 Phase margin = 180 + f (w) at w = w = 180 - 150 = 30c 6.101

s2

Option (A) is correct. Techometer acts like a differentiator so its transfer function is of the form ks .

w1, w2 = ! 3

which gives

7

A I D 6.107

Gain cross over frequency

or

1

There is no sign change in the first column. Thus there is no root lying in the left-half plane.

Option (D) is correct. we have

s3

s1 s (s + 1) s so = lim 2 =0 E (s) = lim s"0 s"0 s + s + K K 1+ s (s + 1) Option (B) is correct. Fig given below shows a unit impulse input given to a zero-order hold circuit which holds the input signal for a duration T & therefore, the output is a unit step function till duration T .

h (t) = u (t) - u (t - T) Taking Laplace transform we have H (s) = 1 - 1 e-sT = 1 61 - e-sT @ s s s 6.110

Option (C) is correct. Phase margin = 180c + qg where qg = value of phase at gain crossover frequency.

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Here so 6.111

6.112

Page 152

Xo2 = dx2 = 0 + x2 + m dt x1 y = [1 1] > H = x1 + x2 x2 dy = dx1 + dx2 dt dt dt dy = x1 + m dt

qg =- 125c P.M = 180c - 125c = 55c

Option (B) is correct. Open loop transfer function is given by K (1 + 0.5s) G (s) H (s) = s (1 + s) (1 + 2s) Close looped system is of type 1. It must be noted that type of the system is defined as no. of poles lies at origin in OLTF. lying Option (D) is correct. Transfer function of the phase lead controller is 1 + (3Tw) j T.F = 1 + 3Ts = 1+s 1 + (Tw) j Phase is

dy dt

= x1 (0) + m (0) t=0

= 1+0 = 0

f (w) = tan-1 (3Tw) - tan-1 (Tw) w f (w) = tan-1 ; 3Tw - T 1 + 3T 2 w2 E f (w) = tan-1 ; 2Tw2 2 E 1 + 3T w For maximum value of phase df (w) =0 dw 1 = 3T 2 w2 Tw = 1 3 So maximum phase is or

6.113

6.114

A I D

fmax = tan-1 ; 2Tw2 2 E at Tw = 1 1 + 3T w 3 R V 1 S 2 W .in 3 W = tan-1 1 = 30c o c = tan-1 S . ; 3E SS1 + 3 # 1 WW dia o 3 .n T X w Option (A) is correct. ww G (jw) H (jw) enclose the (- 1, 0) point so here G (jwp) H (jwp) > 1 wp = Phase cross over frequency 1 Gain Margin = 20 log 10 G (jwp) H (jwp) so gain margin will be less than zero.

O N

Option (B) is correct. The denominator of Transfer function is called the characteristic equation of the system. so here characteristic equation is (s + 1) 2 (s + 2) = 0

6.115

6.116

Option (C) is correct. In synchro error detector, output voltage is proportional to [w (t)], where w (t) is the rotor velocity so here n = 1 Option (C) is correct. By masson’s gain formulae / Dk Pk y = x D Forward path gain

so gain 6.117

P1 = 5 # 2 # 1 = 10 D = 1 - (2 # - 2) = 1 + 4 = 5 D1 = 1 y = 10 # 1 = 2 x 5

Option (C) is correct. By given matrix equations we can have Xo1 = dx1 = x1 - x2 + 0 dt

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UNIT 7

Page 153

2012 7.7

COMMUNICATION SYSTEMS

2013 7.1

The bit rate of a digital communication system is R kbits/s . The modulation used is 32-QAM. The minimum bandwidth required for ISI free transmission is (A) R/10 Hz (B) R/10 kHz (C) R/5 Hz (D) R/5 kHz

7.3

7.4

7.8

TWO MARKS

Let U and V be two independent zero mean Gaussain random variables of variances 1 and 1 respectively. The probability 9 4 P ^3V F 2U h is (A) 4/9 (C) 2/3

The power spectral density of a real process X (t) for positive frequencies is shown below. The values of E [X 2 (t)] and E [X (t)] , respectively, are

ONE MARK

2013 7.2

ONE MARK

(B) 1/2 (D) 5/9

7.9

Consider two identically distributed zero-mean random variables U and V . Let the cumulative distribution functions of U and 2V be F ^x h and G ^x h respectively. Then, for all values of x (A) F ^x h - G ^x h # 0 (B) F ^x h - G ^x h $ 0 (C) ^F (x) - G (x)h .x # 0 (D) ^F (x) - G (x)h .x $ 0

(A) 6000/p, 0

(B) 6400/p, 0

(C) 6400/p, 20/ (p 2 )

(D) 6000/p, 20/ (p 2 )

In a baseband communications link, frequencies upto 3500 Hz are used for signaling. Using a raised cosine pulse with 75% excess bandwidth and for no inter-symbol interference, the maxi mum possible signaling rate in symbols per second is (A) 1750 (B) 2625 (C) 4000 (D) 5250 A source alphabet consists of N symbols with the probability of the first two symbols being the same. A source encoder increases the probability of the first symbol by a small amount e and decreases that of the second by e. After encoding, the entropy of the source (A) increases (B) remains the same (C) increases only if N = 2 (D) decreases

A I D

O N

7.10 Two independent random variables X and Y are uniformly n Let U and V be two independent and identically distributed random o.i distributed in the interval 6- 1, 1@. The probability that max 6X, Y @ c . a 1 i variables such that P ^U =+ 1h = P ^U =- 1h = . The entropy o d is less than 1/2 is 2 .n (B) 9/16 (A) 3/4 H ^U + V h in bits is w w w (A) 3/4 (B) 1 (C) 1/4 (D) 2/3 (C) 3/2 (D) log 2 3

2012

Common Data for Questions 5 and 6:

7.11

Bits 1 and 0 are transmitted with equal probability. At the receiver, the pdf of the respective received signals for both bits are as shown below.

7.12

7.5

7.6

If the detection threshold is 1, the BER will be (B) 1 (A) 1 2 4 (C) 1 (D) 1 8 16 The optimum threshold to achieve minimum bit error rate (BER) is (B) 4 (A) 1 2 5 (C) 1

7.13

TWO MARKS

A BPSK scheme operating over an AWGN channel with noise power spectral density of N 0 /2, uses equiprobable signals s1 (t) = 2E sin (wc t) and s2 (t) =- 2E sin (wc t) over the symbol T T interval (0, T). If the local oscillator in a coherent receiver is ahead in phase by 45c with respect to the received signal, the probability of error in the resulting system is (A) Q c 2E m (B) Q c E m N0 N0 E E (C) Q c (D) Q c 2N 0 m 4N 0 m A binary symmetric channel (BSC) has a transition probability of 1/8. If the binary symbol X is such that P (X = 0) = 9/10, then the probability of error for an optimum receiver will be (A) 7/80 (B) 63/80 (C) 9/10 (D) 1/10 The signal m (t) as shown is applied to both a phase modulator (with k p as the phase constant) and a frequency modulator (with k f as the frequency constant) having the same carrier frequency.

(D) 3 2

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Page 154

(A) (4p 2 f 2 + 1) exp (- pf 2) (C) (4p 2 f 2 + 1) exp (- pf ) The ratio k p /k f (in rad/Hz) for the same maximum phase deviation is (A) 8p (B) 4p (C) 2p (D) p

7.19

Statement for Linked Answer Question 14 and 15 :

7.14

7.15

The transfer function of a compensator is given as Gc (s) = s + a s+b Gc (s) is a lead compensator if (A) a = 1, b = 2 (B) a = 3, b = 2 (C) a =- 3, b =- 1 (D) a = 3, b = 1

6 rad/s

Common Data For Q. 8.5 & 8.6 A four-phase and an eight-phase signal constellation are shown in the figure below.

(D) 1/ 3 rad/s

2011

7.17

ONE MARK

A I D

For the constraint that the minimum distance between pairs of An analog signal is band-limited to 4 kHz, sampled at the Nyquist signal be d for both constellations, the radii r 1 , and r 2 of the in points . o rate and the samples are quantized into 4 levels. The quantized circles are .c levels are assumed to be independent and equally probable. If we o dia (A) r 1 = 0.707d, r2 = 2.782d (B) r 1 = 0.707d, r 2 = 1.932d n . transmit two quantized samples per second, the information rate w is (C) r 1 = 0.707d, r 2 = 1.545d (D) r 1 = 0.707d, r 2 = 1.307d ww (A) 1 bit/sec (B) 2 bits/sec 7.21 Assuming high SNR and that all signals are equally probable, the (C) 3 bits/sec (D) 4 bits/sec additional average transmitted signal energy required by the 8-PSK The Column -1 lists the attributes and the Column -2 lists the signal to achieve the same error probability as the 4-PSK signal is modulation systems. Match the attribute to the modulation system (A) 11.90 dB (B) 8.73 dB that best meets it. (C) 6.79 dB (D) 5.33 dB Column -1 Column -2

O N

P.

Power efficient transmission of 1. signals

Conventional AM

Q.

Most bandwidth efficient 2. transmission of voice signals

FM

R.

Simplest receiver structure

S.

Bandwidth efficient transmission 4. of signals with significant dc component

3.

2011

7.20

2010 7.22

VSB SSB-SC

ONE MARK

Suppose that the modulating signal is m (t) = 2 cos (2pfm t) and the carrier signal is xC (t) = AC cos (2pfC t), which one of the following is a conventional AM signal without over-modulation (A) x (t) = AC m (t) cos (2pfC t) (B) x (t) = AC [1 + m (t)] cos (2pfC t) (C) x (t) = AC cos (2pfC t) + AC m (t) cos (2pfC t) 4 (D) x (t) = AC cos (2pfm t) cos (2pfC t) + AC sin (2pfm t) sin (2pfC t)

(A) P-4, Q-2, R-1, S-3 (B) P-2, Q-4, R-1, S-3 (C) P-3, Q-2, R-1, S-4 (D) P-2, Q-4, R-3, S-1

7.18

A message signal m (t) = cos 2000pt + 4 cos 4000pt modulates the carrier c (t) = cos 2pfc t where fc = 1 MHz to produce an AM signal. For demodulating the generated AM signal using an envelope detector, the time constant RC of the detector circuit should satisfy (A) 0.5 ms < RC < 1 ms (B) 1 μs 0.5 ms

The phase of the above lead compensator is maximum at (A) 2 rad/s (B) 3 rad/s (C)

7.16

(B) (4p 2 f 2 - 1) exp (- pf 2) (D) (4p 2 f 2 - 1) exp (- pf )

7.23

x (t) = 6 cos [2p # 106 t + 2 sin (800pt)] + 4 cos (800pt) The average power of x (t) is (A) 10 W (B) 18 W (C) 20 W (D) 28 W

TWO MARKS

X (t) is a stationary random process with auto-correlation function RX (t) = exp (- pt 2). This process is passed through the system shown below. The power spectral density of the output process Y (t) is

Consider an angle modulated signal

7.24

Consider the pulse shape s (t) as shown below. The impulse response h (t) of the filter matched to this pulse is

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Page 155

Let Sy (f ) be the power spectral density of Y (t). Which one of the following statements is correct (A) Sy (f ) > 0 for all f (B) Sy (f ) = 0 for f > 1 kHz (C) Sy (f ) = 0 for f = nf0, f0 = 2 kHz kHz, n any integer (D) Sy (f ) = 0 for f = (2n + 1) f0 = 1 kHz , n any integer 2009 7.29

2010

TWO MARKS

ONE MARK

For a message siganl m (t) = cos (2pfm t) and carrier of frequency fc , which of the following represents a single side-band (SSB) signal ? (A) cos (2pfm t) cos (2pfc t) (B) cos (2pfc t) (C) cos [2p (fc + fm) t] (D) [1 + cos (2pfm t) cos (2pfc t) 2009

7.30

Statement for linked Answer Question : 8.10 & 8.11 :

TWO MARKS

Consider two independent random variables X and Y with identical distributions. The variables X and Y take values 0, 1 and 2 with probabilities 12 , 14 and 1 respectively. What is the conditional 4 probability P (X + Y = 2 X - Y = 0) ?

A I D

Consider a baseband binary PAM receiver shown below. The additive channel noise n (t) is with power spectral density Sn (f ) = N 0 /2 = 10-20 W/Hz . The low-pass filter is ideal with unity (A) 0 (B) 1/16 gain and cut-off frequency 1 MHz. Let Yk represent the random variable y (tk ). (C) 1/6 (D) 1 n i Yk = Nk , if transmitted bit bk = 0 cAo. discrete random variable X takes values from 1 to 5 with 7.31 . a Yk = a + Nk if transmitted bit bk = 1 di probabilities as shown in the table. A student calculates the mean Where Nk represents the noise sample value. The noise sample has .no X as 3.5 and her teacher calculates the variance of X as 1.5. Which ww a probability density function, PNk (n) = 0.5ae- a n (This has mean w 2 of the following statements is true ? zero and variance 2/a ). Assume transmitted bits to be equiprobable and threshold z is set to a/2 = 10-6 V . 1 2 3 4 5 k

O N

P (X = k)

0.1

0.2

0.3

0.4

0.5

(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 7.32

7.25

7.26

7.27

7.28

The value of the parameter a (in V - 1 ) is (A) 1010 (B) 107 (C) 1.414 # 10-10 (D) 2 # 10-20 The probability of bit error is (A) 0.5 # e-3.5 (C) 0.5 # e-7

7.33

(B) 0.5 # e-5 (D) 0.5 # e-10

The Nyquist sampling rate for the signal sin (500pt) sin (700) pt is given by s (t) = # pt pt (A) 400 Hz (B) 600 Hz (C) 1200 Hz (D) 1400 Hz

A message signal given by m (t) = ( 12 ) cos w1 t - ( 12 ) sin w2 t amplitude - modulated with a carrier of frequency wC to generator s (t)[ 1 + m (t)] cos wc t . What is the power efficiency achieved by this modulation scheme ? (A) 8.33% (B) 11.11% (C) 20% (D) 25% A communication channel with AWGN operating at a signal to noise ration SNR >> 1 and bandwidth B has capacity C1. If the SNR is doubled keeping constant, the resulting capacity C2 is given by (A) C2 . 2C1 (B) C2 . C1 + B (C) C2 . C1 + 2B (D) C2 . C1 + 0.3B

Common Data For Q. 8.19 & 8.20 :

X (t) is a stationary process with the power spectral density Sx (f ) > 0 , for all f . The process is passed through a system shown below

The amplitude of a random signal is uniformly distributed between -5 V and 5 V. 7.34

If the signal to quantization noise ratio required in uniformly

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quantizing the signal is 43.5 dB, the step of the quantization is approximately (A) 0.033 V (B) 0.05 V (C) 0.0667 V (D) 0.10 V 7.35

If the positive values of the signal are uniformly quantized with a step size of 0.05 V, and the negative values are uniformly quantized with a step size of 0.1 V, the resulting signal to quantization noise ration is approximately (A) 46 dB (B) 43.8 dB (C) 42 dB (D) 40 dB

Page 156 7.41

7.42

7.43

2008 7.36

Consider the amplitude modulated (AM) signal Ac cos wc t + 2 cos wm t cos wc t . For demodulating the signal using envelope detector, the minimum value of Ac should be (A) 2 (B) 1 (C) 0.5 (D) 0 2008

7.37

ONE MARK

TWO MARKS

The signal cos wc t + 0.5 cos wm t sin wc t is (A) FM only (B) AM only (C) both AM and FM (D) neither AM nor FM

A speed signal, band limited to 4 kHz and peak voltage varying between +5 V and - 5 V, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.

The probability density function (pdf) of random variable is as shown below

If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is (A) 64 kHz (B) 32 kHz (C) 8 kHz (D) 4 kHz

A I D 7.45

O N

The corresponding commutative distribution function CDF has the form

w

Consider the frequency modulated signal 10 cos [2p # 105 t + 5 sin (2p # 1500t) + 7.5 sin (2p # 1000t)] with carrier frequency of 105 Hz. The modulation index is (A) 12.5 (B) 10 (C) 7.5 (D) 5

Common Data For Q. 8.29, 8.30 and 8.31 :

7.44

Assuming the signal to be uniformly distributed between its peak to peak value, the signal to noise ratio at the quantizer output is (A) 16 dB (B) 32 dB (C) 48 dB (D) 4 kHz

.in oAssuming c . the signal to be uniformly distributed between its peak to ia

7.46

d .no

ww

Four messages band limited to W, W, 2W and 3W respectively are to be multiplexed using Time Division Multiplexing (TDM). The minimum bandwidth required for transmission of this TDM signal is (A) W (B) 3W (C) 6W (D) 7W

peak value, the signal to noise ratio at the quantizer output is (A) 1024 (B) 512 (C) 256 (D) 64 2007

7.47

7.38

7.39

7.40

A memory less source emits n symbols each with a probability p. The entropy of the source as a function of n (A) increases as log n (B) decreases as log ( n1 ) (C) increases as n (D) increases as n log n Noise with double-sided power spectral density on K over all frequencies is passed through a RC low pass filter with 3 dB cut-off frequency of fc . The noise power at the filter output is (A) K (B) Kfc (C) kpfc (D) 3 Consider a Binary Symmetric Channel (BSC) with probability of error being p. To transmit a bit, say 1, we transmit a sequence of three 1s. The receiver will interpret the received sequence to represent 1 if at least two bits are 1. The probability that the transmitted bit will be received in error is (A) p3 + 3p2 (1 - p) (B) p3 (C) (1 - p3) (D) p3 + p2 (1 - p)

7.48

If R (t) is the auto correlation function of a real, wide-sense stationary random process, then which of the following is NOT true (A) R (t) = R (- t) (B) R (t) # R (0) (C) R (t) =- R (- t) (D) The mean square value of the process is R (0) If S (f) is the power spectral density of a real, wide-sense stationary random process, then which of the following is ALWAYS true? (A) S (0) # S (f) (B) S (f) $ 0 (C) S (- f) =- S (f)

7.49

ONE MARK

(D)

3

#- 3 S (f) df = 0

If E denotes expectation, the variance of a random variable X is given by (A) E [X2] - E2 [X] (B) E [X2] + E2 [X] (C) E [X2] (D) E2 [X] 2007

7.50

A Hilbert transformer is a (A) non-linear system (C) time-varying system

TWO MARKS

(B) non-causal system (D) low-pass system

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 7.51

7.52

equiprobable. Let N0 denote the power spectral density of white 2 Gaussian noise.

In delta modulation, the slope overload distortion can be reduced by (A) decreasing the step size (B) decreasing the granular noise (C) decreasing the sampling rate (D) increasing the step size The raised cosine pulse p (t) is used for zero ISI in digital communications. The expression for p (t) with unity roll-off factor is given by sin 4pWt p (t) = 4pWt (1 - 16W2 t2) The value of p (t) at t = 1 is 4W (A) - 0.5 (C) 0.5

7.53

Page 157

The if ratio or the average energy of Constellation 1 to the average energy of Constellation 2 is (A) 4a2 (B) 4 (C) 2 (D) 8

7.57

(B) 0 (D) 3

In the following scheme, if the spectrum M (f) of m (t) is as shown, then the spectrum Y (f) of y (t) will be

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 (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 N0 will determine which of the constellations has a lower probability of symbol error

7.58

Statement for Linked Answer Question 8.44 & 8.45 :

A I D

O N

no

. ww

An input to a 6-level quantizer has the probability density function f (x) as shown in the figure. Decision boundaries of the quantizer are chosen so as to maximize the entropy of the quantizer output. It is given that 3 consecutive decision boundaries are’ - 1'.'0' and '1'..in

co

. dia

w

7.54

During transmission over a certain binary communication channel, bit errors occur independently with probability p. The probability of AT MOST one bit in error in a block of n bits is given by (A) pn (B) 1 - pn (C) np (1 - p) n - 1 + (1 + p) n

7.55

7.56

(D) 1 - (1 - p) n

In a GSM system, 8 channels can co-exist in 200 kHz bandwidth using TDMA. A GSM based cellular operator is allocated 5 MHz bandwidth. Assuming a frequency reuse factor of 1 , i.e. a five-cell 5 repeat pattern, the maximum number of simultaneous channels that can exist in one cell is (A) 200 (B) 40 (C) 25 (D) 5 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 # 103 bits per sec (B) must be greater than 12.288 # 103 bits per sec (C) must be exactly equal to 12.288 # 103 bits per sec (D) can take any value less than 122.88 # 103 bits per sec

7.59

7.60

2006 7.61

ONE MARK

A low-pass filter having a frequency response H (jw) = A (w) e jf (w) does not produce any phase distortions if (A) A (w) = Cw3, f (w) = kw3 (B) A (w) = Cw2, f (w) = kw (C) A (w) = Cw, f (w) = kw2

Common Data For Q. 8.41 & 8.42 : Two 4-array signal constellations are shown. It is given that f1 and f2 constitute an orthonormal basis for the two constellation. Assume that the four symbols in both the constellations are

The values of a and b are (A) a = 1 and b = 1 (B) a = 1 and b = 3 6 12 5 40 (C) a = 1 and b = 1 (D) a = 1 and b = 1 4 16 3 24 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 (A) 152 (B) 64 9 3 (C) 76 (D) 28 3

(D) A (w) = C, f (w) = kw- 1

2006 7.62

TWO MARKS

A signal with bandwidth 500 Hz is first multiplied by a signal g (t) where

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g (t) =

3

Page 158 7.69

/(- 1) k d (t - 0.5 # 10- 4 k)

R =- 3

In the following figure the minimum value of the constant "C" , which is to be added to y1 (t) such that y1 (t) and y2 (t) are different , is

The resulting signal is then passed through an ideal lowpass filter with bandwidth 1 kHz. The output of the lowpass filter would be (A) d (t) (B) m (t) (C) 0 (D) m (t) d (t) 7.63

7.64

The minimum sampling frequency (in samples/sec) required to reconstruct the following signal from its samples without distortion 3 2 x (t) = 5` sin 2p100t j + 7` sin 2p100t j would be pt pt (A) 2 # 103 (B) 4 # 103 (C) 6 # 103 (D) 8 # 103

7.70

The minimum step-size required for a Delta-Modulator operating at 32k samples/sec to track the signal (here u (t) is the unit-step function) x (t) = 125[ u (t) - u (t - 1) + (250t)[ u (t - 1) - u (t - 2)] so that slope-overload is avoided, would be (A) 2 - 10 (B) 2 - 8 (C) 2 - 6

7.65

7.66

7.67

7.68

(B) 3 2

(A) 3

2 (C) 3 (D) 3 12 L A message signal with bandwidth 10 kHz is Lower-Side Band SSB modulated with carrier frequency fc1 = 106 Hz. The resulting signal is then passed through a Narrow-Band Frequency Modulator with carrier frequency fc2 = 109 Hz. The bandwidth of the output would be (A) 4 # 10 4 Hz (B) 2 # 106 Hz (C) 2 # 109 Hz (D) 2 # 1010 Hz

(D) 2 - 4

A zero-mean white Gaussian noise is passes through an ideal lowpass filter of bandwidth 10 kHz. The output is then uniformly sampled with sampling period ts = 0.03 msec. The samples so obtained would be (A) correlated (B) statistically independent (C) uncorrelated (D) orthogonal

Common Data For Q. 8.56 & 8.57 : Let g (t) = p (t)*( pt), where * denotes convolution & p (t) = u (t) - u (t - 1) lim with u (t) being the unit step function

A I D 7.71

O N

z"3

The impulse response of filter matched s (t) = g (t) - d (1 - 2)* g (t) is given as : (A) s (1 - t) (B) - s (1 - t) (C) - s (t) (D) s (t)

to

the

signal

A source generates three symbols with probabilities 0.25, 0.25, 0.50 at a rate of 3000 symbols per second. Assuming independent 7.72 An .inAmplitude Modulated signal is given as o c . generation of symbols, the most efficient source encoder would have xAM (t) = 100 [p (t) + 0.5g (t)] cos wc t dia o average bit rate is n in the interval 0 # t # 1. One set of possible values of modulating w. (A) 6000 bits/sec (B) 4500 bits/sec w w signal and modulation index would be (C) 3000 bits/sec (D) 1500 bits/sec (A) t, 0.5 (B) t, 1.0 The diagonal clipping in Amplitude Demodulation (using envelop (C) t, 2.0 (D) t2, 0.5 detector) can be avoided it RC time-constant of the envelope detector satisfies the following condition, (here W is message bandwidth and Common Data For Q. 8.58 & 8.59 : w is carrier frequency both in rad/sec) 1 1 (A) RC < (B) RC > The following two question refer to wide sense stationary stochasW W tic process (C) RC < 1 (D) RC > 1 w w 7.73 It is desired to generate a stochastic process (as voltage process) with power spectral density S (w) = 16/ (16 + w2) by driving a A uniformly distributed random variable X with probability density Linear-Time-Invariant system by zero mean white noise (As voltage function process) with power spectral density being constant equal to 1. The fx (x) = 1 pu (x + 5) - u (x - 5)] 10 system which can perform the desired task could be (A) first order lowpass R-L filter where u (.) is the unit step function is passed through a transfor(B) first order highpass R-C filter mation given in the figure below. The probability density function (C) tuned L-C filter of the transformed random variable Y would be (D) series R-L-C filter 7.74

(A) fy (y) = 1 [u (y + 2.5) - u (y - 2.25)] 5 (B) fy (y) = 0.5d (y) + 0.5d (y - 1) (C) fy (y) = 0.25d (y + 2.5) + 0.25d (y - 2.5) + 5d (y)

The parameters of the system obtained in previous Q would be (A) first order R-L lowpass filter would have R = 4W L = 1H (B) first order R-C highpass filter would have R = 4W C = 0.25F (C) tuned L-C filter would have L = 4H C = 4F (D) series R-L-C lowpass filter would have R = 1W , L = 4H , C = 4F

(D) fy (y) = 0.25d (y + 2.5) + 0.25d (y - 2.5) + 1 [u (y + 2.5) - u (y - 2.5)] 10

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Page 159

Common Data For Q. 8.60 & 8.61 : Consider the following Amplitude Modulated (AM) signal, where fm < B XAM (t) = 10 (1 + 0.5 sin 2pfm t) cos 2pfc t 7.75

7.76

The average side-band power for the AM signal given above is (A) 25 (B) 12.5 (C) 6.25 (D) 3.125 The AM signal gets added to a noise with Power Spectral Density Sn (f) given in the figure below. The ratio of average sideband power to mean noise power would be : Noise with uniform power spectral density of N0 W/Hz is passed though a filter H (w) = 2 exp (- jwtd ) followed by an ideal pass filter of bandwidth B Hz. The output noise power in Watts is (A) 2N0 B (B) 4N0 B (C) 8N0 B (D) 16N0 B

7.81

An output of a communication channel is a random variable v with the probability density function as shown in the figure. The mean square value of v is

7.82

(A) 25 8N0 B (C) 25 2N0 B

(B) 25 4N0 B (D) 25 N0 B

2005 7.77

ONE MARK

Find the correct match between group 1 and group 2. Group 1 Group 2 P. {1 + km (t) A sin (wc t)} W. Phase modulation Q. km (t) A sin (wc t) X. Frequency modulation R. A sin {wc t + km (t)} Y. Amplitude modulation t S. A sin ; wc t + k m (t) dt E Z. DSB-SC modulation

#- 3

(A) P - Z, Q - Y, R - X, S - W (B) P - W, Q - X, R - Y, S - Z (C) P - X, Q - W, R - Z, S - Y (D) P - Y, Q - Z, R - W, S - X 7.78

Which of the following analog minimum transmitted power and (A) VSB (C) SSB 2005

7.79

7.80

N

D O

IA

d

ww

modulation scheme requires the minimum channel bandwidth ? (B) DSB-SC (D) AM

(B) 6 (D) 9

in Ao.carrier is phase modulated (PM) with frequency deviation of 10 c . ia

7.83

no w.

(A) 4 (C) 8

kHz by a single tone frequency of 1 kHz. If the single tone frequency is increased to 2 kHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is (A) 21 kHz (B) 22 kHz (C) 42 kHz (D) 44 kHz

Common Data For Q. 8.69 and 8.70 : Asymmetric three-level midtread quantizer is to be designed assuming equiprobable occurrence of all quantization levels.

TWO MARKS

A device with input X (t) and output y (t) is characterized by: Y (t) = x2 (t). An FM signal with frequency deviation of 90 kHz and modulating signal bandwidth of 5 kHz is applied to this device. The bandwidth of the output signal is (A) 370 kHz (B) 190 kHz (C) 380 kHz (D) 95 kHz

7.84

A signal as shown in the figure is applied to a matched filter. Which of the following does represent the output of this matched filter ? 7.85

If the probability density function is divide into three regions as shown in the figure, the value of a in the figure is (A) 1 (B) 2 3 3 (C) 1 (D) 1 2 4 The quantization noise power for the quantization region between - a and + a in the figure is (A) 4 (B) 1 81 9 (C) 5 (D) 2 81 81

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2004 7.86

7.87

7.88

7.89

7.90

ONE MARK

In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor (A) 8 (B) 12 6 (C) 16

Page 160

7.93

(D) 8

An AM signal is detected using an envelop detector. The carrier frequency and modulating signal frequency are 1 MHz and 2 kHz respectively. An appropriate value for the time constant of the envelop detector is (A) 500m sec (B) 20m sec (C) 0.2m sec (D) 1m sec An AM signal and a narrow-band FM signal with identical carriers, modulating signals and modulation indices of 0.1 are added together. The resultant signal can be closely approximated by (A) broadband FM (B) SSB with carrier (C) DSB-SC (D) SSB without carrier

(A) B1 = 20 kHz, B2 = 20 kHz (C) B1 = 20 khz, B2 = 10 kHz

IA

D O

The distribution function Fx (x) of a random variable x is shown in the figure. The probability that X = 1 is

(B) 0.25 (D) 0.30

N

7.92

TWO MARKS

A 1 mW video signal having a bandwidth of 100 MHz is transmitted to a receiver through cable that has 40 dB loss. If the effective oneside noise spectral density at the receiver is 10 - 20 Watt/Hz, then the signal-to-noise ratio at the receiver is (A) 50 dB (B) 30 dB (C) 40 dB (D) 60 dB Consider the signal x (t) shown in Fig. Let h (t) denote the impulse response of the filter matched to x (t), with h (t) being non-zero only in the interval 0 to 4 sec. The slope of h (t) in the interval 3 < t < 4 sec is

7.97

7.98

(A) 1 sec - 1 2

(B) - 1 sec - 1

(A) constant

(B)

1 + sin (2p # 106 t)

5 - sin (2p - 106 t) 5 + cos (2p # 106 t) (D) 4 4 co . a 7.95 di Two sinusoidal signals of same amplitude and frequencies 10 kHz o n and 10.1 kHz are added together. The combined signal is given to an w. w w ideal frequency detector. The output of the detector is (A) 0.1 kHz sinusoid (B) 20.1 kHz sinusoid (C) a linear function of time (D) a constant 7.96

2004

(B) B1 = 10 kHz, B2 = 20 kHz (D) B1 = 10 kHz, B2 = 10 kHz

A 100 MHz carrier of 1 V amplitude and a 1 MHz modulating signal of 1 V amplitude are fed to a balanced modulator. The ourput of the modulator is passed through an ideal high-pass filter with cutoff frequency of 100 MHz. The output of the filter is added with 100 MHz signal of 1 V amplitude and 90c phase shift as shown in the figure. The envelope of the resultant signal is

7.94

In the output of a DM speech encoder, the consecutive pulses are of opposite polarity during time interval t1 # t # t2 . This indicates that during this interval (A) the input to the modulator is essentially constant (B) the modulator is going through slope overload (C) the accumulator is in saturation (D) the speech signal is being sampled at the Nyquist rate

(A) zero (C) 0.55

7.91

(C) - 1 sec - 1 (D) 1 sec - 1 2 A source produces binary data at the rate of 10 kbps. The binary symbols are represented as shown in the figure. The source output is transmitted using two modulation schemes, namely Binary PSK (BPSK) and Quadrature PSK (QPSK). Let B1 and B2 be the bandwidth requirements of the above rectangular pulses is 10 kHz, B1 and B2 are

(C) .in

Consider a binary digital communication system with equally likely 0’s and 1’s. When binary 0 is transmitted the detector input can lie between the levels - 0.25 V and + 0.25 V with equl probability : when binary 1 is transmitted, the voltage at the detector can have any value between 0 and 1 V with equal probability. If the detector has a threshold of 0.2 V (i.e., if the received signal is greater than 0.2 V, the bit is taken as 1), the average bit error probability is (A) 0.15 (B) 0.2 (C) 0.05 (D) 0.5 A random variable X with uniform density in the interval 0 to 1 is quantized as follows : If 0 # X # 0.3 , xq = 0 If 0.3 < X # 1, xq = 0.7 where xq is the quantized value of X. The root-mean square value of the quantization noise is (A) 0.573 (B) 0.198 (C) 2.205 (D) 0.266 Choose the current one from among the alternative A, B, C, D after matching an item from Group 1 with the most appropriate item in Group 2. Group 1 Group 2 1. FM P. Slope overload 2. DM Q. m-law

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R. Envelope detector S. Hilbert transform T. Hilbert transform U. Matched filter (A) 1 - T, 2 - P, 3 - U, 4 - S (B) 1 - S, 2 - U, 3 - P, 4 - T (C) 1 - S, 2 - P, 3 - U, 4 - Q (D) 1 - U, 2 - R, 3 - S, 4 - Q

Page 161

3. PSK 4. PCM

7.99

7.100

X (t) is a random process with a constant mean value of 2 and the auto correlation function Rxx (t) = 4 (e - 0.2 t + 1). Let X be the Gaussian random variable obtained by sampling the process at t = ti and let 3 Q (a) = - 1 e dy a 2p The probability that 6x # 1@ is (A) 1 - Q (0.5) (B) Q (0.5) 1 (C) Q c (D) 1 - Q c 1 m m 2 2 2 2 Let Y and Z be the random variable obtained by sampling X (t) at t = 2 and t = 4 respectively. Let W = Y - Z . The variance of W is (A) 13.36 (B) 9.36 (C) 2.64 (D) 8.00

7.105

Three analog signals, having bandwidths 1200 Hz, 600 Hz and 600 Hz, are sampled at their respective Nyquist rates, encoded with 12 bit words, and time division multiplexed. The bit rate for the multiplexed. The bit rate for the multiplexed signal is (A) 115.2 kbps (B) 28.8 kbps (C) 57.6 kbps (D) 38.4 kbps

#

Consider a system shown in the figure. Let X (f) and Y (f) and denote the Fourier transforms of x (t) and y (t) respectively. The ideal HPF has the cutoff frequency 10 kHz.

7.106

A sinusoidal signal with peak-to-peak amplitude of 1.536 V is quantized into 128 levels using a mid-rise uniform quantizer. The quantization-noise power is (A) 0.768 V (B) 48 # 10 - 6 V2

7.107

(B) 12 # 10 - 6 V2

no . w ONE MARK ww

2003 7.101

7.102

7.103

7.109

The noise at the input to an ideal frequency detector is white. The detector is operating above threshold. The power spectral density of the noise at the output is (A) raised - cosine (B) flat (C) parabolic (D) Gaussian At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by. (A) 6 dB (B) 3 dB (C) 2 dB (D) 0 dB

Common Data For Q. 8.90 & 8.91 :

(C) 2.6, 2.7, 3.3, 3.4, 3.6

(B) 3.3, 3.6 (D) 2.7, 3.3

A DSB-SC signal is to be generated with a carrier frequency fc = 1 MHz using a non-linear device with the input-output characteristic V0 = a0 vi + a1 vi3 where a0 and a1 are constants. The output of the non-linear device can be filtered by an appropriate band-pass filter. Let Vi = Aci cos (2pfi ct) + m (t) is the message signal. Then the value of fci (in MHz) is (A) 1.0 (B) 0.333 (B) 0.5 (D) 3.0

Common Data For Q. 8.95 & 8.96 : Let m (t) = cos [(4p # 103) t] be the message signal & c (t) = 5 cos [(2p # 106 t)] be the carrier. 7.110

TWO MARKS

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 #- 1) is (A) zero (B) 1 6 (C) 1 (D) 1 3 12

n o.i 2.7, 3.4 (A) c . ia

d

The input to a coherent detector is DSB-SC signal plus noise. The noise at the detector output is (A) the in-phase component (B) the quadrature - component (C) zero (D) the envelope

2003 7.104

A I D

O N

(D) 3.072 V

Let x (t) = 2 cos (800p) + cos (1400pt). x (t) is sampled with the rectangular pulse train shown in the figure. The only spectral components (in kHz) present in the sampled signal in the frequency range 2.5 kHz to 3.5 kHz are

7.108

The positive frequencies where Y (f) has spectral peaks are (A) 1 kHz and 24 kHz (B) 2 kHz and 244 kHz (C) 1 kHz and 14 kHz (D) 2 kHz and 14 kHz

x2 2

7.111

c (t) and m (t) are used to generate an AM signal. The modulation index of the generated AM signal is 0.5. Then the quantity Total sideband power is Carrier power (A) 1 (B) 1 2 4 (C) 1 (D) 1 3 8 c (t) and m (t) are used to generated an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term cos [2p (1008 # 103 t)] in the FM signal (in terms of the Bessel coefficients) is (B) 5 J8 (3) (A) 5J4 (3) 2

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7.112

7.113

7.114

7.115

7.116

7.117

(C) 5 J8 (4) (D) 5J4 (6) 2 Choose the correct one from among the alternative A, B, C, D after matching an item in Group 1 with most appropriate item in Group 2. Group 1 Group 2 P. Ring modulator 1. Clock recovery Q. VCO 2. Demodulation of FM R. Foster-Seely discriminator 3. Frequency conversion S. Mixer 4. Summing the two inputs 5. Generation of FM 6. Generation of DSB-Sc (A) P - 1; Q - 3; R - 2; S - 4 (B) P - 6; Q = 5; R - 2; S - 3 (C) P - 6; Q - 1; R - 3; S - 2 (D) P - 5; Q - 6; R - 1; S - 3

7.119

-5

If Eb , the energy per bit of a binary digital signal, is 10 wattsec and the one-sided power spectral density of the white noise, N0 = 10 - 6 W/Hz, then the output SNR of the matched filter is (A) 26 dB (B) 10 dB (C) 20 dB (D) 13 dB

Consider a sample signal y (t) = 5 # 10 - 6 # (t)

+3

/ d (t - nTs)

n =- 3

where x (t) = 10 cos (8p # 103) t and Ts = 100m sec. When y (t) is passed through an ideal lowpass filter with a cutoff frequency of 5 KHz, the output of the filter is (A) 5 # 10 - 6 cos (8p # 103) t (b) 5 # 10 - 5 cos (8p # 103) t (C) 5 # 10 - 1 cos (8p # 103) t 7.120

7.121

A superheterodyne receiver is to operate in the frequency range 550 kHz - 1650 kHz, with the intermediate frequency of 450 kHz. Let R = Cmax /Cmin denote 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 (A) R = 4.41, I = 1600 (B) R = 2.10, I - 1150 (C) R = 3.0, I = 600 (D) R = 9.0, I = 1150

(D) 10 cos (8p # 103) t

For a bit-rate of 8 Kbps, the best possible values of the transmitted frequencies in a coherent binary FSK system are (A) 16 kHz and 20 kHz (C) 20 kHz and 32 kHz (C) 20 kHz and 40 kHz (D) 32 kHz and 40 kHz The line-of-sight communication requires the transmit and receive antennas to face each other. If the transmit antenna is vertically polarized, for best reception the receiver antenna should be (A) horizontally polarized (B) vertically polarized (C) at 45c with respect to horizontal polarization (D) at 45c with respect to vertical polarization 2002

7.122

TWO MARKS

An angle-modulated signal is given by s (t) = cos 2p (2 # 106 t + 30 sin 150t + 40 cos 150t). The maximum frequency and phase deviations of s (t) are (A) 10.5 kHz, 140p rad (B) 6 kHz, 80p rad (C) 10.5 kHz, 100p rad (D) 7.5 kHz, 100p rad

A I D

O N

The input to a linear delta modulator having a step-size 3= 0.628 2 sin 2pt , s (t) = cos 200pt and n (t) = sin 199pt is a sine wave with frequency fm and peak amplitude Em . If the 7.123 Ininthe figure m (t) = t t . co. sampling frequency fx = 40 kHz, the combination of the sine-wave . a i The output y (t) will be frequency and the peak amplitude, where slope overload will take no d w. place is w w Em fm (A) 0.3 V 8 kHz (B) 1.5 V 4 kHz (C) 1.5 V 2 kHz (A) sin 2pt (B) sin 2pt + sin pt cos 3pt t t t (D) 3.0 V 1 kHz (C) sin 2pt + sin 0.5pt cos 1.5pt (D) sin 2pt + sin pt cos 0.75pt If S represents the carrier synchronization at the receiver and r t t t t represents the bandwidth efficiency, then the correct statement for 7.124 A signal x (t) = 100 cos (24p # 103) t is ideally sampled with a sampling the coherent binary PSK is period of 50m sec ana then passed through an ideal lowpass filter (A) r = 0.5, S is required (B) r = 1.0, S is required with cutoff frequency of 15 kHz. Which of the following frequencies (C) r = 0.5, S is not required (D) r = 1.0, S is not required is/are present at the filter output ? (A) 12 kHz only (B) 8 kHz only A signal is sampled at 8 kHz and is quantized using 8 - bit uniform (C) 12 kHz and 9 kHz (D) 12 kHz and 8 kHz quantizer. Assuming SNRq for a sinusoidal signal, the correct statement for PCM signal with a bit rate of R is 7.125 If the variance ax2 of d (n) = x (n) - x (n - 1) is one-tenth the (A) R = 32 kbps, SNRq = 25.8 dB variance ax2 of stationary zero-mean discrete-time signal x (n), then (B) R = 64 kbps, SNRq = 49.8 dB R (k) the normalized autocorrelation function xx 2 at k = 1 is (C) R = 64 kbps, SNRq = 55.8 dB (A) 0.95 (B) 0.90 ax (D) R = 32 kbps, SNRq = 49.8 dB (C) 0.10 (D) 0.05 2002

7.118

Page 162

ONE MARK

A 2 MHz sinusoidal carrier amplitude modulated by symmetrical square wave of period 100 m sec . Which of the following frequencies will NOT be present in the modulated signal ? (A) 990 kHz (B) 1010 kHz (C) 1020 kHz (D) 1030 kHz

2001 7.126

ONE MARK

A bandlimited signal is sampled at the Nyquist rate. The signal can be recovered by passing the samples through (A) an RC filter (B) an envelope detector (C) a PLL

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Page 163

(A) Pm cos q 2

(D) an ideal low-pass filter with the appropriate bandwidth 7.127

The PDF of a Gaussian random variable X is given by (x - 4) px (x) = 1 e - 18 . The probability of the event {X = 4} is 3 2p 1 1 (A) (B) 2 3 2p (C) 0 (D) 1 4 2

7.136

7.137

2001 7.128

7.129

7.130

A video transmission system transmits 625 picture frames per second. Each frame consists of a 400 # 400 pixel grid with 64 intensity levels per pixel. The data rate of the system is (A) 16 Mbps (B) 100 Mbps (C) 600 Mbps (D) 6.4 Gbps

7.133

7.135

7.138

ONE MARK

The input to a channel is a bandpass signal. It is obtained by linearly modulating a sinusoidal carrier with a single-tone signal. The output of the channel due to this input is given by y (t) = (1/100) cos (100t - 10-6) cos (106 t - 1.56) The group delay (tg) and the phase delay (t p) in seconds, of the channel are (A) tg = 10-6, t p = 1.56 (B) tg = 1.56, t p = 10-6 (C) tg = 108, t p = 1.56 # 10-6 (D) tg = 108, t p = 1.56

A I D

(D) (1 - p) n + np (1 - p) n - 1

7.139

O N w

w ONE MARK

7.140

The amplitude modulated waveform s (t) = Ac [1 + Ka m (t)] cos wc t is fed to an ideal envelope detector. The maximum magnitude of K0 m (t) is greater than 1. Which of the following could be the detector output ? (A) Ac m (t) (B) Ac2 [1 + Ka m (t)] 2 (C) [Ac (1 + Ka m (t)] (D) Ac [1 + Ka m (t)] 2 The frequency range for satellite communication is (A) 1 KHz to 100 KHz (B) 100 KHz to 10 KHz (C) 10 MHz to 30 MHz (D) 1 GHz to 30 GHz 2000

7.134

In a FM system, a carrier of 100 MHz modulated by a sinusoidal signal of 5 KHz. The bandwidth by Carson’s approximation is 1 MHz. If y (t) = (modulated waveform) 3 , than by using Carson’s approximation, the bandwidth of y (t) around 300 MHz and the and the spacing of spectral components are, respectively. (A) 3 MHz, 5 KHz (B) 1 MHz, 15 KHz (C) 3 MHz, 15 KHz (D) 1 MHz, 5 KHz

A modulated signal is given by s (t) = m1 (t) cos (2pfc t) + m2 (t) sin (2pfc t) where the baseband signal m1 (t) and m2 (t) have bandwidths of The PSD and the power of a signal g (t) are, respectively, Sg (w) and Pg . The PSD and the power of the signal ag (t) are, respectively, 10 kHz, and 15 kHz, respectively. The bandwidth of the modulated 2 2 2 (A) a Sg (w) and a Pg (B) a Sg (w) and aPg n signal, o.i in kHz, is c 2 . (C) aSg (w) and a Pg (D) aSg (w) and aPs (B) 15 dia (A) 10 o n (C) 25 (D) 30 w. 2000

7.132

2 2 (C) Pm sin q (D) Pm cos q 4 4 The Hilbert transform of cos w1 t + sin w2 t is (A) sin w1 t - cos w2 t (B) sin w1 t + cos w2 t (C) cos w1 t - sin w2 t (D) sin w1 t + sin w2 t

1999

The Nyquist sampling interval, for the signal sin c (700t) + sin c (500t) is (A) 1 sec (B) p sec 350 350 (C) 1 sec (D) p sec 700 175 During transmission over a communication channel, bit errors occur independently with probability p. If a block of n bits is transmitted, the probability of at most one bit error is equal to (A) 1 - (1 - p) n (B) p + (n - 1)( 1 - p) (C) np (1 - p) n - 1

7.131

TWO MARKS

(B) Pm 4

1999 7.141

TWO MARKS

In a digital communication system employing Frequency Shift Keying (FSK), the 0 and 1 bit are represented by sine waves of 10 KHz and 25 KHz respectively. These waveforms will be orthogonal for a bit interval of (A) 45m sec (B) 200m sec (C) 50m sec (D) 250m sec A message m (t) bandlimited to the frequency fm has a power of Pm . The power of the output signal in the figure is

A modulated signal is given by s (t) = e-at cos [(wc + Dw) t] u (t), where a wc and Dw are positive constants, and wc >> Dw . The complex envelope of s (t) is given by (A) exp (- at) exp [j (wc + Dw) t] u (t) (B) exp (- at) exp (jDwt) u (t) (C) exp (jDwt) u (t) (D) exp [jwc + Dw) t]

7.142

7.143

TWO MARKS

The Nyquist sampling frequency (in Hz) of a signal given by 6 # 10 4 sin c2 (400t) * 106 sin c3 (100t) is (A) 200 (B) 300 (C) 500 (D) 1000 The peak-to-peak input to an 8-bit PCM coder is 2 volts. The signal power-to-quantization noise power ratio (in dB) for an input of 0.5 cos (wm t) is (A) 47.8 (B) 49.8 (C) 95.6 (D) 99.6 The input to a matched filter is given by 6 -4 "10 sin (2p # 10 t) 0 < 1 < 10 sec s (t) = 0 otherwise The peak amplitude of the filter output is (A) 10 volts (B) 5 volts (C) 10 millivolts (D) 5 millivolts

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Four independent messages have bandwidths of 100 Hz, 200 Hz and 400 Hz , respectively. Each is sampled at the Nyquist rate, and the samples are time division multiplexed (TDM) and transmitted. The transmitted sample rate (in Hz) is (A) 1600 (B) 800 (C) 400 (D) 200 1998

7.145

7.146

Page 164 7.155

7.156

ONE MARK

The amplitude spectrum of a Gaussian pulse is (A) uniform (B) a sine function (C) Gaussian (D) an impulse function

7.148

7.149

7.150

7.151

7.152

7.153

7.154

The probability density function of the envelope of narrow band Gaussian noise is (A) Poisson (B) Gaussian (C) Rayleigh (D) Rician 1997

7.157

The ACF of a rectangular pulse of duration T is (A) a rectangular pulse of duration T (B) a rectangular pulse of duration 2T (C) a triangular pulse of duration T (D) a triangular pulse of duration 2T 7.158

7.147

The spectral density of a real valued random process has (A) an even symmetry (B) an odd symmetry (C) a conjugate symmetry (D) no symmetry

The image channel selectivity of superheterodyne receiver depends upon (A) IF amplifiers only (B) RF and IF amplifiers only (C) Preselector, RF and IF amplifiers (D) Preselector, and RF amplifiers only

7.159

ONE MARK

The line code that has zero dc component for pulse transmission of random binary data is (A) Non-return to zero (NRZ) (B) Return to zero (RZ) (C) Alternate Mark Inversion (AM) (D) None of the above 2

A probability density function is given by p (x) = Ke-x /2 - 3 < x < 3 . The value of K should be 2 (B) (A) 1 p 2p (C) 1 (D) 1 2 p p 2 A deterministic signal has the power spectrum given in the figure is, The minimum sampling rate needed to completely represent this signal is

A I D

In a PCM system with uniform quantisation, increasing the number of bits from 8 to 9 will reduce the quantisation noise power by a factor of (A) 9 (B) 8 (C) 4 (D) 2

O N

.in

Flat top sampling of low pass signals .no w (A) gives rise to aperture effect (B) implies oversampling ww (C) leads to aliasing (D) introduces delay distortion A DSB-SC signal is generated using the carrier cos (we t + q) and modulating signal x (t). The envelope of the DSB-SC signal is (A) x (t) (B) x (t) (C) only positive portion of x (t) (D) x (t) cos q

co ia.

d

(A) 1 kHz (C) 3 kHz 7.160

Quadrature multiplexing is (A) the same as FDM (B) the same as TDM (C) a combination of FDM and TDM (D) quite different from FDM and TDM

(B) 2 kHz (D) None of these

A communication channel has first order low pass transfer function. The channel is used to transmit pulses at a symbol rate greater than the half-power frequency of the low pass function. Which of the network shown in the figure is can be used to equalise the received pulses?

The Fourier transform of a voltage signal x (t) is X (f). The unit of X (f) is (A) volt (B) volt-sec (C) volt/sec (D) volt 2 Compression in PCM refers to relative compression of (A) higher signal amplitudes (B) lower signal amplitudes (C) lower signal frequencies (D) higher signal frequencies For a give data rate, the bandwidth B p of a BPSK signal and the bandwidth B 0 of the OOK signal are related as (A) B p = B 0 (B) B p = B 0 2 4 (C) B p = B 0

(D) B p = 2B 0

7.161

The power spectral density of a deterministic signal is given by [sin (f) /f 2] where f is frequency. The auto correlation function of this signal in the time domain is

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(A) a rectangular pulse (C) a sine pulse 1996 7.162

7.163

7.166

ONE MARK

The image channel rejection in a superheterodyne receiver comes from (A) IF stages only (B) RF stages only (C) detector and RF stages only (D) detector RF and IF stages TWO MARKS

The number of bits in a binary PCM system is increased from n to n + 1. As a result, the signal to quantization noise ratio will improve by a factor (B) 2(n + 1)/n (A) n + 1 n (C) 22 (n + 1)/n

7.165

(B) a delta function (D) a triangular pulse

A rectangular pulse of duration T is applied to a filter matched to this input. The out put of the filter is a (A) rectangular pulse of duration T (B) rectangular pulse of duration 2T (C) triangular pulse (D) sine function

1996 7.164

Page 165

(D) which is independent of n

The auto correlation function of an energy signal has (A) no symmetry (B) conjugate symmetry (C) odd symmetry (D) even symmetry

A I D

O N

An FM signal with a modulation index 9 is applied to a frequency .in o c . tripler. The modulation index in the output signal will be ia d o (A) 0 (B) 3 .n w (C) 9 (D) 27 ww

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Page 166

SOLUTIONS 7.1

7.2

Option (B) is correct. In ideal Nyquist Channel, bandwidth required for ISI (Inter Symbol reference) free transmission is W = Rb 2 Here, the used modulation is 32 - QAM (Quantum Amplitude modulation i.e., q = 32 or 2v = 32 v = 5 bits So, the signaling rate (sampling rate) is Rb = R 5 (R " given bit rate) Hence, for ISI free transmission, minimum bandwidth is W = Rb = R kHz 2 10 Option (B) is correct.

P ^X - Y $ 0h = P ^X - Y # 0h = 1 2 or, we can say P ^2U - 3V # 0h = 1 2 Thus, P ^3V $ 2U h = 1 2 7.3

7.4

Option (C) is correct. The mean of random variables U and V are both zero i.e., U =V=0 Also, the random variables are identical i.e., fU ^u h = fV ^v h or, FU ^u h = FV ^v h i.e., their cdf are also same. So, FU ^u h = F2V ^2v h i.e., the cdf of random variable 2V will be also same but for any instant 2V $ U Therefore, G ^x h = F ^x h but, x G ^x h $ xF ^x h or, 6F ^x h - G ^x h@x # 0 Option (C) is correct.

Given, P ^U =+ 1h = P ^U =- 1h = 1 1 2 Given, random variables U and V with mean zero and variances 4 where U is a random variable which is identical to V i.e., and 1 9 P ^V =+ 1h = P ^V =- 1h = 1 i.e., U =V=0 2 1 2 su = So, random variable U and V can have following values 4 U =+ 1, - 1; V =+ 1, - 1 and sv2 = 1 Therefore the random variable U + V can have the following values, 9 n U+V o.i so, P ^U $ 0h = 1 c . 2 2 When U = V =1 a i d o = *0 When U = 1,V = 1 or u =- 1, v = 1 .n and P ^V $ 0h = 1 w 2 2 When U = V = 1 ww The distribution is shown in the figure below Hence, we obtain the probabilities for U + V as follows

A I D

O N

U+V

P ^U + V h

-2

1 1=1 2#2 4 1 1 1 1 1 b2 # 2l+b2 # 2l = 2

0

1 1=1 2#2 4 Therefore, the entropy of the ^U + V h is obtained as 2

fu ^u h = fv ^v h =

H ^U + V h =

2 u

1 e -v 2s 2p sv2 We can express the distribution in standard form by assuming X = u - 0 = u = 2U su Y2 v 0 and Y = = v = 3V sv Y3 2 v

for which we have X = 2U = 0 Y = 2V = 0 and X2 = 4U2 = 1 also, Y2 = 9V2 = 1 Therefore, X - Y is also a normal random variable with X-Y = 0 Hence,

/ P^U + V h log

1 ' P ^U + V h 1 = 1 log 2 4 + 1 log 2 2 + 1 log 2 4 2 4 4 2 1 2 = + + 4 2 4 =3 2

1 e -u 2s 2p su2

7.5

2

Option (D) is correct. For the shown received signal, we conclude that if 0 is the transmitted signal then the received signal will be also zero as the threshold is 1 and the pdf of bit 0 is not crossing 1. Again, we can observe that there is an error when bit 1 is received as it crosses the threshold. The probability of error is given by the area enclosed by the 1 bit pdf (shown by shaded region)

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Page 167

X (w) 2w = 0 = 0 X (w) w = 0 = 0 Option (C) is correct. For raised cosine spectrum transmission bandwidth is given as BT = W (1 + a) a " Roll of factor BT = Rb (1 + a) Rb " Maximum signaling rate 2 3500 = Rb (1 + 0.75) 2 Rb = 3500 # 2 = 4000 1.75

7.8

P (error when bit 1 received) = 1 # 1 # 0.25 = 1 2 8 or P b received 1 l = 1 8 transmitted 0 Since, the 1 and 0 transmission is equiprobable: i.e., P ^ 0 h = P ^1 h = 1 2 Hence bit error rate (BER) is BER received 0 received 1 P 1 = Pb l P 0 + P b transmitted transmitted 1 ^ h 0l ^ h = 0+1 #1 8 2 1 = 16 7.6

Option (D) is correct. Entropy function of a discrete memory less system is given as

7.9

H =

or,

#S 3

-3

E [X 2 (t)] = 1 2p

X

(f ) df

#S 3

-3

X

(w) dw

3 E [X 2 (t)] = 2 # 1 SX (w) dw (Since the PSD is even) 2p 0 = 1 [area under the triangle + integration of delta function] p = 1 ;2 b 1 # 1 # 103 # 6 l + 400E p 2 = 1 66000 + 400@ = 6400 p p E [X (t)] is the absolute value of mean of signal X (t) which is also equal to value of X (w) at (w = 0). From given PSD

#

or,

k

N-1

H = P1 log b 1 l + P2 log b 1 l + Pk log b 1 l P1 P2 Pk k=3

/

=-e P1 log P1 + P2 log P2 +

N-1

/ P log P o k

k

k=3

IA

no w.

BER ww = P b received 1 l P ^0 h + P b received 0 l P ^1 h transmitted 0 transmitted 1 = b1 # 1 # 1l# 1 +b1 # 4 # 1l# 1 2 5 2 2 2 5 5 2 = 1 0 b-a > 0 b >a Note: For phase lead compensator zero is nearer to the origin as compared to pole, so option (C) can not be true. 7.15

r1 = d/ 2 = 0.707d

Option (A) is correct. f = tan-1 a w k - tan-1 a w k a b 1/a 1/b df = =0 2 2 w dw 1 +a k 1 +awk a b 2 2 1 + w = 1+1w a ab2 b b a2 1 - 1 = w2 1 - 1 a b ab b a b l w = ab = 1 # 2 =

7.16

7.17

7.18

d2 = r 12 + r 12 d2 = 2r 12

Now

q = 2p = 2p = p 8 4 M Applying Cooine law we have

2 rad/ sec

Option (D) is correct. Quantized 4 level require 2 bit representation i.e. for one sample 2 bit are required. Since 2 sample per second are transmitted we require 4 bit to be transmitted per second.

d2 = r 22 + r 22 - 2r 22 cos p 4

A I D

= 2r 22 - 2r 22 1/ 2 = (2 d r2 = = 1.3065d 2- 2

or

O N

2 ) r 22

Option (B) is correct. n 7.21 Option o.i (D) is correct. c In FM the amplitude is constant and power is efficient transmitted. . dia Here Pe for 4 PSK and 8 PSK is same because Pe depends on d . No variation in power. o Since Pe is same, d is same for 4 PSK and 8 PSK. .n w There is most bandwidth efficient transmission in SSB- SC. because ww we transmit only one side band. Simple Diode in Non linear region ( Square law ) is used in conventional AM that is simplest receiver structure. In VSB dc. component exists. Option (A) is correct. We have Sx (f) = F {Rx (t)} = F {exp (- pt2)} 2

= e- pf The given circuit can be simplified as

Additional Power SNR = (SNR) 2 - (SNR) 1 = 10 log b ES2 l - 10 log b ES1 l No No = 10 log b ES2 l ES1

Power spectral density of output is Sy (f) = G (f) 2 Sx (f) = j2pf - 1 2 e- pf

2

= ( (2pf) 2 + 1) 2 e- pf Sy (f) = (4p2 f 2 + 1) e- pf

or 7.19

Additional SNR = 5.33 dB

2

Option (B) is correct. Highest frequency component in m (t) is fm = 4000p/2p = 2000 Hz Carrier frequency fC = 1 MHz For Envelope detector condition 1/fC 2pfm Rc 1 < RC < 1 2pfc 2pfm 1 < RC < 1 2pfc 2pfm 1 1 < RC < 2p106 2 # 103 1.59 # 10 - 7 < RC < 7.96 # 10 - 5 so, 20 msec sec best lies in this interval. 7.88

From graph it may be easily seen that slope between 3 < t < 4 is - 1. 7.93

Option (B) is correct.

A I D

SAM (t) = Ac [1 + 0.1 cos wm t] cos wm t sNBFM (t) = Ac cos [wc t + 0.1 sin wm t] s (t) = SAM (t) + SNB fm (t)

O N

= Ac [1 + 0.1 cos wm t] cos wc t + Ac cos (wc t + 0.1 sin wm t) = Ac cos wc t + Ac 0.1 cos wm t cos wc t + Ac cos wc t cos (0.1 sin wm t) - Ac sin wc t. sin (0.1 sin wm t) As 0.1 sin wm t ,+ 0.1 to - 0.1 so, cos (0.1 sin wm t) . 1 As when q is small cos q . 1 and sin q , q, thus

Thus it is SSB with carrier. Option (A) is correct. Consecutive pulses are of same polarity when modulator is in slope overload. Consecutive pulses are of opposite polarity when the input is constant. F (x1 # X < x2) = p (X = x2) - P (X = x1) or 7.91

P (X = 1) = P (X = 1+) - P (X = 1 -) = 0.55 - 0.25 = 0.30

Option (A) is correct. The SNR at transmitter is SNRtr = Ptr NB

7.94

Thus

Option (C) is correct. We have MHz

M = 4 = 2n $ n = 2 B2 = 2Rb = 10 kHz 2

fc = 100 MHz = 100 # 106 and fm = 1

= 1 # 106 The output of balanced modulator is

USB

Option (D) is correct.

no

o .cFor a QPSK, i d

w

- Ac 0.1 sin wm t sin wc t = 2Ac cos wc t + 0.1Ac cos (wc + wm) t 1 44 2 44 3 1 4444 4 2 4444 43 cosec

7.90

where 2n = M and Rb is bit rate For BPSK, M = 2 = 2n $ n = 1 Thus B1 = 2Rb = 2 # 10 = 20 kHz 1 i. n

. ww

sin (0.1 sin wm t) = 0.1 sincos wc t cos wm t + Ac cos wc t

7.89

Option (C) is correct. The required bandwidth of M array PSK is BW = 2Rb n

VBM (t) = [cos wc t][ cos wc t] = 1 [cos (wc + wm) t + cos (wc - wm) t] 2 If VBM (t) is passed through HPF of cut off frequency fH = 100 # 106 , then only (wc + wm) passes and output of HPF is VHP (t) = 1 cos (wc + wm) t 2 Now

V0 (t) = VHP (t) + sin (2p # 100 # 106) t

= 1 cos [2p100 # 106 + 2p # 1 # 106 t] + sin (2p # 100 # 106) t 2 = 1 cos [2p108 + 2p106 t] + sin (2p108) t 2 = 1 [cos (2p108 t) t cos (2p106 t)] - sin [2p108 t sin (2p106 t) + sin 2p108 t] 2 = 1 cos (2p106 t) cos 2p108 t + `1 - 1 sin 2p106 t j sin 2p108 t 2 2 This signal is in form

10 - 3 109 - 20 6 = 10 # 100 # 10

= A cos 2p108 t + B sin 2p108 t

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ELECTRONICS & COMMUNICATION by RK Kanodia & Ashish Murolia in 3 Volumes 7.98

7.99

7.95

A2 + B2

2 6 6 2 1 1 ` 2 cos (2p10 t)j + `1 - 2 sin (2p10 t j

1 cos2 (2p106 t) + 1 + 1 sin2 (2p106 t) - sin (2p106 t) 4 4 1 + 1 - sin (2p106 t) = 4 5 - sin (2p106 t) = 4

Option (A) is correct. s (t) = A cos [2p10 # 10 t] + A cos [2p10.1 # 103 t] 1 Here = 100m sec T1 = 10 # 103 1 and = 99m sec T2 = 10.1 # 103 Period of added signal will be LCM [T1, T2] Thus T = LCM [100, 99] = 9900m sec Thus frequency f = 1 = 0.1 kHz 9900m

7.96

7.101

7.103

Option (C) is correct. FM DM PSK PCM

P (0 # X # 0.2) = 0.2 Probability of error of 0 : P (0.2 # X # 0.25) = 0.05 # 2 = 0.1 P (0 # X # 0.2) + P (0.2 # X # 0.25) 2 = 0.2 + 0.1 = 0.15 0

3

s2 =

#- 3 (x - xq) 2 f (x) dx

=

#0 (x - xq) 2 f (x) dx

$ Capture effect $ Slope over load $ Matched filter $ m - law

Option (C) is correct. Since fs = 2fm , the signal frequency and sampling frequency are as follows fm1 = 1200 Hz $ 2400 samples per sec fm2 = 600 Hz $ 1200 samples per sec fm3 = 600 Hz $ 1200 samples per sec Thus by time division multiplexing total 4800 samples per second will be sent. Since each sample require 12 bit, total 4800 # 12 bits per second will be sent Thus bit rate Rb = 4800 # 12 = 57.6 kbps

Option (A) is correct. The input is a coherent detector is DSB - SC signal plus noise. The noise at the detector output is the in-phase component as the quadrature component nq (t) of the noise n (t) is completely rejected by the detector. Option (C) is correct. The noise at the input to an ideal frequency detector is white. The PSD of noise at the output is parabolic Option (B) is correct. We have

Probability of error of 1

Option (B) is correct. The square mean value is

0.039 = 0.198

A I D 7.102

7.97

s2 =

Option (B) is correct. The input signal X (f) has the peak at 1 kHz and - 1 kHz. After balanced modulator the output will have peak at fc ! 1 kHz i.e. : 10 ! 1 $ 11 and 9 kHz 10 ! (- 1) $ 9 and 11 kHz in . o c ia. 9 kHz will be filtered out by HPF of 10 kHz. Thus 11 kHz will red main. After passing through 13 kHz balanced modulator signal will .no w have 13 ! 11 kHz signal i.e. 2 and 24 kHz. ww Thus peak of Y (f) are at 2 kHz and 24 kHz.

Option (A) is correct. The pdf of transmission of 0 and 1 will be as shown below :

Average error =

s2 = 0.039

7.100

O N

3

0. 1

#0.3 (x - 0.7) 2 f (x) dx

RMS =

The envelope of this signal is

=

(x - 0) 2 f (x) dx +

or

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=

0. 3

3 0. 3 3 2 1 = ; x E + ; x + 0.49x - 14 x E 3 0 3 2 0. 3

Fully Revised

=

#0

Pe = 1 erfc c 2

Ed 2h m

Since Pe of Binary FSK is 3 dB inferior to binary PSK 7.104

Option (D) is correct. The pdf of Z will be convolution of pdf of X and pdf of Y as shown below. Now

p [Z # z] = p [Z #- 2] =

z

#- 3 fZ (z) dz -2

#- 3fZ (z) dz

= Area [z #- 2] = 1 # 1 #1 = 1 2 6 12

1

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Cn = A sin ` np j pn 3 From Cn it may be easily seen that 1, 2, 4, 5, 7, harmonics are present and 0, 3, 6, 9,.. are absent. Thus p (t) has 1 kHz, 2 kHz, 4 kHz, 5 kHz, 7 kHz,... frequency component and 3 kHz, 6 kHz.. are absent. The signal x (t) has the frequency components 0.4 kHz and 0.7 kHz. The sampled signal of x (t) i.e. x (t)* p (t) will have 1 ! 0.4 and 1 ! 0.7 kHz 2 ! 0.4 and 2 ! 0.7 kHz 4 ! 0.4 and 4 ! 0.7 kHz Thus in range of 2.5 kHz to 3.5 kHz the frequency present is 2 + 0.7 = 2.7 kHz 4 - 0.7 = 3.3 kHz or

Option (C) is correct.

7.109

7.105

vi = Ac1 cos (2pfc t) + m (t) v0 = ao vi + avi3 v0 ' ' = a0 [Ac cos (2pfc t) + m (t)] + a1 [Ac' cos (2pfc' t) + m (t)] 3

Option (D) is correct. We have

RXX (t) = 4 (e - 0.2 t + 1) RXX (0) = 4 (e - 0.2 0 + 1) = 8 = s2

or mean Now

s =2 2 m =0 P (x # 1) = Fx (1)

Given

= a0 Ac' cos (2pfc' t) + a0 m (t) + a1 [(Ac' cos 2pfc' t) 3

X-m at x = 1 s m = 1 - Qc 1 - 0 m = 1 - Qc 1 m 2 2 2 2

= 1 - Qc

7.106

+ (Ac' cos (2pfc') t) 2 m (t) + 3Ac' cos (2pfc' t) m2 (t) + m3 (t)]

IA

D O

Option (C) is correct.

7.108

+ 3a1 Ac'2 ;

1 + cos (4pfc' t) E m (t) 2

W = Y-Z n = 3a1 Ac' cos (2pfc' t) m2 (t) + m3 (t) o.i c E [W2] = E [Y - Z] 2 = E [Y2] + E [Z2] - 2E [YZ] . ia The term 3a1 Ac' ( cos 42pf t ) m (t) is a DSB-SC signal having carrier d 2 o = sw .n frequency 1. MHz. Thus 2fc' = 1 MHz or fc' = 0.5 MHz w 2 We have E [X (t)] = Rx (10) ww 7.110 Option (D) is correct. = 4 [e - 0.2 0 + 1] = 4 [1 + 1] = 8 2 PT = Pc c1 + a m 2 2 E [Y ] = E [X (2)] = 8 2 2 E [Z2] = E [X2 (4)] = 8 Pc a2 = Pc (0.5) P = sb 2 2 E [YZ] = RXX (2) = 4 [e-0.2 (4 - 2) + 1] = 6.68 Psb = 1 E [W2] = sw2 = 8 + 8 - 2 # 6.68 = 2.64 or Pc 8 7.111 Option (C) is correct. Option (D) is correct. 2mp AM Band width = 2fm Step size d = = 1.536 = 0.012 V L 128 Peak frequency deviation = 3 (2fm) = 6fm 2 2 ( 0 . 012 ) 6f Quantization Noise power = d = Modulation index b = m = 6 12 12 fm = 12 # 10-6 V2 The FM signal is represented in terms of Bessel function as

N

7.107

= a0 Ac' cos (2pfc' t) + a0 m (t) + a1 (Ac' cos 2fc' t) 3

Option (D) is correct. The frequency of pulse train is f 1- 3 = 1 k Hz 10 The Fourier Series coefficient of given pulse train is -T /2 Cn = 1 Ae-jnw t dt To -T /2

#

xFM (t) = Ac

#

o

-To /6

7.112

Ae-jhw t dt o

A [e-jw t] --TT //66 To (- jhwo) A = (e-jw t - e jhw T /6) (- j2pn) = A (e jhp/3 - e-jhp/3) j2pn =

o

o o

o

o

/Jn (b) cos (wc - nwn) t

wc + nwm = 2p (1008 # 103) 2p106 + n4p # 103 = 2p (1008 # 103), n = 4 Thus coefficient = 5J4 (6)

o

-To /6

3

n =- 3

o

= 1 To

' c

o

7.113

Option (B) is correct. Ring modulation $ VCO $ Foster seely discriminator $ mixer $

Generation of DSB - SC Generation of FM Demodulation of fm frequency conversion

Option (A) is correct. fmax = 1650 + 450 = 2100 kHz

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fmin = 550 + 450 = 1000 kHz 1 or f = 2p LC frequency is minimum, capacitance will be maximum R = Cmax Cmin or 7.114

Page 178

For best reception, if transmitting waves are vertically polarized, then receiver should also be vertically polarized i.e. transmitter and receiver must be in same polarization. 7.122

f2 = max = (2.1) 2 2 fmin

s (t) = cos 2p (2 # 106 t + 30 sin 150t + 40 cos 150t) = cos {4p106 t + 100p sin (150t + q)} Angle modulated signal is

R = 4.41 fi = fc + 2fIF = 700 + 2 (455) = 1600 kHz

s (t) = A cos {wc t + b sin (wm t + q)} Comparing with angle modulated signal we get Phase deviations b = 100p Frequency deviations 3 f = bfm = 100p # 150 = 7.5 kHz 2p

Option (D) is correct. Eb = 10 - 6 watt-sec No = 10 - 5 W/Hz o (SNR) matched filler = E = N

106 = .05 2 # 10 - 5 2 (SNR)dB = 10 log 10 (0.05) = 13 dB o

7.115

7.123

Option (B) is correct. 3 fs 2pfm This is satisfied with Em = 1.5 V and fm = 4 kHz For slopeoverload to take place Em $

7.116

7.117

Option (A) is correct. If s " carrier synchronization at receiver r " represents bandwidth efficiency then for coherent binary PSK r = 0.5 and s is required.

7.118

7.119

Option (*) is correct. We have m (t) s (t) = y1 (t) 2 sin (2pt) cos (200pt) = t sin (202pt) - sin (198pt) = t y1 (t) + n (t) = y2 (t) = sin 202pt - sin 198pt + sin 199pt t t

Option (B) is correct. Bit Rate = 8k # 8 = 64 kbps (SNR)q = 1.76 + 6.02n dB = 1.76 + 6.02 # 8 = 49.8 dB

Option (D) is correct.

A I D =

O N

y2 (t) s (t) = u (t)

[sin 202pt - sin 198pt + sin 199pt] cos 200pt t

Option (C) is correct. n o=.i 1 [sin (402pt) + sin (2pt) - {sin (398pt) - sin (2pt)} c The frequency of message signal is . 2 dia o fc = 1000 kHz .n + sin (399pt) - sin (pt)] w w 1 The frequency of message signal is w After filtering 1 = 10 kHz fm = -6 sin (2pt) + sin (2pt) - sin (pt) 100 # 10 y (t) = 2t Here message signal is symmetrical square wave whose FS has only sin (2pt) + 2 sin (0.5t) cos (1.5pt) = odd harmonics i.e. 10 kHz, 30 kHz 50 kHz. Modulated signal con2t tain fc ! fm frequency component. Thus modulated signal has = sin 2pt + sin 0.5pt cos 1.5pt 2t t fc ! fm = (1000 ! 10) kH = 1010 kHz, 990 kHz 7.124 Option (B) is correct. fc ! 3fm = (1000 ! 10) kH = 1030 kHz, 970 kHz The signal frequency is Thus, there is no 1020 kHz component in modulated signal. 24p103 = 12 kHz = f m Option (C) is correct. 2p +3 Ts = 50m sec " fs = 1 = 1 # 106 = 20 We have y (t) = 5 # 10 - 6 x (t) d (t - nTs) Ts 50 n =- 3 kHz 3 x (t) = 10 cos (8p # 10 ) t After sampling signal will have fs ! fm frequency component i.e. 32 Ts = 100m sec and 12 kHz The cut off fc of LPF is 5 kHz At filter output only 8 kHz will be present as cutoff frequency is We know that for the output of filter x (t) y (t) 15 kHz. = Ts 7.125 Option (A) is correct.

/

10 cos (8p # 103) t # 5 # 10 - 6 100 # 10 - 6 = 5 # 10 - 1 cos (8p # 103) t =

7.120

7.121

Option (C) is correct. Transmitted frequencies in coherent BFSK should be integral of bit rate 8 kHz. Option (B) is correct.

d (n) = x (n) - x (n - 1) E [d (n)] 2 = E [x (n) - x (n - 1)] 2 or E [d (n)] 2 = E [x (n)] 2 + E [x (n - 1)] 2 - 2E [x (n) x (n - 1)] or sd2 = sx2 + sx2 - 2Rxx (1) 2 As we have been given sd2 = sx , therefore 10

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as k = 1

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

sx2 = s2 + s2 - 2R (1) x x xx 10 2Rxx (1) = 19 sx2 10 Rxx = 19 = 0.95 20 sx2

or or 7.126

7.127

Page 179

Rb = HCF (f1, f2) = HCF (10k, 25k) = 5 kHz Tb = 1 = 1 = 0.2 msec = 200 msec Rb 5k

Bit rate

Thus bit interval is Option (D) is correct.

7.135

Option (A) is correct. An ideal low - pass filter with appropriate bandwidth fm is used to recover the signal which is sampled at nyquist rate 2fm .

We have The input to LPF is

Pm = m2 (t) x (t) = m (t) cos wo t cos (wo t + q)

Option (A) is correct. For any PDF the probability at mean is 1 . Here given PDF is 2 Gaussian random variable and X = 4 is mean.

7.128

7.129

m (t) [cos (2wo t + q) + cos q] 2 m (t) cos (2wo t + q) m (t) cos q = + 2 2

=

Option (C) is correct. We require 6 bit for 64 intensity levels because 64 = 26 Data Rate = Frames per second # pixels per frame # bits per pixel = 625 # 400 # 400 # 6 = 600 Mbps sec

The output of filter will be

Option (C) is correct. We have

Power of output signal is

y (t) =

2 Py = y2 (t) = 1 m2 (t) cos2 q = Pm cos q 4 4

sin (700pt) sin (500pt) sin c (700t) + sin c (500t) = + 700pt 500pt

Option (D) is correct. Probability of error = p Probability of no error = q = (1 - p) Probability for at most one bit error = Probability of no bit error + probability of 1 bit error

A I D

O N

Thus

no

. ww

dia

w

FT

g (t) then PSD of g (t) is

G (w) 7.138

Sg (w) = G (w) 2

7.139

and power is Pg = 1 2p Now PSD of ag (t) is

or Similarly 7.132

7.133

7.134

ag (t)

FT

3

#- 3Sg (w) dw aG (w) 7.140

Sag (w) = a (G (w)) = a2 G (w) 2 Sag (w) = a2 Sg (w) Pag = a2 Pg

2

Option (C) is correct. The envelope of the input signal is [1 + ka m (t)] that will be output of envelope detector. Option (D) is correct. Frequency Range for satellite communication is 1 GHz to 30 GHz, Option (B) is correct. Waveform will be orthogonal when each bit contains integer number of cycles of carrier.

x (t) = Ac cos {wc t + b sin wm t} y (t) = {x (t)} 3

= Ac2 cos (3wc t + 3b sin wm t) + 3 cos (wc t + b sin wm t) Thus the fundamental frequency doesn’t change but BW is three times. BW = 2 (3 f') = 2 (3 f # 3) = 3 MHz

Option (A) is correct. If

cos wt " sin wt sin wt " cos wt cos w1 t + sin w2 t " sin w1 t - cos w2 t

Option (A) is correct. n o.i have .cWe

7.137

= (1 - p) n + np (1 - p) n - 1 7.131

Option (A) is correct. Hilbert transformer always adds - 90c to the positive frequency component and 90c to the negative frequency component. Hilbert Trans form

7.136

Here the maximum frequency component is 2pfm = 700p i.e. fm = 350 Hz Thus Nyquist rate fs = 2fm = 2 (350) = 700 Hz Thus sampling interval = 1 sec 700 7.130

m (t) cos q 2

Option (C) is correct. Option (C) is correct. This is Quadrature modulated signal. In QAM, two signals having bandwidth. B 1 & B 2 can be transmitted simultaneous over a bandwidth of (B 1 + B 2) Hz so B.W. = (15 + 10) = 25 kHz Option (B) is correct. A modulated signal can be expressed in terms of its in-phase and quadrature component as S (t) = S1 (t) cos (2pfc t) - SQ (t) sin (2pfc t) Here S (t) = [e-at cpsDwt cos wc t - eat sin Dwt sin wc t] m (t) = [e-at cos Dwt] cos 2pfc t - [e-at sin Dwt] sin 2pfc t = S1 (t) cos 2pfc t - SQ (t) sin 2pfc t Complex envelope of s (t) is S (t) = S1 (t) + jSQ (t) = e-at cos Dwt + je-at sin Dwt = e-at [cos Dwt + j sin Dwt]

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Page 180

= exp (- at) exp (jDwt) m (t) 7.141

Option (B) is correct. Given function g (t) 4

2

6

= 6 # 10 sin c (400t) ) 10 sin c3 (100t) Let g1 (t) = 6 # 10 4 sin c2 (400t) g2 (t) = (106) sin c3 (100t) We know that g1 (t) ) g2 (t) ? G1 (w) G2 (w) occupies minimum of Bandwidth of G1 (w) or G2 (w) Band width of G1 (w)

Auto correlation function is given by T/2 Rxx (t) = 1 x (t) x (t - t) dt T -T/2 When x (t) is shifted to right (t > 0), x (t - t) will be shown as dotted line.

#

= 2 # 400 = 800 rad/ sec or = 400 Hz Band width of G2 (w) = 3 # 100 = 300 rad/ sec or 150 Hz Sampling frequency = 2 # 150 = 300 Hz 7.142

7.143

Option (B) is correct. For a sinusoidal input SNR (dB) is PCM is obtained by following formulae. SNR (dB) = 1.8 + 6n n is no. of bits Here n =8 So, SNR (dB) = 1.8 + 6 # 8 = 49.8 Option (D) is correct. We know that matched filter output is given by g 0 (t) = t = T0

6g 0 (t)@max = = [g 0 (t)] max 7.144

3

0

# g (l) g (l) dl = # 3

-3

#

1 # 10-4

0

3

g 2 (t) dt

O N -3

[10 sin (2p # 106) 2] dt

#

T +t 2

A2 dt

T - +t 2

2 2 = A :T + T - tD = A :T - tD 2 T 2 T 2 (t) can be negative or positive, so generalizing above equations 2 Rxx (t) = A :T - t D T 2 Rxx (t) is a regular pulse of duration T .

A I D

# g (l) g (T - t + l) dl at -3

Rxx (t) = 1 T

.in

co ia.

od

.n = 1 # 100 # 10-4 = 5 mV ww 2 w

Option (B) is correct. Sampling rate must be equal to twice of maximum frequency. f s = 2 # 400 = 800 Hz

7.145

Option (C) is correct. The amplitude spectrum of a gaussian pulse is also gaussian as shown in the fig. -y 2 fY (y) = 1 exp c 2 m 2p

7.147

7.148

7.149

7.146

Option (C) is correct. Let the rectangular pulse is given as 7.150

Option (B) is correct. Selectivity refers to select a desired frequency while rejecting all others. In super heterodyne receiver selective is obtained partially by RF amplifier and mainly by IF amplifier. Option (C) is correct. In PCM, SNR a 22n so if bit increased from 8 to 9 2#8 (SNR) 1 = 22 # 9 = 22 = 1 4 (SNR) 2 2 so SNR will increased by a factor of 4 Option (A) is correct. In flat top sampling an amplitude distortion is produced while reconstructing original signal x (t) from sampled signal s (t). High frequency of x (t) are mostly attenuated. This effect is known as aperture effect. Option (A) is correct. Carrier C (t) = cos (we t + q) Modulating signal = x (t) DSB - SC modulated signal = x (t) c (t) = x (t) cos (we t + q) envelope = x (t)

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 7.151

7.152

7.153

7.154

7.155

7.156 7.157

Page 181

Option (D) is correct. In Quadrature multiplexing two baseband signals can transmitted or modulated using I 4 phase & Quadrature carriers and its quite different form FDM & TDM. Option (A) is correct. Fourier transform perform a conversion from time domain to frequency domain for analysis purposes. Units remain same.

output of matched filter y (t) = g (t) ) h (t)

Option (A) is correct. In PCM, SNR is depends an step size (i.e. signal amplitude) SNR can be improved by using smaller steps for smaller amplitude. This is obtained by compressing the signal. Option (C) is correct. Band width is same for BPSK and APSK(OOK) which is equal to twice of signal Bandwidth.

if we shift g (- t) for convolution y (t) increases first linearly then decreases to zero.

Option (A) is correct. The spectral density of a real value random process symmetric about vertical axis so it has an even symmetry. Option (A) is correct. Option (C) is correct. It is one of the advantage of bipolar signalling (AMI) that its spectrum has a dc null for binary data transmission PSD of bipolar signalling is

Option (C) is correct. The difference between incoming signal frequency (fc) and its image frequency (fc) is 2I f (which is large enough). The RF filter may provide poor selectivity against adjacent channels separated by a small frequency differences but it can provide reasonable selectivity against .in a station separated by 2I f . So it provides adequate suppression o c . ia of image channel.

A I D 7.163

O N

no w.

ww

7.158

Option (A) is correct. Probability Density function (PDF) of a random variable x defined as Px (x) = 1 e-x /2 2p so here K = 1 2p

d

7.164

if no. of bits is increased from n to (n + 1) SNR will increase by a factor of 22 (n + 1)/n

2

7.159

7.160

7.161

Option (C) is correct. In PCM SNR is given by SNR = 3 22n 2

7.165

Option (C) is correct. Here the highest frequency component in the spectrum is 1.5 kHz [at 2 kHz is not included in the spectrum] Minimum sampling freq. = 1.5 # 2 = 3 kHz

Option (D) is correct. The auto correlation of energy signal is an even function. auto correlation function is gives as R (t) = put Let

Option (B) is correct. We need a high pass filter for receiving the pulses. Option (D) is correct. Power spectral density function of a signal g (t) is fourier transform of its auto correlation function

-3 3

R (- t) = # x (t) x (t - t) dt -3 t-t = a dt = da

R (- t) = change variable a " t

F

Sg (w) Rg (t) here Sg (w) = sin c2 (f) so Rg (t) is a triangular pulse. f [triang.] = sin c2 (f) 7.162

3

# x (t) x (t + t) dt

R (- t) =

3

# x (a + t) x (a) da -3

3

# x (t) x (t + t) dt = R (t) -3

R (- t) = R (t) even function 7.166

Option (D) is correct.

Option (C) is correct. For a signal g (t), its matched filter response given as h (t) = g (T - t) so here g (t) is a rectangular pulse of duration T .

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Page 182

UNIT 8

p # 10 z (B) - E 0 ^atx + atz h e j 3 V/m 2 4

ELECTROMAGNETICS

p # 10 ^x - z h (C) 0.44 E 0 ^atx + atz h e-j 3 2 V/m 2 4

p # 10 ^x + z h 3 (D) E 0 ^atx + atz h e-j V/m 2 4

2013 8.1

v ^rvh. The closed loop line integral A v : dlv Consider a vector field A can be expressed as vh : dsv over the closed surface bounded by the loop (A) ^d # A vh dv over the closed volume bounded by the loop (B) ^d : A (D)

8.3

2012

#

(C)

8.2

ONE MARK

A plane wave propagating in air with j (wt + 3x - 4y) E = (8ax + 6ay + 5az ) e V/m is incident on a perfectly conducting slab positioned at x # 0 . The E field of the reflected wave is (A) (- 8ax - 6ay - 5az ) e j (wt + 3x + 4y) V/m

8.6

## ### ### ^d : Avhdv over the open volume bounded by the loop ## ^d # Avh : dsv over the open surface bounded by the loop

-

(B) (- 8ax + 6ay - 5az ) e j (wt + 3x + 4y) V/m (C) (- 8ax - 6ay - 5az ) e j (wt - 3x - 4y) V/m (D) (- 8ax + 6ay - 5az ) e j (wt - 3x - 4y) V/m

v = xatx + yaty + zatz is The divergence of the vector field A (A) 0 (B) 1/3 (C) 1 (D) 3

TWO MARKS

A I D

O N

Statement for Linked Answer Questions 52 and 53:

The electric field of a uniform plane electromagnetic wave in free space, along the positive x direction is given by E = 10 (ay + jaz ) e-j 25x . The frequency and polarization of the wave, respectively, are (A) 1.2 GHz and left circular (B) 4 Hz and left circular (C) 1.2 GHz and right circular (D) 4 Hz and right circular

8.7

The return loss of a device is found to be 20 dB. The voltage standing wave ratio (VSWR) and magnitude of reflection coefficient are respectively (A) 1.22 and 0.1 (B) 0.81 and 0.1 (C) – 1.22 and 0.1 (D) 2.44 and 0.2 2013

ONE MARK

8.8

.in o(A) c . 330 W ia

od

A monochromatic plane wave of wavelength l = 600 mm is propa-w.n ww 8.9 gating in the direction as shown in the figure below. Evi , Evr and Evt denote incident, reflected, and transmitted electric field vectors associated with the wave.

A coaxial-cable with an inner diameter of 1 mm and outer diameter of 2.4 mm is filled with a dielectric of relative permittivity 10.89. -9 Given m0 = 4p # 10-7 H/m, e0 = 10 F/m , the characteristic 36p impedance of the cable is (C) 143.3 W

(B) 100 W (D) 43.4 W

The radiation pattern of an antenna in spherical co-ordinates is given by F (q) = cos 4 q ; 0 # q # p/2 . The directivity of the antenna is (A) 10 dB (B) 12.6 dB (C) 11.5 dB (D) 18 dB 2012

8.10

8.4

The angle of incidence qi and the expression for Evi are p # 10 ^x + 2h (A) 60c and E 0 ^atx - atz h e-j 3 2 V/m 2 4

8.11

TWO MARKS

A transmission line with a characteristic impedance of 100 W is used to match a 50 W section to a 200 W section. If the matching is to be done both at 429 MHz and 1 GHz, the length of the transmission line can be approximately (A) 82.5 cm (b) 1.05 m (C) 1.58 cm (D) 1.75 m The magnetic field among the propagation direction inside a rectangular waveguide with the cross-section shown in the figure is Hz = 3 cos (2.094 # 102 x) cos (2.618 # 102 y) cos (6.283 # 1010 t - bz)

p # 10 z (B) 45c and E 0 ^atx + atz h e-j 3 V/m 2 4

p # 10 ^x + z h (C) 45c and E 0 ^atx - atz h e-j 3 2 V/m 2 4

p # 10 z (D) 60c and E 0 ^atx - atz h e-j 3 V/m 2 The expression for Evr is p # 10 ^x - z h (A) 0.23 E 0 ^atx + atz h e-j 3 2 V/m 2 4

8.5

4

The phase velocity v p of the wave inside the waveguide satisfies (A) v p > c (B) v p = c (C) 0 < v p < c (D) v p = 0

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Page 183

2011

Statement for Linked Answer Question 7 and 8 : An infinitely long uniform solid wire of radius a carries a uniform dc current of density J 8.12

8.13

8.17

The magnetic field at a distance r from the center of the wire is proportional to (A) r for r < a and 1/r 2 for r > a (B) 0 for r < a and 1/r for r > a (C) r for r < a and 1/r for r > a (D) 0 for r < a and 1/r 2 for r > a

TWO MARKS

A current sheet Jv = 10uty A/m lies on the dielectric interface x = 0 between two dielectric media with er 1 = 5, mr 1 = 1 in Region-1 (x < 0) and er2 = 2, mr2 = 2 in Region-2 (x 2 0). If the magnetic field in Region-1 at x = 0- is Hv1 = 3utx + 30uty A/m the magnetic field in Region-2 at x = 0+ is

A hole of radius b (b < a) is now drilled along the length of the wire at a distance d from the center of the wire as shown below. (A) Hv2 = 1.5utx + 30uty - 10utz A/m (B) Hv2 = 3utx + 30uty - 10utz A/m (C) Hv2 = 1.5utx + 40uty A/m (D) Hv2 = 3utx + 30uty + 10utz A/m 8.18

The magnetic field inside (A) uniform and depends (B) uniform and depends (C) uniform and depends (D) non uniform 2011 8.14

the hole is only on d only on b on both b and d

The electric and magnetic fields for a TEM wave of frequency 14 GHz in a homogeneous medium of relative permittivity er and relative permeability mr = 1 are given by Ev = E p e j (wt - 280py) utz V/m and Hv = 3e j (wt - 280py) utx A/m . Assuming the speed of light in free space to be 3 # 108 m/s , the intrinsic impedance of free space to be 120p , the relative permittivity er of the medium and the electric field amplitude E p are in . o (B) er = 3, E p = 360p c(A) er = 3, E p = 120p ia. d (C) er = 9, E p = 360p (D) er = 9, E p = 120p no 8.19

A I D

ONE MARK

Consider the following statements regarding the complex Poynting vector Pv for the power radiated by a point source in an infinite homogeneous and lossless medium. Re(Pv ) denotes the real part of Pv, S denotes a spherical surface whose centre is at the point source, and nt denotes the unit surface normal on S . Which of the following w. statements is TRUE? w w (A) Re(Pv ) remains constant at any radial distance from the source (B) Re(Pv ) increases with increasing radial distance from the

O N

source

2010 8.20

(C) ## Re (Pv) : nt dS remains constant at any radial distance from s the source (D)

##s Re (Pv) : nt dS

decreases with increasing radial distance from

the source 8.15

8.16

A transmission line of characteristic impedance 50 W is terminated in a load impedance ZL . The VSWR of the line is measured as 5 and the first of the voltage maxima in the line is observed at a distance of l/4 from the load. The value of ZL is (A) 10 W (B) 250 W (C) (19.23 + j 46.15) W (D) (19.23 - j 46.15) W

A transmission line of characteristic impedance 50 W is terminated by a 50 W load. When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be p/4 radians. The phase velocity of the wave along the line is (A) 0.8 # 108 m/s (B) 1.2 # 108 m/s (C) 1.6 # 108 m/s (D) 3 # 108 m/s The modes in a rectangular waveguide are denoted by TE mn where TM mn m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is TRUE? (A) The TM 10 mode of the waveguide does not exist (B) The TE 10 mode of the waveguide does not exist (C) The TM 10 and the TE 10 modes both exist and have the same cut-off frequencies (D) The TM 10 and the TM 01 modes both exist and have the same cut-off frequencies

8.21

8.22

ONE MARK

If the scattering matrix [S ] of a two port network is 0.2+0c 0.9+90c , then the network is [S ] = > 0.9+90c 0.1+90cH (A) lossless and reciprocal (B) lossless but not reciprocal (C) not lossless but reciprocal (D) neither lossless nor reciprocal A transmission line has a characteristic impedance of 50 W and a resistance of 0.1 W/m . If the line is distortion less, the attenuation constant(in Np/m) is (A) 500 (B) 5 (C) 0.014 (D) 0.002 The electric field component of a time harmonic plane EM wave traveling in a nonmagnetic lossless dielectric medium has an amplitude of 1 V/m. If the relative permittivity of the medium is 4, the magnitude of the time-average power density vector (in W/m2 ) is (B) 1 (A) 1 30p 60p (C) 1 (D) 1 120p 240p 2010

8.23

v = xyatx + x 2 aty , then If A is

TWO MARKS

# Av $ dlv over the path shown in the figure

o

C

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Page 184

(A) Only P has no cutoff-frequency (B) Only Q has no cutoff-frequency (C) Only R has no cutoff-frequency (D) All three have cutoff-frequencies (B) 2 3 (D) 2 3

(A) 0 (C) 1 8.24

8.25

2009

(C)

#S #C #S #C #C #C #C D # V $ dl = #S #C D # A $ d S (D) #C D # V $ dl = #S #CV $ d S

A transmission line terminates in two branches, each of length l , 4 as shown. The branches are terminated by 50W loads. The lines are lossless and have the characteristic impedances shown. Determine the impedance Zi as seen by the source.

8.29

A I D

In the circuit shown, all the transmission line sections are lossless. The Voltage Standing Wave Ration(VSWR) on the 60 W line is

O N w ww

(A) 1.00 (C) 2.50

(B) 1.64 (D) 3.00

2009

8.30

ONE MARK

Two infinitely long wires carrying current are as shown in the figure below. One wire is in the y - z plane and parallel to the y - axis. The other wire is in the x - y plane and parallel to the x - axis. Which components of the resulting magnetic field are non-zero at the origin ?

(A)n 200W .i o .c (C) 50W

ia

d .no

8.26

If a vector field V is related to another vector field A through V = 4# A , which of the following is true? (Note : C and SC refer to any closed contour and any surface whose boundary is C . ) (A) V $ dl = (B) A $ dl = A $ dS V $ dS

8.28

A plane wave having the electric field components Evi = 24 cos ^3 # 108 - by h atx V/m and traveling in free space is incident normally on a lossless medium with m = m0 and e = 9e0 which occupies the region y $ 0 . The reflected magnetic field component is given by (A) 1 cos (3 # 108 t + y) atx A/m 10p (B) 1 cos (3 # 108 t + y) atx A/m 20p (C) - 1 cos (3 # 108 t + y) atx A/m 20p (D) - 1 cos (3 # 108 t + y) atx A/m 10p

TWO MARKS

(B) 100W (D) 25W

A magnetic field in air is measured to be y B = B0 c 2 x 2 yt - 2 xt m x +y x + y2 What current distribution leads to this field ? [Hint : The algebra is trivial in cylindrical coordinates.] t t (B) J =- B0 z c 2 2 2 m, r ! 0 (A) J = B0 z c 2 1 2 m, r ! 0 m0 x + y m0 x + y t (C) J = 0, r ! 0 (D) J = B0 z c 2 1 2 m, r ! 0 m0 x + y 2008

8.31

8.32

ONE MARK

For a Hertz dipole antenna, the half power beam width (HPBW) in the E -plane is (A) 360c (B) 180c (C) 90c (D) 45c For static electric and magnetic fields in an inhomogeneous sourcefree medium, which of the following represents the correct form of Maxwell’s equations ? (A) 4$ E = 0 , 4# B = 0 (B) 4$ E = 0 , 4$ B = 0 (C) 4# E = 0 , 4# B = 0 (D) 4# E = 0 , 4$ B = 0 2008

(A) x, y, z components (C) y, z components 8.27

(B) x, y components (D) x, z components

Which of the following statements is true regarding the fundamental mode of the metallic waveguides shown ?

8.33

TWO MARKS

A rectangular waveguide of internal dimensions (a = 4 cm and b = 3 cm) is to be operated in TE11 mode. The minimum operating frequency is (A) 6.25 GHz (B) 6.0 GHz (C) 5.0 GHz

(D) 3.75 GHz

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 8.34

8.35

8.36

8.37

One end of a loss-less transmission line having the characteristic impedance of 75W and length of 1 cm is short-circuited. At 3 GHz, the input impedance at the other end of transmission line is (A) 0 (B) Resistive (C) Capacitive (D) Inductive A uniform plane wave in the free space is normally incident on an infinitely thick dielectric slab (dielectric constant e = 9 ). The magnitude of the reflection coefficient is (A) 0 (B) 0.3 (C) 0.5 (D) 0.8

ONE MARK

A plane wave of wavelength l is traveling in a direction making an angle 30c with positive x - axis and 90c with positive y - axis. The E field of the plane wave can be represented as (E0 is constant) 3p p p 3p t 0 e j c wt - l x - l z m t 0 e jc wt - l x - l z m (B) E = yE (A) E = yE

(B) (C) (D)

(D) > 43 -4

jc wt +

3 p x+ p z l l m

t 0e (D) E = yE

O N

A I D

jc wt - p x + 3 p z m l l

1

- 43 1 4

H

##

#S H $ d l = ##S c j + 22Dt m $ dS

od

8.44

##S H $ dS = #C c j + 22Dt m $ d t

#C H $ d l # = ##S c j + 22Dt m $ ds

2007

TWO MARKS

8.45

The E field in a rectangular waveguide of inner dimension a # b is given by 2 wm E = 2 ` l j H0 sin ` 2px j sin (wt - bz) yt a h 2 Where H0 is a constant, and a and b are the dimensions along the x - axis and the y - axis respectively. The mode of propagation in the waveguide is (A) TE20 (B) TM11 (C) TM20 (D) TE10 A load of 50 W is connected in shunt in a 2-wire transmission line of Z0 = 50W as shown in the figure. The 2-port scattering parameter matrix (s-matrix) of the shunt element is

8.46

(B) - 5 7 (D) 5 7

The nH field (in A/m) of a plane wave propagating in free space is o.i c . ia given by H = xt 5 3 cos (wt - bz) + yt wt - bz + p .

8.43

c

8.41

2 -1 3 (C) > 23 1H 3 3

(A) - j 7 5 (C) j 5 7

If C is code curve enclosing a surface S , then magnetic field intensityw.n H , the current density j and the electric flux density D are related ww by 2D (A) H $ ds = c j + 2t m $ d t S c

##

8.40

0 1 (B) = 1 0G

At 20 GHz, the gain of a parabolic dish antenna of 1 meter and 70% efficiency is (A) 15 dB (B) 25 dB (C) 35 dB (D) 45 dB

t 0e (C) E = yE 8.39

1 -1 2 (A) > 12 1H 2 2

The parallel branches of a 2-wirw transmission line re terminated in 100W and 200W resistors as shown in the figure. The characteristic impedance of the line is Z0 = 50W and each section has a length of l . The voltage reflection coefficient G at the input is 4

8.42

In the design of a single mode step index optical fibre close to upper cut-off, the single-mode operation is not preserved if (A) radius as well as operating wavelength are halved (B) radius as well as operating wavelength are doubled (C) radius is halved and operating wavelength is doubled (D) radius is doubled and operating wavelength is halved

2007 8.38

Page 185

` h0 2j The time average power flow density in Watts is h (A) 0 (B) 100 100 h0 (C) 50h20 (D) 50 h0

An air-filled rectangular waveguide has inner 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) 308 W (B) 355 W (C) 400 W (D) 461 W A l dipole is kept horizontally at a height of l0 above a perfectly 2 2 conducting infinite ground plane. The radiation pattern in the lane of the dipole (E plane) looks approximately as

A right circularly polarized (RCP) plane wave is incident at an angle 60c to the normal, on an air-dielectric interface. If the reflected wave is linearly polarized, the relative dielectric constant xr2 is.

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

Page 186

incident normally on the interface from region I . The electric file E2 in region II at the interface is (A) E2 = E1 (B) 4atx + 0.75aty - 1.25atz (C) 3atx + 3aty + 5atz (D) - 3atx + 3aty + 5atz (A) 2 (C) 2 2006 8.47

8.48

8.50

8.51

8.52

ONE MARK

The electric field of an electromagnetic wave propagation in the positive direction is given by E = atx sin (wt - bz) + aty sin (wt - bz + p/2). The wave is (A) Linearly polarized in the z -direction (B) Elliptically polarized (C) Left-hand circularly polarized (D) Right-hand circularly polarized A transmission line is feeding 1 watt of power to a horn antenna having a gain of 10 dB. The antenna is matched to the transmission line. The total power radiated by the horn antenna into the free space is (A) 10 Watts (B) 1 Watts (C) 0.1 Watts (D) 0.01 Watt 2006

8.49

8.53

(B) 3 (D) 3

A mast antenna consisting of a 50 meter long vertical conductor operates over a perfectly conducting ground plane. It is base-fed at a frequency of 600 kHz. The radiation resistance of the antenna is Ohms is 2 2 (A) 2p (B) p 5 5 2 (C) 4p 5

(D) 20p2

2005 8.54

The magnetic field intensity vector of a plane wave is given by H (x, y, z, t) = 10 sin (50000t + 0.004x + 30) aty where aty , denotes the unit vector in y direction. The wave is propagating with a phase velocity. (A) 5 # 10 4 m/s (B) - 3 # 108 m/s (C) - 1.25 # 107 m/s (D) 3 # 108 m/s

8.55

Refractive index of glass is 1.5. Find the wavelength of a beam of light with frequency of 1014 Hz in glass. Assume velocity of light is 3 # 108 m/s in vacuum (A) 3 mm (B) 3 mm (C) 2 mm (D) 1 mm

A I D

TWO MARKS

ONE MARK

When a planes wave traveling in free-space is incident normally on a medium having the fraction of power transmitted into the medium 2005 TWO MARKS n i is given by . 8.56 coWhich one of the following does represent the electric field lines . 8 1 a (A) (B) i d for the mode in the cross-section of a hollow rectangular metallic 9 2 .no w waveguide ? w (C) 1 (D) 5 w 3 6 A medium of relative permittivity er2 = 2 forms an interface with free - space. A point source of electromagnetic energy is located in the medium at a depth of 1 meter from the interface. Due to the total internal reflection, the transmitted beam has a circular crosssection over the interface. The area of the beam cross-section at the interface is given by (A) 2p m 2 (B) p2 m 2 (C) p m 2 (D) p m 2 2 8.57 Characteristic impedance of a transmission line is 50 W. Input impedance of the open-circuited line when the transmission line a A rectangular wave guide having TE10 mode as dominant mode is short circuited, then value of the input impedance will be. having a cut off frequency 18 GHz for the mode TE30 . The inner (A) 50 W (B) 100 + j150W broad - wall dimension of the rectangular wave guide is (C) 7.69 + j11.54W (D) 7.69 - j11.54W (A) 5 cm (B) 5 cm 3 8.58 Two identical and parallel dipole antennas are kept apart by a (C) 5 cm (D) 10 cm 2 distance of l in the H - plane. They are fed with equal currents 4 but the right most antenna has a phase shift of + 90c. The radiation A medium is divide into regions I and II about x = 0 plane, as pattern is given as. shown in the figure below.

O N

An electromagnetic wave with electric field E1 = 4atx + 3aty + 5atz is

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Page 187

2004 8.64

8.65

TWO MARKS

A parallel plate air-filled capacitor has plate area of 10 - 4 m 2 and plate separation of 10 - 3 m. It is connect - ed to a 0.5 V, 3.6 GHz source. The magnitude of the displacement current is ( e = 361p 10 - 9 F/m) (A) 10 mA (B) 100 mA (C) 10 A (D) 1.59 mA Consider a 300 W, quarter - wave long (at 1 GHz) transmission line as shown in Fig. It is connected to a 10 V, 50 W source at one end and is left open circuited at the other end. The magnitude of the voltage at the open circuit end of the line is

Statement of Linked Answer Questions 9.46 & 9.47 : Voltage standing wave pattern in a lossless transmission line with characteristic impedance 50 and a resistive load is shown in the figure. (A) 10 V (C) 60 V 8.66

8.59

8.60

8.61

O N

If E = (atx + jaty) e jkz - kwt and H = (k/wm) (aty + katx ) e jkz - jwt , the timeaveraged Poynting vector is n (A) (B) (k/wm) atz o.i null vector c . (D) (k/2wm) atz dia (C) (2k/wm) atz

8.67

no

Many circles are drawn in a Smith Chart used for transmission linew. 8.68 ww calculations. The circles shown in the figure represent

(A) Unit circles (B) Constant resistance circles (C) Constant reactance circles (D) Constant reflection coefficient circles. 2004 8.62

8.63

In a microwave test bench, why is the microwave signal amplitude modulated at 1 kHz (A) To increase the sensitivity of measurement (B) To transmit the signal to a far-off place (C) To study amplitude modulations (D) Because crystal detector fails at microwave frequencies

A I D

The value of the load resistance is (A) 50 W (B) 200 W (C) 12.5 W (D) 0 The reflection coefficient is given by (A) - 0.6 (B) - 1 (C) 0.6 (D) 0

(B) 5 V (D) 60/7 V

(A) adding an inductance in series with Z (B) adding a capacitance in series with Z (C) adding an inductance in shunt across Z (D) adding a capacitance in shunt across Z

ONE MARK

The phase velocity of an electromagnetic wave propagating in a hollow metallic rectangular waveguide in the TE10 mode is (A) equal to its group velocity (B) less than the velocity of light in free space (C) equal to the velocity of light in free space (D) greater than the velocity of light in free space Consider a lossless antenna with a directive gain of + 6 dB. If 1 mW of power is fed to it the total power radiated by the antenna will be (A) 4 mW (B) 1 mW (C) 7 mW (D) 1/4 mW

Consider an impedance Z = R + jX marked with point P in an impedance Smith chart as shown in Fig. The movement from point P along a constant resistance circle in the clockwise direction by an angle 45c is equivalent to

8.69

8.70

A plane electromagnetic wave propagating in free space is incident normally on a large slab of loss-less, non-magnetic, dielectric material with e > e0 . Maxima and minima are observed when the electric field is measured in front of the slab. The maximum electric field is found to be 5 times the minimum field. The intrinsic impedance of the medium should be (A) 120p W (B) 60p W (C) 600p W (D) 24p W A lossless transmission line is terminated in a load which reflects a part of the incident power. The measured VSWR is 2. The percentage

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Page 188

(A) (0.01 - j0.02) mho (C) (0.04 - j0.02) mho

of the power that is reflected back is (A) 57.73 (B) 33.33 (C) 0.11 (D) 11.11 8.78

2003 8.71

8.72

The unit of 4# H is (A) Ampere (C) Ampere/meter 2

8.74

8.75

(B) Ampere/meter (D) Ampere-meter

8.79

The depth of penetration of electromagnetic wave in a medium having conductivity s at a frequency of 1 MHz is 25 cm. The depth of penetration at a frequency of 4 MHz will be (A) 6.25 dm (B) 12.50 cm (C) 50.00 cm (D) 100.00 cm 2003

8.73

ONE MARK

8.77

A rectangular metal wave guide filled with a dielectric material of relative permittivity er = 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 Two identical antennas are placed in the q = p/2 plane as shown in Fig. The elements have equal amplitude excitation with 180c polarity difference, operating at wavelength l. The correct value of the magnitude of the far-zone resultant electric field strength normalized with that of a single element, both computed for f = 0 , is

TWO MARKS

Medium 1 has the electrical permittivity e1 = 1.5e0 farad/m and occupies the region to the left of x = 0 plane. Medium 2 has the electrical permittivity e2 = 2.5e0 farad/m and occupies the region to the right of x = 0 plane. If E1 in medium 1 is E1 = (2ux - 3uy + 1uz ) volt/m, then E2 in medium 2 is (A) (2.0ux - 7.5uy + 2.5uz ) volt/m (B) (2.0ux - 2.0uy + 0.6uz ) volt/m (C) (2.0ux - 3.0uy + 1.0uz ) volt/m (D) (2.0ux - 2.0uy + 0.6uz ) volt/m

(A) 2 cos b 2ps l l (C) 2 cos a ps k l

(B) 2 sin b 2ps l l (D) 2 sin a ps k l

A I D

If the electric field intensity is given by E = (xux + yuy + zuz ) volt/m, the potential difference between X (2, 0, 0) and Y (1, 2, 3) is (A) + 1 volt (B) - 1 volt (C) + 5 volt (D) + 6 volt

O N

2002

8.80

no w.

The VSWR can have any value between (A) 0 and 1 (B) - 1 and + 1 (C) 0 and 3 (D) 1 and 3

n oIn.i in impedance Smith movement along a constant resistance circle c . ia

d

w A uniform plane wave traveling in air is incident on the w plane boundary between air and another dielectric medium with er = 4 . The reflection coefficient for the normal incidence, is (A) zero (B) 0.5+180c (B) 0.333+0c (D) 0.333+180c If the electric field intensity associated with a uniform plane electromagnetic wave traveling in a perfect dielectric medium is given by E (z, t) = 10 cos (2p107 t - 0.1pz) V/m, then the velocity of the traveling wave is (A) 3.00 # 108 m/sec (B) 2.00 # 108 m/sec (C) 6.28 # 107 m/sec (D) 2.00 # 107 m/sec

A short - circuited stub is shunt connected to a transmission line as shown in fig. If Z0 = 50 ohm, the admittance Y seen at the junction of the stub and the transmission line is

ONE MARK

8.81

8.82 8.76

(B) (0.02 - j0.01) mho (D) (0.02 + j0) mho

gives rise to (A) a decrease in the value of reactance (B) an increase in the value of reactance (C) no change in the reactance value (D) no change in the impedance

The phase velocity for the TE10 waveguide is (c is the velocity of (A) less than c (C) greater than c

-mode in an air-filled rectangular plane waves in free space) (B) equal to c (D) none of these

2002 8.83

8.84

TWO MARKS

t jp/2) e jwt - jkz . This A plane wave is characterized by E = (0.5xt + ye wave is (A) linearly polarized (B) circularly polarized (C) elliptically polarized (D) unpolarized Distilled water at 25c C is characterized by s = 1.7 # 10 - 4 mho/m and e = 78eo at a frequency of 3 GHz. Its loss tangent tan d is ( e = 10 36p F/m) -9

(A) 1.3 # 10-5 (C) 1.3 # 10-4 /78 8.85

(B) 1.3 # 10-3 (D) 1.3 # 10-5 /78e0

The electric field on the surface of a perfect conductor is 2 V/m. The conductor is immersed in water with e = 80eo . The surface charge density on the conductor is ( e = 10 36p F/m) (A) 0 C/m 2 (B) 2 C/m 2 -9

(C) 1.8 # 10 - 11 C/m 2

(D) 1.41 # 10 - 9 C/m 2

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 8.86

ONE MARK

8.96

(D) RG = LC 2

8.89

8.95

A transmission line is distortonless if (A) RL = 1 (B) RL = GC GC (C) LG = RC

8.88

2000

A person with receiver is 5 Km away from the transmitter. What is the distance that this person must move further to detect a 3-dB decrease in signal strength (A) 942 m (B) 2070 m (C) 4978 m (D) 5320 m 2001

8.87

Page 189

8.97

2

If a plane electromagnetic wave satisfies the equal d E2x = c2 d E2x , dZ dt the wave propagates in the (A) x - direction (B) z - direction (C) y - direction (D) xy plane at an angle of 45c between the x and z direction

8.98

The plane velocity of wave propagating in a hollow metal waveguide is (A) grater than the velocity of light in free space (B) less than the velocity of light in free space (C) equal to the velocity of light free space (D) equal to the velocity of light in free

8.93

8.94

If the diameter of a l dipole antenna is increased from l to l 2 100 50 , then its (B) bandwidth decrease (D) gain decreases TWO MARKS

A uniform plane wave in air impings at 45c angle on a lossless dielectric material with dielectric constant dr . The transmitted wave propagates is a 30c direction with respect to the normal. The value of dr is (A) 1.5 (B) 1.5

A I D (C) 2

(D)

2

A rectangular waveguide has dimensions 1 cm # 0.5 cm. Its cut-off frequency is (A) 5 GHz (B) 10 GHz (C)in15 GHz (D) 12 GHz

The dominant mode in a rectangular waveguide is TE10 , because this mode has (A) the highest cut-off wavelength (B) no cut-off co. . a 8.100 i d Two coaxial cable 1 and 2 are filled with different dielectric constants (C) no magnetic field component o n . er1 and er2 respectively. The ratio of the wavelength in the cables (D) no attenuation ww (l /l ) is 2001

8.92

A TEM wave is incident normally upon a perfect conductor. The E and H field at the boundary will be respectively, (A) minimum and minimum (B) maximum and maximum (C) minimum and maximum (D) maximum and minimum

2000

O N w

8.91

The magnitudes of the open-circuit and short-circuit input impedances of a transmission line are 100 W and 25 W respectively. The characteristic impedance of the line is, (A) 25 W (B) 50 W (C) 75 W (D) 100 W

(A) bandwidth increases (C) gain increases

8.99

8.90

ONE MARK

1

er1 /er2 (C) er1 /er2 (A)

TWO MARKS

A material has conductivity of 10 - 2 mho/m and a relative permittivity of 4. The frequency at which the conduction current in the medium is equal to the displacement current is (A) 45 MHz (B) 90 MHz (C) 450 MHz (D) 900 MHz A uniform plane electromagnetic wave incident on a plane surface of a dielectric material is reflected with a VSWR of 3. What is the percentage of incident power that is reflected ? (A) 10% (B) 25% (C) 50% (D) 75% A medium wave radio transmitter operating at a wavelength of 492 m has a tower antenna of height 124. What is the radiation resistance of the antenna? (A) 25 W (B) 36.5 W (C) 50 W (D) 73 W In uniform linear array, four isotropic radiating elements are spaced l apart. The progressive phase shift between required for forming 4 the main beam at 60c off the end - fire is : (A) - p (B) - p2 radians (C) - p4 radians (D) - p8 radians

2

8.101

(B) er2 /er1 (D) er2 /er1

For an 8 feet (2.4m) parabolic dish antenna operating at 4 GHz, the minimum distance required for far field measurement is closest to (A) 7.5 cm (B) 15 cm (C) 15 m (D) 150 m 1999

8.102

ONE MARK

An electric field on a place is described by its potential V = 20 (r-1 + r-2) where r is the distance from the source. The field is due to (A) a monopole (B) a dipole (C) both a monopole and a dipole (D) a quadruple

8.103

8.104

Assuming perfect conductors of a transmission line, pure TEM propagation is NOT possible in (A) coaxial cable (B) air-filled cylindrical waveguide (C) parallel twin-wire line in air (D) semi-infinite parallel plate wave guide Indicate which one of the following will NOT exist in a rectangular resonant cavity. (B) TE 011 (A) TE110 (C) TM110 (D) TM111

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 8.105

Identify which one of the following will NOT satisfy the wave equation. (A) 50e j (wt - 3z) (B) sin [w (10z + 5t)] 2 (C) cos (y + 5t) (D) sin (x) cos (t) 1999

8.106

8.107

8.108

8.109

8.111

8.112

1998 8.113

TWO MARKS

In a twin-wire transmission line in air, the adjacent voltage maxima are at 12.5 cm and 27.5 cm . The operating frequency is (A) 300 MHz (B) 1 GHz (C) 2 GHz (D) 6.28 GHz A transmitting antenna radiates 251 W isotropically. A receiving antenna, located 100 m away from the transmitting antenna, has an effective aperture of 500 cm2 . The total received by the antenna is (A) 10 mW (B) 1 mW (C) 20 mW (D) 100 mW

8.114

8.115

In air, a lossless transmission line of length 50 cm with L = 10 mH/m , C = 40 pF/m is operated at 25 MHz . Its electrical path length is (A) 0.5 meters (B) l meters (C) p/2 radians (D) 180 deg rees

8.116

A plane wave propagating through a medium [er = 8, vr = 2, and s = 0] t - (z/3) sin (108 t - bz) V/m . The has its electric field given by Ev = 0.5Xe wave impedance, in ohms is (A) 377 (B) 198.5+180c (C) 182.9+14c (D) 133.3

8.117

ONE MARK

The intrinsic impedance of copper at high frequencies is (A) purely resistive (B) purely inductive (C) complex with a capacitive component (D) complex with an inductive component

8.118

The Maxwell equation V # H = J + 2D is based on 2t (A) Ampere’s law (B) Gauss’ law (C) Faraday’s law (D) Coulomb’s law

8.119

8.120

All transmission line sections shown in the figure is have a characteristic impedance R 0 + j 0 . The input impedance Zin equals

8.121

8.122

(A) 2 R 0 3 (C) 3 R 0 2

(B) R 0 (D) 2R 0

The time averages Poynting vector, Ev = 24e j (wt + bz) avy V/m in free space is (B) (A) - 2.4 avz p (C) 4.8 avz (D) p

in W/m2 , for a wave with 2.4 av p z - 4.8 avz p

The wavelength of a wave with propagation constant (0.1p + j0.2p) m-1 is 2 m (B) 10 m (A) 0.05 (C) 20 m (D) 30 m The depth of penetration of wave in a lossy dielectric increases with increasing (A) conductivity (B) permeability (C) wavelength (D) permittivity The polarization of Ev = E 0 e j^wt + bz h ^avx + avy h is (A) linear (C) left hand circular

wave

with

electric

field

vector

(B) elliptical (D) right hand circular

The vector H in the far field of an antenna satisfies (A) d $ Hv = 0 and d # Hv = 0 (B) d $ Hv ! 0 and d # Hv ! 0 (C) d $ Hv = 0 and d # Hv ! 0 (D) d $ Hv ! 0 and d # Hv = 0

The radiation resistance of a circular loop of one turn is 0.01 W. The radiation resistance of five turns of such a loop will be (A) 0.002 W (B) 0.01 W (C) 0.05 W (D) 0.25 W

.in oAn c . antenna in free space receives 2 mW ia

d .no

w

ww

TWO MARKS

A I D

O N

1998 8.110

Page 190

of power when the incident electric field is 20 mV/m rms. The effective aperture of the antenna is (A) 0.005 m2 (B) 0.05 m2 (C) 1.885 m2 (D) 3.77 m2 The maximum usable frequency of an ionospheric layer at 60c incidence and with 8 MHz critical frequency is (A) 16 MHz (B) 16 MHz 3 (C) 8 MHz (D) 6.93 MHz A loop is rotating about they y -axis in a magnetic field Bv = B 0 cos (wt + f) avx T. The voltage in the loop is (A) zero (B) due to rotation only (C) due to transformer action only (D) due to both rotation and transformer action The far field of an antenna varies with distance r as (B) 12 (A) 1 r r (C) 13 (D) 1 r r

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1997 8.123

8.124

8.125

8.127

ONE MARK

8.130

8.132

8.133

TWO MARKS

A very lossy, l/4 long, 50 W transmission line is open circuited at the load end. The input impedance measured at the other end of the line is approximately (A) 0 (B) 50 W (C) 3 (D) None of the above

A metal sphere with 1 m radius and a surface charge density of 10 Coulombs/m2 is enclosed in a cube of 10 m side. The total outward electric displacement normal to the surface of the cube is (A) 40p Coulombs (B) 10p Coulombs (C) 5p Coulombs (D) None of these 1996

A parabolic dish antenna has a conical beam 2c wide, the directivity of the antenna is approximately (A) 20 dB (B) 30 dB (C) 40 dB (D) 50 dB

TWO MARKS

A uniform plane wave in air is normally incident on infinitely thick slab. If the refractive index of the glass slab is 1.5, then the percentage of incident power that is reflected from the air-glass interface is (A) 0% (B) 4% (C) 20% (D) 100% The critical frequency of an ionospheric layer is 10 MHz. What is the maximum launching angle from the horizon for which 20 MHz wave will be reflected by the layer ? (A) 0c (B) 30c (C) 45c (D) 90c

O N

A I D

A 1inkm long microwave link uses two antennas each having 30 dB o. If the power transmitted by one antenna is 1 W at 3 GHz, the cgain. . a di power received by the other antenna is approximately The skin depth at 10 MHz for a conductor is 1 cm. The phase velocity .no (A) 98.6 mW of an electromagnetic wave in the conductor at 1, 000 MHz is about ww w 6 (B) 76.8 mW (A) 6 # 10 m/ sec 7 (C) 63.4 mW (B) 6 # 10 m/ sec 8 (D) 55.2 mW (C) 3 # 10 m/ sec

1996

8.129

8.131

A rectangular air filled waveguide has cross section of 4 cm #10 cm . The minimum frequency which can propagate in the waveguide is (A) 0.75 GHz (B) 2.0 GHz (C) 2.5 GHz (D) 3.0 GHz

(D) 6 # 108 m/ sec

8.128

(C) 25% (D) 0%

A transmission line of 50 W characteristic impedance is terminated with a 100 W resistance. The minimum impedance measured on the line is equal to (A) 0 W (B) 25 W (C) 50 W (D) 100 W

1997 8.126

Page 191

8.134

8.135

ONE MARK

A lossless transmission line having 50 W characteristic impedance and length l/4 is short circuited at one end and connected to an ideal voltage source of 1 V at the other end. The current drawn from the voltage source is (A) 0 (B) 0.02 A (C) 3 (D) none of these

Some unknown material has a conductivity of 106 mho/m and a permeability of 4p # 10-7 H/m . The skin depth for the material at 1 GHz is (A) 15.9 mm (B) 20.9 mm (C) 25.9 mm (D) 30.9 mm

The capacitance per unit length and the characteristic impedance of a lossless transmission line are C and Z 0 respectively. The velocity of a travelling wave on the transmission line is (B) 1 (A) Z 0 C Z0 C (C) Z 0 (D) C Z0 C A transverse electromagnetic wave with circular polarization is received by a dipole antenna. Due to polarization mismatch, the power transfer efficiency from the wave to the antenna is reduced to about (A) 50% (B) 35.3%

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Page 192

SOLUTION

Also, direction of propagation is v v avk = ax + az 2 x + z So, k = 2 Substituting it in equation (1), we get p # 10 ^x + z h Evi = Eo _avx - avz i e-j 3 2 2 4

8.1

Option (D) is correct. v around Stoke’s theorem states that the circulation a vector field A v over a closed path l is equal to the surface integral of the curl of A the open surface S bounded by l . v : dlv = vh : dsv ^d # A i.e., A

Option (A) is correct. We obtain the reflection coefficient for parallel polarized wave (since, electric field is in the plane of wave propagation) as h cos qt - h1 cos qi Gz = 2 h2 cos qt + h1 cos qi ...(1)

8.5

##

#

Here, line integral is taken across a closed path which is denoted by a small circle on the integral notation where as, the surface invh is taken over open surface bounded by the loop. tegral of ^d # A 8.2

As we have already obtained qi = 45c, qt = 19.2c m 1 = h0 Also, h2 = = h0 e 4.5 4.5 m 1 and h1 = = h0 = h0 e 1 Substituting these in eq. (1) we get G z = cos 19.2c - 4.5 cos 45c cos 19.2c + 4.5 cos 45c =- 0.227 .- 0.23 Therefore, the reflected field has the magnitude given by Ero = T 11' Eio or Ero = G z Eio =- 0.23 Eio Hence, the expression of reflected electric field is p # 10 Evr =- 0.23 Eo _- avx - avz i e-j 3 k n i . 2 co(2)

Option (D) is correct. Given, the vector field v = xavx + yavy + zavz A so, v (Divergence of A v ) = 2Ax + 2Ay + 2Az d$A 2x 2y 2z = 1+1+1 = 3

8.3

Option (A) is correct. Given, the return loss of device as 20 dB i.e., G in dB =- 20 dB (loss) ^

or,

20 log G =- 20

& G = 10-1 = 0.1 Therefore, the standing wave ration is given by 1+ G VSWR = 1- G = 1 + 0.1 = 1.1 = 1.22 1 - 0.1 0.9 8.4

A I D

h

O N

.

ia od

n

. ww

w

Option (C) is correct. For the given incidence of plane wave, we have the transmitting angle qt = 19.2c From Snell’s law, we know n1 sin qi = n2 sin qt

Now, the wavelength of EM wave is l = 600 mm So,

b = 2p = p # 10 4 3 l

Again, we have the propagation vector of reflected wave as v v avk = ax - az 2 x z or, k = 2 Substituting it in Eq. (2), we get p # 10 x - z Evr =- 0.23 Eo _- avx - avz i e-j 3 b 2 l 2 jp # 10 ^x - z h V m Evr = 0.23 Eo _avx + avz i e- 3 2 2 4

4

c m1 e1 sin qi = c m2 e2 sin qt ...(1) For the given interfaces, we have m1 = m2 = 1 e1 = 1, e2 = 4.5 So, from Eq. (1) sin qi = 4.5 sin 19.2 or, qi . 45c v Now, the component of Ei can be obtained as Evi = _Eox avx - Eoz avz i e-jbk (observed from the shown figure) Since, the angle qi = 45c so, Eox = Eoz = Eo 2 E v o v Therefore, Ei = _ax - avz i e-jbk 2 ...(1)

4

8.6

Option (C) is correct. Electric field of the propagating wave in free space is given as Ei = (8ax + 6ay + 5az ) e j (wt + 3x - 4y) V/m So, it is clear that wave is propagating in the direction (- 3ax + 4ay) . Since, the wave is incident on a perfectly conducting slab at x = 0 . So, the reflection coefficient will be equal to - 1. i.e. Er = (- 1) Ei =- 8ax - 6ay - 5az Again, the reflected wave will be as shown in figure. 0

0

i.e. the reflected wave will be in direction 3ax + 4ay . Thus, the electric field of the reflected wave will be.

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Page 193

Ex = (- 8ax - 6ay - 5az ) e j (wt - 3x - 4y) V/m 8.7

8.8

8.9

Option (A) is correct. The field in circular polarization is found to be Es = E 0 (ay ! jaz ) e-jbx propagating in + ve x -direction. where, plus sign is used for left circular polarization and minus sign for right circular polarization. So, the given problem has left circular polarization. b = 25 = w c 8 2pf & f = 25 # c = 25 # 3 # 10 = 1.2 GHz 25 = c 2p 2 # 3.14

Substituting above values, b = - j261

8.12

Option ( ) is correct. For r > a , Ienclosed = (pa2) J

# H : dl

For r < a , So,

Fmax = 1 Favg = 1 F (q, f) dW 4p 2p 2p F (q, f) sin qdq dfE = 1 ; 4p 0 0

#

A I D

# # 0

p/2

cos 4 q sin qdqdfE

= Ienclosed

H # 2pr = (pa2) J H = Io 2pr H \ 1 , for r > a r

Option (A) is correct. The directivity is defined as D = Fmax Favg

# #

6.283 # 1010 2 - (2.0942 + 2.6182) 10 4 # c 3 108 m #

b is imaginary so mode of operation is non-propagating. vp = 0

Option (B) is correct. Let b " outer diameter a " inner diameter Characteristic impedance, m0 4p # 10-7 # 36p ln 2.4 = 100 W Z0 = ln b b l = b 1 l a e0 er 10-9 # 10.89

2p = 1 ; 4p 0

w 2 - (b 2 + b 2) x y c2

b =

8.13

O N

Ienclosed =

Io = (pa2) J

J (pr 2) Jr 2 = 2 a pa 2

# H : dl

= Ienclosed 2 H # 2pr = Jr2 a H = Jr 2 2pa H \ r , for r < a

Option (B) is correct. Assuming the cross section of the wire on x -y plane as shown in figure.

in

p/2 5 co. . = 1 ;2p b- cos q lE = 1 # 2p :- 0 + 1 D a 5 5 4p 4p di 0 o n w. = 1 # 2p = 1 w 5 10 4p w D = 1 = 10 10

or, 8.10

D (in dB) = 10 log 10 = 10 dB

Option (C) is correct. Z0 =

Since

Z1 Z 2

100 = 50 # 200 This is quarter wave matching. The length would be odd multiple of l/4 . l = (2m + 1) l 4 c = 3 # 108 = 0.174 m f1 # 4 429 # 106 # 4 8 f2 = 1 GHz , l2 = c = 3 # 10 = 0.075 m 9 f2 # 4 1 # 10 # 4 Only option (C) is odd multiple of both l1 and l2 . (2m + 1) = 1.58 = 9 l1 (2m + 1) = 1.58 - 21 l2 f1 = 429 MHz,

8.11

l1 =

Option (D) is correct. Hz = 3 cos (2.094 # 102 x) cos (2.618 # 102 y) cos (6.283 # 1010 t - bz) bx = 2.094 # 102 by = 2.618 # 102 w = 6.283 # 1010 rad/s For the wave propagation,

Since, the hole is drilled along the length of wire. So, it can be assumed that the drilled portion carriers current density of - J . Now, for the wire without hole, magnetic field intensity at point P will be given as Hf1 (2pR) = J (pR2) Hf1 (2pR) = JR 2 Since, point o is at origin. So, in vector form H1 = J (xax + yay) 2 Again only due to the hole magnetic field intensity will be given as. (Hf2) (2pr) =- J (pr 2) Hf2 = - Jr 2 Again, if we take Ol at origin then in vector form

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H2 = - J (xlax + ylay) 2 where xl and yl denotes point ‘P ’ in the new co-ordinate system. Now the relation between two co-ordinate system will be. x = xl + d y = yl So, H2 = - J [(x - d) ax + yay] 2 So, total magnetic field intensity = H1 + H2 = J dax 2 So, magnetic field inside the hole will depend only on ‘d ’. 8.14

8.15

Page 194

or Now 8.20

Option (C) is correct. We have d = 2 mm and f = 10 GHz Phase difference = 2p d = p ; 4 l

8.17

8.21

= l = 8d = 8 # 2 mm = 16 mm

8.18

8.19

8.22

Now

f = 14 GHz l = C = er f

3 # 108 = 3 9 er 14 # 10 140 er

.in (C) is correct. oOption c . ia

8.23

no w.

d

v = xyatx + x 2 aty A v = dxatx + dyaty dl v = # (xyatx + x 2 aty) : (dxatx + dyaty) # Av : dl

ww

Option (A) is correct. Since voltage maxima is observed at a distance of l/4 from the load and we know that the separation between one maxima and minima equals to l/4 so voltage minima will be observed at the load, Therefore load can not be complex it must be pure resistive. Now G = s-1 s+1 also RL = R 0 (since voltage maxima is formed at the load) s RL = 50 = 10 W 5

again,

Option (C) is correct. Intrinsic impedance of EM wave m m0 = h = = 120p = 60p e 2 4e0 Time average power density 2 1 Pav = 1 EH = 1 E = = 1 2 # 60p 120p 2 2 h

A I D

O N

Option (D) is correct. From the expressions of Ev & Hv , we can write, b = 280 p 2p = 280 p & l = 1 or 140 l v E Wave impedance, Zw = E = p = 120 p 3 v er H

RG a = R = 0.1 = 0.002 50 Z0

So,

Option (A) is correct. From boundary condition

10ut (3utx + 30uty) utx = (1.5utx + Auty + Butz ) xt + v y J =- 30utz =- Autz + Buty + 10uty Comparing we get A = 30 and B =- 10 So H z = 1.5utx + 30uty - 10utz A/m

Option (D) is correct. For distortion less transmission line characteristics impedance Z0 = R G Attenuation constant a =

Option (A) is correct. TM11 is the lowest order mode of all the TMmn modes.

Bn1 = Bn2 m1 Hx1 = m2 Hx2 or Hx2 = Hx1 = 1.5 2 or Hx2 = 1.5utx Further if H z = 1.5utx + Auty + Buz Then from Boundary condition

(not lossless)

S12 = S21 = 0.9 90c (Reciprocal)

v = fl = 10 # 109 # 16 # 10-3 = 1.6 # 108 m/ sec 8.16

Option (C) is correct. For a lossless network S11 2 + S21 2 = 1 For the given scattering matrix S11 = 0.2 0c , S12 = 0.9 90c S21 = 0.9 90c , S22 = 0.1 90c Here, (0.2) 2 + (0.9) 2 ! 1 Reciprocity :

Option (C) is correct. Power radiated from any source is constant.

or

3 = 1 140 140 er er = 9 Ep = 120p = E p = 120p 3 9

or

= # (xydx + x 2 dy)

C

C

C

=

2/ 3

#1/

3

xdx +

1/ 3

#2/

3

3xdx +

#1

3

4 dy + 3

#3

1

1 dy 3

= 1 : 4 - 1 D + 3 :1 - 4 D + 4 [3 - 1] + 1 [1 - 3] = 1 2 3 3 2 3 3 3 3 8.24

Option (A) is correct. In the given problem

Reflection coefficient h - h1 t = 2 = 400p - 120p =- 1 h2 + h 1 2 40p + 120p t is negative So magnetic field component does not change its direction Direction of incident magnetic field atE # atH = atK atZ # atH = aty atH = atx ( + x direction)

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So, reflection magnetic field component Hr = t # 24 cos (3 # 108 + by) atx , y $ 0 h = 1 # 24 cos (3 # 108 + by) atx , y $ 0 2 # 120p 8 b = w = 3 # 108 = 1 vC 3 # 10 1 So, Hr = cos (3 # 108 + y) atx , y $ 0 10p

Page 195

8.25

Option (D) is correct.

8.29

The transmission line are as shown below. Length of all line is l 4

Option (B) is correct. For length of l/4 transmission line Z + jZo tan bl Z in = Zo ; L Zo + jZL tan bl E

2 2 Zi1 = Z01 = 100 = 200W ZL1 50 2 2 Zi2 = Z02 = 100 = 200W ZL2 50

Zo = 30 W, b = 2p , l = l 4 l So, tan bl = tan b 2p : l l = 3 4 l R ZL V S tan bl + jZo W 2 W = Z 0 = 60 W Z in = Zo S S Zo + jZL W ZL S tan bl W T X For length of l/8 transmission line Z + jZo tan bl Z in = Zo ; L Zo + jZL tan bl E ZL = 30 W ,

ZL3 = Zi1 Zi2 = 200W 200W = 100W 2 2 Zi = Z0 = 50 = 25W ZL3 100

Option (C) is correct. We have Bv = B0

8.30

x a - y a y xm x + y2 x2 + y2 To convert in cylindrical substituting x = r cos f and y = r sin f

Zo = 30 W, ZL = 0 (short) tan bl = tan b 2p : l l = 1 8 l

IA

D O

N

w

ww

8.32

Bv = Bv0 af v Bv a Hv = B = 0 f m0 m0 v v J = 4# H = 0

Now

n

.i .co

ia

d .no 8.31

2

...(1)

ax = cos far - sin faf ay = sin far + cos faf

and In (1) we have

Z in = jZo tan bl = 30j Circuit is shown below.

c

constant since H is constant

Option (C) is correct. The beam-width of Hertizian dipole is 180c and its half power beamwidth is 90c. Option (D) is correct. Maxwell equations 4- B = 0

Reflection coefficient 60 + 3j - 60 = t = ZL - Zo = 60 + 3j + 60 ZL + Zo 1+ t VSWR = = 1 + 17 = 1.64 1- t 1 - 17 8.26

8.27

8.28

4$ E = r/E 1 17

Option (D) is correct. Due to 1 A current wire in x - y plane, magnetic field be at origin will be in x direction. Due to 1 A current wire in y - z plane, magnetic field be at origin will be in z direction. Thus x and z component is non-zero at origin.

4# E =- B 4# Ht = D + J For static electric magnetic fields 4$ B = 0 4$ E = r/E 4# E = 0 S 4# H = J 8.33

Option (A) is correct. Rectangular and cylindrical waveguide doesn’t support TEM modes and have cut off frequency. Coaxial cable support TEM wave and doesn’t have cut off frequency. Option (B) is correct. We have V = 4# A By Stokes theorem

# A $ dl

## (4 # A) $ ds

= From (1) and (2) we get

# A $ dl

=

##V $ ds

...(1) ...(2)

Option (A) is correct. Cut-off Frequency is fc = c 2 For TE11 mode,

m 2 n 2 ` a j +`b j

3 # 1010 fc = 2 8.34

1 2 1 2 ` 4 j + ` 3 j = 6.25 GHz

Option (D) is correct. Zin = Zo

ZL + iZo tan (bl) Zo + iZL tan (bl)

For ZL = 0 , Zin = iZo tan (bl) The wavelength is

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8.35

8 l = c = 3 # 109 = 0.1 m or 10 cm f 3 # 10 bl = 2p l = 2p # 1 = p l 10 5 Thus Zin = iZo tan p 5 Thus Zin is inductive because Zo tan p is positive 5 Option (C) is correct. m We have h = e Reflection coefficient h - h1 G= 2 h2 + h1

Page 196

S22

Option (D) is correct. The input impedance is

8.42

2 Zin = Zo ; ZL

mo eo er mo eo er

er = 9

+

mo eo mo eo

= 11+

2 2 Zin2 = Zo2 = 50 = 12.5 ZL2 200

Now er = 1 er 1+

9 9

since

ZL = Zin1 Zin2 25 12.5 = 25 3 (50) 2 = 300 25/3 G = ZS - Zo = 300 - 50 = 5 ZS + Zo 300 + 50 7

Zs = =- 0.5

8.36

Option (C) is correct. In single mode optical fibre, the frequency of limiting mode increases as radius decreases Hence r \ 1 f

8.38

l = c = 3 # 10 9 = 3 f 200 20 # 10 2 2 Gp = hp2 ` D j = 0.7 # p2 c 13 m = 30705.4 l 100

Option (A) is correct. g = b cos 30cx ! b sin 30cy = 2p 3 x ! 2p 1 y l 2 l 2

O N

8.39

fc = c 2

n o.i

.c Since the mode is TE

ia od

.n

w

ww

8.45

= p 3 x! py l l E = ay E0 e j (wt - g) = ay E0 e j;wt - c

H

= 2 and n = 0 8 fc = c m = 3 # 10 # 2 = 10 GHz 2 2 2 # 0.03 ho 377 = h' = = 400W 10 2 fc 2 10 1-c m 1-c m f 3 # 1010 Option (B) is correct. Using the method of images, the configuration is as shown below 20, m

p 3 x! p y l l mE

Maxwell Equations

## 4# H $ ds = ## `J + 22Dt j .ds

Integral form

## `J + 22Dt j .ds

Stokes Theorem

Here d = l, a = p, thus bd = 2p

s

s

# H $ dl

=

bd cos y + a E 2 2p cos y + p = cos ; E = sin (p cos y) 2 = cos ;

Array factor is

s

Option (A) is correct. 2 wm p H sin ` 2px j sin (wt - bz) yt 2 `2j 0 a h This is TE mode and we know that mpy Ey \ sin ` mpx j cos ` a b j

E =

8.46

Option (D) is correct. The Brewster angle is tan qn =

Thus m = 2 and n = 0 and mode is TE20 8.41

m 2 n 2 ` a j +`b j

Option (D) is correct. 4# H = J + 2D 2t

8.40

2

Option (C) is correct. The cut-off frequency is

8.44

8

= 44.87 dB

2 2 2 = Hx2 + Hy2 = c 5 3 m + c 5 m = c 10 m ho ho ho 2 2 ho H E 2 h = = o c 10 m = 50 P = 2 ho ho 2ho 2

We have

A I D

Option (D) is correct.

Gain

Option (D) is correct.

8.43

For free space watts

So. if radius is doubled, the frequency of propagating mode gets halved, while in option (D) it is increased by two times. 8.37

if l = l 4

2 2 Zin1 = Zo1 = 50 = 25 ZL1 100

Substituting values for h1 and h2 we have t =

2 (ZL Zo) 2 (50 50) = =2 (ZL Zo) + Zo (50 50) + 50 3 (Z Z ) - Zo (50 50) - 50 = =- 1 = L o (ZL Zo) + Zo (50 50) + 50 3

S12 = S21 =

Option (C) is correct. The 2-port scattering parameter matrix is S11 S12 S == S21 S22 G (Z Z ) - Zo (50 50) - 50 S11 = L 0 = =- 1 (ZL Z0) + Zo (50 50) + 50 3

tan 60c = or 8.47

er2 er1 er2 1

er2 = 3

Option (C) is correct. We have E = atxx sin (wt - bz) + aty sin (wt - bz + p/2) Here Ex = Ey and fx = 0, fy = p2

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Phase difference is 8.48

p 2

, thus wave is left hand circularly polarized.

2 2 4p2 W 50 Rrad = 80p2 ` dl j = 80p2 c = 5 l 0.5 # 103 m

Option (C) is correct.

8.54

Option (C) is correct. Refractive index of glass m = 1.5 Frequency f = 1014 Hz

8.55

Option (A) is correct. = 1 + er = 1 - 4 =- 1 3 1 + er 1+ 4 The transmitted power is

h2 - h1 = h2 + h1

mo eo er mo eo er

+

mo eo mo eo

c = 3 # 108 m/sec 8 l = c = 3 # 10 = 3 # 10 - 6 14 f 10 wavelength in glass is

Pt = (1 - G2) Pi = 1 - 1 = 8 9 9 Pt = 8 Pi 9

or

-6 lg = a = 3 # 10 = 2 # 10 - 6 m 1.5 m

Option (D) is correct.

8.56

Option (D) is correct.

8.57

Option (D) is correct.

sin q = 1 = 1 er 2 q = 45c = p 4

or

ZZC

A I D

The configuration is shown below. Here A is point source.

8.51

AO = 1 m BO = 1 m = pr2 = p # OB = p m 2

Option (C) is correct. The cut-off frequency is fc = c 2

O N w

Thus

or 8.52

A = cos b

bd sin q + a l 2

= 2p , d = l and a = 90c l 4 2p l sin q + A = cos c l 4 2

p 2

The option (A) satisfy this equation. 8.59

m 2 m 2 ` a j +` b j

Since the mode is TE30 , m = 3 and n = 0 fc = c m 2 a or

n o.i c . ia Here b

od

n

. ww

Zo2 = ZOC .ZSC 2 = Zo = 50 # 50 = 50 ZOC 100 + j150 2 + 3j 50 (2 - 3j) = 7.69 - 11.54j = 13

Option (A) is correct. The array factor is

8.58

Now From geometry Thus area

w = 50, 000 and b =- 0.004 4 vP = w = 5 # 10 - 3 = 1.25 # 107 m/s b - 4 # 10

Phase Velocity is

Prad = 10 Watts

G=

8.50

Option (C) is correct. Since antenna is installed at conducting ground,

8.53

Option (A) is correct. We have 10 log G = 10 dB or G = 10 Now gain G = Prad Pin or 10 = Prad 1W or

8.49

Page 197

p p m = cos ` sin q + j 4 2

Option (C) is correct. From the diagram, VSWR is s = Vmax = 4 = 4 Vmin 1 When minima is at load ZO = s.ZL or ZL = Zo = 50 = 12.5W s 4

8 18 # 109 = 3 # 10 3 2 a a = 1 m = 5 cm 40 2

8.60

Option (C) is correct. We have E1 = 4ux + 3uy + 5uz Since for dielectric material at the boundary, tangential component of electric field are equal E21 = E1t = 3aty + 5atz at the boundary, normal component of displacement vector are equal i.e. Dn2 = Dn1 or e2 E2n = e1 E1n or 4eo E2n = 3eo 4atz or E2n = 3atx Thus E2 = E2t + E2a = 3atx + 3aty + 5atz

8.61

8.62

8.63

Option (A) is correct. The reflection coefficient is G = ZL - ZO = 12.5 - 50 =- 0.6 ZL + ZO 125. + 50 Option (C) is correct. The given figure represent constant reactance circle. Option (D) is correct. We know that vp > c > vg . Option (A) is correct. We have

GD (q, f) =

4pU (q, f) Prad

For lossless antenna Prad Here we have Prad and 10 log GD (q, f) or GD (q, f)

= Pin = Pin = 1 mW = 6 dB = 3.98

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Page 198

Thus the total power radiated by antenna is 4pU (q, f) = Prad GD (q, f) = 1 m # 3.98 = 3.98 mW 8.64

i.e. 11.11% of incident power is reflected. 8.71

Option (D) is correct. The capacitance is - 12 -4 C = eo A = 8.85 # 10 - 3 # 10 = 8.85 # 10 - 13 d 10

4# H = 2D + J 2t Thus 4# H has unit of current density J that is A/m2

The charge on capacitor is Q = CV = 8.85 # 10 - 13 = 4.427 # 10 - 13 Displacement current in one cycle Q I = = fQ = 4.427 # 10 - 13 # 3.6 # 109 = 1.59 mA T 8.65

8.72

Option (D) is correct.

8.67

Option (A) is correct.

8.68

E # H* = (atx + jaty) e jkz - jwt # k (- jatx + aty) e-jkz + jwt wm = atz ; k - (- j) (j) k E = 0 wm wm Ravg = 1 Re [E # H*] = 0 2

f1 f2 1 4 1 # 25 = 12.5 cm 4

d2 = d1 d2 = 25 d2 =

or 8.73

Option (A) is correct. Suppose at point P impedance is

d \ 1 f

Thus

Ravg = 1 Re [E # H*] 2

Thus

Option (B) is correct. We know that

Option (C) is correct. VL = ZO Zin Vin or VL = ZO Vin = 10 # 300 = 60 V Zin 50

8.66

Option (C) is correct. By Maxwells equations

Option (C) is correct. We have E1 = 2ux - 3uy + 1uz E1t = - 3uy + uy and E1n = 2ux Since for dielectric material at the boundary, tangential component of electric field are equal (x = 0 plane) E1t =- 3uy + uy = E2t E1n = 2ux At the boundary the for normal component of electric field are

A I D or or

O N

D1n = D2n e1 E1n = e2 E2n 1.5eo 2ux = 2.5eo E2n E2n = 3 ux = 1.2ux 2.5

n Z = r + j (- 1) oor.i c . If we move in constant resistance circle from point P in clockwise dia o Thus E2 = E2t + E2n =- 3uy + uz + 1.2ux .n direction by an angle 45c, the reactance magnitude increase. Let us w w Option (C) is correct. consider a point Q at 45c from point P in clockwise direction.wIt’s 8.74 We have E = xux + yuy + zuz impedance is dl = utx dx + uty dy + utz dz Z1 = r - 0.5j or Z1 = Z + 0.5j Thus movement on constant r - circle by an +45c in CW direction is the addition of inductance in series with Z . 8.69

=

Option (D) is correct. We have or Thus Now or

or Thus or

2

xdxutx +

=-= x 2

8.75

0

0

2 0 y2 +z G 2 2 2 3

Option (D) is correct. h =

m e

Reflection coefficient t =

h2 = 24p

h2 - h1 h2 + h1

Substituting values for h1 and h2 we have t =

1- G 2= 1+ G G =1 3 Pref = G2= 1 9 Pinc Pref = Pinc 9

1

+

0

#2 ydyutz + #3 zdzuzt

=- 1 [22 - 12 + 02 - 22 + 02 - 32] = 5 2

Option (D) is correct. The VSWR

#1

Y

#XE.dl 2 2

1- G VSWR = Emax = 5 = Emin 1+ G G =2 3 G =- 2 3 h - h1 G= 2 h2 + h1 h - 120p -2 = 2 3 h2 + 120p

or 8.70

VXY =-

er = 4

mo eo er mo eo er

+

m0 eo mo eo

= 11+

er = 1 er 1+

= - 1 = 0.333+180c 3 8.76

Option (B) is correct. We have E (z, t) = 10 cos (2p # 107 t - 0.1pz)

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4 4

since

GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia

w = 2p # 107 t b = 0.1p 7 u = w = 2p # 10 = 2 # 108 m/s b 0.1p

where Phase Velocity 8.77

Page 199

constant resistance circle. Option (C) is correct. Phase velocity

8.82

VC

VP =

Option (A) is correct. The fig of transmission line is as shown below. [Z + jZo tan bl] We know that Zin = Zo L [Zo + jZL tan bl] For line 1, l = l and b = 2p , ZL1 = 100W l 2 [Z + jZo tan p] Thus Zin1 = Zo L = ZL = 100W [Zo + jZL tan p] For line 2, l = l and b = 2p , ZL2 = 0 (short circuit) 8 l [0 + jZo tan p4 ] Thus Zin2 = Zo = jZo = j50W [Zo + 0] Y = 1 + 1 = 1 + 1 = 0.01 - j0.02 Zin1 Zin2 100 j50

f 2 1-c c m f

When wave propagate in waveguide fc < f $ VP > VC Option (C) is correct.

8.83

We have

p 2

t j ) e j (wt - kz) E = (0.5xt + ye Ex = 0.5e j (wt - kz) p

Ey = e j 2 e j (wt - kz) Ex p = 0.5e- 2 Ey Since

Ex ! 1, it is elliptically polarized. Ey

Option (A) is correct.

8.84

1.7 # 10 - 4 tan a = s = we 2p # 3 # 109 # 78eo

Loss tangent

-4 9 = 1.7 # 10 9# 9 # 10 = 1.3 # 10 - 5 3 # 10 # 39

Option (D) is correct. The flux density is

8.85

D O 8.86

8.78

N

Option (A) is correct.

no w.

Option (D) is correct. Normalized array factor = 2 cos

Now

d f d y 2 cos 2

= 2 cos ;

8.88

s = 1+G 1-G

bd sin q cos f + d E 2

8.89

where G varies from 0 to 1 8.90

Thus s varies from 1 to 3. 8.81

Option (C) is correct. A transmission line is distortion less if LG = RC Option (B) is correct. d2 Ex = c2 d2 Ex dz2 dt2 This equation shows that x component of electric fields Ex is traveling in z direction because there is change in z direction.

Option (D) is correct. VSWR

P \ 12 r P1 = r22 Thus P2 r12 3 dB decrease $ Strength is halved P1 = 2 Thus P2

We have

= 2 cos 8 2p 2 s cos 45c + 180 B l. 2 2 s p p = 2 cos 8 + 90cB = 2 sin ` s j l l 8.80

.in

2 2 = r22 5 or r2 = 5 2 kM = 7071 m Distance to move = 7071 - 5000 = 2071 m 8.87

= 2 s, = 45c, = 180c

Option (B) is correct.

Substituting values we have

y 2

y = bd sin q cos f + d q = 90c,

s = 1.41 # 10 - 9 C/m 2

co ia.

d

8 u = c = 3 # 10 = 1.5 # 108 ww 2 e0 In rectangular waveguide the dominant mode is TE10 and fC = v ` m j2 + ` n j2 2 a b 8 1 2 + 0 2 = 1.5 # 108 = 2.5 GHz = 1.5 # 10 ` 2 0.03 j ` b j 0.06 8.79

IA or

s = eE = e0 er E = 80 # 8.854 # 10 - 12 # 2

Option (B) is correct. Reactance increases if we move along clockwise direction in the

8.91

Option (A) is correct. In wave guide vp > c > vg and in vacuum vp = c = vg where vp $ Phase velocity c $ Velocity of light vg $ Group velocity Option (A) is correct. In a wave guide dominant gives lowest cut-off frequency and hence the highest cut-off wavelength. Option (A) is correct.

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or or

8.92

Ic = Id sE = jw d E s = 2pfeo er

Page 200 8.99

or

f = 45 # 106 = 45 MHz

Option (B) is correct.

or or Now

8.100

8.96

8.97

l = 492 m 8.101

e2 e1

Option (D) is correct. 2 l ` 2 jd = l

8 l = c = 3 # 10 9 = 3 m f 40 4 # 10

Option (C) is correct. The array factor is y = bd cos q + d d =l 4

A I D

Distance between elements

or

2 3 ` 40 # 2 j d = (2.4)

d =

80 # (2.4) 2 . 150 m 3

Because of end fire 8.102 Option (C) is correct. y =0 q = 60c We know that for a monopole its electric field varies inversely with p 1 2 p l r.2inwhile its potential varies inversely with r . Similarly for a dipole Thus 0= # cos 60c + d = # + d o c 2 2 l 4 ia. its electric field varies inversely as r 3 and potential varies inversely d no as r 2 . or d =- p . w 4 In the given expression both the terms a _ r1 + r1 i are present, so ww Option (B) is correct. this potential is due to both monopole & dipole. Zo = ZOC .ZSC = 100 # 25 = 10 # 5 = 50W 8.103 Option (D) is correct.

O N

-1

Option (B) is correct.

8.104

1 (Diameter)

8.105

As diameter increases Bandwidth decreases. Option (C) is correct. The fig is as shown below :

As per snell law sin qt = 1 sin qi er sin 30c = 1 or sin 45c er 1 2 = 1 1 er 2 or er = 2

-2

In TE mode Ez = 0 , at all points within the wave guide. It implies that electric field vector is always perpendicular to the waveguide axis. This is not possible in semi-infine parallel plate wave guide.

Option (C) is correct. As the impedance of perfect conductor is zero, electric field is minimum and magnetic field is maximum at the boundary. BW \

8.98

l1 = l2

we get

25 W.

8.95

l \ 1 e

Thus

= 124 m . l 4 It is a quarter wave monopole antenna and radiation resistance is

where

b = 2p = w me l l = 2p w me

or

and height of antenna

8.94

For air

Option (B) is correct. Phase Velocity

G = 0.5 Pr = G2 = 0.25 Pi

Option (A) is correct. We have

m 2 n 2 ` a j +`b j

= 15 GHz

VSWR = 1 + G 1-G 3 = 1+G 1-G

Thus 25% of incident power is reflected. 8.93

vp 2

For rectangular waveguide dominant mode is TE01 8 v Thus fc = p = 3 # 10- 2 = 15 # 109 2a 2 # 10 vp = 3 # 108

-2 9 s = 2s = 9 # 10 # 2 # 10 2p # eo er 4peo er 4

f =

fc =

Cutoff frequency

w = 2pf and e = er e0

or

Option (C) is correct.

Option (A) is correct. Option (C) is correct. A scalar wave equation must satisfy following relation 2 2 E - m 22 2 E = 0 ...(1) 2t 2 2z 2 Where m = w (Velocity) b Basically w is the multiply factor of t and b is multiply factor of z or x or y . In option (A) E = 50e j (wt - 3z) m =w=w 3 b We can see that equations in option (C) does not satisfy equation (1)

8.106

Option (B) is correct. We know that distance between two adjacent voltage maxima is equal to l/2 , where l is wavelength. l = 27.5 - 12.5 2 l = 2 # 15 = 30 cm

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Frequency 8.107

8.108

Page 201

10 u = C = 3 # 10 = 1 GHz 30 l

d\ 1 \ l f so depth increases with increasing in wavelength.

Option (D) is correct. Power received by antenna -4 PR = PT 2 # (apeture) = 251 # 500 # 102 = 100 mW 4p r 4 # p # (100) Option (C) is correct. Electrical path length = bl Where b = 2p , l = 50 cm l We know that l =u =1# 1 au= f f LC 1 1 = # 25 # 106 10 # 10-6 # 40 # 10-12

Option (A) is correct. Given j (wt + bz) v a x + e0 e j (wt + bz) avy E (z, t) = Eo e

8.116

Generalizing ...(2) E (z) = avx E1 (z) + avy E2 (z) Comparing (1) and (2) we can see that E1 (z) and E2 (z) are in space quadrature but in time phase, their sum E will be linearly polarized along a line that makes an angle f with x -axis as shown below.

1 LC

Option (C) is correct. v v Hv = 1 4 A m # v is auxiliary potential function. where A So 4: H = 4: (4 # A) = 0

8.117

7 = 5 # 10 6 = 2 m 25 # 10 Electric path length = 2p # 50 # 10-2 = p radian 2 5 8.109

8.110

A I D

Option (C) is correct. We have

8.119

O N

#l H $ dl

= I enclosed

#l H $ dl

=

Power Re ceived Polynting vector of incident wave A =W P =

n

. ww

w

2 P = E h0 = 120p is intrinsic impedance h0

of space So

(ampere's law)

A -6

= 2 # 10 E2 c h0 m

#s Jds

Applying curl theorem = # Jds s 4# H = J then it is modified to 4# H = J + 2D 2t

n Aperture Area o.i

.c

ia od

Option (A) is correct. This equation is based on ampere’s law as we can see

or

Option (D) is correct. Radiation resistance of a circular loop is given as Rr = 8 hp3 :ND2 S D 3 l 2 Rx \ N N " no. of turns So, Rr 2 = N 2 # Rr 1 = (5) 2 # 0.01 = 0.25 W

8.118

Option (D) is correct. Impedance is written as jwm s + jwe Copper is good conductor i.e. s >> we jwm wm So = 45c h = s s Impedance will be complex with an inductive component.

8.111

4# H = 4# (4 # A) = Y 0

Option (D) is correct. In a lossless dielectric (s = 0) median, impedance is given by m m0 mr mr h = 0c = = 120p # e er e0 er = 120p # 2 = 188.4 W 8

h =

=

-6

2 # 10 # 120 # 3.14 (20 # 10-3) 2 -6 3.14 = 1.884 m2 = 2 # 10 # 12 -# 6 400 # 10

#s (4 # H) $ ds

8.112

Option (A) is correct.

8.113

Option (B) is correct.

8.114

8.115

8.120

Based on continuity equation

Option (B) is correct. Propagation constant here

...(1)

r = a + ib = 0.1p + j0.2p b = 2p = 0.2p l l = 2 = 10 m 0.2

Option (C) is correct. The depth of penetration or skin depth is defined as – 1 d= pfms

8.121

8.122

Option (B) is correct. Maximum usable frequency fo fm = sin Ae fm = 8MHz = sin 60c

8 = 16 MHz 3 3 c 2 m

Option (D) is correct. When a moving circuit is put in a time varying magnetic field educed emf have two components. One for time variation of B and other turn motion of circuit in B . Option (A) is correct. Far field \ 1 r

8.123

Option (B) is correct. Z in min = Z 0 S where S = standing wave ratio

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S =

Page 202

1 + GL 1 - GL

and

GL = reflection coefficient

so

GL = ZL - Z 0 = 100 - 50 = 50 = 1 ZL + Z 0 100 + 50 150 3

Thus infinite impedance, and current will be zero.

1+1 3 =2 S = 1 13 50 Z in min = = 25 W 2 8.124

8.129

Option (A) is correct. The cutoff frequency is given by ml m 2+ n 2 fc = a 2 a k a2k Here a < b , so minimum cut off frequency will be for mode TE 01 m = 0, n = 1 8 fc = 3 # 10 2#2

1 (10 # 10-12) 3 # 108 = 0.75 GHz = 2 # 2 # 10 # 10-2

8.125 8.126

a ml = c 2 *

Option (B) is correct. For lossless transmission line, we have Velocity V =w= 1 b LC Characteristics impedance for a lossless transmission line Z0 = L C From eqn. (1) and (2) 1 V = = 1 C (Z 0 C ) Z 0 C

8.130

Option (C) is correct.

8.131

Option (A) is correct.

8.132

c = 3 # 108

Option (A) is correct. For any transmission line we can write input impedance Z + jZ 0 tanh lg Zin = Z 0 ; L Z 0 + jZL tanh lg E R jZ tanh lg V W S1 + 0 Z0 ZL W= so Zin = Z 0 lim S Z " 3S Z 0 j tanh lg W S ZL + j tanh lg W X T Option (A) is correct. We know that skin depth is given by 1 s = = 1 # 10-2 m pf1 ms 1 or = 10-2 6 p # 10 # 10 # m s -3 or ms = 10 p

A I D So

O N

L

f 2 = 1000 MHz is 4p f 2 V = ms -3

ms = 10 in above equation p

20 # 106 = 10 # 10 sin Ae sin Ae = 1 2

6

Ae = 30c 8.134

Option (C) is correct.

8.135

Option (A) is correct. Skin depth

d=

1 pfms

Putting the given value d= = 15.9 mm

1 3.14 # 1 # 10 # 4p # 10-7 # 106 9

V = 8.128

6

4 # p # 1000 # 10 # p - 6 106 m/ sec # 10-3

Option (A) is correct. Input impedance of a lossless transmission line is given by Z + jZ 0 tan bl Zin = Z 0 ; L Z 0 + jZL tan bl E where

so here

...(2)

Ei " Incident power

8.133

f 1 = 10 MHz

Now phase velocity at another frequency

Put

Er = GEi G = Reflection coefficient h - h1 G = 2 = 1.5 - 1 = 1 h 2 + h1 1.5 + 1 5 Er = 1 # Ei 5 Er = 20% Ei

Option (B) is correct. .inhave maximum usable frequency formulae as oWe c . f0 dia fm = o sin Ae n .

w

ww

...(1)

Option (C) is correct. Reflected power

Option (B) is correct.

Here given ZL = 3 (open circuited at load end)

8.127

ZL = 0 (Short circuited) Z 0 = 50 W 0 + j50 tan p/2 Zin = 50 = =3 50 + j0 tan p/2G

Z 0 = Charateristic impedance of line ZL = Load impedance b = 2p l = length l bl = 2p l = p 2 l 4

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Page 203

UNIT 9

(A) The emergence of nationalism in colonial India led to our Independence (B) Nationalism in India emerged in the context of colonialism (C) Nationalism in India is homogeneous (D) Nationalism in India is heterogeneous

GENERAL APTITUDE 9.9

YEAR 2013 9.1

9.2

9.3

9.4

9.5

9.1

ONE MARK

Choose the grammatically CORRECT sentence: (A) Two and two add four (B) Two and two become four (C) Two and two are four (D) Two and two make four Statement: You can always give me a ring whenever you need. Which one of the following is the best inference from the above statement? (A) Because I have a nice caller tune. (B) Because I have a better telephone facility (C) Because a friend in need is a friend indeed (D) Because you need not pay towards the telephone bills when you give me a ring In the summer of 2012, in New Delhi, the mean temperature of Monday to Wednesday was 41°C and of Tuesday to Thursday was 43cC . If the temperature on Thursday was 15% higher than that of Monday, then the temperature in cC on Thursday was (A) 40 (B) 43 (C) 46 (D) 49

9.11

9.12

ONE MARK

If (1.001) 1259 = 3.52 and (1.001) 2062 = 7.85, then (1.001) 3321 (A) 2.23 (B) 4.33 (C) 11.37 (D) 27.64 Choose the most appropriate alternate from the options given below to complete the following sentence : If the tired soldier wanted to lie down, he..................the mattress out on the balcony. (A) should take (B) shall take (C) should have taken (D) will have taken

A I D 9.13

9.15

TWO MARKS

A car travels 8 km in the first quarter of an hour, 6 km in the second quarter and 16 km in the third quarter. The average speed of the car in km per hour over the entire journey is (A) 30 (B) 36 (C) 40 (D) 24

9 ^9n - 1h (D) + n2 8

Statement: There were different streams of freedom movements in colonial India carried out by the moderates, liberals, radicals, socialists, and so on. Which one of the following is the best inference from the above statement?

One of the parts (A, B, C, D) in the sentence given below contains an ERROR. Which one of the following is INCORRECT ? I requested that he should be given the driving test today instead of tomorrow. (A) requested that (B) should be given (C) the driving test (D) instead of tomorrow 2012

9.16

Find the sum to n terms of the series 10 + 84 + 734 + ... 9 ^9n + 1h 9 ^9n - 1h (A) (B) +1 +1 10 8 9 ^9n - 1h (C) +n 8

9.8

2012

O N

YEAR 2013

9.7

What is the chance that a leap year, selected at random, will contain 53 Sundays? (A) 2/7 (B) 3/7 (C) 1/7 (D) 5/7

Choose the most appropriate word from the options given below to n complete the following sentence : o.i c . Give the seriousness of the situation that he had to face, his........ a di was impressive. o .n w (A) beggary (B) nomenclature ww They were requested not to quarrel with others. (C) jealousy (D) nonchalance Which one of the following options is the closest in meaning to the word quarrel? 9.14 Which one of the following options is the closest in meaning to the (A) make out (B) call out word given below ? (C) dig out (D) fall out Latitude (A) Eligibility (B) Freedom Option (D) is correct. (C) Coercion (D) Meticulousness They were requested not to quarrel with others. Complete the sentence: Dare .................. mistakes. (A) commit (B) to commit (C) committed (D) committing

Quarrel has a similar meaning to ‘fall out’

9.6

9.10

The set of values of p for which the roots of the equation 3x2 + 2x + p ^p - 1h = 0 are of opposite sign is (A) ^- 3, 0h (B) ^0, 1h (C) ^1, 3h (D) ^0, 3h

TWO MARKS

One of the legacies of the Roman legions was discipline. In the legious, military law prevailed and discipline was brutal. Discipline on the battlefield kept units obedient, intact and fighting, even when the odds and conditions were against them. Which one of the following statements best sums up the meaning of the above passage ? (A) Through regimentation was the main reason for the efficiency of the Roman legions even in adverse circumstances. (B) The legions were treated inhumanly as if the men were animals (C) Disciplines was the armies inheritance from their seniors

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Page 204

(D) The harsh discipline to which the legions were subjected to led to the odds and conditions being against them. 9.17

9.18

9.19

Raju has 14 currency notes in his pocket consisting of only Rs. 20 notes and Rs. 10 notes. The total money values of the notes is Rs. 230. The number of Rs. 10 notes that Raju has is (A) 5 (B) 6 (C) 9 (D) 10 There are eight bags of rice looking alike, seven of which have equal weight and one is slightly heavier. The weighing balance is of unlimited capacity. Using this balance, the minimum number of weighings required to identify the heavier bag is (A) 2 (B) 3 (C) 4 (D) 8

Category

Amount (Rs.)

Food

4000

Clothing

1200

Rent

2000

Savings

1500

Other Expenses

1800

9.22

9.23

9.25

(B) most (D) available

Choose the word from the from the options given below that is most opposite in meaning to the given word : Frequency (A) periodicity (B) rarity (C) gradualness (D) persistency Choose the most appropriate word from the options given below to complete the following sentence : It was her view that the country’s had been ............. by foreign techno-crafts, so that to invite them to come back would be counter-productive. (A) identified (B) ascertained (C) exacerbated (D) analysed 2011

9.26

TWO MARKS

The fuel consumed by a motor cycle during a journey while travelling at various speed is indicated in the graph below.

A I D

O N

.in distance covered during four laps of the journey are listed in oThe A and B are friends. They decide to meet between 1 PM and 2 PM c . a the table below on a given day. There is a conditions that whoever arrives first will o di .n Lap Distance (km) Average speed (km/hour) not wait for the other for more than 15 minutes. The probability w w w that they will meet on that days is P 15 15 (A) 1/4 (B) 1/16 Q 75 45 (C) 7/16 (D) 9/16 R 40 75 2011

9.21

9.24

The data given in the following table summarizes the monthly budget of an average household.

The approximate percentages of the monthly budget NOT spent on savings is (A) 10% (B) 14% (C) 81% (D) 86%

9.20

(A) similar (C) uncommon

S 10 10 From the given data, we can conclude that the fuel consumed per kilometre was least during the lap (A) P (B) Q (C) R (D) S

ONE MARK

There are two candidates P and Q in an election. During the campaign, 40% of voter promised to vote for P , and rest for Q . However, on the day of election 15% of the voters went back on their promise to vote for P and instead voted for Q . 25% of the voter went back on their promise to vote for Q and instead voted for P . Suppose, P lost by 2 votes, then what was the total number of voters ? (A) 100 (B) 110 (C) 90 (D) 95 The question below consists of a pair of related words followed by four pairs of words. Select the pair that best expresses the relations in the original pair : Gladiator : Arena (A) dancer : stage (B) commuter : train (C) teacher : classroom (D) lawyer : courtroom Choose the most appropriate word from the options given below to complete the following sentence : Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which...................treatments are unsatisfactory.

9.27

9.28

9.29

The horse has played a little known but very important role in the field of medicine. Horses were injected with toxins of disease until their blood build up immunities. Then a serum was made from their blood. Serums to fight with diphteria and tetanus were developed this way. It can be inferred from the passage, that horses were (A) given immunity to diseases (B) generally quite immune to diseases (C) given medicines to fight toxins (D) given diphtheria and tetanus serums The sum of n terms of the series 4 + 44 + 444 + ........ (A) (4/81) [10n + 1 - 9n - 1] (B) (4/81) [10n - 1 - 9n - 1] (C) (4/81) [10n + 1 - 9n - 10] (D) (4/81) [10n - 9n - 10] Given that f (y) = y /y, and q is any non-zero real number, the value of f (q) - f (- q) is (A) 0 (B) - 1

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(C) 1 9.30

Page 205

(C) 1623

(D) 2

Three friends R, S and T shared toffee from a bowl. R took 1/3 rd of the toffees, but returned four to the bowl. S took 1/4 th of what was left but returned three toffees to the bowl. T took half of the remainder but returned two back into the bowl. If the bowl had 17 toffees left, how many toffees were originally there in the bowl ? (A) 38 (B) 31 (C) 48 (D) 41

5 skilled workers can build a wall in 20 days; 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall ? (A) 20 days (B) 18 days (C) 16 days (D) 15 days

9.38

Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how much distinct 4 digit numbers greater than 3000 can be formed ? (A) 50 (B) 51 (C) 52 (D) 54

9.39

2010 9.31

9.32

9.33

9.34

9.35

ONE MARK

Which of the following options is the closest in meaning to the word below ? Circuitous (A) Cyclic (B) Indirect (C) Confusing (D) Crooked

O N

9.37

A I D

***********

.in

co ia.

Choose the most appropriate word from the options given below to d no . complete the following sentence : w His rather casual remarks on politics..................his lack of seriousww ness about the subject. (A) masked (B) belied (C) betrayed (D) suppressed 25 persons are in a room 15 of them play hockey, 17 of them play football and 10 of them play hockey and football. Then the number of persons playing neither hockey nor football is (A) 2 (B) 17 (C) 13 (D) 3 2010

9.36

Hari (H), Gita (G), Irfan (I) and Saira (S) are siblings (i.e. brothers and sisters.) All were born on 1st January. The age difference between any two successive siblings (that is born one after another) is less than 3 years. Given the following facts : 1. Hari’s age + Gita’s age > Irfan’s age + Saira’s age. 2. The age difference between Gita and Saira is 1 year. However, Gita is not the oldest and Saira is not the youngest. 3. There are no twins. In what order were they born (oldest first) ? (A) HSIG (B) SGHI (C) IGSH (D) IHSG

9.40

The question below consist of a pair of related words followed by four pairs of words. Select the pair that best expresses the relation in the original pair. Unemployed : Worker (A) Fallow : Land (B) Unaware : Sleeper (C) Wit : Jester (D) Renovated : House Choose the most appropriate word from the options given below to complete the following sentence : If we manage to ........ our natural resources, we would leave a better planet for our children. (A) unhold (B) restrain (C) cherish (D) conserve

(D) 1531

TWO MARKS

Modern warfare has changed from large scale clashes of armies to suppression of civilian populations. Chemical agents that do their work silently appear to be suited to such warfare ; and regretfully, their exist people in military establishments who think that chemical agents are useful fools for their cause. Which of the following statements best sums up the meaning of the above passage ? (A) Modern warfare has resulted in civil strife. (B) Chemical agents are useful in modern warfare. (C) Use of chemical agents in ware fare would be undesirable. (D) People in military establishments like to use chemical agents in war. If 137 + 276 = 435 how much is 731 + 672 ? (A) 534 (B) 1403

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Page 206

SOLUTIONS 9.1

9.2

9.3

9.4

9.5

9.6

S2 =

9.8

Option (D) is correct Two and two make four Option (C) is correct. You can always given me a ring whenever you need. Because a friend is need is a friend indeed

9.9

S3 9 ^93 - 1h = + 32 = 828 8 Option (D) is correct. Nationalism in India is heterogeneous Option (B) is correct. Given, the quadratic equation 3x2 + 2x + P ^P - 1h = 0 It will have the roots with opposite sign if

Option (C) is correct. Let the temperature on Monday, Tuesday, Wednesday and Thursday be respectively as TM , TTU , TW , TTH So, from the given data we have TH + TTU + TW = 41 ....(1) 3 TTU + TW + TTH = 43 and ....(2) 3

P ^P - 1h < 0 So it can be possible only when P < 0 and P - 1 > 0 or P > 0 and P - 1 < 0 The 1 st condition tends to no solution for P . Hence, from the second condition, we obtain

also, as the temperature on Thursday was 15% higher than that of Monday i.e. ....(3) TTH = 1.15 TM solving eq (1), (2) and (3), we obtain

0 B1 B 2 B 3 it means one of the A1 A2 A 3 will be heavier So we will perform next = 60 # 60 - 2 b 1 # 45 # 45 l 2 weighting as: 2 nd weighting " A1 is kept on one side of the balance and A2 on the = 1575 other. So, the required probability = 1575 = 7 3600 16 it means A 3 will be heavier if A1 = A 2 9.21 Option (A) is correct. then A1 will be heavier A1 > A 2 Let us assume total voters are 100. Thus 40 voter (i.e. 40 %) promised then A2 will be heavier A1 < A 2 to vote for P and 60 (rest 60 % ) promised to vote fore Q. Case 3 : A1 A 2 A 3 < B1 B 2 B 3 Now, 15% changed from P to Q (15 % out of 40) This time one of the B1 B2 B 3 will be heavier, So again as the above 15 Changed voter from P to Q 40 = 6 case weighting will be done. 100 # 2 nd weighting " B1 is kept one side and B2 on the other Now Voter for P 40 - 6 = 34 if B1 = B 2 B 3 will be heavier Also, in 25% changed form Q to P (out of 60%) . o B1 > B 2 B1 will be heavier c 25 # 60 = 15 ia. Changed voter from Q to P 100 d B1 < B 2 B2 will be heavier o .n w Now Voter for P 34 + 15 = 49 So, as described above, in all the three cases weighting is done w w only two times to give out the result so minimum no. of weighting Thus P P got 49 votes and Q got 51 votes, and P lost by 2 votes, required = 2. which is given. Therefore 100 voter is true value.

A I D

O N

9.19

Option (D) is correct. Total budget = 4000 + 1200 + 2000 + 1500 + 1800 = 10, 500 The amount spent on saving = 1500 So, the amount not spent on saving

9.23

= 10, 500 - 1500 = 9000 So, percentage of the amount = 9000 # 100% = 86% 10500 9.20

9.22

9.24

Option (S) is correct. The graphical representation of their arriving time so that they met is given as below in the figure by shaded region.

9.25

9.26

9.27

9.28

Option (A) is correct. A gladiator performs in an arena. Commutators use trains. Lawyers performs, but do not entertain like a gladiator. Similarly, teachers educate. Only dancers performs on a stage. Option (D) is correct. Available is appropriate because manipulation of genes will be done when other treatments are not useful. Option (B) is correct. Periodicity is almost similar to frequency. Gradualness means something happening with time. Persistency is endurance. Rarity is opposite to frequency. Option (C) is correct. The sentence implies that technocrats are counterproductive (negative). Only (C) can bring the same meaning. Option (B) is correct. Since fuel consumption/litre is asked and not total fuel consumed, only average speed is relevant. Maximum efficiency comes at 45 km/ hr, So least fuel consumer per litre in lap Q Option (B) is correct. Option B fits the sentence, as they built up immunities which helped humans create serums from their blood. Option (C) is correct.

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Page 208

4 + 44 + 444 + .............. 4 (1 + 11 + 111 + .......) = 4 (9 + 99 + 999 + ............) 9

n (A , B) = n (A) + n (B) - n (A + B) = 15 + 17 - 10 = 22 Thus people who play neither hockey nor football = 25 - 22 = 3 9.36

= 4 [(10 - 1) + (100 - 1) + ........] 9

9.37

= 4 [10 (1 + 10 + 102 + 103) - n] 9 n = 4 :10 # 10 - 1 - nD 9 10 - 1 = 4 610n + 1 - 10 - 9n@ 81 9.29

9.38

Option (D) is correct. y y -y f (- y) = =- f (y) y f (q) - f (- q) = 2f (q) = 2

9.30

Bowl Status

R

= x -4 3

= 2x + 4 3

S

= 1 :2x + 4D - 3 4 3 = x +1-3 = x -2 6 6

T

= 1 a x + 6k - 2 2 2 = x +1 4

Option (D) is correct. Let W be the total work. Per day work of 5 skilled workers

Option (C) is correct. Let total no of toffees be x . The following table shows the all calculations. Friend

Option (C) is correct. Since 7 + 6 = 13 but unit digit is 5 so base may be 8 as 5 is the remainder when 13 is divided by 8. Let us check. 137 8 731 8 276 8 672 8 435 Thus here base is 8. Now 1623

=W 20 Per day work of one skill worker = W =W 5 # 20 100 Similarly per day work of 1 semi-skilled workers = W = W 8 # 25 200 Similarly per day work of one semi-skill worker = W = W 10 # 30 300 Thus total per day work of 2 skilled, 6 semi-skilled and 5 unskilled workers is = 2W + 6W + 5W = 12W + 18W + 10W = W 100 200 300 600 15 Therefore time to complete the work is 15 days.

f (y) =

Now or

Option (D) is correct.

A I D

Option (B) is correct. As the number must be greater than 3000, it must be start with 3 or 4. Thus we have two case: Case (1) If left most digit is 3 an other three digits are any of 2, 2, 3, 3, 4, 4, 4, 4. .inUsing 2, 2, 3 we have 3223, 3232, 3322 i.e. 3! = 3 no. = x +6-x -1 o(1) c 2 4 . 2! dia o x .n = +5 (2) Using 2, 2, 4 we have 3224, 3242, 3422 i.e. 3! = 3 no. w 4 2! ww (3) Using 2, 3, 3 we have 3233, 3323, 3332 i.e. 3! = 3 no. 2! 9.39

Now, or

x + 5 = 17 4 x = 17 - 5 = 12 4

= 2x + 4 - x + 2 3 6 = x +6 2

O N

x = 12 # 4 = 48 9.31

9.32

9.33

9.34

9.35

Option (B) is correct. Circuitous means round about or not direct. Indirect is closest in meaning to this circuitous (A) Cyclic : Recurring in nature (B) Indirect : Not direct (C) Confusing : lacking clarity of meaning (D) Crooked : set at an angle; not straight Option (B) is correct. A worker may by unemployed. Like in same relation a sleeper may be unaware. Option (D) is correct. Here conserve is most appropriate word. Option (C) is correct. Betrayed means reveal unintentionally that is most appropriate. Option (D) is correct. Number of people who play hockey n (A) = 15 Number of people who play football n (B) = 17 Persons who play both hockey and football n (A + B) = 10 Persons who play either hockey or football or both :

(4) Using 2, 3, 4 we have 3! = 6 no. (5) Using 2, 4, 4 we have 3244, 3424, 3442 i.e. 3! = 3 no. 2! (6) Using 3, 3, 4 we have 3334, 3343, 3433 i.e. 3! = 3 no. 2! (7) Using 3, 4, 4 we have 3344, 3434, 3443 i.e. 3! = 3 no. 2! (8) Using 4, 4, 4 we have 3444 i.e. 3! = 1 no. 3! Total 4 digit numbers in this case is 1 + 3 + 3 + 3 + 6 + 3 + 3 + 3 + 1 = 25 Case 2 : If left most is 4 and other three digits are any of 2, 2, 3, 3, 3, 4, 4, 4. (1) Using 2, 2, 3 we have 4223, 4232, 4322 i.e. . 3! = 3 no 2! (2) Using 2, 2, 4 we have 4224, 4242, 4422 i.e. . 3! = 3 no 2! (3) Using 2, 3, 3 we have 4233, 4323, 4332 i.e. . 3! = 3 no 2! (4) Using 2, 3, 4 we have i.e. . 3! = 6 no

(5) Using 2, 4, 4 we have 4244, 4424, 4442 i.e. . 3! = 3 no 2! (6) Using 3, 3, 3 we have 4333 i.e 3! = 1. no. 3! (7) Using 3, 3, 4 we have 4334, 4343, 4433 i.e. . 3! = 3 no 2!

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(8) Using 3, 4, 4 we have 4344, 4434, 4443 i.e. . 3! = 3 no 2! (9) Using 4, 4, 4 we have 4444 i.e. 3! = 1. no 3! Total 4 digit numbers in 2nd case = 3 + 3 + 3 + 6 + 3 + 3 + 1 + 3 + 1 = 26 Thus total 4 digit numbers using case (1) and case (2) is = 25 + 26 = 51 9.40

Option (B) is correct. Let H , G , S and I be ages of Hari, Gita, Saira and Irfan respectively. Now from statement (1) we have H + G > I + S Form statement (2) we get that G - S = 1 or S - G = 1 As G can’t be oldest and S can’t be youngest thus either GS or SG possible. From statement (3) we get that there are no twins (A) HSIG : There is I between S and G which is not possible (B) SGHI : SG order is also here and S > G > H > I and G + H > S + I which is possible. (C) IGSH : This gives I > G and S > H and adding these both inequalities we have I + S > H + G which is not possible. (D) IHSG : This gives I > H and S > G and adding these both inequalities we have I + S > H + G which is not possible.

A I D

O N

no w.

.in

co ia.

d

ww

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Page 210

UNIT 10

(C) x

(D) 1

2012

ENGINEERING MATHEMATICS

TWO MARKS

d 2 y (t) dy (t) Consider the differential equation +2 + y (t) = d (t) 2 dt dt dy with y (t) t = 0 =- 2 and =0 dt t = 0 dy The numerical value of is dt t = 0 (A) - 2 (B) - 1 (C) 0 (D) 1

10.8

-

2013 10.1

10.2

10.3

ONE MARK

10.9

The minimum eigen value of the following matrix is R3 5 2V S W S5 12 7W SS2 7 5WW T X (A) 0 (B) 1 (C) 2 (D) 3

10.10

TWO MARKS

10.12

N

2

2011

ONE MARK

Consider a closed surface S surrounding volume V . If rv is the position vector of a point inside S , with nt the unit normal on S , the value of the integral ## 5rv $ nt dS is S

(A) 3V (C) 10V

(B) 5V (D) 15V dy = ky, y (0) = c is dx (B) x = kecy (D) y = ce-kx

The solution of the differential equation

10.15

The value of the integral z = 1 is given by (A) 0 (C) 4/5

# c

- 3z + 4 dz where c is the circle (z 2 + 4z + 5) (B) 1/10 (D) 1

t (D) x = 2

(C) x = t 2

1 - 2 . z+1 z+3 If C is a counter clockwise path in the z -plane such that z + 1 = 1, the value of 1 f (z) dz is 2p j C (A) - 2 (B) - 1 (C) 1 (D) 2

2011

Given f (z) =

10.16

#

10.7

IA

Given that

(A) x = ce-ky (C) y = cekx

ONE MARK

With initial condition x (1) = 0.5 , the solution of the differential equation t dx + x = t , is dt (B) x = t 2 - 1 (A) x = t - 1 2 2

The maximum value of f (x) = x3 - 9x2 + 24x + 5 in the interval [1, 6] is (A) 21 (B) 25 (C) 41 (D) 46

-5 -3 1 0 , the value of A3 is A=> and I = > 2 0H 0 1H (A)n 15A + 12I (B) 19A + 30I o.i c . (D) 17A + 21I ia (C) 17A + 15I

Let A be an m # n matrix and B an n # m matrix. It is given that d determinant ^Im + AB h = determinant ^In + BAh, where Ik is the.no w k # k identity matrix. Using the above property, the determinant ww of the matrix given below is R V 10.13 S2 1 1 1W S1 2 1 1W S1 1 2 1W S W S1 1 1 2W T X (A) 2 (B) 5 (C) 8 (D) 16 2012

10.6

A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is (A) 1/3 (B) 1/2 (C) 2/3 (D) 3/4

10.11

10.14

10.5

The direction of vector A is radially outward from the origin, with A = krn . where r2 = x2 + y2 + z2 and k is a constant. The value of n for which d:A = 0 is (A) - 2 (B) 2 (C) 1 (D) 0

D O

2013 10.4

+

The maximum value of q until which the approximation sin q . q holds to within 10% error is (A) 10c (B) 18c (C) 50c (D) 90c

A polynomial f (x) = a 4 x 4 + a 3 x3 + a2 x2 + a1 x - a 0 with all coefficients positive has (A) no real roots (B) no negative real root (C) odd number of real roots (D) at least one positive and one negative real root

-

If x = - 1, then the value of xx is (A) e- p/2 (B) e p/2

10.17

TWO MARKS

A numerical solution of the equation f (x) + x - 3 = 0 can be obtained using Newton- Raphson method. If the starting value is x = 2 for the iteration, the value of x that is to be used in the next step is (A) 0.306 (B) 0.739 (C) 1.694 (D) 2.306 The system of equations

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x+y+z = 6 x + 4y + 6y = 20 x + 4y + l z = m has NO solution for values of l and μ given by (A) l = 6, m = 20 (B) l = 6, m = Y 20 (C) l = (D) l = Y 6, m = 20 Y 6, m = 20 10.18

A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is (A) 2/36 (B) 2/6 (C) 5/12 (D) 1/2 2010

10.19

10.20

(C) 1 10.25

ONE MARKS

The eigen values of a skew-symmetric matrix are (A) always zero (B) always pure imaginary (C) either zero or pure imaginary (D) always real The trigonometric Fourier series for the waveform f (t) shown below contains 10.26

10.21

10.22

The residues of a complex function 1 - 2z x (z) = z (z - 1) (z - 2) at its poles are (B) 1 , - 1 and - 1 (A) 1 , - 1 and 1 2 2 2 2 (C) 1 , 1 and - 3 (D) 1 , - 1 and 3 2 2 2 2 dy (x) Consider differential equation - y (x) = x , with the initial dx condition y (0) = 0 . Using Euler’s first order method with a step size of 0.1, the value of y (0.3) is (A) 0.01 (B) 0.031 (C) 0.0631 (D) 0.1

A I D 10.27

(A) only (B) only (C) only (D) only

(B) 2 3 (D) 2 3

(A) 0

O N

3s + 1 Given f (t) = L-1 ; 3 . If lim f (t) = 1, then the value t"3 s + 4s2 + (k - 3) s E of k is (A) 1 (B) 2 (C) (D) 4 .in3

cosine terms and zero values for the dc components .co a i cosine terms and a positive value for the dc components od n . cosine terms and a negative value for the dc components w 2009 ONE MARK w w sine terms and a negative value for the dc components 10.28 The order of the differential equation d2y dy 3 -t 4 A function n (x) satisfied the differential equation + c dt m + y = e 2 2 dt d n (x) n (x) is - 2 =0 dx 2 L (A) 1 (B) 2 where L is a constant. The boundary conditions are : n (0) = K (C) 3 (D) 4 and n (3) = 0 . The solution to this equation is 10.29 (A) n (x) = K exp (x/L) (B) n (x) = K exp (- x/ L ) A fair coin is tossed 10 times. What is the probability that only the 2 first two tosses will yield heads? (C) n (x) = K exp (- x/L) (D) n (x) = K exp (- x/L) 2 2 (A) c 1 m (B) 10C2 b 1 l 2 2 10 10 2010 TWO MARKS (C) c 1 m (D) 10C2 b 1 l 2 2 If ey = x1/x , then y has a 1 + f (z) (A) maximum at x = e (B) minimum at x = e 10.30 If f (z) = c 0 + c1 z-1 , then dz is given by -1 -1 z (C) maximum at x = e (D) minimum at x = e unit circle (A) 2pc1 (B) 2p (1 + c0) A fair coin is tossed independently four times. The probability of the (C) 2pjc1 (D) 2p (1 + c0) event “the number of time heads shown up is more than the number of times tail shown up” 2009 TWO MARKS (A) 1/16 (B) 1/3 (C) 1/4 (D) 5/16 10.31 The Taylor series expansion of sin x at x = p is given by x-p v = xyatx + x 2 aty , then # A v $ dlv over the path shown in the figure If A (x - p) 2 (x - p) 2 (A) (B) 1 + + ... 1 + ... C is 3! 3!

#

10.23

10.24

(C) 1 10.32

(x - p) 2 + ... 3!

(D) - 1 +

(x - p) 2 + ... 3!

Match each differential equation in Group I to its family of solution

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curves from Group II Group I dy y A. = dx x dy y B. =dx x dy C. =x dx y dy D. =- x dx y (A) A - 2, B - 3, C - 3, D - 1 (B) A - 1, B - 3, C - 2, D - 1 (C) A - 2, B - 1, C - 3, D - 3 (D) A - 3, B - 2, C - 1, D - 2 10.33

2008

Group II

10.35

10.37

10.38

10.39

2. Straight lines

The recursion relation to solve x = e - x using Newton - Raphson method is (A) xn + 1 = e-x (B) xn + 1 = xn - e-x -x x 2 - e-x (1 - xn) - 1 (C) xn + 1 = (1 + xn) e -x (D) xn + 1 = n xn - e-x 1+e n

n

n

n

3. Hyperbolas 10.41

n

The residue of the function f (z) = (A) - 1 32 (C) 1 16

10.42

The Eigen values of following matrix are V R S- 1 3 5 W S- 3 - 1 6 W SS 0 0 3 WW X T (A) 3, 3 + 5j, 6 - j (B) - 6 + 5j, 3 + j, 3 - j (C) 3 + j, 3 - j, 5 + j (D) 3, - 1 + 3j, - 1 - 3j

10.43

p11 p12 are nonzero, p21 p22 G and one of its eigenvalue is zero. Which of the following statements is true? (A) p11 p12 - p12 p21 = 1 (B) p11 p22 - p12 p21 =- 1 (C) p11 p22 - p12 p21 = 0 (D) p11 p22 + p12 p21 = 0 All the four entries of the 2 # 2 matrix P = =

The system of linear equations

0 1 Consider the matrix P = = . The value of e p is - 2 - 3G 2e-2 - 3e-1 e-1 - e-2 e-1 + e-1 2e-2 - e-1 (B) > -1 (A) > -2 H H 2e - 2e-1 5e-2 - e-1 2e - 4e2 3e-1 + 2e-2

O N w

ww

The equation sin (z) = 10 has (A) no real or complex solution (B) exactly two distinct complex solutions (C) a unique solution (D) an infinite number of complex solutions

10.45

points P and Q in the x - y plane, with P = (1, 0) and

#

Q

Q = (0, 1). The line integral 2 (xdx + ydy) along the semicircle P with the line segment PQ as its diameter (A) is - 1 (B) is 0 (C) is 1 (D) depends on the direction (clockwise or anit-clockwise) of the semicircle 2007

10.46

Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x = 0 ? (A) sin (x3) (B) sin (x2) (C) cos (x3) (D) cos (x2)

(A) 1.0 (C) 4.0 10.47

(D) x (t) = 3t2 10.48

ONE MARK

The following plot shows a function which varies linearly with x . The value of the integral I =

For real values of x , the minimum value of the function f (x) = exp (x) + exp (- x) is (A) 2 (B) 1 (C) 0.5 (D) 0

Which of the following is a solution to the differential equation dx (t) + 3x (t) = 0 ? dt (A) x (t) = 3e - t (B) x (t) = 2e - 3t

The value of the integral of the function g (x, y) = 4x3 + 10y 4 along the straight line segment from the point (0, 0) to the point (1, 2) in the x - y plane is (A) 33 (B) 35 (C) 40 (D) 56

.in oConsider c . ia

d .no

2e-1 - e-2 e-1 - e-2 (D) > H - 2e-1 + 2e-2 - e-1 + 2e-2

In the Taylor series expansion of exp (x) + sin (x) about the point x = p , the coefficient of (x - p) 2 is (A) exp (p) (B) 0.5 exp (p) (C) exp (p) + 1 (D) exp (p) - 1

A I D 10.44

1 at z = 2 is (z + 2) (z - 2) 2 2

(B) - 1 16 (D) 1 32

5e-2 - e-1 3e-1 - e-2 (C) > -2 H 2e - 6e-1 4e-2 + 6-1

ONE MARKS

(C) x (t) =- 23 t2

TWO MARKS

n

4x + 2y = 7 2x + y = 6 has (A) a unique solution (B) no solution (C) an infinite number of solutions (D) exactly two distinct solutions 10.36

10.40

1. Circles

2008 10.34

Page 212

2

#1 ydx

is

(B) 2.5 (D) 5.0

For x 0 , ex > 1 and 0 < e-x < 1 For x < 0 , 0 < ex < 1 and e-x > 1 Thus f (x) have minimum values at x = 0 and that is e0 + e-0 = 2 . Option (A) is correct. 3 5 sin x = x + x + x + ... 3! 5!

Substituting x - p = y ,we get sin (y + p) sin y f (y + p) = == - 1 (sin y) y y y

2 4 cos x = 1 + x + x + ... 2! 4!

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10.39

Thus only sin (x3) will have odd power of x .

Now

Option (B) is correct. dx (t) We have + 3x (t) = 0 dt

or, Now

Thus only (B) may be

10.45

Option (B) is correct.

x = e-x f (x) = x - e - x

or

10.46

-x

f'( x) = 1 + e The Newton-Raphson iterative formula is f (xn) xn + 1 = xn f'( xn) f (xn) = xn - e

Now

- xn

f'( xn) = 1 + e - x

xn + 1 = xn - xn - e- x 1+e

Thus

n

(1 + xn) e - x = 1 + e-x

#1

= L- 1 f>

s+3 (s + 1)( s + 2) -2 (s + 1)( s + 2)

Now

I =

1 (s + 1)( s + 2) s (s + 1)( s + 2)

2

#1 ydx

=

2

#1 (x + 1) dx

(x + 1) 2 2 9 4 E = - = 2.5 2 2 2

O N w

.no

Option .in (D) is correct.

co ia.

10.49

d

We have,

lim 12 = 3 x"0 x lim x2 = 3 x"3

lim e - x = 3

x"3

lim e - x = 0 2

x"3

lim e - x = 1

Thus e - x is strictly bounded.

2

2

x"0

Hp

10.50

Option (A) is correct. We have

f (x) = e - x = e - (x - 2) - 2 = e - (x - 2) e - 2 (x - 2) 2 = ;1 - (x - 2) + ...E e - 2 2! = 61 - (x - 2)@ e - 2

Option (B) is correct. Taylor series is given as

= (3 - x) e f (x) = f (a) + x - a f'( a) + 1!

(x - a) 2 f"( a) + ... 2!

f (x) = f (p) + x - p f'( p) + 1!

(x - p) 2 f"( x)... 2!

For x = p we have

= ex + sin x = ex + cos x = ex - sin x = e p - sin p = e p f"( p) Thus the coefficient of (x - p) 2 is 2! Now

1

#0 ydy = 0

Option (A) is correct. q sin ^ q2 h sin ^ q2 h 1 lim sin ^ 2 h = 1 = 0.5 lim = lim = q"0 q " 0 2^ q h 2 q 2 q " 0 ^ q2 h 2

2e - 1 - e - 2 e-1 - e-2 == G - 2e - 1 + 2e - 2 - e - 1 + 2e - 2

Thus

Q

#P ydy

Option (C) is correct.

A I D 10.48

ww

eP = L- 1 6(sI - A) - 1@ s 0 0 1 -1 -1 = L e= 0 s G =- 2 - 3Go s - 1 -1 -1 o = L e= 2 s + 3G

xdx + 2

as x (a + 2) F (s) = e5 ; s - (a + 2) E

Option (D) is correct. Let d " defective and y " supply by Y P (y + d) y pa k = d P (d)

A ==

or or

4 2 2 4G

A I D

O N

no

. ww

w

10.73

(- 4 - l)(3 - l) - 8 = 0 - 12 + l + l2 - 8 = 0 l2 + l - 20 = 0

or or l =- 5, 4 Eigen vector for l =- 5 (A - lI) Xi = 0 1 - (- 5) 2 x1 0 == G G G = = 4 8 - 4 x2 0

Option (D) is correct. Probability of coming odd number is 12 and the probability of coming even number is 12 . Both the events are independent to each other, thus probability of coming odd number after an even number is 12 # 12 = 14 . Option (B) is correct. d2 y dy -5 + 6y = 0 2 dx dx

Eigen values

R2 - 4R1

x1 + 2x2 = 0 Let - x1 = 2 & x2 =- 1, X ==

x

Option (B) is correct. Order is the highest derivative term present in the equation and degree is the power of highest derivative term. Order = 2 , degree = 1

We have

or

Thus

e 1 + ex For x " 3, the value of f (x) monotonically increases.

10.72

4-l 2 =0 4 3-l

1 2 x1 0 =0 0G=x G = = 0 G 2

Option (A) is correct. f (x) =

A - lI = 0

or

in coor.

. dia

/

We have

-4 2 A == 4 3G

Characteristic equation is

d2 y + k2 y = 0 2 dx or D2 y + k2 y = 0 The AE is m2 + k2 = 0 The solution of AE is m = ! ik Thus y = A sin kx + B cos kx From x = 0 , y = 0 we get B = 0 and x = a, y = 0 we get A sin ka = 0 or sin ka = 0 k = mpx a Thus y = Am sin ` mpx j a m We have

Method-Solving nonlinear eq. Solving ordinary differential eq. Numerical Integration Solving linear simultaneous eq.

We have

(101)( 4 - l) + 2 (101) = 0 l =6

Option (A) is correct.

" " " "

Option (C) is correct.

10.77

4 - l 2 101 0 = 2 4 - l G=101G = = 0 G

or

10.71

Option (C) is correct. Newton - Raphson Runge - kutta Method Simpson’s Rule Gauss elimination

10.76

6A - lI @ [X] = 0

Now

10.70

Option (C) is correct. For x > 0 the slope of given curve is negative. Only (C) satisfy this condition.

10.75

Option (C) is correct. We have

10.69

Option (A) is correct.

10.74

P (y + d) = 0.3 # 0.02 = 0.006 P (d) = 0.6 # 0.1 + 0.3 # 0.02 + 0.1 # 0.03 = 0.015 y P a k = 0.006 = 0.4 d 0.015 10.68

=0 = 3, 2 = C1 e3x + C2 e2x = C1 e3x + C2 e2x

10.78

2 -1G

Eigen vector

Option (A) is correct. We have A == Now

1 2 - 0.1 a and A - 1 = = 2 G G 0 3 0 b

AA - 1 = I

or

2 - 0. 1 1 a 1 0 =0 3 G= 2 G = =0 1G 0 b

or

1 2a - 0.1b 1 0 == G =0 3b 0 1G

2a - 0.1 = 0 and 3b = 1 Thus solving above we have b = 1 and a = 1 3 60 Therefore a+b = 1 + 1 = 7 3 60 20 or

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GATE Electronics and Communication Topicwise Solved Paper by RK Kanodia & Ashish Murolia 10.79

Option (A) is correct. Gaussian PDF is f (x) = 1 2p s

#

3 - (x - m)2 2s2

-3

e

# f (x) dx 3

and

-3

dx

Page 223

for - 3 # x # 3

=1

Substituting m = 0 and s = 2 in above we get 3 1 e dx = 1 2p 2 - 3

10.80

#

x2 8

or

1 2 2p 2

#

3 - x2 8

e

dx = 1

or

1 2p

#

3 - x2 8

dx = 1

0

0

e

Option (C) is correct. From orthogonal matrix [AAT ] = I Since the inverse of I is I , thus [AAT ] -1 = I-1 = I

A I D

O N

no w.

.in

co ia.

d

ww

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