3320100XZ8L_Control System Engineering_Solution Manual
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control system...
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3
Transfer Function and Impulse Response
Solutions of Examples for Practice Example 3.3.7 Solution :
Taking Laplace transform of the network, R1
R2 1 sC1
Ei(t)
1 sC2
Eo(s)
Fig. 3.1 (a)
Divide the network into two parts, Part 1 : E i (s) = I(s) R 1 +
1 1 ù é I(s) = I(s) êR 1 + ú sC 1 sC 1û ë ... (1)
but,
V1 (s) =
R1
Ei(s)
1 I(s) sC 1
1 sC1
I(s)
V1(s)
Fig. 3.1 (b)
I(s) = V1(s) (sC1) \
1 ù é E i (s) = V1 (s) (s C 1 ) êR 1 + ú sC 1û ë
\
V1 ( s) = E i ( s)
Port 2
1 (R 1 C 1 s + 1)
Part 2 : V2(s) =
V2(s)
1 ö æ ç R2 + ÷ I(s) sC 2 ø è
R2
I(s)
1 sC2
Fig. 3.1 (c) (3 - 1) TM
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Eo(s)
Control System Engineering
E o (s) = I(s)
3-2
Transfer Function and Impulse Response
1 sC 2
1 sC 2 E o (s) 1 1 = = = V2 (s) 1 ö R 2 C2 s + 1 1 ö æ æ sC 2 ç R 2 + ç R2 + ÷ ÷ sC 2 ø sC 2 ø è è Amplifier gain = 10 hence Eo(s) (1+R2C2s) =
V2(s) = 10 V1(s)
10E i (s) i.e. 1 + R 1C1s
E o (s) = E i (s)
10
(1 + sR 1 C1 )(1 + sR 2 C2 )
Example 3.3.8 Solution : The s - domain network of the given circuit is shown in the Fig. 3.2 (a). Applying KVL to the loop, 1 I(s) - I(s) R + Vi (s) = 0 sC \
+ R –
Vo(s)
–
Fig. 3.2 (a)
Vi (s) 1 = é + Rù = ê úû I (s) ë sC
And, Vo (s) = I(s) R i.e.
Vi(s) I(s)
1 + Rù Vi (s) = I(s) é êë sC úû
\
+
1 sC + –
1 + sCR sC
Vo (s) = R I(s)
Example 3.3.9 Solution : Applying we get the equations as, di 1 ... (1) + idt E i = iR + L dt C ò Input = E i ; Output = E o F(s) Laplace transform of ò f(t) dt = , … Neglecting initial conditions s df(t) … Neglecting initial conditions and Laplace transform of = sF(s) dt Take Laplace transform, I(s) 1 1 ù = ... (2) i.e. \ E i (s) = I(s) éR + sL + êë E i ( s) é sC úû 1 ù R + sL + êë sC úû
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3-3
Transfer Function and Impulse Response
1 1 idt i.e. E o (s) = I(s) i.e. I(s) = sCE o (s) Cò sC Substituting value of I(s) in equation (2), sCE o (s) 1 = \ E i (s) éR + sL + 1 ù êë sC úû Now
\
Eo =
E o (s) = E i (s)
... (3)
1 1 = 1 ù RsC + s 2 LC + 1 é sC R + sL + êë sC úû
So we can represent the system as in the Fig. 3.3 (a).
1
Ei(s)
2
s LC + sRC + 1
Eo(s)
Fig. 3.3 (a) Transfer function model
Example 3.3.10 Solution : The s-domian network is shown in the Fig. 3.4 (a). 1 R1 || ––– sC1 R1 = ––––––– 1+sR1C1 1 ––– sC1
R1
Z1(s)
Ei(s)
Ei(s) 1 ––– sC2
R2
Z2(s)
Eo(s)
Eo(s)
1 R2 || ––– sC2 R2 = ––––––– 1+sR2C2 (a)
(b)
Fig. 3.4
\
\
Eo(s) Z2 T.F. = = = E1(s) Z1 + Z 2
T.F. =
R2 1 + s R 2 C2 R1 R2 + 1 + sR1C1 1 + sR 2 C2
R 2 (1 + s R 1 C 1 ) (R 1 + R 2 ) + sR 1 R 2 (C 1 + C 2 )
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Example 3.4.2
3-4
Transfer Function and Impulse Response
Kept this unsolved example for student's practice.
Example 3.5.2 Solution : From pole-zero plot given, the transfer function has 3 poles at s = – 1, – 2+j and – 2– j. And it has one zero at s = – 3. K(s + 3) K(s + 3) K(s + 3) = = \ T(s) = 2 2 (s + 1) (s + 2 + j) (s + 2 - j) (s + 1) [(s + 2) - (j) ] (s + 1) [s 2 + 4s + 5] Now d.c. gain is value of T(s) at s = 0 which is given as 10. K´ 3 50 i.e. K = = 16.667 d.c. gain = T(s)|at s= 0 i.e. 10 = \ 1´5 3 \
T(s) =
16.667 (s + 3) ... Transfer function
( s + 1) ( s 2 + 4s + 5)
Example 3.5.3 Solution : From the pole-zero plot shown in the Fig. 1, there is 1 pole at origin, s= 0 , 1 pole at s = –1, 1 pole at s = –2, 1 zero at s = –3. Hence the overall transfer function with gain 2.5 is, \
T(s) =
2.5( s + 3) s( s + 1)( s + 2)
qqq
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4
Solutions of Examples for Practice Example 4.6.1 Solution : As discussed earlier work done by the gears remains same. T1 N1 q2 = \ = T2 N2 q1 Now T2 is supplying torque to load. d2 q2 dq 2 + B2 \ T2 (t) = J 2 2 dt dt \
As T2 =
\
T1 =
Now
q2
and
T1 , Substituting in equation (1)
N1 d2 q2 N1 dq 2 J2 + B2 N2 N2 dt dt 2 N1 = q 1 , Substituting in equation (2) N2
T1 = Hence
N2 N1
... (1)
2 N1 N1 æ N ö d q1 J2 ç 1 ÷ + N2 N2 è N 2 ø dt 2
æ N ö dq 1 B2 ç 1 ÷ è N 2 ø dt
2
J2
é N1 ù ê N ú = Equivalent M.I. referred to primary = J eq ë 2û
B2
é N1 ù ê N ú = Equivalent friction referred to primary = B eq ë 2û
2
\ Taking Laplace of equation (3) and assuming initial conditions zero. T1 (s) = J eq s 2 q 1 (s) + B eq s q 1 (s) = q 1 (s) [s 2 J eq + s B eq ] \ where
q 1 (s) = T1 (s)
1 s [s J eq + B eq ]
J eq = Equivalent M.I., B eq = Equivalent friction. (4 - 1) TM
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... (2)
... (3)
Control System Engineering
4-2
Mathematical Models of Control Systems
Example 4.9.10
At node 2,
0 = K1 (x 2 - x 1 ) + K2 x 2 + M2
i) F-V analogy :
d2 x 2 dt
2
1 C1
ò (i 1 - i 2 ) dt
0 =
1 C1
ò (i 2 - i 1 ) dt + L 2
x2
M2
F
K2 B2
Equivalent system
+ B2
dx 2 dt
M ® L, B ® R, K ® 1/C, x ® ò i dt,
v(t) =
K1
x1
Solution : Two displacements : No element under x 1 (t) alone as force is directly applied to a spring K1 . So it will store energy and hence is the cause to change the force applied to M2 . Hence displacement of M2 is x 2 and as B 2 and K2 are connected to fixed supports both are under x 2 (t) only as shown in the equivalent system. At node 1, f(t) = K1 (x 1 - x 2 ) ... (1)
... (2) dx d2 x di ® i, ® 2 dt dt dt ... (3) Loop (1)
di 2 1 + dt C2
ò i 2 dt + R 2 i 2
... (4) Loop (2)
The F-V analogous network is shown in the Fig. 4.1. L2
V(s)
I1
C1 (1/K1)
C2 (1/K2) I2
R2
Same displacement same current.
Fig. 4.1
ii) F - I analogy : M ® C, B ® 1/R, K ® 1/L, x ® ò v dt, \
i(t) =
1 L1
ò (v 1 - v 2 ) dt
0 =
1 L1
ò (v 2 - v 1 ) dt + C2
dx d2 x dv ® v, ® 2 dt dt dt ... (5) Node (1)
dv 2 v 2 1 + + dt R2 L2
TM
ò v 2 dt
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... (6) Node (2)
Control System Engineering
4-3
Mathematical Models of Control Systems
The F-I analogous system is shown in the Fig. 4.1 (a). v1
L1(1/K1)
v2
2
1 I(s)
L2 (1/K2)
Same displacement same voltage.
R2 (1/B2)
C2
Fig. 4.1 (a)
Example 4.9.11 Solution : There are three displacements x i , x o and x y . The input is x i and output is x o . So transfer function of the mechanical system is X o (s)/X i (s). The equilibrium equations are, d(x o - x y ) d(x o - x i ) = 0. ... (1) B1 + K1 (x o - x i ) + B 2 dt dt d(x y - x o ) ... (2) B2 + K2 x y = 0 dt Taking Laplace transform of both the equations, neglecting initial conditions, B 1 sX o (s) - B 1 sX i (s) + K1 X o (s) - K1 X i (s) + B 2 sX o (s) - B 2 sX y (s) = 0 \ X o (s)[sB 1 + K1 + sB 2 ] - X i (s) [sB 1 + K1 ] - B 2 sX y (s) = 0 B 2 sX y (s) - B 2 sX o (s) + K2 X y (s) = 0
i.e. X y (s) =
... (3)
B2 s X (s) sB 2 + K2 o
... (4)
Substituting in equation (3), sB 2 sB 2 X o (s) =0 sB 2 + K2 (sB 1 + K1 ) (sB 2 + K2 )
X o (s)[sB 1 + K1 + sB 2 ] - X i (s)[sB 1 + K1 ] Solving,
X o (s) = X i (s) s 2 B 1 B 2 + sB 1 K2 + sK1 B 2 + K1 K2 + s 2 B 22 + sK2 B 2 - s 2 B 22
\
B öæ B ö æ ç1 + s 1 ÷ ç1 + s 2 ÷ K K X o (s) 1ø è 2 ø è = X i (s) B öæ B ö B æ ç1 + s 1 ÷ ç1 + s 2 ÷ + s 2 K K K1 1ø è 2 ø è
TM
... Required transfer function
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4-4
Mathematical Models of Control Systems
The s domain network for the given electric network is, R2 Z(s) R1
1 sC2
Ei(s)
Eo(s)
R1 +
Ei(s)
1 sC1
I(s)
(a)
1 sC1
Eo(s)
(b)
Fig. 4.2
Now
E i (s)
I(s) =
Z(s) + R 1 + and
\
1 sC 1
1 ù 1 é with Z(s) = R 2|| E o (s) = I(s) êR 1 + ú sC sC 1û 2 ë 1 ù é E i (s) êR 1 + sC 1 úû E i (s)(1 + sR 1 C 1 ) ë = E o (s) = 1 sC 1 Z(s) + sC 1 R 1 + 1 Z(s) + R 1 + sC 1 1 R2 sC 2 = Z(s) = 1 + sR 2 C 2 1 R2 + sC 2 R2 ´
Substituting
\
E o (s) = E i (s)
(1 + sR 1 C 1 ) (1 + sR 1 C 1 ) (1 + sR 2 C 2 ) = sC 1 R 2 (1 + sR 1 C 1 ) (1 + sR 2 C 2 ) + sC 1 R 2 + (1 + sR 1 C 1 ) (1 + sR 2 C 2 )
1 , B ® R the two transfer functions are identical hence the two systems are C analogous in nature.
As K ®
Example 4.9.12 Solution : The displacements are q 1 and q as shown in the Fig. 4.3 (a). The J 1 is under q 1 . The displacement changes to q due to K. While J2 is under q. Thus the equivalent system is as shown in the Fig. 4.3 (a). (See Fig. 4.3 (a) on next page) The equilibrium equations are, d2 q1 T = J1 + K( q 1 - q) dt 2
…(1)
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4-5
Mathematical Models of Control Systems fv3 x1(t)
K J1 T
q1
J2 q1
q
(Same)
K2
f(t)
q
M1
K1
fv1
(Same)
(a)
K3
M2
fv2
(b)
Fig. 4.3
0 = K( q - q 1 ) + J 2
2
d q
…(2)
dt 2 Taking Laplace transform of equation (1) and equation (2), T(s) = J 1 s 2 q 1 (s) + Kq 1 ( s) - Kq( s) …(3) \ \
0 = Kq(s) - Kq 1 (s) + J 2 s 2 q(s)
é K+ J 2 From equation (4), q 1 (s) = ê K êë
s2
…(4)
ù ú q(s) úû
q1
T
K
J1
q
J2
…(5) Fig. 4.3 (c)
Using equation (5) in equation (1),
é K+ J 2 s 2 ù T(s) = q 1 (s) [ J 1 s 2 + K] - Kq(s) = ê ú ( J 1 s 2 + K) q(s) - Kq(s) K êë úû \
ìï (K + J 2 s 2 ) (J 1 s 2 + K) üï T(s) = í - Ký q(s) K ïî ïþ
\
q(s) K K = = 2 2 2 4 T(s) (K + s J 2 ) (K + s J 1 ) - K s J 1 J 2 + Ks 2 J 2 + Ks 2 J 1
\
q(s) K = 2 2 T(s) s [s J 1 J 2 + K( J 1 + J 2 )]
Example 4.9.13 Solution : The mass M1, K1 and friction fv1 are under the displacement x1(t) while mass M2, K3 and friction fv2 are under the displacement x2(t). The spring K2 and the friction fv3 are between x1(t) and x2(t). The equivalent mechanical system is shown in the Fig. 4.4 (a). The equilibrium equations are,
fv3 x1(t) K2
f(t)
K1
M1
fv1
K3
Fig. 4.4 (a) TM
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M2
fv2
Control System Engineering
f(t) = M1
4-6
d2 x 1 dt 2
dx 1 d (x 1 - x 2 ) + K2 (x1 – x2) + fv3 dt dt
…(1)
d (x 2 - x 1 ) dx 2 d2 x 2 + M2 + fv2 + K3 x2 2 dt dt dt
…(2)
+ K1 x1 + fv1
0 = K2 (x2 – x1) + fv3
Mathematical Models of Control Systems
Taking Laplace transform of both the equations, 2
F(s) = X1(s) [M1s + s(fv1 + fv3) + (K1 + K2)] – X2(s) [sfv3 + K2] 0 = – X1(s) [K2 + sfv3] + X2(s) [M2s2 + s (fv2 + fv3) + (K2 + K3)]
…(3) …(4)
é M s 2 + s(fv2 + fv3 ) + (K2 + K 3 ) ù From equation (4), X1(s) = ê 2 ú X2(s) K2 + sfv3 êë úû Using in equation (3), ì [M s 2 + s(fv1 + fv3 ) + (K1 + K2 )][M2 s 2 + s(fv2 + fv3 ) + (K2 + K3 )] ü F(s) = X 2 (s) í 1 - (sfv3 + K2 ) ý (sfv3 + K2 ) î þ
X 2 (s) (sfv3 + K2 ) = 2 F(s) [M1 s + s(fv1 + fv3 ) + (K1 + K2 )] [M2 s 2 + s(fv2 + fv3 ) + (K2 + K 3 )] - [sfv3 + K2 ] 2
\
Example 4.9.14 Solution : Write the equilibrium equation At 1, At 2,
F(t)
= K(x 1 - x 2 )
... (1)
0 = K(x 2 - x 1 ) + M
d2 x 2 2
+ B
d x2 dt
dt Taking Laplace of both, neglecting initial conditions F(s) = KX 1 (s) - KX 2 (s) 0 = KX 2 (s) - KX 1 (s) + Ms 2 X 2 (s) + BsX 2 (s) from equation (3)
... (2)
... (3) ... (4)
1 X 1 (s) = [F(s) + KX 2 (s)] K
Substituting in (4)
1 0 = KX 2 (s) - Kìí [F(s) + KX 2 (s)]üý Ms 2 X 2 (s) + Bs X 2 (s) îK þ
\
0 = KX 2 (s) - F(s) KX 2 (s) + Ms 2 X 2 (s) + Bs X 2 (s)
\
F(s) = sX 2 (s) [Ms + B] \
X 2 (s) 1 = F(s) s(Ms + B)
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4-7
Mathematical Models of Control Systems
Example 4.9.15 Solution : The equilibrium equation at node X o is, B1
d (x o - x i ) dx o + K1 (x o - x i ) + B 2 + K2 x o = 0 dt dt
Taking Laplace transform and neglecting initial conditions, B 1 s X o (s) - B 1 s X i (s) + K1 X o (s) - K1 X i (s) + B 2 sX o (s) + K2 X o (s) = 0 \ X o (s) [s B 1 + K1 + s B 2 + K2 ] = [K1 + s B 1 ] X i (s) s B 1 + K1 X o (s) = s (B 1 + B 2 ) + (K1 + K2 ) X i (s)
\
For force-current analogy, B ®
1 , R
R1(
· 1 K ® , x ® f, x = V, x = ò V dt L 1 1 \ [V - Vi ] + (Vo - Vi ) dt R1 o L1 ò
+
1 1 V + R2 o L2
ò
Vi L1( Vi
Vo dt = 0
The force-current equivalent analogous electrical network is as shown in the Fig. 4.5 (a).
1 ) B1
Vo
1 ) K1 R2 (1/B2)
L2 (1/K2)
Vo
Fig. 4.5 (a)
Example 4.9.16 Solution : q(t) = Deflection and e(t) is the applied voltage which is to be measured. Torque developed by coil is proportional to current passing through coil T µ i(t) T = KT i(t) dq(t) dt Now torque produced has to overcome spring torque and M. I. J of system. d 2 q(t) T = Ks q (t) + J \ dt 2 Now analysing coil circuit, dq(t) i.e. e(t) = i(t) R + Kb e(t) = i(t) R + e b (t) dt Taking Laplace of all the equations, T(s) = KT I(s) e b = Kb
... (1)
T(s) = Ks q (s) + s 2 J q (s)
... (2)
e(s) = I (s) R + Kb s q (s)
... (3) TM
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Mathematical Models of Control Systems
\ Equating (1) and (2), KT I(s) = Ks q(s) + s 2 J q(s) é K + s2 Jù I(s) = q (s) ê s \ ú êë KT úû \ Substituting in (3), e(s) = \
é R (Ks + s 2 J) + Kb s KT q (s) R ( Ks + s 2 J ) + Kb s q (s) = q (s) ê KT KT êë
ù ú úû
KT q (s) = is the required transfer function 2 e(s) J R s + Kb KT s + Ks R
Example 4.9.17 K2
Solution : Due to force applied, mass M1 will be displaced by x 1 (t). Now K 3 will consume same energy hence one end of B 1 will get displaced by displacement x 2 (t) which is less than x 3 (t). Now due to B 1 and K2 , the force F(t) on M2 will be different and M2 will be displaced by x 1 (t) which is other than x 1 (t) and x 2 (t).
x1
x2 K3
M1
K1
x3 B1 M2
So mass M1 and K1 are under influence of x 1 (t). Spring K 3 under influence of x 1 - x 2 Spring K2 under influence of x 1 - x 3 Friction B 1 under influence of x 2 - x 3 . Hence equivalent mechanical system will be as shown in above figure. The equations of equilibrium are, d2 x 1 At x 1 , F(t) = M1 + K1 x 1 + K 3 (x 1 - x 2 ) + K2 (x 1 - x 3 ) dt 2 d(x 2 - x 3 ) At x 2 , 0 = K 3 (x 2 - x 1 ) + B 1 dt d(x 3 - x 2 ) d2 x 3 + K2 (x 3 - x 1 ) + M2 dt dt 2 Use F-V analogy and the replacements, dx 1 d2 x di M ® L, B ® R, ® i, K® , x ® q, ® 2 dt C dt dt At x 3 ,
0 = B1
and x ® q ® ò i dt TM
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... (1) ... (2) ... (3)
Control System Engineering
V(t) = L 1 0 =
1 C3
di 1 1 + dt C 1
4-9
1
Mathematical Models of Control Systems
1
ò i 1 dt + C 3 ò (i 1 - i 2 )dt + C2 ò (i 1 - i 3 )dt
... (1)
ò (i 2 - i 1 )dt + R 1 (i 2 - i 3 )
0 = R 1 (i 3 - i 2 ) +
1 C2
ò (i 3 - i 1 )dt + L 2
... (2) di 3 dt
... (3)
Simulating using loop basis, L1
C1 (1/K1)
R1
C3 (1/K3) i2 V
+ _
L2 C2 (1/K2)
i1
i3
Currents through various elements, L1 , G1
i1 alone
C3
i1 - i2
R1
i2 - i3
C2
i1 - i3
L2
i3
alone
Note The elements in parallel in equivalent mechanical system drawn based on nodal analysis
appears in series in F-V analogous system based on loop analysis. Similarly elements in series in mechanical system appears in parallel in F-V analogous system. Remember this, while drawing F-V analagous system. In this problem it can be observed that M1 , K1 in parallel
®
K 3 , B 1 in series ®
C 3 , R 1 in parallel
L 1 , C 1 in series
(K 3 , B 1 series) parallel with (K2 ) ® (C 3 , R 1 in parallel) series with (C 2 )
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4 - 10
Mathematical Models of Control Systems
Example 4.9.18 Solution : Due to applied force, M1 will be displaced by x 1 . Spring K1 will be under same displacement. Due to friction B 1 , K 3 will get displaced by x 2 as friction causes change in displacement. Due to spring K2 , friction B 2 will be under influence of x 3 which is different than x 2 . This is due to storing of energy by K2 . Due to B 2 and K 3 , mass M2 will again under different displacement x 4 . Due to friction B 3 , again there will be change in force applied to K4 hence K4 will be under the influence of displacement x5 . So mass M1 and K1 are under x 1 . B 1 between x 1 and x 2 . K2 between x 1 and x 3 . K 3 between x 2 and x 4 . B 2 between x 3 and x 4 . B 3 between x 4 and x5 . K4 under x5 alone. Hence equivalent system is as shown, K2 x2
x1
B2
x3
B3 x4
B1 F(t)
M1
K1
K3
M2
The equations of equilibrium are, d2 x 1 d(x 1 - x 2 ) F(t) = M1 At x 1 , + K1 x 1 + B 1 + K2 (x 1 - x 3 ) 2 dt dt
x5
K4
... (1)
d(x 2 - x 1 ) + K 3 (x 2 - x 4 ) dt
... (2)
d(x 3 - x 4 ) dt
... (3)
At x 2 ,
0 = B1
At x 3 ,
0 = K2 (x 3 - x 1 ) + B 2
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At x 4 ,
4 - 11
0 = K 3 (x 4 - x 2 ) + B 2
Mathematical Models of Control Systems
d(x 4 - x 3 ) d(x 4 - x5 ) d2 x 4 + B3 + M2 dt dt dt 2
d(x5 - x 4 ) dt Using F-V analogy, the replacements are,
At x5 ,
M ® L,
0 = K 4 x5 + B 3
B ®R
K®
and x ® q ® ò i dt as i = V(t) = L 1
1 , C
x ® q,
... (4) ... (5)
dx ® i, dt
d2 x dt
2
®
di dt
dq dt
di 1 1 + dt C 1
0 = R 1 (i 2 - i 1 ) +
1
ò i 1 dt + R 1 (i 1 - i 2 ) + C2 ò (i 1 - i 3 )dt
... (6)
1 C3
ò (i 2 - i 4 )dt
... (7) ... (8)
0 =
1 C2
ò (i 3 - i 1 )dt + R 2 (i 3 - i 4 )
0 =
1 C3
ò (i 1 - i 2 )dt + R 2 (i 4 - i 3 ) + R 3 (i 4 - i5 ) + L 2
0 =
1 C4
ò i5 dt + R 3 (i5 - i 4 )
di 4 dt
... (9) ... (10)
Elements
Currents
L1 , C1
i1
R1
i1 - i2
C2
i1 - i3
C3
i2 - i4
R2
i3 - i4
R3
i4 - i5
C4
i5
L2
i4
Remember parallel elements in series and series elemets in parallel. Mechanical
F-V
M1, K1 parallel
L1 , C1 series
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4 - 12
Mathematical Models of Control Systems
B1, K 3 series
R1, C3 parallel
K 2 , B2 series
C2 , R2 parallel
B1 , K 3 series in parallel with K 2 , B2
R1 C3 parallel in series with C2 ,R2 parallel
Thus the F-V analogous network based on loop basis is,
L1
C1 (1/K1)
R1 V (t)
L3
i2
C3 (1/K3) R3
+ _
i1
C2 (1/K2) i3
i4
R2
C4 (1/K4) i5
Check the elements and currents as per the table made above. Example 4.9.19 Solution : For the mass M 1 , the displacement is x 1 due to force f(t), against the friction force, d 2 x1 dx + B1 1 \ f (t) = M1 K (1) 2 dt dt Let force transmitted to mass M 2 be f2 (t) which will cause the displacement of x 2 of mass M 2 against spring K 2 and friction B2 . d 2 x2 dx2 + K1 x2 +B2 \ f2 (t) = M 2 K (2) 2 dt dt Now taking moments about fulcrum, f (t) ´ L1 = f 2 (t) ´ L 2 \
f 2 (t) =
L1 ´ f (t) L2
... (3)
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Mathematical Models of Control Systems
The equivalent mechanical system is, x2
x1
M1
B1
For F-V analogy,
f(t)
L 1/ L 2
M2
f2(t)
B2
M ® L , B ® R, K ® 1/C, x ® q, dx/dt ® I
Hence equations (1) and (2) get modified as, dI V(t) = L1 1 + I1 R1 dt V2 (t) = L 2 and
V2 (t) =
K2
dI 2
dt 2
+ I2 R 2 +
L1 . V(t) L2
1 C2
K (4)
ò I 2 dt
K (5) K (6)
The equation (6) indicates that it acts as a transformer with turns ratio
L1 . Hence F-V L2
analogous network is, R1
L1
L2
I1
v(t)
V2
R2
I2
C2 (1/K2)
(L1 / L2 )
For obtaining the transfer function X 2 (s) /X1 (s), take laplace transform of the equations (1) and (2),
and
F (s) = M1 s 2 ´ X1 (s) + B1 s X1 (s)
K (7)
F2 (s) = M2 s 2 X 2 (s) + B 2 s X 2 (s) + K2 X 2 (s)
... (8)
F2 (s) =
L1 F (s) L2
... (9)
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Substituting (7) and (8) in (9) we get, L1 M1s 2 X1 (s) + B1s X1 (s) L2
[
M 2 s 2 X 2 (s) + B2 s X 2 (s) + K 2 X 2 (s) = \
]
L1 X ( s) M1s 2 + B1s L2 1
]
L1s (M1s + B1 ) X2 ( s) = X1 ( s) L 2 M 2 s 2 + B2 s + K 2
)
[
X 2 (s) M 2 s 2 + B2 s + K 2
[
]=
(
This is the required transfer function.
qqq
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5
Solutions for Examples for Practice Example 5.3.11 Solution : The blocks G 1 and G 2 are in series. Minor loop
3 blocks in parallel
G3 R(s)
+
+ –
G4
G1 G2
+
G6 +
G5
H1 H2 Series R(s)
+ –
G1 G2 1 + G1 G2 H1
G3 + G4 + G5
G6
C(s)
H2 Minor loop R(s)
+ –
G1 G2 (G3 + G4 + G5) 1 + G1 G2 H1
G6
H2
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C(s)
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Block Diagram Models of Control Systems
G1 G2 (G3 + G4 + G5) 1 + G1 G2 H1
R(s)
G6
1 + G1 G2 (G3 + G4 + G5)H2 1 + G1 G2 H1
C(s)
G1 G2 G6 (G3 + G4 + G5)
R(s)
C(s)
1 + G1 G2 H1 + G1 G2 H2 G3 + G1 G2 H2 G4 + G1 G2 G5 H2
Fig. 5.1
Example 5.3.12 Solution : Shifting the take off point to the right as shown we get, Parallel G3 G2 R(s)
G1
G2
1
+
+
C(s)
–
–
H1
The dotted portion shown has a parallel combination of block of 1 and G 3 / G 2 , so they add up. Then shift summing point to the left of G 1 . R(s) +
G1
G2
–
–
1+
G3 G2
C(s)
H1
R(s)
G1 –
G2
–
G2+G3 G2
1 G1 H1
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Block Diagram Models of Control Systems
Interchanging the positions of summing points using Associative law,
Minor feedback loop
R(s)
G1 –
G2
–
G2 + G3 –––––– G2
... Rule 1
C(s)
H1
G1G2 ––––––––– 1+G1G2H1
1 –– G1
Combining two blocks in forward path we get canonical form from which we can write, C(s) = R(s)
G 1 G 2 (G 2 + G 3 ) G 2 (1 + G 1 G 2 H 1 ) = G G (G + G 3 ) 1 1+ 1 2 2 × G 2 (1 + G 1 G 2 H 1 ) G 1
G 1 (G 2 + G 3 ) 1 + G 1G 2 H 1 + G 2 + G 3
... Ans
Example 5.3.13 Solution : Combine the parallel blocks G 2 and G 3 to give G 2 + G 3 and minor loop of G 1 and H 1 . G4
R(s) –
+
G1 –––––––––– 1 + G1H1
C(s)
G2 + G3
Minor feedback loop
H2
Reduce the minor feedback loop of G 1 and H 1 to give
G1 . 1+ G 1 H 1
G4
R(s)
G1(G2 + G3)
+
1 + G1H1 + G1H2(G2+G3)
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Block Diagram Models of Control Systems
Reducing inner minor loop we get, The two blocks are in parallel, so adding them we get the single equivalent block as
C(s) R(s)
R(s)
=
G4+ G4G1H1+ G4G1G2H2+G4G1G3H2+ G1G2 + G1G3 1 + G1H1 + G1G2H2 + G1G3H2
G1G2 + G1G3
G4 +
… Ans.
C(s)
1 + G1H1 + G1G2H2 + G1G3H2
Example 5.3.14 Solution : The blocks G 2 and G 3 in parallel, use rule 3. Minor loop of 1 and G6
G1
R(s)
–
+
G2 + G3
–
+
G5
G6
C(s)
+
– G4
G7
G2 + G3
G1
R(s)
Minor loop with G = 1, H = G1(G2 + G3)
–
+
G2 + G3
G5
– G4
+ +
Series
G7
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G6 1 + G6
C(s)
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5-5
Block Diagram Models of Control Systems G4 + G5(G2 + G3)
Parallel blocks, Rule 3 R(s)
+
1 1 + G1(G2 + G3)
G5(G2 + G3)
–
+
G6 1 + G6
+
C(s)
G4
G7
+
R(s)
1 1 + G1(G2 + G3) G4 + G5(G2 + G3)
–
G6 1 + G6
C(s)
G7
G 2 G5 G 6 + G 3 G5 G 6 + G 4 G 6 (1 + G 1 G 2 + G 1 G 3 ) (1 + G 6 )
G eq =
\ \
H eq = G 7 Geq
R(s)
C(s)
1 + GeqHeq
Example 5.3.15 Solution : Rearranging the block diagram we get,
R
+
G3 –
+
G8
G5
G2
G1
C
– G7 + G6
G4 +
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Block Diagram Models of Control Systems
Separate the feedback paths through G 4 and G 7 , combined at summing point before G 6 . So G 6 will appear separately in each path of G 4 and G 7 , as shown below. Minor loop
R
G3
G8
–
–
G2G5
G1
C
– G6
G7 G6
G4
Reducing inner minor feedback loop, R
G3 –
–
G8 1 + G6G7G8
G2G5
G1
C
G4G6
Hint : Combine inner blocks, reduce minor feedback loop. Again combine blocks in forward path to get simple form. Using \
G we get, 1 + GH G 1 G 2 G 3 G5 G 8 C(s) = 1 + G 6 G7 G 8 + G 2 G 4 G5 G 6 G 8 + G 1 G 2 G 3 G5 G 8 R(s)
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... Ans.
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Block Diagram Models of Control Systems
Example 5.3.16 Solution : Shifting take off point before G 2 .
... Rule 6
Critical rule R(s)
G3
G1
+
+
G2
C(s)
–
–
H2 H1
G2
Shifting take off point before summing point using critical rule No. 10, G3 +
R(s)
G1 –
G2
C(s)
+ H1G2
H2
–
For this negative sign, include block of '–1' while separating the path.
Separating the paths in the feedback path shown dotted, G3 +
R(s)
G1 –
G2
C(s)
–
–
H2
H1G2 H1G2
–1
H2
Key Point Remember that though the paths through summing point are separated, signs at the
summing points to those paths must be carried as it is. Hence after H 2 , carry the negative sign and then H 1 G 2 to get – H 1 H 2 G 2 .
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Shifting summing point as shown and then interchanging the two summing points using associative Law we get, ... Rule 4 and 1 Parallel
G3
Minor loop
G2 R(s)
+
G1 1 + H1G1G2
–
C(s)
G2
1 –
H2 – H1H2G2
Rule 3 R(s)
G1 1 + H1G1G2
–
1+
Rule 8
G3
G2
G2
1 + G2H2
C(s)
– H1H2G2
Geq
Minor loop R(s)
C(s)
G1 (G2 + G3) –
(1 + G2H2) (1 + H1G1G2) – H1H2G2 Heq
Rule 8 gives,
G eq C(s) = = R(s) 1 + G eq H eq
G 1 (G 2 + G 3 ) 1 + G 2 H2 + H1 G 1 G 2 - G 1 G 2 G 3 H1 H2
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Block Diagram Models of Control Systems
Example 5.3.17 Solution : Shift summing point before G1 and take off point of H2 after the block G4.
–
R
1/G1
H2
1/G4
G1
G2
G3
–
–
Y
G4
H3 H1 Minor G3G4 –––––––– loop 1+G3G4H3
H2 –––– G1G4 –
R
G3G4 –––––––– 1+G3G4H3
G1G2
– H1
R –
Y
G1G2G3G4 –––––––––– 1+G3G4H3 Minor loop = ––––––––––––––––– G1G2G3G4 H2 1+ –––––––––– ´ –––– 1+G3G4H3 G1G4
G1G2G3G4 –––––––––––––––– 1+G3G4H3+G2G3H2
Y
H1
R
G1G2G3G4 –––––––––––––––– G1G2G3G4 1+G3G4H3+G2G3H2 –––––––––––––––––– = ––––––––––––––––––––––––––––– G1G2G3G4 1+G3G4H3+G2G3H2+G1G2G3G4H1 1 + –––––––––––––––––– ´ H1 1+G3G4H3+G2G3H2
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Block Diagram Models of Control Systems
Example 5.3.18 Solution : Shift the take off point of V3(s) after G2. 1/G2
R(s)
+
G1 –
+
G2 –
+ – H3
H2 H1
Parallel 1 1+G2 1+ –– = –––– G2 G2
Minor loop G2 –––––– 1+G2H2
R(s) –
G2 –––––– 1+G2H2
G1
Minor loop G3 –––––– 1+G3H3
G3 –––––– 1+G3H3
1+G2 ––––– G2
H1
C(s)
G1G2 Minor loop ––––––– G1G2 1+G2H2 –––––––––– = –––––––––––––– 1+G G1G2H1 2H2+G1G2H1 1+ ––––––– 1+G2H2
G1G2 –––––––––––––– 1+G2H2+G1G2H1
R(s)
C(s)
G3
1+G2 ––––– G2
G3 –––––– 1+G3H3
C(s)
Series
R(s)
G1G3(1+G2) –––––––––––––––––––––––––––––––––––––––– 1+G2H2+G1G2H1+G3H3+G2G3H2H3+G1G2G3H1H3
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Block Diagram Models of Control Systems
Example 5.3.19 Solution : Shifting take off point before G 2 . G3 R(s)
G1
+
+
G2
C(s)
–
–
H2 H1
G2
Shifting take off point before summing point using critical rule No. 10, G3 +
R(s)
G1 –
G2 –
+ H1G2
C(s)
H2
–
Separating the paths in the feedback path shown dotted, G3
Minor feedback loop R(s)
+
G1 –
G2
C(s)
–
–
H2
H1G2 – H1H2G2
Key Point Remember that though the paths through summing point are separated, signs at
the summing points to those paths must be carried as it is. Hence after H 2 , carry the negative sign and then H 1 G 2 to get – H 1 H 2 G 2 . Shifting summing point as shown and then interchanging the two summing points using Associative Law we get,
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Block Diagram Models of Control Systems
Blocks in parallel G3 G2 R(s)
Minor feedback loop
+
G1 1 + H1G1G2
–
C(s)
G2 – H2 – H1H2G2
G2 + G3 G2 R(s)
G1 –
1 + H1G1G2
1+
C(s)
G3
G2
G2
1 + G2H2
– H1H2G2
R(s)
G1 (G2 + G3) –
(1 + G2H2) (1 + H1G1G2) – H1H2G2
C(s) = R(s)
\
G 1 (G 2 + G 3 ) (1 + G 2 H 2 ) (1 + H 1 G 1 G 2 ) G (G + G 3 ) ( - H 1 H 2 G 2 ) 1+ 1 2 (1 + G 2 H 2 ) (1 + H 1 G 1 G 2 )
G 1 (G 2 + G 3 ) C(s) = 1 + G 2 H2 + H1 G 1 G 2 - G 1 G 2 G 3 H1 H2 R(s)
Example 5.3.20 Solution : Separating the feedback paths,
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R(s)
G1
–
Block Diagram Models of Control Systems
G2
– +
–
H1
–
S
C(s)
G3
H2
Separating the dotted paths from the summing points 'S'. S1
T1
R(s)
G1
–
S2 G2
–
–
4
H1
1
T2
C(s)
3
+
H1
–
2
G3
H2
Separating the paths from T1 to S 1 , T1 to S 2 , T2 to S 1 and T2 to S 2 . Parallel R(s)
G1
–
H1
–
–
H1
G2 1 H1
4
–H2
–
C(s)
G3
–
Minor feedback loop
H1
2
– H2 3 – H1H2
– R(s) –
G1 – H1
G3
G2
1 + G3(–H1H2)
– H1
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Block Diagram Models of Control Systems
–H1H2 – R(s)
G1–H1
–
Interchange
G3
G2
–
C(s)
1 – G3H1H2
Interchange the summing points
1 G2
H1
–H1H2 R(s) –
–
+
G1–H1
C(s)
G2G3
1 – G3H1H2
–
H1
G2
Minor loop
Parallel
G2G3 1 – G3H1H2–G2G3H1H2
G2G3 R(s)
G1 – H1 –
–
1 – G3H1H2
H1 G2
1+
G2G3 1 – G3H1H2
C(s) × –H1H2
G1G2–H1G2–H1 G2
R(s)
G1G2 – H1(1 + G2) –
G2
×
G2G3 1 – G3H1H2 – G2G3H1H2
[G 1 G 2 - H 1 (1 + G 2 ) ] G 3
\
C(s) = R(s)
1 - G 3 H 1H 2 - G 2 G 3 H 1H 2 1+
[G 1 G 2 - H 1 (1 + G 2 ) ] G 3 1 - G 3 H 1H 2 - G 2 G 3 H 1H 2 TM
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\
5 - 15
Block Diagram Models of Control Systems
G 1G 2 G 3 - H 1G 3 - H 1G 2 G 3 C(s) = R(s) 1 - G 3 H 1H 2 - G 2 G 3 H 1H 2 + G 1G 2 G 3 - H 1G 3 - H 1G 2 G 3
Example 5.3.21 Solution : Separating two feedback from second takeoff point which is after block having transfer function G 2 as shown, we get, Minor feedback loop
H2
–
R(s)
G2
G1
C(s)
G3
–
+
+ H1
H1 G4
H2 –
R(s)
G2
G1
1 + G2H1
+
C(s)
G3 +
H1 G4
Shifting summing point behind the block having transfer function ‘G 1 ’ as shown we get, 1 / G1
H2
–
R(s)
G2
G1
1 + G2H1
+ H1 G4
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Block Diagram Models of Control Systems
Use Associative law for the two summing points and interchange their positions, we get, H2 G1 –
R(s)
G2
G1 +
C(s)
G3
1 + G2H1
+ Minor loop
Series H1 G4
G1G2 1 + G2H1 1–
G1G2 1 + G2H1
=
× H1
G1G2G3
Minor loop
1 + G2H1 – G1G2H1 + G2G3H2
H2 G1 R(s)
–
G1G2 1 + G2H1 – G1G2H1
G1G2G3
G1G2G3
C(s)
1 + G2H1 – G1G2H1
1 + G2H1 – G1G2H1
R(s) 1+
+
C(s)
G1G2G3
H2
1 + G2H1 – G1G2H1
G1
G4
+
G4
G4 and other block are in parallel hence, R(s)
\
G4 +
G1G2G3
C(s)
1 + G2H1 – G1G2H1 + G2G3H2
G 4 + G 4 G 2 H1 – G 4 G 1 G 2 H1 + G 2 G 3 G 4 H2 + G 1 G 2 G 3 C(s) = R(s) 1 + G 2 H1 – G 1 G 2 H1 + G 2 G 3 H2
Example 5.3.22 Solution : Shift summing point to the left and take off point to the right as shown in the Fig. 5.3 (a).
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Block Diagram Models of Control Systems R2
sC1 –
+
R(s)
–
1 sC1
+
1 R1
–
1 sC2
1 R2
C(s)
Fig. 5.3 (a)
Interchange the summing points and solve minor feedback loops. sC1R2 R(s)
+
– –
1 sC1R1
1 sC2R2
–
Loop 1
C(s)
Loop 2
Fig. 5.3 (b)
Loop 1 =
Loop 2 =
\
C( s) = R( s)
1 1 = sC 1 R 1 1 + sC 1 R 1 1 1+ sC 1 R 1
sC1R2 R(s)
–
1 1 + sC1R1
1 1 = sC 2 R 2 1 + sC 2 R 2 1 1+ sC 2 R 2
1 1 + sC2R2
C(s)
Fig. 5.3 (c)
1 1 æ ö æ ö ç ÷ ç ÷ 1 sC R 1 sC R + + 1 1ø è 2 2 ø è 1 1 1+ ´ sC 1 R 2 1 sC R 1 sC + + ( 1 1) ( 2 R2 )
C( s) 1 = 2 R( s) s C 1 C 2 R 1 R 2 + s (R 1 C 1 + R 2 C 2 + R 2 C 1 ) + 1 Example 5.3.23 Solution : No blocks are in series or parallel and no minor feedback loop is existing so shifting summing point towards left i.e. behind block with transfer function G 1 as shown, we get,
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G3
Block Diagram Models of Control Systems
1 / G1 +
R(s) –
C(s)
G2
G1 – H1 H2
Use Associative law for the summing points, we get, Parallel G3 1 + —— G1
R(s)
G3 G1
Minor loop
+
–
C(s)
G2
G1 – H1 H2 G1 + G3 G1
Series
G3 1 + —— G1
R(s) –
G1
C(s)
G2
1 + G1H1 H2
R(s) –
G1 + G3
G1 G2
G1
1 + G1 H1
C(s)
R(s) –
(G1 + G3)
G1 G2
G1
1 + G1 H1
H2
C(s)
H2
G2 (G1 + G3) R(s)
G2 (G1 + G3) –
C(s)
1 + G1H1
R(s)
1 + G1H1
1+
G2 (G1 + G3) H2
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1 + G1H1
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5 - 19
Block Diagram Models of Control Systems
G1 G2 + G2 G 3 C(s) = 1 + G 1 H1 + G 1 G 2 H2 + G 2 H2 G 3 R(s)
\ Example 5.3.24
Solution : Separate all the paths linked by take-off point and summing point in the feedback path. Parallel of ‘1’ and G3
Minor loop
R(s)
+
G1
–
–
G2
+
G3
C(s)
+
+
H1
+ Separate the dotted paths Minor loop
G2 ––––––– 1 + G2
+
G1
R(s)
–
–
C(s)
1 + G3
H1 H1
R(s)
G1
G2 –––––––––––– 1 + G2 + G2H1
G2 ––––––– 1 + G2
+ –
G2H1 1 + ––––––– 1 + G2
1 + G3
H1
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G2 (1 + G3) ––––––––––– 1 + G2+ G2H1
G1 +
R(s)
5 - 20
Block Diagram Models of Control Systems
C(s)
–
G2 (1 + G3) ––––––––––– 1 + G2+ G2H1
G1
R(s)
C(s)
G2 (1 + G3)H1 1 + ––––––––––– 1 + G2+ G2H1 H1 Minor loop
G 1 G 2 (1 + G 3 ) C(s) = 1 + G 2 + 2G 2 H 1 + G 2 G 3 H 1 R(s)
\ Example 5.4.3
Solution : In this case there are two inputs and two outputs. Consider one input at a time assuming other zero and one output at a time. Consider R 1 acting, R 2 = 0 and C 2 not considered R 2 = 0 and C 2 is suppressed (not considered). C 2 suppressed does not mean that C 2 = 0. Only it is not the focus of interest while C 1 is considered. As R 2 = 0, summing point at R 2 can be removed but block of '–1' must be introduced in series with the signal which is shown negative at that summing point. Forward path R1
G1
–
Minor loop
C1
G3
G1
R1
»
– – G2G3G4
G4 G2
–1
C1
Feedback path
C1 G1 G1 = = R1 1 + [G 1 ] [- G 2 G 3 G 4 ] 1 - G 1 G 2 G 3 G 4 For
C2 , assume C 1 suppressed. R1 R1
G1 – G3
Forward path
»
G4 –1
G2
– G1G3G2
R1 –
Feedback path
G4
C2
Minor loop
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C2
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Block Diagram Models of Control Systems
C2 - G1 G 3 G2 - G1 G2 G 3 = = R1 1 + [- G 1 G 3 G 2 ] [G 4 ] 1 - G 1 G 2 G 3 G 4
\
For
C1 , R 1 = 0 and C 2 is suppressed. R2 C1
G1
–1 Forward path
Feedback path
G3
R2
–
» G4
C1
G3
–
Minor loop
G2
R2
C1 - G1 G2 G4 -G1 G2 G4 = = R2 1 + [- G 1 G 2 G 4 ] [G 3 ] 1- G1 G 2 G 3 G4
\
For
– G1G2G4
C2 , R 1 = 0 and C 1 is suppressed. R2 G1
–1
G3 Feedback path
G4 –
R2
G2
G2
R2
»
– – G1G3G4 Minor loop
C2 Forward path
C2 G2 G2 = = R2 1 + [G 2 ] [- G 1 G 3 G 4 ] 1 - G 1 G 2 G 3 G 4 Example 5.4.4 Solution : i) With N(s) = 0 block diagram becomes
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Block Diagram Models of Control Systems Minor loop
3 R(s)
+
E(s) –
C(s)
10 s(s + 1)
s+4 –
0.5 s
10 s (s + 1) 10 10 = Minor feedback loop = = 2 2 10 1+ 0.5s s + s + 5s s + 6s s (s + 1) 3 R(s)
E(s)
+ s+4
C(s)
10 2
s + 6s
–
3 s+4 R(s)
E(s)
+ X(s)
–
s+4
10
C(s)
2
s + 6s
Assume output of second summing points as X(s), Hence
E(s) = R(s) – C(s)
While
X(s) = E(s) +
…(i)
and
C(s) = X(s)
10 (s + 4) s 2 + 6s
3 R(s) s+ 4
... (ii)
... (iii)
Substituting value of X(s) and R(s) from (i) and (ii) in (iii) we get, s 2 + 6s 3 3 C(s) = E(s) + E(s) + C(s) s+ 4 s+ 4 10 (s + 4) C(s) 10 (s + 7) = 2 E(s) s + 6s - 30
Solving,
ii) To find
C(s) , R(s)
when N(s) = 0
we have to reduce block diagram solving minor feedback loop and
shifting summing point to the left as shown earlier in (i). TM
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Block Diagram Models of Control Systems
So referring to block diagram after these two steps i.e. 3 s+4 R(s)
+
E(s)
C(s)
10
s+4
2
s + 6s
–
Exchanging two summing points using associative law, 3 s+4 +
R(s)
C(s)
10 (s + 4) 2
3 Parallel of '1' and –––– s+4 3 = 1 + –––– s+4 s+7 = –––– s+4
\
s + 6s
–
G Minor loop –––––– = 1 + GH
10 (s + 4) –––––––– s2 + 6s 10 (s + 4) 1 + –––––––– s2 + 6s
10(s+4) = ––––––––––– 2 s +16s+40
C(s) 10(s + 7) æ s + 7 ö æç 10 (s + 4) ö÷ = = ç ÷ ´ 2 2 R(s) s + 4 ç ÷ è ø è s + 16s + 40 ø s + 16s + 40
iii) With R(s) = 0 block diagram becomes, + s+4 –
10 s(s + 1)
+ –
N(s)
+
C(s)
0.5 s
The block of ‘3’ will not exist as R(s) = 0. Similarly first summing point will also vanish but student should note that negative sign of feedback must be considered as it is, though summing point gets deleted.
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In general while deleting summing point, it is necessary to consider the signs of the different signals at that summing points and should not be disturbed. So introducing block of ‘–1’ to consider negative sign.
+
–
+
+
10 s(s + 1)
s+4
–1
–
+
C(s)
+
+
N(s) C(s)
0.5 s
N(s)
+ 10 s(s + 1)
Block Diagram Models of Control Systems
C(s)
+
10 s(s + 1)
+
0.5 s
N(s)
– (s + 4) – 0.5 s
– (s + 4)
After simplification ( – 1.5 s – 4)
Parallel
Removing summing point, as sign is positive no need of adding a block. + 10 s(s + 1)
+
N(s) C(s)
N(s)
+
C(s) + – 10 (1.5 s + 4) s(s + 1)
– (1.5 s + 4)
Minor loop with G = 1
\
C(s) N(s)
=
1 1 = 15s + 40 é -10 (1.5s + 4) ù 1+ 1-ê ú s(s + 1) ë s(s + 1) û
=
s(s + 1) s2
+ 16s + 40
Example 5.4.5 Solution : Consider R 1 alone R 2 , R 3 , R 4 are zero. Note Whenever R 1 is zero, as the sign of feedback at R 1 is negative, while removing
summing point at R 1 , do not forget to insert a block of '–1' to consider effect of negative sign.
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Block Diagram Models of Control Systems
Minor loop R1
G1 –
R1
C
G2
G2
G1
–
– H2
C
1 + G2H2
H1
H1
G 1G 2 G 1G 2 1 + G 2 H2 C = = \ 1 + G 2 H 2 + G 1G 2 H 1 R1 G 1G 2 H 1 1+ 1 + G 2 H2
C =
i.e.
G 1G 2 R 1 1 + G 2 H 2 + G 1G 2 H 1
Consider R 2 alone, with R 3 = R 1 = R 4 = 0 Note this
R2
–1
G1
R2
Minor loop
+ +
C
G2
+
+
G2 1 + G2H2
– H2 –G1H1
H1
G2 + 1 G C 2 H2 = \ R2 G2 æ ö 1 -ç ÷ ( -G 1 H 1 ) + 1 G H 2 2 ø è
C=
i.e.
G 2 R2 1 + G 2 H 2 + G 1G 2 H 1
Consider R 3 alone, R 1 = R 2 = R 4 = 0 Note this
R3 –1
G1
+
+
– G2
– H2 H1
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C
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Block Diagram Models of Control Systems
Combining two summing points we get, Minor loop R3
–
R3
C
G2
–
+ –
G2
C
1 + G2H2
+ H2
–G1H1
–G1H1
C -R3
\
\
C
G2 G2 1 + G 2 H2 = = + 1 G H G æ ö 2 2 + G 1G 2 H 1 2 1 -ç ÷ ( -G 1 H 1 ) è 1 + G 2 H2 ø =
- R3G2 1 + G 2 H 2 + G 1G 2 H 1
Consider R 1 alone, with R 1 = R 2 = R 3 = 0 Note this –1
G1
+
C
G2 – H2
R4 +
+
H1
–G1H1
+ R4
\
\
C R4
-G 1 G 2 H 1 -G 1 G 2 H 1 1 + G 2 H2 = = 1 + G G æ ö 2 H 2 + G 1G 2 H 1 2 1 - ( -G 1 H 1 )ç ÷ è 1 + G 2 H2 ø
C
=
- G 1G 2 H 1R 4 1 + G 2 H 2 + G 1G 2 H 1
Combining all the values of C, we get C
=
G 1 G 2 R 1 + G 2 (R 2 - R 3 ) - G 1 G 2 H 1 R 4 1 + G 2 H 2 + G 1G 2 H 1 TM
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G2 1 + G2H2
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Block Diagram Models of Control Systems
Example 5.4.6 Solution : i) Input R at station I R(s)
H3
+
–
+
G1
+
G2
C(s)
G3 –
–
Shift the takeoff point
H2 H1 R(s)
1 G3
H3
+
G1
–
–
+
G2
+
G3 –
C(s) Minor loop
H2 H1 R(s)
H3
Minor loop
G3 + –
–
+
G1
G3
G2
C(s)
1 + G3H2
H1
G2G3 1 + G3H2 + G2H3
R(s) G2G3 +
1 + G3H2
G1 1+
–
G2G3 1 + G3H2
C(s) ×
H3 G3
H1 TM
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G 1G 2 G 3 1 + G 3 H2 + G 2 H 3 C(s) = = R(s) G 1G 2 G 3 ´ H1 1+ 1 + G 3 H2 + G 2 H 3
Block Diagram Models of Control Systems
G1 G2G 3 1 + G 3 H 2 + G 2 H 3 + G 1G 2 G 3 H 1
ii) Input R at station II Note that at station I, through input is zero while removing the summing point, the negative sign of H 1 as it is. Thus add a block of '– 1' as shown in the system. Note this
R(s)
H3 –
+
G1
–1
+
+
G2
C(s)
G3
–
H2 H1
Shift the summing point to the left of G 2 and interchange the two summing points. Minor loop –
+
G1
–1
R(s)
H3
+ G2
+
–
H2
C(s)
G3
1 G2
H1 R(s)
+
–
G2 1+G2H3
+ G3
+
H2 G2
– G1H1 Blocks in parallel
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C(s)
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Block Diagram Models of Control Systems
R(s) + G2 1+G2H3
+
C(s)
G3
H – G1H1 – 2 G2 – G1G2H1– H2 G2
C(s) = R(s)
\
\
G3 G 3 (- G 1G 2 H 1 - H 2 ) 1– 1+ G2H3
R(s) + G3
+
G 3 (1 + G 2 H 3 ) C(s) = 1 + G 2 H 3 + G 3 H 2 + G 1G 2 G 3 H 1 R(s)
C(s)
– G1G2H1– H2 1 + G2H3
Example 5.4.7
Solution : Assume R( s) = 0 but to consider negative sign at the summing point introduce block of '–1' as shown in the Fig. 5.5 (a). Rearranged block diagram is shown in the Fig. 5.5 (b).
–1
W(s)
4
Note this 2
1 s+1
Y(s) W(s)
+
Y(s)
4
+ – 14 s+1
7 (a)
(b)
Fig. 5.5
é ù ê ú Y( s) 1 = 4´ ê ú W( s) ê1 - 1 ´ é -14 ù ú êë s + 1 úû úû êë
=
4( s + 1) s + 15
Example 5.4.8 Solution :
Solving the minor feedback loop of ‘G 4 ’ and ‘1’.
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Block Diagram Models of Control Systems
– G1
G2
G3
C1
+ H2 H1 + R1
G4
G5
1 + G4
+
As C 2 is not the focus of interest, G 6 becomes meaningless in the block diagram. Shifting take off point to the right of G 2 . Minor loop
– G1
G2
H2
1 G2
G3
C1
+ Shift
H1 +
R1
G4G5 1 + G4 G1G2
G3
1 + G1G2 1 G2
H2
1 G3
H1 + R1
G4G5 1 + G4
+
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Block Diagram Models of Control Systems
Combining all the blocks in series. G1G2G3G4G5H1
R1
C1
(1 + G1G2) (1 + G4)
+
H2 G2G3
C1 = R1
\
G 1 G 2 G 3 G 4 G5 H1 (1 + G 1 G 2 ) (1 + G 4 ) H2 G G G G G H 1- 1 2 3 4 5 1 × (1 + G 1 G 2 ) (1 + G 4 ) G 2 G 3
C1 G 1 G 2 G 3 G 4 G5 H1 = R1 1 + G 1 G 2 + G 4 + G 1 G 2 G 4 - G 1 G 4 G5 H1 H2
\
1 sC1
Example 5.5.2 Solution : The s-domain network, currents and voltages are shown in the Fig. 5.6 V (s) - V1 (s) I 1 (s) = i 1 sC 1
1 sC2
V1(s)
Vi(s) I1(s)
Vo(s)
R2
R1 I2(s)
Fig. 5.6
= sC 1 [Vi (s) - V1 (s)]
…(1)
V1 (s) = R 1 [I 1 (s) - I 2 (s)]
…(2)
I 2 (s) = sC 2 [V1 (s) - Vo (s)]
…(3)
Vo (s) = R 2 I 2 (s) The block diagrams for equations are, Vi(s)
sC1
I1(s)
I1(s)
Equation (1)
R1
V1(s)
–
– sC1
…(4)
V1(s)
R1
I2(s)
sC2
I2(s)
–
I2(s)
Equation (2)
TM
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sC2 Equation (3)
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Vo(s)
Vo(s)
Equation (4)
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Block Diagram Models of Control Systems
The complete block diagram is shown in the Fig. 5.7. Vi(s)
I1
sC1
V1
R1
I2
sC2 R1
Shift
Vo(s)
R2 Shift sC2
sC1
Fig. 5.7
Interchange Vi(s)
sC1
–
–
Interchange R1
sC2
1 –– R1
R2
–
Vo(s)
1 –– R2
R1
sC1
sC2 Minor loops
Vi(s)
sC1
R1
sC2
R2
–
Vo(s)
– sC2
sC1 1 1 1 –– x R1 x –– = –– R2 R 2 R1
R1 –––––– 1+s R1C1
R2 –––––– ––– 1+s R2 C2
Series sC1
Vi(s)
–
R1 ––––––– 1+sC1R1
sC2
R2 ––––––– 1+s R2 C 2
Vo(s)
1 –– R2
Vi(s)
sC1
s R1R2C2 (1 + sC1R1)(1 + sC2R2) sR1R2C2 x1 1+ (1 + sC1R1)(1 + sC2R2) R1
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Minor loop
Vo(s)
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Block Diagram Models of Control Systems
sR 1 R 2 C 2 Vo (s) = sC 1 ´ Vi (s) (1 + sC 1 R 1 ) (1 + sC 2 R 2 ) + sR 1 C 2
=
s 2 R 1R 2 C1C2
s 2 R 1 R 2 C 1 C 2 + s[C 1 R 1 + C 2 R 2 + R 1 C 2 ]+ 1
qqq
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6
Solutions of Examples for Practice Example 6.5.14 Solution : T1 = G 1 G 2 G 3 G 4
... Only one forward path
No combination of nontouching loops. G3
G2
G1
G3
G2
G4
– H2
–H1
L2 = –G3H2
L1 = –G2H1
– H3 L3 = –G1G2G3G4H3
D = 1– [L 1 + L 2 + L 3 ] = 1 + G 2 H 1 + G 3 H 2 + G 1 G 2 G 3 G 4 H 3
\
D1 = 1 \
... All loops touching to T1
TD G 1G 2 G 3 G 4 C(s) = 1 1 = D 1 + G 2 H 1 + G 3 H 2 + G 1G 2 G 3 G 4 H 3 R(s)
Example 6.5.15 Solution : Name all the summing and take off points as shown below and representing each separately as a node draw the signal flow graph. G4 R(s)
s2
s1 –
t1
s3 G1
G2
–
t2
s4 G3
– H2
H1
(6 - 1) TM
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t3
C(s)
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Signal Flow Graph Models of Control Systems G4
s1
1
1
s2
1
t1
G1
s3
G2
t2
G3
s4
t3
1
1 C(s)
R(s) –H2
–H1 –1
Number of forward paths are K = 2 Forward path gains are T1 = G 1 G 2 G 3 T2 = G 4 Individual feedback loops are G4 G1
1
G2
G2
G3
G2
1
1
–H2
–H1 L1 = –G1 G2 H1
–H1
L2 = –G2 G3 H2
1
–H2
L3 = +G2 G4 H1 H2
G4 1
G1
G3
G2
1
1
1
1
–1
–1
L4 = –G1 G1 G3
L5 = –G4
There are no combination of nontouching loops \
D = 1 - [L 1 + L 2 + L 3 + L 4 + L5 ]
All the loops are touching to both the forward paths \
D1 = D2 = 1
\ Using Mason’s gain formula T D + T2 D 2 C(s) = 1 1 D R(s)
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Signal Flow Graph Models of Control Systems
C(s) G1 G 2 G 3 + G 4 = 1 + G1 G 2 H1 + G 2 G 3 H 2 + G1 G 2 G 3 + G 4 - G 2 G 4 H1 H 2 R(s)
Example 6.5.16 Solution : Name the various summing and take-off points to draw the signal flow graph as shown, G3
+ s3
s1
R(s)
+
Input
H1
–
+ –
G2
G1
t1
C(s)
t2
s2
Output
H2
The corresponding signal flow graph is, +G3 1
1
Input
s1
s3
1 G1
–H1
s2
t1
1
G2 t2
Output
–H2
Forward path gains are, T1 = G 1 G 2
T2 = G 3 G 2
The various loops are, +G3 1
G1
G2
1
1
1
G2
–H1
G2 –H2
–H2
L1 = – G 1 G 2 H2
L2 = - G 2 G 3 H2
L 3 = – G 2 H1
No combinations of non toucting loops. \D = 1 – [L 1 + L 2 + L 3 ] = 1 + G 1 G 2 H 2 + G 2 G 3 H 2 + G 2 H 1 All loops are touching, \ D 1 = 1 For T1 , For T2 ,
All loops are touching, \ D 2 = 1 TM
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According to Mason's gain formula, T D + T2 D 2 C(s) = 1 1 R(s) D G1 G2 + G2 G 3 C(s) = 1 + G 1 G 2 H2 + G 2 G 3 H2 + G 2 H1 R(s) Example 6.5.17 Solution : Representing each take off point and summing point by separate node, the signal flow graph is as shown in the Fig. 6.1. G6
1
R(s)
G5
G1
1
G2
G3
G4
1
1
C(s)
–H1 –H2
Fig. 6.1
Forward paths :
T1 = G1G2G3G4 ,
T2 = G 6G 2G 3G4,
T3 = G1G2G5 , T 4 = G 6 G2 G5 Individual feedback loops : L1 = – G2G3H1, L2 = – G1G2G3G4H2, L3 = – G1G2G5H2 No combination of two non touching loops. \ D = 1– [L 1 + L 2 + L 3 ] = 1 + G2G3H1 + G1G2G3G4H2 + G1G2G5H2 All loops are touching to T1, T2, T3, T4 hence D 1 = D 2 = D 3 = D 4 = 1 \
C( s) T1 D 1 + T2 D 2 + T3 D 3 + T4 D 4 = = D R( s)
G 1G 2 G 3 G 4 + G 6 G 2 G 3 G 4 + G 1G 2 G5 + G 6 G 2 G5 1 + G 2 G 3 H 1 + G 1G 2 G 3 G 4 H 2 + G 1G 2 G5 H 2
Example 6.5.18 Solution : The various forward paths are, G1
G2
T1= G1G2
G3
G4
G3
G6
G5 G2
T2 = G3G4
G1
T4 = G3G5G2
G7 T3 = G7
G4
T5 = G1G6G4
Fig. 6.2
The various individual loop gains are, The combinations of two non-touching loops are, L 1 L2 = H 1 H 2 No combination of three non-touching loops.
H2 L 1 = H2
H1 L 2 = H1
Fig. 6.2 (a) TM
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G5
G6
L3 = G5G6
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H2
All loops are non-touching G6 to T3
H2 G5
L1 non-touching to T1 G1
Signal Flow Graph Models of Control Systems
G3
G4
H1
G2
H1 L2 non-touching to T2
G7
Fig. 6.2 (b)
\ D = 1 – [L1 + L2 + L3] + [L1 L2] All loops are touching to T4 and T5. \ D 1 = 1 – L1 , D 2 = 1 – L2 , D 3 = 1 – [L1 + L2 + L3] + [L1 L2] , D 4 = D5 = 1 T D + T2 D 2 + T3 D 3 + T4 D 4 + T5 D5 C(s) = 1 1 \ D R(s)
\
G 1 G 2 (1 - H 2 ) + G 3 G 4 (1 - H 1 ) + G 7 [1 - H 1 -H 2 - G 5 G 6 + H 1 H 2 ] + G 3 G5 G 2 + G 1 G 6 G 4 1 - H 1 - H 2 - G5 G 6 + H 1H 2
C(s) = R(s)
Example 6.5.19 Solution : System node variables are Y1 , Y2 , Y 3 , Y4 . Consider equation 1 : This indicates Y2 depends on Y1 and Y 3 Consider equation 2 : This indicates Y 3 depends on Y1 , Y 2 and Y 3 G1 Y2
G4
Y3
G5
G3
Y1
S.F.G. for equation (1)
Y2
S.F.G. for equation (2)
Consider equation 3 : This indicates Y4 depends on, Y 3 and Y2 G6
Y2
Y3
Y3
G2
G4 G5
G7
Y1
Y4
G1
G2 Y2
G7 Y3
G3
S.F.G. for equation (3)
G6
Fig. 6.3 TM
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Combining all three we get, complete S.F.G. as shown in Fig. 6.3, No. of forward paths = K = 4 4 T D T D + T2 D 2 + T3 D 3 + T4 D 4 \ T.F. = å K K = 1 1 D D
... Mason's gain formula
K= 1
T1 = G 1 G 2 G 7 ,T2 = G 4 G 7 ,
T3 = G 1 G 6 , T4 = G 4 G 3 G 6
Individual loops are,
L1
G2
G5
L2
SELF LOOP
G3
L1 = G2 G3
L2 = G5
\ D = 1 - [L 1 + L 2 ] = 1 - G 2 G 3 - G 5 No nontouching loop combinations. \ D1 = D2 = D4 = 1
For T1 , T2 and T4 , all loops are touching.
\ D3 = 1 -G
And for T3 , `G 5 ' self loop is nontouching, Y4 T D + T2 D 2 + T3 D 3 + T4 D 4 = 1 1 Y1 D =
L2
5
G5 G1
G 1 G 2 G 7 × 1 + G 4 G 7 × 1 + G 1 G 6 (1 - G 5 ) + G 4 G 3 G 6 × 1 D
T3
G6
L2 nontouching to T3
\
Y4 G G G + G 4 G 7 + G 1 G 6 (1 - G 5 ) + G 4 G 3 G 6 = 1 2 7 Y1 1 - G 2 G 3 - G5
Example 6.5.20 Solution : There are two inputs and two R1 outputs hence there are four gains possible as C1 C2 C2 C C , , and 2 . Let us find the gain 1 R1 R2 R1 R2 R1
G1
1 G3
The forward paths are, T1 = G1 (only one forward path).
H1
G4
by Mason's gain formula. The input R2 = 0 and C2 is suppressed as shown in the Fig. 6.4 (a).
1
H2 G2
Fig. 6.4 (a)
The individual feedback loops are, The loops L1 and L2 are nontouching to each other, hence L1 L2 = G3G4H1H2 \ D = 1 – [L1 + L2 + L3] + [L1 L2] TM
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Signal Flow Graph Models of Control Systems
All loops are touching to T1 hence D1 = 1. G1 G3
H2
G4
H1
H1
H2
G2 L1 = G3 H2
L3 = G1 G2 H1 H2
L2 = G4 H1
Fig. 6.4 (b)
According to Mason's gain formula, C1 T D = 1 1 = R1 D
G1 1 - G 3 H 2 - G 4 H 1 - G 1G 2 H 1H 2 + G 3 G 4 H 1H 2
Similarly other gains can be obtained. Example 6.5.21 Solution : Forward paths : T1 = G1 G2 G3 G4 G5 G7,
T2 = G1 G2 G3 G4 G6 G7
Individual feedback loops are, G6 G2
G4
H1
H2
L1 = G2H1
L2 = G4H2
G4
G5
H3 L3 = G4G5H3
G4
H3 L4 = G4G6H3
Fig. 6.5
Two nontouching loop combinations : L1L2, L1L3, L1L4 \ D = 1 – [L1 + L2 + L3 + L4] + [L1L2 + L1L3 + L1L4] All the loops are touching to T1 and T2 hence D1 = D2 = 1. ì ü ï ï G G G G G G G G G G G G + C(s) T1 D 1 + T2 D 2 ï ï 1 2 3 4 5 7 1 2 3 4 6 7 =í = \ D R(s) 1 - G 2 H 1 - G 4 H 2 - G 4 G5 H 3 - G 4 G 6 H 3 + G 2 G 4 H 1H 2 ý ï ï ïî ïþ + G 2 G 4 G5 H 1H 3 + G 2 G 4 G 6 H 1H 3 Example 6.5.22 Solution : Only one forward path, T1 = G 1 G 2 G 3 G 4 G 5
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Signal Flow Graph Models of Control Systems
The individual feedback loops are, G1
G3
G1
–H1
–H3
L1 = – G1H1
L2 = – G3H3
G2
G3
G4
–H4
–H5
L4 = – G4H4
L5 = – G5H5
–H2 L3 = – G1G2G3H2
G5
Fig. 6.6
The nontouching loops : L 1 L 2 , L 1 L 4 , L 1 L5 , L 2 L5 , L 3 L5 Three nonotouching loops :
L 1 L 2 L5
\D = 1 - [L 1 + L 2 + L 3 + L 4 + L5 ] + [L 1 L 2 + L 1 L 4 + L 1 L5 + L 2 L5 + L 3 L5 ] - [L 1 L 2 L5 ] D 1 = 1 as all the loops are touching to T1 \
TD G 1G 2 G 3 G 4 G5 C(s) = 1 1 = D {1 + G 1 H 1 + G 3 H 3 + G 1 G 2 G 3 H 2 + G 4 H 4 + G 5 H5 + G 1 G 3 H 1 H 3 R(s) + G 1 G 4 H 1 H 4 + G 1 G 5 H 1 H5 + G 3 G 5 H 3 H5 + G 1 G 2 G 3 G 5 H 2 H5 + G 1 G 3 G 5 H 1 H 3 H5 }
Example 6.5.23 Solution : Number of forward paths K = 2 2
\
C(s) = T.F. = R(s)
S
K= 1
TK D K D
=
T1 = 5 × 4 × 2 × 10 = 400
T1 D 1 + T2 D 2 D
... Mason's gain formula T2 = 1 × 10 × 2 × 10 = 200
and
Key Point Self loops cannot be part of forward path or another individual loop as node
containing self loop gets traced twice which is not allowed. Individual loops are, 4
2
–1
–2
L1 = – 4
–1 SELF LOOP
L2 = – 4
L 3 = –1
Combinations of two non-touching loops are, i) L 1 and L 3 and
ii) L 2 and L 3
No combination of three non-touching loops : \ D = 1 -[L 1 + L 2 + L 3 ] + [L 1 L 3 + L 2 L 3 ] = 1 – [– 4 – 4 – 1] + [4 + 4] = 1 + 9 + 8 = 18 Consider T1 , L3 loop is non-touching. TM
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5
1
T1
4
Signal Flow Graph Models of Control Systems –1
10
2
L3
\ D 1 = 1 – L 3 = 1 – [ –1] = 2 Consider T2 , L 1 is non-touching, –1 T2 10
2
5
1
4 –1
2
10
L1
D 2 = 1 – L 1 = 1 – [ – 4] = 5 C(s) R(s)
\
=
T1 D 1 + T2 D 2 400 ´ 2 + 200 ´ 5 = = 100 D 18
Example 6.5.24 Solution : The various forward path gains are, T1 = G 1 G 2 G 3 G 6 G 7 G 8 ,
T2 = G 1 G 4 G 6 G 7 G 8
T3 = G 1 G 4 G 5 G 8 ,
T4 = G 1 G 2 G 3 G 5 G 8
The individual feedback loop gains are, G6
G3
G2
G7
G3
– H2
G8
G7
– H1
L1 = – G3G6G7H2
L2 = – G2G3G6G7G8H1
G4
G4 G6
G6
G7
– H1 L3 = – G4G6G7G8H1
G5
G5 G8
G8
G2
– H1 L4 = – G4G5G8H1
Fig. 6.7 TM
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G3
– H1 L5 = – G2G3G5G8H1
G8
Control System Engineering
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Signal Flow Graph Models of Control Systems
All loops are touching to each other, no combination of non touching loops. \ D = 1 – [L 1 + L 2 + L 3 + L 4 + L5 ] All loops are touching to all the forward paths hence, D1 = D2 = D 3 = D4 = 1 According to Mason's gain formula, x2 T D + T2 D 2 + T3 D 3 + T4 D 4 = 1 1 1 – [L 1 + L 2 + L 3 + L 4 + L5 ] x1 =
G1G 2 G 3 G 6G 7G 8 + G1G 4 G 6G 7G 8 + G1G 4 G5 G 8 + G1G 2 G 3 G5 G 8 1 + G 3 G 6G 7H2 + G 2 G 3 G 6G 7G 8 H1 + G 4 G 6G 7G 8 H1 + G 4 G5 G 8 H1 + G 2 G 3 G5 G 8 H1
Example 6.5.25 Solution : The forward path gains are, T1 = G 1 G 2 G 3 G 4 ,
T2 = G 1 G 6 G 4 ,
T3 = G 1 G 7
The individual feedback loop gains are, L 1 = G 2H1
L2 = G5 (self loop)
Both loops are nontouching to each other hence \ D = 1 – [L 1 + L 2 ] + [L 1 L 2 ]
L1L2 = G 2 G5 H 1
For T1 , both loops touching hence D 1 = 1 For T2 , both loops touching hence D 2 = 1 For T3 , L2 is nontouching hence \
D 3 = 1 – L2 = 1 – G5
x5 T D + T2 D 2 + T3 D 3 = = 1 1 x1 D
G 1 G 2 G 3 G 4 + G 1 G 6 G 4 + G 1 G 7 (1 – G 5 ) 1 – G 2 H1 – G5 + G 2 G5 H1
Example 6.7.4 Solution : Laplace transform of the given network is as shown in following figure. R1
Vi(s)
I1(s)
R2
V1(s) 1 sC
sL I2(s)
Vo(s)
Key Point Assume the network variables alternately as the loop current and node voltage and
then write the equations by analysing the horizontal and vertical branches alternately. I 1 (s)
=
(Vi - V1 ) R1
... (I) TM
V1 = (I 1 - I 2 )
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1 sC
... (II)
Control System Engineering
I2 =
6 - 11
( V1 - Vo ) R2
... (III)
S.F.G. for equation (I) + Vi
1 R1
Signal Flow Graph Models of Control Systems
Vo = I 2 sL
... (IV)
S.F.G. for equation (II) V1
I1
1 sC
I1
V1
I2
– 1 R1
–1 sC
S.F.G. for equation (III) V1
+
1 R2
S.F.G. for equation (IV)
I2
Vo I2
sL
Vo
–1 R2
Total Signal Flow Graph for the network is as follows. V Use Mason's gain formula to find o . Vi
Vi
1 R1
1 +1 I1 sC V1 R2 –1 R1
S TK D K Vo = ; Number of forward paths = 1 D Vi \
T D1 Vo = 1 D Vi L T1 = R1 R2 C
Individual feedback loops are, 1 , L1 = – sR 1 C
L2 = –
1 , sR 2 C
L3 = –
sL R2
L 1 and L 3 are non-touching. \ D = 1 – [L 1 + L 2 + L 3 ] + [L 1 L 3 ] D = 1+
1 1 sL L + + + sR 1 C sR 2 C R2 R1 R2 C
As all loops are touching to T1 , D 1 = 1, L R1 R2 C Vo = \ Vi 1 1 sL L 1+ + + + sR 1 C sR 2 C R 2 R 1 R 2 C
TM
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–1 sC
I2
sL –1 R2
Vo
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Signal Flow Graph Models of Control Systems
Vo sL = Vi sR 1 R 2 C + R 2 + R 1 + s 2 L R 1 C + sL
\
\
6 - 12
Vo (s) sL = 2 Vi (s) s LR 1 C + s[L + R 1 R 2 C] + (R 1 + R 2 )
Example 6.7.5 Solution : Laplace transform of the given network is, R1
Vi(s)
I1(s)
R3
V1 R2
R4
I2(s)
Vo(s)
Equations for different currents and voltages are
S.F.G. (I)
Vi
1 R1
I1
V1
I1 = ( Vi - V1) ´
1 R1
... (1)
–1 R1
S.F.G. (II)
I1
R2
V1
I2
V1 = (I1 - I 2 ) R 2
... (2)
–R2 V1
1 R3
I2
Vo
S.F.G. (III)
S.F.G. (IV)
I 2 = (V1 - Vo ) ´
–1 R3 I2
R4
Vo
Vo = I 2 R 4
Total S.F.G. is as shown in the following figure.
Vi
1 R1
I1
1 R 2 V1 R 3 –1 R1
–R2
TM
I2
R4
Vo
–1 R3
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1 R3
... (3)
... (4)
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Use Mason's gain formula to find S TK D K Vo = D Vi
Signal Flow Graph Models of Control Systems
Vo Vi
Number of forward paths = K = 1 =
T1 D 1 D
Individual feedback loops are, R2 R4 \ T1 = R1 R 3 L1 = -
R2 , R1
L2 = –
R2 , R3
R4 R3
L3 = –
L 1 and L 3 are non-touching D = 1 – [L 1 + L 2 + L 3 ]+ [L 1 L 3 ] R R R R R = 1+ 2 + 2 + 4 + 2 4 R1 R 3 R 3 R1 R 3 \
…All loops are touching to T1
D1 = 1 R1 R 3 + R2 R 3 + R1 R2 + R1 R4 + R2 R4 D = R1 R 3 T D Vo = 1 1 D Vi \
R2 R4 Vo = R2 R4 + R1 R4 + R1 R2 + R2 R 3 + R1 R 3 Vi
Example 6.7.6 Solution : The Laplace domain representation of the given network is shown below. The various branch currents are shown, I2(s)
Vi(s)
I2(s)
V1(s)
(I1+I2) Vi(s)
R1
I1(s)
1 sC
R
(I1+I2)
1 sC
I2(s)
Vo(s)
Vo(s)
(I1+I2)
\
I 1 (s) =
Then,
I 2 (s) =
Vi (s) - V1 (s) = sC Vi (s) - sC V1 (s) æ 1 ö ç sC ÷ è ø Vi (s) – Vo (s) 1 1 = V (s) – Vo (s) R R i R TM
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... (1)
... (2)
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Signal Flow Graph Models of Control Systems
V1 (s) = [I 1 (s) + I 2 (s)]R = R I 1 (s) + R I 2 (s)
... (3)
Vo (s) - V1 (s) = sC Vo (s) - sC V1 (s) 1 sC From this obtain the equation for Vo (s) as I 2 (s) equation is already obtained.
and also,
I 2 (s) =
Note Write the separate equation for separate branch and each element must be
considered at least once. 1 I (s) sC 2 Hence the signal flow graph is,
\
Vo (s) = Vi (s) +
... (4)
1 R sC
Vi(s)
–1 R
R
R
V1(s)
I1(s)
1 sC
Vo(s)
I2(s)
–sC +1
The forward path gains are, T1
=
sCR
T2 =
1 sCR
T3 =
1 ´ R´ 1 = 1 R
The various loop gains are, L1
=
L2 = -
– sCR
1 sCR
L3 = -
1 ´ R´ 1 = – 1 R
The loops L 1 and L 2 are non touching. \ L1L2 = 1 Hence system determinant is, D = 1 - [L 1 + L 2 + L 3 ] + [L 1 L 2 ] = 1 + sCR +
1 3sCR + s 2 C 2 R 2 + 1 +1 +1 = sCR sCR
For T1 ,
D1
=
1
For T2 ,
D2
=
1 – L1
=
(1 + sCR)
… All loops touching to T1 … As L 1 is non touching to T2
For T3 , D3 = 1 According to Mason’s gain formula, T D + T2 D 2 + T3 D 3 Vo (s) = 1 1 D Vi (s)
… All loops touching to T3
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Signal Flow Graph Models of Control Systems
æ s 2 C 2 R 2 + 1 + sCR + sCR ö ÷ ç ÷ ç sCR ø è æ 3sCR + s 2 C 2 R 2 + 1 ö ç ÷ ç ÷ sCR è ø
1 ö sCR + æç ÷(1 + sCR) + 1 è sCR ø = = æ 3sCR + s 2 C 2 R 2 + 1 ö ç ÷ ç ÷ sCR è ø Vo (s) s 2 C 2 R 2 + 2sCR + 1 = Vi (s) s 2 C 2 R 2 + 3sCR + 1
\
Example 6.7.7 Solution : For the given current distributions, V - V2 1 1 For branch r1 , i1 = 1 = V V r1 r1 1 r1 2
… (1)
For branch r 3 , V2 = (i 1 - i 2 ) r 3 = i 1 r 3 - i 2 r 3 Now current through r2 is (i 2 - a i 1 ) . Hence according to Ohm's law, V2 - V 3 i2 - a i1 = r2 \
i2 =
… (2)
1 1 V V + a i1 r2 2 r2 3
...(3)
And V 3 = i 2 r4 The simulations of all the equations are,
… (4) a
V1
1 r1
i1
V2
i1
r3
i2
V2
V3
1 r2
– r3
1 r1 Equation (1) –
i2
V2
i1
–
Equation (2)
Equation (3)
There are two forward paths, r r T1 = 3 4 r1 r2
V1
1 r1
i1
r3
a r4 r1
–
The various loops and loop gains are, r3
–
1 r1
– r3
1 r1
a
1 r2
r4
– TM
r4
V3
Equation (4)
a
Thus the overall signal flow graph is as shown,
T2 =
1 r2
i2
1 r2
–
1 r1
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– r3
1 V2 r2
– r3
i2 r4
–
1 r2
V3
Control System Engineering
L1 =
- r3 r1
6 - 16
L2 =
- r3 r2
L3 =
Signal Flow Graph Models of Control Systems
- r4 r2
L4 =
a r3 r1
One combination of two non-touching loops is L 1 L3 . \
D = 1 - [L 1 + L 2 + L 3 + L 4 ] + [L 1 L 3 ]
All loop are touching to all the forward paths, \
D1 = D2 = 1
\
r 3 r4 a r4 + V3 T1 D 1 + T2 D 2 r1 r2 r1 = = V1 D 1 - [L 1 + L 2 + L 3 + L 4 ]+ [L 1 L 3 ] r 3 r4 a r4 + r1 r2 r1 = r r r r r ar 1+ 3 + 3 + 4 - 3 + 3 4 r1 r2 r1 r2 r2 r1
\
V3 r 3 r4 + a r4 r2 = V1 r1 r2 + r 3 r2 + r 3 r1 + r1 r4 - a r2 r 3 + r 3 r4
qqq
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State Variable Models
Solutions of Examples for Practice Example 7.4.4 Solution : Applying KVL to the two loops, di di 1 R 1 1 e – L1 1 – i1 R1 – eT + VC = 0 i.e. = – 1 i1 + V dt dt L1 L1 C L1 T – L2
di 2 – i 2 R2 + V C = 0 dt
And
(i1 + i2) = – C
i.e.
di 2 R 1 = – 2 i2 + V dt L2 L2 C
dVC dt
dVC 1 1 = – i1 - i2 dt C C Selecting state variables as X1 = i1, X2 = i2, X3 = VC, U = eT · R 1 1 U X \ X 1 = – 1 X1 + L1 L1 3 L1 ·
R2 1 X + X L2 2 L2 3
·
1 1 X - X C 1 C 2
X2 = –
\
X3 = –
\
é R1 ê- L ê 1 ê A=ê 0 ê ê ê-1 êë C
1 ù é- 1 L1 ú ú ê L1 ê 1 ú ú,B=ê 0 L2 ú ê ú ê 0 ë 0 ú úû
0 -
R2 L2
-
…(2)
… Polarities of VC are opposite
\
\
…(1)
1 C
ù ú ú ú ú ú û
·
where state mode is X = AX + BU
(7 - 1) TM
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…(3)
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State Variable Models
Example 7.4.5 Solution : i) For transfer function model, obtain its s-domain network as shown in the Fig. 7.1. Applying KVL +
1 V(s) = I( s) éR + + sLù sC ëê ûú
1/sC
R –
=
=
I(s)
V( s) s 2 LC s 2 LC + sRC + 1 s2 R 1 s2 + s + L LC
… T.F. model
While
X1 and e(t) = X2 , v(t) = U
· di( t ) R 1 1 = - X1 - X2 + U = X1 dt L L L
\ and
=
C
· de( t ) 1 de( t ) = X1 = X2 = i( t ) i.e. dt C dt ·
y(t) = L X 1 = -RX 1 - X 2 + U ·
Hence the state variable model is X = AX + BU and Y = CX + DU
where,
–
Fig. 7.1
ii) For state variable model, apply KVL in time domain L di( t ) di( t ) and y(t) = L v( t ) = i( t )R + e( t ) + dt dt Let i(t)
+ sL Y(s)
–
V( s) ´ sL éR + 1 + sLù sC ûú ëê
Y( s) = V( s)
\
–
V(s) +
Y(s) = I( s) ´ sL
and
+
é- R ê A=ê L 1 ê ë C
1 - ù é1 Lú , B = êL ú ê0 0 ú ë û
ù ú , C = [-R -1] , D = [1] ú û
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State Variable Models
Example 7.4.6 Solution : i)
For transfer function model example 3.3.1 (Chapter - 3).
–
+
vC
1 + s CR 2 E o ( s) = E i ( s) 1 + sC [R 1 + R 2 ]
\
R1
refer
ei(t) +
+ –
C eo(t)
+
–
i(t)
ii) For state variable model consider network shown in the Fig. 7.2. Applying KVL,
R2
–
Fig. 7.2
– i( t )R 1 - v C ( t ) - i( t )R 2 + e i ( t ) = 0 i( t ) =
\
C Let
ei ( t) - v C ( t) R1 + R2
dv C ( t ) = i( t ) dt
i.e.
X1 = v C ( t ) ·
X1 =
… (1)
dv C ( t ) i( t ) -v C ( t ) ei ( t) = = + dt C C (R 1 + R 2 ) C (R 1 + R 2 )
and e i ( t ) = U
–X 1
C(R 1 + R 2 )
+
U C(R 1 + R 2 )
i.e.
·
X = AX+BU
\
é e (t) - v C (t)ù e o ( t ) = v C ( t ) + i( t )R 2 = v C ( t ) + ê i ú R2 ë R1 + R2 û
\
eo ( t) =
\
… (2)
… (3)
R1 R2 R1 R2 v C (t) + ei ( t) = X1 + U R1 + R2 R1 + R2 R1 + R2 R1 + R2
… (4)
Y = CX + DU
The equations (3) and (4) give the state variable model. Example 7.4.7 Solution : Assume the loop currents as shown in following figure Applying KVL to first loop,
+
R = 1 M W V2(t)
R = 1 MW
+
+
- I 1 R - V2 + u( t ) = 0 u( t ) - V2 … (1) i.e. I1 = R Applying KVL to second loop,
– +
u(t) I1(t)
–
TM
–
–
+
C = 1 mF +
–
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V1(t)
I2(t)
–
C = 1 mF
Control System Engineering
- I 2 R - V1 + V2 = 0
7-4
i.e.
I2 =
State Variable Models
V2 - V1 R
… (2)
Now current through capacitors are I1 – I2 and I2, \
I1 – I 2 = C
dV2 … (3) dt
and
I2 = C
dV1 dt
… (4)
Substituting I1 and I2, C
dV2 u( t ) - V2 V2 - V1 u( t ) - 2V2 + V1 = = dt R R R
… (5)
C
V2 - V1 dV1 = R dt
… (6)
Now
RC = 1 ´ 10 6 ´ 1 ´ 10 -6 = 1
Using
u(t) = U, ·
U + X 1 - 2X 2 R
·
X2 - X1 R
\
CX 2 =
and
CX 1 =
Ans
X1(t) = V1, X2(t) = V2, i.e.
i.e.
·
X1 =
dV1 · dV2 , X2 = dt dt
·
U + X 1 - 2X 2 = U + X1 – 2X2 RC
… (7)
·
-X 1 + X 2 = – X 1 + X2 RC
… (8)
X2 = X1 =
Y(t) = V1(t) = X1(t)
The equations (7), (8) and (9) gives the required state model.
\
é· ù -1 1 ù é X 1 ù é0ù êX 1 ú = é U + ê · ê ú ë 1 -2úû êëX 2 úû êë1 úû ëX 2 û
and
éX ù Y(t) = [1 0] ê 1 ú ëX 2 û
Example 7.4.8 Solution : Applying KVL to the first loop, di (t) di 1 (t) 1 1 =– V (t) + e (t) – L1 1 - VC (t) + e i (t) = 0 i.e. dt dt L1 C L1 i
…(1)
Applying KVL to the second loop, – L2
di 2 (t) – i2(t) R2 + VC(t) = 0 dt i1(t) – i2(t) = C
dVC (t) dt
i.e.
i.e.
di 2 (t) R 1 = – 2 i2(t) + V (t) dt L2 L2 C
dVC (t) 1 1 i (t) – i (t) = dt C 1 C 2
eo(t) = i2(t) R2
…(2) …(3) ... (4)
Use the state variables as, X1 = i1(t), X2 = i2(t), X3 = VC(t) Using in the equations (1) to (4), we get TM
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é 0 ê0 é· ù 1ú ê êX · êX 2 ú = ê 0 - R 2 ê L2 ê· ú ê1 êX 3 ú 1 ê ë û C C êë
1 ù L1 ú ú 1 ú L2 ú ú 0 ú úû
-
State Variable Models
é ù ê1 ú éX1 ù êL ú êX ú + ê 1 ú U ê 2ú ê 0 ú êëX 3 úû ê ú ê 0 ú ë û
éX1 ù Y = [0 1 0] ê X 2 ú ê ú êëX 3 úû
… U = ei(t)
… Y = eo(t) ·
This is the required state model in the form, X = AX + BU
and Y = CX.
Example 7.5.4
Kept this unsolved example for student's practice.
Example 7.5.5
Kept this unsolved example for student's practice.
Example 7.5.6
Kept this unsolved example for student's practice.
Example 7.5.7
Kept this unsolved example for student's practice.
Example 7.7.4 Solution : System is 3 rd order, n = 3 3 integrators and variables are required. Select y = X 1 and then successive differentiation of y as next variable. \
·
X 1 = X 2 = dy/dt ·
X2 = X3 =
... (1)
d2 y dt 2
... (2) ·
·
Now as 3 variables are defined, X 3 ¹ X 4 but X 3 must be obtained by substituting all selected variables in original differential equation. ·
\ \
X 3 + 6X 3 + 11X 2 + 10 X 1 = 3U ·
as
·
X3 =
d3y dt 3
X 3 = 3U – 10X 1 – 11X 2 – 6X 3
... (3) which is output equation.
y = X1 \ State model can be written as, ·
X = AX + BU and
Y = CX
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State Variable Models
0ù é 0 1 é0 ù A=ê 0 0 1 ú , B = ê0 ú , C = [ 1 0 0 ] ê ú ê ú êë– 10 –11 –6úû êë 3úû
where
State diagram : U(s)
X3
3 –
X3
X1
X2
Y(s)
–
–
6 11 10
Example 7.7.5 Solution : Dividing both N and D by s 2 , 3 4 + s s2 Y (s) = U (s) 5 6 1+ + s s2 6 ù 5 6 é 5 \ L1 = - , L2 = D = 1 - ê- 2ú s s s û s2 ë The numerator is T1 D 1 + T2 D 2 , i.e. K = 2 = Forward paths T1 =
3 , s
T2 =
U(s)
1
4 s2
,
X2
D1 = D2 = 1 3 1 __ s
1 __ s
X2
4
1
X1
X1 –5 –6
Fig. 7.3
Complete signal flow graph is, \
·
X 1 = X2 ·
X 2 = U (s) - 6 X 1 - 5 X 2 Y = 4 X1 + 3 X2
TM
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Y(s)
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\Model is,
7-7
State Variable Models
·
X = AX + BU and Y = CX + DU 1ù é0ù é0 A=ê ú , B = ê1 ú , 6 5 ë û û ë
where
C = [4
3],
D = [0]
Example 7.7.6 Solution : The closed loop transfer function of the system is, K (s + z) Y(s) s (s + a)(s + p) Ks + Kz = = 3 U(s) K(s + z) s + (a + p) s 2 + (ap + K) s + Kz 1+ s (s + a)(s + p) Decompose the denominator as, {([s + a + p] s + (ap + K)) s + Kz} For numerator shift take off point once for Ks. The entire state diagram is shown in the Fig. 7.4. K U(s)
X3 – –
1/s
X2 X3
–
1/s
X1
1/s
X2
X1
+
Kz
+
a+p ap + K Kz
Fig. 7.4 ·
\ X 1 = X 2,
·
·
X 2 = X3, X 3 = U – Kz X1 – (ap + K) X2 – (a + p) X3
while
Y = Kz X1 + KX2
Hence,
X = AX + BU
·
and
Y = CX + DU
where,
1 0 ù é 0 é0ù ê ú A= 0 , B = ê0ú , C = [Kz 0 1 ê ú ê ú êë-Kz -(ap + K) -(a + p) úû êë1 ûú Example 7.7.7 Solution : By direct decomposition, y( s) 5 = u( s ) s + 0 s {[( ) + 6]s + 7} TM
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K
0], D = [0]
Y(s)
Control System Engineering
7-8
Hence the state diagram is as shown in the Fig. 7.5. Assigning output of each integrator as the state variable,
X3 u(s)
State Variable Models X3 = X 2
1/s –
X2 = X 1 1/s
1/s
X1
5
y(s)
– 6 7
Fig. 7.5 ·
·
·
X 1 = X 2 , X 2 = X 3 , X 3 = – 6 X 2 – 7 X 1 + U, Y = 5X 1 \
·
X = AX + BU
and Y = CX + DU where é0 A = ê0 ê êë–7
0ù é0ù 1 ú , B = ê0ú , C = [5 0 0], D = [0] ú ê ú êë1 úû –6 0úû 1 0
Example 7.8.3 Solution : Finding factors of denominator Y ( s) 6 A B C = = + + U ( s) ( s + 1)( s + 2)( s + 3) s + 1 s + 2 s + 3 Taking partial fractions, Y ( s) 3 6 3 = + U ( s) s + 1 s + 2 s + 3
X1
U(s)
ò
–
X1
3
Hence the state diagram is as shown in the Fig. 7.6. X2
This is Foster's form of representation. From Fig. 7.6, we get,
ò
–
X2
6
2 ·
X 1 = U ( s) - X 1 ,
X3
·
ò
–
X 2 = U ( s) - 2 X 2
X3
3
·
X 3 = U ( s) - 3 X 3
Fig. 7.6
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3
+ –
+
Y(s)
Control System Engineering
7-9
State Variable Models
Y (s) = 3X 1 - 6X 2 + 3X 3
and
Hence the state space model is, ·
X = AX + BU
where
and
Y = CX
é1ù é-1 0 0 ù ú ê A = 0 -2 0 , B = ê1ú , ê ú ú ê êë1ûú êë 0 0 -3úû
C = [3 -6
3]
Example 7.10.3 Solution : T.F. = C [sI – A] –1 B Adj [sI – A] [sI – A ]–1 = |sI – A| é1 0ù é-2 -3ù é s 0ù é -2 -3ù [sI – A] = s ê ú -ê ú=ê ú -ê ú ë0 1 û ë 4 2 û ë0 s û ë+ 4 2 û és + 2 [sI – A] = ê ë–4
3 ù s – 2úû
C 12 ù T és – 2 éC Adj = ê 11 ú =ê– 3 ë ëC 21 C 22 û
4 ùT és – 2 =ê ú s + 2û ë 4
–3ù s + 2úû
|sI – A| = (s + 2) (s – 2) + 12 = s 2 – 4 + 12 = s 2 + 8 és – 2 –3 ù ê 4 s + 2úû [sI – A] –1 = ë s2 + 8
\
é 3s – 21ù és – 2 –3 ù é 3ù [1 1] ê ú ê ú ê5s + 22 ú s + 2û ë5 û û = ë 4 =[1 1] ë T.F. = s2 + 8 s2 + 8
[8s + 1] s2 + 8
Example 7.10.4 Solution : The transfer function is given by, T(s) = C [sI - A] - 1 B 0 0 ù 0 ù és + 1 é1 0 0ù é- 1 0 ê ú ê ú ê [sI – A] = s 0 1 0 - 0 - 1 0 = 0 s+ 1 0 ú ú ú ê ú ê ê êë0 0 1 úû êë 0 0 s + 3úû 0 - 3úû êë 0 Adj [sI – A] = [co factor [sI - A]] T
és 2 + 4s + 3 0 0 ê 2 0 s + 4s + 3 =ê 0 2 + ê 0 0 s 2s + ë
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ù ú ú 1ú û
T
Control System Engineering
7 - 10
State Variable Models
ù ú ú 1ú û Adj [sI - A] = |sI - A|
és 2 + 4s + 3 0 0 ê 2 = ê 0 s + 4s + 3 0 2 ê 0 0 s + 2s + ë |sI – A| = (s + 1) 2 (s + 3)
and
[sI - A] - 1
és 2 + 4s + 3 0 0 ê 2 [1 1 1] ê 0 s + 4s + 3 0 2 ê 0 0 s + 2s + ë T(s) = (s + 1) 2 (s + 3)
\
ù ú ú 1ú û
é 3ù ê- 6ú ê ú êë 3 úû
é 3 (s 2 + 4s + 3) ù ê ú [1 1 1] ê- 6 (s 2 + 4s + 3) ú ê 3 (s 2 + 2s + 1) ú 2 2 2 ë û = [3s + 12s + 9 - 6s - 24s - 18 + 3s + 6s + 3] = (s + 1) 2 (s + 3) (s + 1) 2 (s + 3) T(s) =
\
- 6s - 6 (s +
1) 2 (s +
3)
=
- 6 (s + 1) (s + 1) 2 (s + 3)
Example 7.10.5 Solution : The T.F. is given by,
és + 5 6ù s úû
T.F. = C[sI - A ] -1 B + D, éC Adj [sI - A ] = ê 11 ëC 21
[sI - A ] = ê ë -1
C 12 ù T 1 ùT és -6 ù és =ê = ú ú ú ê C 22 û ë1 s + 5û ë-6 s + 5û
|sI - A| = s (s + 5) + 6 = s 2 + 5s + 6 = ( s + 2)( s + 3)
\
C Adj [sI - A ]B = T.F. = |sI - A|
\
T.F. =
[1
és -6 ù é1 ù ésù 1 ]ê [1 1]ê1ú ú ê0ú 1 s + 5 ûë û = ë ë û ( s + 2)( s + 3) ( s + 2)( s + 3)
( s + 1) s + ( 2)( s + 3)
Example 7.16.6 Solution : Use Laplace transform method, - 1ù é1 0ù é 0 [sI – A] = s ê -ê ú= ú ë0 1 û ë+ 2 -3 û
+ 1ù é s ê-2 s + 3ú û ë
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\
7 - 11
State Variable Models
T és + 3 2ù és + 3 -1ù Adj [sI – A] = ê =ê ú s úû ë -1 s û ë 2
+ 1ù é s 2 |sI – A| = ê ú = s + 3s + 2 = (s + 1) (s + 2) 2 3 s + û ë
\
\
\
–1
[sI – A]
és + 3 -1ù ê 2 s úû Adj [sI - A] = = ë ( s + 1) ( s + 2) |sI - A|
–1 f (s) = [sI – A]
At
e
= L
–1
s+ 3 é ê(s + 1) (s + 2) =ê ê 2 ê (s + 1)(s + 2) ë
[sI – A]
–1
=L
–1
-1 ù (s + 1) (s + 2) ú ú ú s (s + 1)(s + 2) úû
s+ 3 é ê(s + 1) (s + 2) ê ê 2 ê(s + 1) (s + 2) ë
-1 ù (s + 1) (s + 2) ú ú ú s (s + 1) (s + 2) úû
Using partial fraction expansion for all the elements. At
e
= L
–1
é 2 - 1 ês + 1 s + 2 ê ê 2 - 2 êë s + 1 s + 2
-1 1 ù + s +1 s + 2ú ú= -1 2 ú + s + 1 s + 2 úû
é 2e - t - e -2t ê -t -2t êë2e - 2e
Example 7.16.7 3ù é- 2 Solution : From the given state equation, A = ê ú ë 0 - 3û eAt = L- 1
{(sI - A) - 1} = State transition matrix
é1 0ù é- 2 [sI – A] = s ê ú -ê ë0 1 û ë 0
[sI - A] - 1 =
Adj [sI - A] = sI - A
3ù és + 2 - 3 ù = - 3 úû êë 0 s + 3úû éC 11 êC ë 21
C 12 ù T C 22 úû
sI - A
3 ù és + 3 é 1 ê 0 s + 2úû ê s + 2 ë = =ê (s + 2) (s + 3) ê 0 ë
0 ùT és + 3 ê 3 s + 2úû = ë s+2 - 3 0 s+ 3
3 ù (s + 2) (s + 3) ú ú 1 ú (s + 3) û TM
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-e - t + e -2t ù ú -e - t + 2e -2t úû
Control System Engineering
\
At
e
-1
= L
7 - 12
[(sI - A)
-1
-1
]= L
State Variable Models
ìé 1 ïï ê(s + 2) íê ïê 0 ïî ë
3 ùü (s + 2) (s + 3) ú ïï úý 1 úï (s + 3) û ïþ
é 1 3 öù æ 3 ç ÷ú ê = L- 1 ê(s + 2) è (s + 2) (s + 3) øú 1 ê 0 ú s+ 3 ë û \
ée - 2t eAt = ê êë 0
3e - 2t - 3e - 3t ù ú e - 3t úû
Example 7.16.8 Solution : From the given state model 0ù é- 1 1 ê A = 0 -1 1 ú ê ú 0 - 1úû ëê 0 ì Adj [sI - A ] ü eAt = L- 1 { [sI - A - 1 ] } = L- 1 í ý î sI - A þ 0 ù és + 1 - 1 0 ù é1 0 0ù é- 1 1 ê ú ê ú ê [sI – A] = s 0 1 0 - 0 - 1 1 = 0 s +1 - 1 ú ê ú ê ú ê ú êë0 0 1 úû êë 0 0 - 1úû êë 0 0 s + 1úû Adj [sI – A] = [Cofactor matrix]
T
é(s + 1) 2 ê = ê (s + 1) ê 1 ë
0 (s + 1) 2 (s + 1)
0 ù ú 0 ú (s + 1) 2 ú û
sI - A = (s + 1) 3
\
[sI - A ]
-1
é 1 ê(s + 1) ê = ê 0 ê ê ê 0 êë
1 ö -t L- 1 æç ÷ = e , s è +1 ø
1 1) 2
(s + 1 (s + 1) 0
ù ú (s + ú 1 ú (s + 1) 2 ú ú 1 ú (s + 1) úû 1
1) 3
é 1 ù L- 1 ê ú = t e- t , êë( s + 1) 2 úû
TM
é 1 ù 1 L- 1 ê ú = t 2 e- t êë( s + 1) 3 úû 2
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\eAt = L- 1 [sI - A ] - 1
ée - t ê =ê 0 ê ê 0 ë
t e- t e- t 0
7 - 13
State Variable Models
1 2 -tù t e ú 2 t e- t ú ú e- t ú û
... State transition matrix
Example 7.16.9 Solution : The matrix A is, é-3 1 ù A = ê ú ë 0 -1û The state transition matrix is, eAt = L-1 [sI - A ] -1 é1 0ù é-3 1 ù és + 3 -1 ù sI – A = s ê ú=ê ú -ê s + 1úû ë0 1 û ë 0 -1û ë 0 0 ù T és + 1 1 ù és + 1 Adj [sI – A] = ê =ê ú s + 3û s + 3úû ë 0 ë 1 |sI – A| = (s + 1) (s + 3) é ( s + 1) ù é 1 1 Adj [sI - A ] ê( s + 1)( s + 3) ( s + 1)( s + 3) ú ê s + 3 -1 ú= =ê [sI - A ] = ( s + 3) ú ê s I -A ê 0 ê 0 êë ( s + 1)( s + 3) úû ë Finding the partial fractions,
[sI - A ] -1
eAt
0.5 0.5 ù é 1 ê( s + 3) ( s + 1) ( s + 3) ú = ê ú 1 ê 0 ú ( s + 1) êë úû 3t ée 0.5 e - t - 0.5 e -3t ù = L-1 [sI - A ] -1 = ê ú e- t úû êë 0
Example 7.16.10 Solution : System is homogeneous so ZSR = 0 \ Let
Now
X(t) = eAt X(0) éA A = ê 1 ëA 3
A2ù A 4 úû
é e – 2t ù X(t) = ê ú êë– 2 e –2 t úû
é – 2e – 2 t ù · hence X(t) = ê ú êë 4e – 2 t úû
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1
ù
1 s +1
ú ú û
( s + 1)( s + 3) ú
Control System Engineering
7 - 14
and at t = 0
é– 2ù X(t) = ê ú ë 4 û
So
· é– 2ù X(0) = ê ú ë 4 û
Similarly
ée–t ù X(t) = ê ú, êë– e – t úû
So
· é+ 1ù é– 1ù X(0) = ê ú when X(0) = ê ú ë– 1û ë+ 1û
State Variable Models
é 1ù when X(0) = ê ú ë– 2û é– e – t ù · X(t) = ê ú êë e – t úû
· é– 1ù at t = 0 X(t) = ê ú ë+ 1û
·
Substituting in X(t) = A X(t) éA 1 é–2ù ê 4ú = êA ë û ë 3 A 1 – 2A 2 = – 2,
\ \
+ A 1 – A 2 = – 1,
A 2 ù é 1ù éA é–1ù and ê ú = ê 1 1 A 4 úû êë–2úû ë û ëA 3 A 3 – 2A 4 = + 4
A 2 ù é 1ù A 4 úû êë–1úû ... (1 and 2)
A 3 – A4 = 1
... (3 and 4)
Solving these 4 equations simultaneously, A 2 = 1, A 1 = 0,
A 4 = – 3,
A3 = – 2
1 ù é 0 A = ê ú ë– 2 – 3û
\ Example 7.16.11 Solution :
é s 0ù é 0 1 ù é s –1 ù [sI – A] = ê ú ú=ê ú -ê ë0 s û ë-2 0û ë2 s + 3û
C 12 ù T éC és + 3 Adj [sI – A] = ê 11 =ê ú ë 1 ëC 21 C 22 û
–2ù T és + 3 =ê ú sû ë –2
1ù súû
|sI – A| = s 2 + 3s + 2 = (s +1) (s + 2) \
eAt = L–1 [(sI – A) –1 ] s+ 3 é ê(s + 1) (s + 2) = ê –2 ê ëê(s + 1) (s + 2)
1 ù (s + 1) (s + 2) ú ú = f (s) s ú (s + 1) (s + 2) úû
Finding partial fractions, eAt = L–1
é 2 – 1 ê s+1 s+ 2 ( f(s)) = ê ê –2 + 2 s+ 2 êë s + 1 TM
1 1 ù – s+1 s+ 2 ú ú –1 2 ú + s+1 s + 2 úû
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Control System Engineering
7 - 15
State Variable Models
é 2e – t – e – 2 t = ê êë– 2e – t + 2e – 2 t ZIR = eAt To find
\
= L–1
ZSR = L–1
e – t – e – 2t ù ú – e – t + 2e – 2t úû –t – 2t ù é1 ù é 2e – e X(0) = eAt ê ú = ê ú ë0û êë – 2e - t + 2e – 2 t úû { f(s) B U(s) }
U(t) = Unit step \U(s) = 1/s s+ 3 1 ü ìé ù ï ï ê(s + 1) (s + 2) (s + 1) (s + 2) ú é0ù ï ï ZSR = L–1 í ê ú ê ú [1 / s]ý 5 s –2 ú ë û ï ïê ïþ ïî êë(s + 1) (s + 2) (s + 1) (s + 2) úû 5 ì ì 2.5 – 5 + 2.5 ü ü ïï s (s + 1) (s + 2) ïï ï s + 1 s + 2 ïï é2.5 – 5e – t + 2.5 e – 2 t ù –1 ï s L = ú í í ý ý =ê –t – – 2t 5 5 5 ê úû 5 e 5e ï ï ï ï – ë ïî (s + 1) (s + 2) ïþ ïî s + 1 s + 2 ïþ é2.5 – 3e –t + 1.5 e – 2 t ù X(t) = ZIR +ZSR = ê ú êë 3 e – t – 3e – 2 t úû é 0 1ù Y(t) = ê ú X(t) ë–2 –3û
\ \
–t – 2t ù é Y1 (t) ù é 0 1ù é2.5 – 3e + 1.5 e ú êY (t) ú = ê–2 –3ú ê úû û êë 3 e – t – 3e – 2 t ë ë 2 û
Y1 (t) = 3 (e – t – e
\
– 2t )
and
é 3e – t – 3e – 2t =ê êë – 5 + 6e – 2t – 3e – t
ù ú úû
Y2 (t) = – 5 + 3 (2e –2t – e –t )
These are the outputs for unit step input applied. Example 7.16.12 Solution : Given system is homogeneous whose solution is X(t) = eAt X(0) Now
eAt = L–1
{(sI – A) – 1}
é 0 1 ù é s –1ù é1 0ù [sI – A] = s ê ú – ê– 2 0ú = ê2 s ú 0 1 û û ë ë û ë (sI – A) –1 =
Adj (sI – A) |sI – A|
éC Adj (sI – A) = ê 11 ëC 21
T C 12 ù T é s 1ù és –2ù = ê1 s ú = ê–2 sú C 22 úû û û ë ë
TM
and
|sI – A| = s 2 + 2
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L–1 L–1 L–1
\
\
7 - 16
1 ù é s ês 2 + 2 2 s + 2ú ú i.e. eAt (sI – A) –1 = ê ê –2 s ú ê 2 ú s2 + 2û ës + 2 üï ìï s üï ì s –1 ï = cos 2t ý = L í 2 í 2 2ý îï s + 2 þï îï s + ( 2) ïþ üï ìï 1 üï ì 1 2 –1 ï 1 × ý = L í ý = í 2 2 2 2 s + ( 2) ïþ ïî s + 2 ïþ ïî 2 üï ìï –2 üï ì 2 –1 ï ý = L í – 2 × 2 ý =– í 2 s + ( 2) 2 þï îï s + 2 þï îï 1 é sin 2 cos 2 t At ê 2 = e ê – 2 sin 2 t cos 2 t êë At
X(t) = e
At
X(0) = e
State Variable Models
= L–1
é s ês 2 + 2 ê ê –2 ê 2 ës + 2
ù s + 2ú ú s ú ú s2 + 2û 1
2
sin 2 t 2 sin
2t
tù ú ú úû
1 é sin 2 t ù é1ù ê cos 2 t + ú 2 ê1ú = ê ú ë û êë cos 2 t – 2 sin 2 t úû
\Output response écos 2 t + 1 sin 2 t ù ú 2 Y(t) = [1 – 1] X(t) = [1 – 1] ê ú ê êë cos 2 t – 2 sin 2 t ûú Y(t) =
3 2
sin 2 t
is the required response.
qqq
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The Performance of Feedback Control Systems
9
Solutions of Examples For Practice Example 9.5.7 Solution : G( s)H( s) =
K s ( s + 1)( s + 4) 2
Lim Kp = s ® 0 G( s)H( s)
=
Lim s®0
Lim Kv = s ® 0 sG( s)H( s) =
Lim s®0
Lim Ka = s ® 0 s 2 G( s)H( s) =
Lim s®0
r( t ) = 1 + 8t +
s ( s + 1)( s + 4) sK s ( s + 1)( s + 4) 2
s2K s2
( s + 1)( s + 4)
=¥ =¥
=
K 4
A 18 2 t = A 1 + A 2t + 3 t2 2 2
e ss = e ss1 + e ss2 + e ss3 =
\
K 2
But e ss is to be limited to 0.8 72 i.e. 0.8 = \ K
A1 A A 1 8 18 72 = 0+0 + + 2 + 3 = + + 1 + Kp Kv Ka 1+¥ ¥ K K 4
K = 90
Example 9.5.8 Solution : The static error coefficients are, Kp =
Lim s®0
G(s)H(s) =
Lim s®0
Kv =
Lim s®0
sG(s)H(s) =
Ka =
Lim s®0
s 2 G(s)H(s) =
Lim s®0
10K 2
s(s + 4)(s + 2s + 5)
=
s ´ 10K 2
s(s + 4)(s + 2s + 5)
Lim s®0
s 2 10K s(s + 4)(s 2
(9 - 1) TM
+ 2s + 5)
10K = ¥ 0´ 4´5 10K = 0.5 K 4´5
=
=
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0 ´ 10K = 0 4´5
Control System Engineering
9-2
The Performance of Feedback Control Systems
Example 9.5.9 Solution : Using superposition principle, consider inputs separately. a) R acting, TL = 0 +
R
5 1 + 0.1s
–
\
C(s)
100 (s + 1) (1 + 0.2s)
500 (1 + 0.1s) (s + 1) (1 + 0.2s)
G(s)H(s) =
For step input Kp = Lim G(s)H(s) = 500 s®0
\
A 10 10 = where A = magnitude of step = 1 + Kp 1 + 500 501
e ss1 =
b) TL acting, R = 0 E –
hence E(s) = – C(s) 5 1 + 0.1s
C(s)
100 (s + 1) (1 + 0.2s)
– TL
As system is not in standard form, error coefficient method cannot be used. –1
5 1 + 0.1s
+
C(s)
100 (s + 1) (1 + 0.2s)
– TL
TL
– +
100 (s + 1) (1 + 0.2s) –5 1 + 0.1s
C(s) = T(s)
100 (1 + s) (1 + 0.2s) 100 -5 ö 1´æ (1 + s) (1 + 0.2s) çè 1 + 0.1s ÷ø TM
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C(s)
Control System Engineering
9-3
The Performance of Feedback Control Systems
C(s) 100(1 + 0.1s) = T(s) (1 + s) (1 + 0.2s) ´ (1 + 0.1s) + 500 Now
T(s) =
-4 –ve sign as TL applied with –ve sign. s
\
C(s) =
– 400(1 + 0.1s) s[(1 + s) (1 + 0.2s) ´ (1 + 0.1s) + 500]
but
E(s) = – C(s) =
\
e ss2 = Lim sE(s) = Lim
+400(1 + 0.1s) s[(1 + s) (1 + 0.2s) ´ (1 + 0.1s) + 500]
s®0
s®0
400(1 + 0.1s) [(1 + s) (1 + 0.2s) ´ (1 + 0.1s) + 500]
+ 400 400 ù = é = êë1 + 500 úû 501 \Total error e ss = e ss1 + e ss2 =
10 400 410 = 0.8183 = + 501 501 501
Example 9.5.10 Solution : i) This is Type 0 system. It will produce finite steady state error for step input. 20 Kp = Lim G( s)H( s) = Lim \ = 3.3333 s ® 0 ( s + 2 )( s + 3 ) s®0 \
e ss =
A 1 = = 0.2307 1 + Kp 1 + 3. 333
… Assuming unit step
ii) This is Type 2 system. It will produce finite steady state error for parabolic input. \
Ka = Lim s 2 G(s)H(s) = Lim
\
e ss =
s®0
s®0
s 2 20( s + 1) s 2 ( s + 2)( s + 4)
A 1 = = 0.4 Ka 2.5
= 2.5 … Assuming unit parabolic input
iii) This is Type 1 system. It will produce finite steady state error for ramp input. \
Kv = Lim
\
e ss =
s®0
( ) = 0.1 2 s ® 0 s( s + 1 ) s + 5s + 25 ( )
sG(s)H(s) = Lim
s( 2.5) s 2 + 2s + 1
A 1 = = 10 Kv 0.1
… Assuming unit ramp input
Example 9.5.11 Solution : G(s)H(s) =
K s( s + 1)( 0.5s + 1)
… H(s) =1
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9-4
sK =K s( s + 1)( 0.5s + 1)
Kv = Lim s G( s)H( s) = Lim s®0
s®0
i) K = 1.5 , r(t) = 5 t \
i.e.
ess =
ess =
A=5
A 5 = = 3.333 Kv 1.5
ii) ess £ 0.1 , unit ramp \
The Performance of Feedback Control Systems
i.e. A = 1 A 1 = Kv K
0.1 £
i.e.
1 K
K ³ 10
i.e.
Example 9.5.12 Solution : G(s)H(s)
=
(
20
… H(s) =1
)
s( s + 2) s 2 + 2s + 20
i) r(t) = 5, A = 5 Kp = Lim G( s )H( s) = Lim s®0
\
ess =
s®0
(
20
)
=¥
)
= 0.5
s( s + 2) s 2 + 2s + 20
A 5 = =0 1 + Kp 1 + ¥
ii) r(t) = 5 t , A = 5 Kv = Lim sG( s)H( s) = Lim s®0
\ iii) r(t) =
ess =
s®0
s ´ 20
A 5 = = 10 Kv 0.5
3 2 t ,A=3 2 Ka = Lim s 2 G( s)H( s) = Lim s®0
\
(
s( s + 2) s 2 + 2s + 20
ess =
s®0
s 2 ´ 20
(
)
s( s + 2) s 2 + 2s + 20
=0
A 3 = =¥ Ka 0
Example 9.5.13 Solution : Unity feedback system \ H(s) = 1 To determine type of the system it is required to bring G(s) H(s) into its time constant form. K(s + 2) G(s) H(s) = s(s 3 + 7s 2 + 12s)
TM
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=
=
=
9-5
The Performance of Feedback Control Systems
2K(1 + 0.5s) s2
(s 2 + 7s + 12)
2K(1 + 0.5s) s2
(s + 4) (s + 3) 2K(1 + 0.5s)
s 2 × 4 (1 + 0.25 s) 3 × (1 + 0.33s)
2K (1 + 0.5s) 12 G(s) H(s) = s 2 (1 + 0.25s) (1 + 0.33s) Comparing this with, G(s) H(s) =
A(1 + T1 s) (1 + T2 s) ......... s j (1 + Ta s) (1 + Tb s) .......
Where j = Type of the system
j = 2, So system is Type 2 system. \ Error coefficients calculation : Kp =
Lim G(s) H(s) s®0
K (1 + 0.5s) 6 =¥ = Lim s ® 0 s 2 (1 + 0.25s) (1 + 0.33s) Kv =
Lim sG(s) H(s) s®0
K (1 + 0.5s) 6 = Lim =¥ s ® 0 s (1 + 0.25s) (1 + 0.33s) Ka =
Lim 2 s G(s) H(s) s®0
= Lim
s®0
K (1 + 0.5s) K 6 = (1 + 0.25s) (1 + 0.33s) 6
To find error use, e ss = where
Lim
s ® 0
s R(s) 1 + G(s) H(s)
R(s) = Laplace Transform of input r(t) to the system r(t) =
R 2 t 2
, R(s) =
R s3 TM
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9-6
s× \
e ss =
=
Lim
s ® 0
Lim
s ® 0
e ss =
Lim
s ® 0
R s3
K (1 + 0.5 s) 6 1+ s 2 (1 + 0.25s) (1 + 0.33s)
s2
=
The Performance of Feedback Control Systems
R K é ù (1 + 0.5s) ê1 + ú 6 ê s 2 (1 + 0.25s) (1 + 0.33s) ú êë ûú
R K (1 + 0.5s) 6 s2 + (1 + 0.25s) (1 + 0.33s)
6R K
Steady state error can also be determined as e ss =
R as system is type 2 system, where Ka
R is magnitude of the input. R 6R e ss = = K K 6 Example 9.5.14 Solution : To determine type of the system arrange G(s) H(s) its time constant form as, 10(1 + s) G(s) H(s) = 2 s × 2(1 + 0.5s) (10) (1 + 0.1s) =
0.5 (1 + s) s2
(1 + 0.5s) (1 + 0.1s)
Comparing this with standard form K(1 + T1 s) (1 + T2 s) ....... where j = Type G(s) H(s) = s j (1 + Ta s) (1 + Tb s) ........ Type of system is 2. Error coefficients are, Lim G(s) H(s) Kp = s®0 Lim 0.5 (1 + s) = 2 s ® 0 s (1 + 0.5s) (1 + 0.1s) \
Kp = ¥
TM
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Kv =
Lim s®0
Kv = Ka =
The Performance of Feedback Control Systems
s G(s) H(s)
Lim 0.5 (1 + s) s ® 0 s (1 + 0.5s) (1 + 0.1s)
= \
9-7
¥ Lim s®0
s 2 G(s) H(s)
Lim
0.5 (1 + s) (1 + 0.5s) (1 + 0.1s) s®0
=
\ Ka = 0.5 The steady state error for step input is, A1 e ss1 = 1+ Kp where
A 1 = magnitude of step input part of total input
Here
A1 = 1
\
e ss1 =
and Kp = ¥
1 1+¥
= 0 The steady state error for ramp input is, A2 e ss2 = Kv where
A 2 = magnitude of ramp input = 4
\
e ss2 =
A ¥
= 0 The steady stade error for parabolic input is, A3 e ss 3 = Ka where
A 3 = magnitude of parabolic input 1 The parabolic input is expressed as t 2 in R(t) 2 \ A3 = 1 Remember that for parabolic input, A is the magnitude when input is expressed as \
e ss 3 =
1 = 2 0.5 TM
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A 2 t . 2
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9-8
The Performance of Feedback Control Systems
As the input is combination of all the three inputs, the total error is also the algebraic sum of the steady state errors due to the individual type of inputs. \ e ss = e ss1 + e ss2 + e ss3 = 0+0+2 \
e ss = 2
Example 9.5.15 Solution : Simplifying the given system by eliminating minor feedback loop in between we get, R(s)
G1
25
1 + G1H1
–
250
R(s)
C(s)
1/2
C(s)
2
(s +214s + 40)
–
G1 Now = 1 + G 1 H1
20 (s + 4) (s + 10) 20 20 = = (s + 4) (s + 10) + 200s s 2 + 214s + 40 20 1+ 10s (s + 4) (s + 10)
Combining all blocks in series and simplifying. 250 So G(s) = , H(s) = 1 2 s + 214s + 40 Now comparing G(s) H(s) with standard time constant form i.e.
K(1 + s T1 ) (1 + s T2 ) ...... s j (1 + s Ta ) (1 + s Tb ) .......
it is clear that quadratic denominator can be expressed as
( 1 + sTa ) (1 + sTb ) hence value of `j' for given system is zero. Hence it is Type Zero system. Error coefficients : Kp =
Lim Lim 250 25 G(s) H(s) = = 2 4 s®0 s ® 0 s + 214s + 40
Kv =
Lim sG(s) H(s) s®0
Ka =
Lim Lim 2 æ 250 ö s 2 G(s) H(s) = s ç ÷=0 2 s®0 s®0 è s + 214s + 40 ø
=
Lim æ 250 ö sç ÷=0 2 s ® 0 è s + 214s + 40 ø
TM
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9-9
The Performance of Feedback Control Systems
Now for first input, i)
r(t) = 10 \ R(s) =
\
e ss =
10 , s
so
A = 10
A 1 + Kp
10 40 = = 1.379 25 29 1+ 4 5 , so A = 5 ramp r(t) = 5t, \ R(s) = s2 A 5 = =¥ e ss = Kv 0 6 2 r(t) = 10 + 5t + t 2 =
ii) \ iii)
A 2 = 5 ramp, A 3 = 6 parabolic A 1 = 10 step, A1 A2 A3 10 5 6 + = + + + =¥ \ e ss = 1 + Kp Kv Ka 25 0 0 1+ 4 Thus as system is type 0 system, error is finite only for step input. i.e.
Example 9.5.16 Solution : From the system shown we can write, K G(s) = H(s) = 1 s(s + 1) The input is r(t) = 0.1 t i.e. ramp of magnitude 0.1. For ramp input K V controls the error. Lim Lim s. K sG(s)H(s) = =K \ Kv = s®0 s ® 0 s(s + 1) \
e ss =
A 0.1 = Kv K
Maximum e ss allowed is 0.005 0.1 0.005 = \ K 0.1 = 20 0.005 For any value of K greater than 20, e ss will be less than 0.005. Hence the range of value of K for e ss £ 0.005 is, 20 £ K < ¥
\
K =
Example 9.5.17 Solution : The input is 2t i.e. ramp of magnitude 2, so Kv will control the error. Lim Lim 200 sG(s) H(s) = \ Kv = s× ×1 = 25 s®0 s ® 0 s(s + 8) TM
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\
e ss =
9 - 10
The Performance of Feedback Control Systems
A 2 = 0.08 = Kv 25
This error is to be reduced by 5% of existing value, with new gain of G(s) as K2 instead of 200. 5 ... reduce by 5% ´ e ss ö÷ \ e ss1 = e ss - æç 100 è ø 5 ´ 0.08 = 0.076 100
= 0.08 – New error is 0.076. New
G(s) =
\
Kv = =
\
e ss1 = =
\
0.076 =
K2 and s(s + 8)
H(s) = 1 with same input
Lim sG(s) H(s) s®0 s × K2 K = 2 s(s + 8) 8 A Kv 2 16 = æ K2 ö K2 ç 8 ÷ ø è 16 K2
\ K2 = 210.52 So new gain is 210.52 Example 9.5.18 Solution : The Kp, Kv, Ka are given by, Kp = Lim G(s)H(s), s®0
i) Kp = ¥,
Kv = ¥,
ii) Kp = 0.7,
Kv = 0
iii) Kp = ¥,
Kv = ¥,
Kv = Lim s G(s)H(s), s®0
Ka = Lim s 2 ´ s®0
10 s2
Ka = Lim s 2 G(s)H(s) s®0
´ 0.7 = 7
Ka = 0 10 Ka = ´ 0.8 = 1.6 5
Example 9.7.2 Solution : According to generalized error coefficient method, K ess(t) = K0 r(t) + K1r ¢(t) + 2 r ¢¢(t) + … 2! TM
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The Performance of Feedback Control Systems
2
Now
r(t) = 1 + 2t + t , r ¢(t) = 2 + 2t, r ¢¢(t) = 2
\
K0 = F1(s) =
= \
Lim s®0
F1(s), K1 =
1 = 1 + G(s)
=
0 . 1 s 2 + s + 50 Lim s®0
d F1 (s) , K2 = ds
d 2 F1 (s)
Lim s®0
ds 2
( 0 . 1s + 1) s 1 = s (1 + 0 . 1s) + 50 50 1+ s (1 + 0 . 1s)
0 .1 s 2 + s
K0 =
Lim s®0
s 2 + 10 s s 2 + 10 s + 500
F1(s) = 0
dF1 (s) (s 2 + 10s + 500) (2s + 10) - (s 2 + 10s) (2s + 10) = ds (s 2 + 10s + 500) 2 = \
(2s + 10) [s 2 + 10 s + 500 - s 2 - 10s] (s 2 Lim s®0
K1 = d 2 F1 (s) ds 2
\
=
=
1000 s + 5000 ( s 2 + 10s + 500) 2
d F1 (s) 5000 = 0.02 = ds (500) 2
(s 2 + 10s + 500) 2 [1000] - (1000 s + 5000) (2) (s 2 + 10 s + 500) (2 s + 10)
K2 =
\
+ 10
s + 500) 2
(s 2 + 10 s + 500) 4 Lim s®0
d 2 F1 (s) ds
2
=
ess(t) = 0.02 (2 + 2t) +
(500) 2 (1000) - (5000)( 2)(500)(10) (500) 4
= 3.2 ´ 10– 3
3 . 2 ´ 10 -3 ( 2) = 0.04t + 0.0032 2!
Example 9.9.4 Solution :
Let the first order system has transfer function
Input r(t) = Unit ramp input = t \ \
hence
C(s) =
1 s2
2
A (s + T) + Bs (s + T) + C s = 1
\ AT = 1
\ A=
1 T
R(s) =
(s + T)
=
C(s) 1 = R(s) ( s + T)
1 s2 A s2
+
B C + s s+T
2
i.e. (B + C) s + (A + BT) s + AT = 1 \ B= -
and A + BT = 0
TM
1 T2
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B+C =0
\ C(s) =
9 - 12
1
\ C=
c(t), r(t)
T2
T2
1 T 1 s s2
+
The Performance of Feedback Control Systems
r(t) = t
T2
1 s+T
S.S. error
6
c(t) for T = 1
4
Taking Laplace inverse, 1 1 1 -T t \ c(t) = æç ö÷ t + e 2 T è ø T T2
2 0
The sketch of c(t) and r(t) is shown in Fig. 9.1. Take T = 1 for plotting c(t).
2
4
6
Fig. 9.1
Example 9.9.5 Solution : Let the transfer function of a first order system is, C(s) 1 1 2 1 and r(t) = = t i.e. R(s) = R(s) (s + T) 2 s3 R (s) 1 A B C D C(s) = = = + + + \ 3 3 2 (s + T) s s+T s (s + T) s s \ A (s + T) + B s (s + T) + C s 2 (s + T) + D s 3 = 1 C + D = 0, B + C T = 0, A + B T = 0, A T = 1 -1 -1 1 \ A = 1 , B= , C= , D= 2 3 T T3 T T \
C (s) =
(1 / T) (1 / T 2 ) (1 / T 3 ) (1 / T 3 ) + s s+T s3 s2
\ C(t) = L-1 [C(s)] =
1 2 1 1 1 - Tt t t+ e 2 3 2T T T T3
Example 9.9.6 Solution : R(s) = T(s) =
1 s C(s) K K hence C(s) = = R(s) s+ a s(s + a)
Finding partial fractions, we get A1 A2 K K , A2 = and A 1 = + C(s) = s s+ a a a
\
æ Kö æ Kö ça÷ ça÷ C(s) = è ø - è ø s s+ a
i.e. c(t) =
K K - at e a a
TM
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Time t
Control System Engineering
9 - 13
The steady state of c(t) is 2 hence
Lim t®¥
The Performance of Feedback Control Systems
c(t) = 2
K = 2 a
\ While slope at t = 0 is \
–
\
K ´ ( -a) e - at a K = 1
dc(t) 2.0 = 1 i.e. 2.0 dt
=1 t= 0
= 1 t= 0
and
a = 1/2
Example 9.14.15 Solution :
Comparing the T.F. with the standard form w2n
=
25 and
s 2 + 2x wn s + w2n
2x wn = 6
\
wn = 5
w2n
x = 0.6
é 1 - x2 q = tan -1 ê x ê ë
ù ú = 0.9272 radians ú û
wd = wn 1 - x 2 = 5 1 - (0.6) 2 = 4 rad/sec. Tr =
p - q p - 0.9272 = = 0.5535 sec. , wd 4
% Mp = e
1 - x2
-px
´ 100 = 9.48 % , e- x wn t
Tp =
p p = = 0.785 sec. wd 4
Ts =
4 = 1.33 sec. x wn
sin ( wd t + q) = 1 –
And
c(t) = 1 –
\
c(t) = 1 – 1.5625 e - 3t sin (4t + 0.9272)
1 - x2
e - 3t 1 - (0.6) 2
sin(4t + 0.9272)
Example 9.14.16
Solution :
C(s) R(s)
=
20 (s + 1) (s + 4) 20 = 2 20 s + 5s + 24 1+ (s + 1) (s + 4)
Key Point Now though T.F. is not in standard form, denominator always reflect
wn2
from middle term and the last term respectively.
\ Comparing, s 2 + 5s + 24 with s 2 + 2x wn s + wn2 TM
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2x wn and
Control System Engineering
\
9 - 14
The Performance of Feedback Control Systems
w2n = 24 \ wn = 4.8989 rad/sec. 2 x wn = 5 \ x = 0.51031 1 - x 2 = 4.2129 rad/sec.
wd = wn
Now, for c(t) we can use standard expression for
C(s) in standard form. So writing R(s)
20 ïì 24 ïü × 24 íï s 2 + 5s + 24 ýï þ î For the bracket term use standard expression, and then c(t) can be obtained by multiplying 20 this expression by constant . 24 é ù 20 e - x wn t ê1 c(t) = sin ( wd t + q) ú \ 24 ê ú 1 - x2 ë û C(s) R(s)
=
1 - x2 radians = 1.03 radians x
q = tan - 1 \
c(t) =
[
]
20 1 - 1.1628 e - 2.5 t sin (4.2129 t + 1.03) 24
Example 9.14.17 Solution : G(s) = K1 , H(s) = 1 + K s 2 s2 K1 G(s) T.F. = = 1 + G(s)H(s)
s2 1+
K1 s2
=
(1 + K2 s)
K1 s 2 + K1 K2 s + K1
\ Comparing with standard form, wn =
2x wn = K1 × K2 ,
K1 ,
\
x=
1 2
K1 × K2
Now, Mp is function of x alone, \
% Mp = e - p x
1 - x2
\
0.25 = e - p x
1 - x2
\
Now,
– 1.3862 =
Tp =
-px 1 - x2
´ 100 i.e.
25 = e - p x
i.e.
1 - x2
´ 100
-px
ln (0.25) =
1 - x2
and solving x = 0.4037
p = 4 sec. wd
i.e.
p wn
1 - x2 TM
=4
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\
p wn
1 - (0.4037) 2
x =
and,
The Performance of Feedback Control Systems
=4
i.e.
wn = 0.8584 rad/sec.
K1
i.e.
K 1 = w2n = 0.7369
wn =
Now,
9 - 15
1 2
K1 × K2 = 0.4037
i.e.
K 2 = 0.9405
Example 9.14.18 Solution : System differential equation is, d2 y
dy + 8y = 8x dt Y(s) To find T.F. , take Laplace transform from above equation and neglect initial X(s) dt 2
+ 4
conditions. s 2 Y(s) + 4s Y(s) + 8 Y(s) = 8 X(s) i.e. Y(s) [s 2 + 4s + 8] = 8 X(s) \ \
T.F.
Y(s) w2n 8 = = X(s) s 2 + 4s + 8 s 2 + 2x w s + w2 n n
w2n = 8
i.e.
wn = 2.83 rad/sec.
2 x wn = 4
\ x = 0.7067 1 - x 2 = 2.83 1 - (0.7067) 2 = 2.002 rad/sec.
\
wd = wn
\
Tp = Time for peak overshoot =
p p = = 1.57 sec. wd 2.002
1 - x2
1 - (0.706) 2 ´ 100
% Mp = e - p x
´ 100 = e - p ´ 0.706
Ts = Settling time =
4 4 = 2 sec. = 0.7067 ´ 2.83 x wn
Example 9.14.19 Solution : The characteristic equation is 1 + G(s)H(s) = 0 \ 1+
16 =0 s (s + 6)
i.e.
s2 + 6s + 16 = 0
2
Comparing with s + 2 x wn s + w2n = 0, wn = 4 rad /s, 2 xwn = 6
x = 0.75
i.e.
wd = wn 1 - x 2 = 2.6457 rad/s
\
Tr =
p-q where wd
Tr =
p - 0.7227 = 0.9142 sec 2.6457
q = tan
1 - x2 = 0.7227 rad x
–1
TM
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= 4.33 %
Control System Engineering
TP =
9 - 16
p p = 1.1874 sec = wd 2.6457
% MP = 100 e - px / Ts =
The Performance of Feedback Control Systems
1- x 2
= 2.8375 %
4 4 = 1.333 sec = x wn 0.75 ´ 4
Example 9.14.20 Solution : The applied force f(t) gets consumed in moving a mass through x(t) against spring and friction force. d 2x dx f(t) = M \ + B + K x( t ) K (1) 2 dt dt Taking Laplace and neglecting initial conditions, F(s) = M s 2 X(s ) + B s X (s ) + K X(s ) = X(s) [M s 2 + B s + K] X(s ) 1 1/M = = 2 F (s ) 2 M s + B s +K s + B s + K M M
\
K (2)
Comparing denominator with s 2 + 2 x wn s + w2n , w2n = Now \
K M
i.e.
K M
wn =
f(t) = 8.9 N X(s) =
8.9 . s
and
i.e.
2 x wn =
F(s) =
B B i.e. x = M 2 MK
8.9 s
(1 M)
B K s+ M M The final value of x(t) is 0.03 m.
\
Final value
K (3)
s2 +
= Lim s . X(s)
i.e.
s® 0
Lim s .
s® 0
8.9 . s
(1 M) s2 +
B K s+ M M
= 0.03
8.9 = 0.03 i.e. K = 296.67 N/m K The overshoot is 0.0029. But this is for the final value of 0.03. Hence the % M p can be calculated as, 2 0.0029 % Mp = ´ 100 = 9.67 % = e - p x 1 - x ´ 100 0.03
\
Solving, Now
x = 0.5966 Tp = 2 sec =
p wn
1 - x2
wn =
i.e.
TM
p 2 1 - x2
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\
wn =
9 - 17
p 2 1 -( 0 . 5966) 2
The Performance of Feedback Control Systems
= 1.957 rad/sec
Equating to equation (3), K M
wn = And,
x =
i.e.
B 2 MK
1.957 =
296.67 M
i.e. 0.5966 =
i.e.
M = 77.44 kg
B 2 77.44 ´ 296.67
i.e.
B = 180.855 N/M/sec
Example 9.14.21 Solution : From the T.F. the poles are at s = – 3 ± 7j and there are no zeros. \ Pole-zero plot is as shown. The transfer function is, C ( s) 1 1 = = 2 R ( s) (s + 3 + j7)(s + 3 - j7) ( s + 3) - (j7) 2 =
j7
X
–3
0
1 s2
– j7
X
+ 6s + 58
Comparing denominator with s 2 + 2 xwn s + w2n , w2n = 58 hence 2 xwn = 6
wn = 7.6157 rad/sec x =
hence
6 = 0.3939 2 ´ 7.6157
wd = wn 1 - x 2 = 7.6157 × 1 - ( 0. 3939) 2 = 7 rad/sec \
Tp = Peak time = % Mp = e - px
1- x 2
p p = = 0.4487 sec wd 7
´ 100 = 23.11%
Example 9.14.22 Solution : The closed loop T.F. is, (s + 2) C(s) G(s) s(s + 1) s+2 = = = 2 R(s) 1 + G(s) (s + 2) s + 2s + 2 1+ s(s + 1) Comparing denominator with s 2 + 2 xwn s + w2n , w2n = 2 2 xwn = 2
i.e. wn = i.e.
x =
2 = 1.414 rad/sec 2 = 0.7071 2´ 2 TM
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Fig. 9.2
Control System Engineering
9 - 18
The Performance of Feedback Control Systems
wd = wn 1 - x 2 = 1.414 1 - (0.7071) 2 = 1 rad/s é 1 - x2 q = tan -1 ê êë x
ù p ú = 45° = rad. 4 úû
\
Tr =
p-q = 2.3561 sec, wd
\
Td =
1 + 0.7 x = 1.0572 sec, wn
%
Mp = 100 e
-p x
1 - x2
Tp =
p = 3.1416 sec wd
Ts =
4 = 4 sec xwn
= 4.32 %
Key Point As [C(s) / R(s)] is not in the standard form and numerator contains s term, the
standard expression of c(t) cannot be used. \
C(s) = R(s) ´
s+2 2
s + 2s + 2
=
(s + 2) s(s 2
…R(s) =
+ 2s + 2)
1 s
Find the partial fractions, C(s) = \ i.e. \
A + B = 0,
2A + C = 1, 2A = 2
A = 1,
B = –1, C = –1 (s + 1) 1 1 ì s +1 ü C(s) = = -í 2 2 s s + 2s + 2 s î s + 2s + 1 + 1 ýþ =
\
A(s 2 + 2s + 2) + s(Bs + C) A Bs + C = + s s 2 + 2s + 2 s(s 2 + 2s + 2)
üï 1 ìï s +1 -í s ï (s + 1) 2 + (1) 2 ýï þ î
c(t) = L-1 [C(s)] = 1 - e - t cost
Example 9.14.23 Solution : The closed loop transfer function is, K (s + a) C(s) G(s) = = R(s) 1 + G(s)H(s)
K (s + a) (s + b) 2 = 2 + K (s + a) s s (2 b + K) + (ka + b 2 ) 1+ 2 (s + b)
Comparing denominater with s 2 + 2xwn s + w2n w2n = K a b 2
i.e.
wn =
2 x wn = 2 b + K
i.e.
x
= TM
K a + b2 2b + K 2´
K a +b 2
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… (1) … (2)
Control System Engineering
9 - 19
For unit step input, A = 1. Kp = Lim G( s) H( s ) = Lim s®0
\
e ss =
s®0
K( s + a )
(s + b ) 2
The Performance of Feedback Control Systems
=
Ka b2
b2 A 1 = = Kp Ka Ka
… (3)
b2 But e ss is 10 % hence, b2 0.1 = Ka From (1),
10 =
From (2),
0.5 =
i.e. K a = 10 b 2
10 b 2 + b 2
i.e. b 2 =
2 ´ (0.9534) + K
10 11
… (4) i.e. b = 0.9534
10 ´ (0.9534) 2 + (0.9534) 2
2´
From (4), 1.2552 ´ a = 10 ´ (0.9534) 2
i.e. K = 1.2552.
i.e. a = 7.2416
Example 9.14.24 Solution : Substituting value of e, d2 c dt d2 c
\
d
t2
2
+ 6.4
+ 6.4
dc = 160 (r - 0.4 c) dt
dc + 64 c = 160 r dt
Taking Laplace transform of both the sides, and neglecting initial conditions,
\
s 2 C (s) + 64 s C(s) + 64 C (s) = 160 R(s) C(s) 160 = 2 R (s) s + 6.4 s + 64
… Transfer function
Comparing denominater with s 2 + 2 x wn s + w2n , wn 2 = 64
and
2 x wn = 6.4
6.4 = 0.4 2´ 8 As the damping ratio x is less than unity, the system is underdamped and will produce the time response with damped oscillatory transients for the step input as shown in the figure
i.e.
wn = 8 rad/sec.
and
x=
c(t)
t
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9 - 20
The Performance of Feedback Control Systems
Example 9.14.25 Solution:
The closed loop transfer function is, C( s) G( s) 1 + 0.4s 1 + 0.4s = = = R( s) 1 + G( s)H( s) s( s + 0.6) s2 + s +1 (1 + 0.4s) 1+ s( s + 0.6)
It is not standard second order system hence standard expression of c(t) cannot be used. Use partial fractions. 1 + 0.4s A Bs + C é 1 + 0.4s ù = = + C( s) = R ( s) ê 2 + + ú 2 + + 2 s s s 1 s s 1 ë û s s + s +1
(
\
)
)
A s 2 + s + 1 + ( B s + C) s = 1 + 0.4s
\ \
(
A + B = 0, A + C = 0.4, A = 1 C(s) =
=
=
i.e.
B = –1, C = – 0.6
1 ( s + 0.6) 1 s + 0.6 – = – s s2 + s +1 s 1 3 s2 + s + + 4 4 1 – s
s + 0.6 2
æ s + 1 ö + æç 3 ö÷ ç 2 ÷ø è 2 ø è
2
=
1 s + 0.5 + 0.1 – s ( s + 0.5) 2 + ( 0.866) 2
( s + 0.5) 1 0.1 0.866 ´ – – s ( s + 0.5) 2 + ( 0.866) 2 0.866 ( s + 0.5) 2 + ( 0.866) 2
c(t) = 1 – e –0.5t cos 0.866t – 0.1154 e –0.5t sin ( 0.866t )
\
From denominator of C(s)/R(s), w2n = 1 , 2xwn = 1 \ \
wn = 1, Mp
=
x = 0.5 , wd = wn 1 – x 2 = 0.866
100 e – px /
1– x 2
= 16.3 %,
Tp =
p = 3.627 sec wd
Example 9.14.26 K , H(s) = 1 + bs s(s + 2) K C( s) G(s) s(s + 2) K = = = 2 R( s) 1+ G(s)H(s) K(1 + bs) s + s( 2 + Kb) + K 1+ s( s + 2)
Solution : For the given system, G(s) =
\
Comparing denominator with s 2 + 2xwn s + wn2 , w2n = K
i.e.
wn =
... (1)
K
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9 - 21
2xwn = 2 + Kb
x=
i.e.
The Performance of Feedback Control Systems
(2 + Kb)
...(2)
2 K
Given specifications are Mp = 20 % and Tp = 1 sec \ Solving, and \
1- x 2
20 = e - px /
Tp = wn
p wn 1 - x
2
p wn 1 - ( 0.4559) 2
= 3.5297 rad/sec.
From equation (2),
0.4559 =
wd
1=
i.e.
3.5297 =
K
i.e.
K = 12.4587
i.e.
b = 0.0978.
(2 + 12.4587 ´ b) 2 ´ 3.5297
= wn 1 - x 2 = 3.1415 rad / sec
æ 1 - x2 q = tan -1 ç x ç è
\
1 – x2
x = 0.4559
From equation (1),
\
– px
´ 100 i. e. ln(0.2) =
ö ÷ = 62.8771º = 1.0974 rad ÷ ø
Tr =
p - q p - 1.0974 = 0.651 sec. = wd 3.1415
Ts =
4 4 = 2.4857 sec. = xwn 0.4559 ´ 3.5297
Example 9.14.27 Solution : i) All the three elements M, K and B are under same displacement x(t). Thus the differential equation is, d 2 x( t ) dx( t ) f(t) = M +B + K x( t ) 2 dt dt Taking Laplace of both sides, ... Neglect initial conditions F(s) = Ms 2 X( s) + Bs X( s) + K X( s) \
X( s) 1M 1 = = 2 F ( s) Ms + Bs + K s 2 + B s + K M M
ii) For K = 33, B = 15, M = 3 X( s) 1 3 0.333 = = 2 2 F( s) s + 5s + 11 s + 5s + 11 Comparing denominator with s 2 + 2xwn s + wn2 , TM
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w2n = 11
9 - 22
wn = 3.3166 rad/sec.
hence
2xwn = 5
x = 0.7537
hence
% Mp = e - px
1- x 2
The Performance of Feedback Control Systems
´ 100 = 2.7234 %
wd = wn 1 - x 2 = 2.1797 rad/sec \
Tp =
p = 1.4412 sec wd
and
Ts =
4 = 1.6 sec xwn
Example 9.14.28 Solution : ei(t) = R i(t) + L \
di(t) 1 + ò i(t) dt dt C
and eo(t) =
1 i(t) dt Cò
I(s) 1 Ei(s) = I(s) éR + sL + ù and Eo(s) = sC sC ûú ëê
E o (s) 1 1 / LC = = 2 E i (s) 2 s LC + sRC + 1 s + R s + 1 L LC 2 2 Comparing denominator with s + 2x wn s + wn , 1 R and 2x wn = w2n = LC L
\
\
wn =
1 LC
and
x=
R 2
L/ C
qqq
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The Stability of Linear Feedback Systems
10
Solutions of Examples for Practice Example 10.6.2 Solution : The Routh's array is, s4
1
26
80
s3
6
56
0
2
s
16.67
80
0
s1
27.2
0
s0
80
No sign change in the first column of the Routh's array hence the system is stable in nature.
Example 10.6.3 Solution : The Routh's array is, s4
1
23
50
s3
6
40
0
s2
16.333
50
0
s1
21.633
0
s0
50
There are no sign changes in the first column of the Routh's array, hence the system is stable in nature.
Example 10.6.4 Solution :
a 0 = 1,
a 1 = 6, s3 s2 s1 s0
a 2 = 11, a 3 = 6, n = 3 1 6 11 ´ 6 - 6 = 10 6 6
11 6 0
As there is no sign change in first column, system is stable. (10 - 1) TM
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10 - 2
The Stability of Linear Feedback Systems
Example 10.6.5 a 0 = 1,
Solution :
a 1 = 4, a 2 = 1,
a 3 = 16
s3 s
1
2
1
+4 4 - 16 = -3 4 + 16
s1 s0
16 0
As there are two sign changes, system is unstable. Number of roots located in the right half of s-plane = Number of sign changes = 2. Example 10.6.6 Solution : a) The characteristic equation is, 1 + G( s)H( s) = 0. 1 = 0 i.e. 1+ s 2 + 6s + 9 = 0 s + 2 ( )( s + 4) Routh's array is, s
2 1
s s
b)
1+
9 2
s ( s + 2)
0
1
9
6
0
No sign change in the first column, hence system is stable.
9
i.e. s3 + 2s2 + 9 = 0
= 0
Routh's array is, s s s s
3 2 1 0
1
0
2
9
– 4.5
0
There are two sign changes in the first column, hence system is unstable.
9
Example 10.7.6 Solution : s5
1
2
3
s4
1
2
15
s3
0
– 12
0
Replace 0 by small positive number e.
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Lim e®0 Lim e®0
10 - 3
=
( 2 e+ 12 ) ( -12) - 15 e e 2 e+ 12 e
=
s5
1
2
3
s4
1
2
15
s3
e
–12
0
s2
2e + 12 e
15
0
s1
æ 2e + 12 ö ÷( -12) - 15 e ç è e ø 2e + 12 e
0
0
s0
15
2e + 12 12 = 2+ = +¥ e e =
Lim -24e - 144 - 15e 2 2e + 12 e®0
0 - 144 - 0 = – 12 0 + 12
s5
1
2
3
s4
1
2
15
3
s
e
– 12
0
s2
+¥
15
0
s1
–12
0
s
0
The Stability of Linear Feedback Systems
There are two sign changes, so system is unstable.
15
Example 10.7.8 Solution : s 4 + 0s 3 + 2s 2 + 0s + 1 = 0 Routh’s array is, s4
1
2
s3
0
0
\
Row of zeros, special case 2,
\
Auxiliary equation is,
1
A(s) = s 4 + 2s 2 + 1 = 0 dA(s) = 4s 3 + 4s ds
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\
10 - 4
The Stability of Linear Feedback Systems
Routh’s array is, s4
1
2
1
s3
4
4
0
s2
1
1
0
s1
0
0
\
Routh’s array s4
1
2
1
s3
4
4
0
s2
1
1
0
s1
2
0
s0
1
No sign change so no root in right half. Solving A(s) = 0 i.e. s 4 + 2s 2 + 1 = 0 s2 =
\ \
-2 ± 4 - 4 = –1, –1 2
s = ± j, \
±j
Number of roots on imaginary axis = 4 Number of roots in left half = 0 = Number of roots in right half.
Example 10.7.9 Solution : s6
1
3
– 16
s5
4
0
– 64
0
s4
3
0
– 48
0
s3
0
0
0
– 48
Ü Special case 2 A(s) = 3s4 - 48 = 0 dA = 12s3 ds
s6
1
3
– 16
– 48
s5
4
0
– 64
0
s4
3
0
– 48
0
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The Stability of Linear Feedback Systems
s3
12
0
0
0
s2
[ e] 0
– 48
0
0
s1
576 e
0
0
s0
– 48
Ü Special case 2 Lim 576 =+¥ e® 0 e
\ One sign change and system is unstable. With one root in right half of s-plane. 4 Solve, A(s) = 3s 4 - 48 = 0 i.e. s = 16 s2 = 4
s2 = – 4
i.e. s = ± 2 and
i.e s = ± 2j
So two roots (± j2) located on jw axis and remaining in left half of s-plane. Roots with positive real part ® One Roots with negative real part ® Three Roots with zero real part
®
Two.
Example 10.7.10 Solution : The characteristic equation is , s 6 + s5 + 3s 4 + 3s 3 + 2s 2 + s 1 + 1 = 0, The Routh array is , For s 3 , Lim e® 0
3 e -1 ® Negative e
æ 3 e -1 ö ç e ÷ - ( e - 1) ø 2 For s , Lim è e® 0 æ 3 e -1 ö ç e ÷ ø è ® Positive i.e. X ® positive e -1 ö æ 3 e -1 ö Xæç ÷ ÷ -ç è e ø è e ø 1 For s , Lim X e® 0 ® Negative
6
1
3
2
1
5
1
3
1
0
4
0 +e
1
1
3
3e – 1 e 3e – 1 – ( e – 1) e
e–1 e
0
s
s s s
2
s
(3e – 1) e 1
s
X
0 Thus there are four sign changes in the s first column of the Routh's array.
e–1 3e – 1 – e e X
1 X
0
1
Hence the system is unstable with four roots lying in the right hand side of s plane. Example 10.8.7 Solution : The characteristic equation is 1 + G(s)H(s) = 0 TM
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\
1+
10 - 6
The Stability of Linear Feedback Systems
K = 0 s (s + 2) (s +3) (s + 4)
...H(s) = 1
s 4 + 9s 3 + 26s 2 + 24s + K = 0
\
The Routh's array is, s4
1
26
K
From the row of s 0 , K > 0
s3
9
24
0
From the row of s 1 , 560 – 9K > 0
s2
23.333
K
\ 560 > 9K
s1
560 – 9 K 23. 333
0
\ K < 62.222
s0
Hence the range of K for stability is,
K
0 < K < 62.222
Example 10.8.8 Solution : F(s) = s [s 3 + 5s 2 + 5s + 4] + K = 0 s4
1
5
K
s3
5
4
0
s2
4.2
K
0
s1
16.8 - 5K 4.2
0
s0
K
i.e. s 4 + 5s 3 + 5s 2 + 4s + K = 0 For system to be stable there should not be sign change in the first coumn. \ K>0
from s0
0 \ 16.8 – 5 K > 0 from s
\ 16.8 > 5 K \ 3.36 > K \ K < 3.36 \ Range of K is 0 < K < 3.36
Example 10.9.5 Solution :
The characteristic equation is 1 + G(s) H(s) = 0 K = 0 i.e. s 4 + 4s 3 + 4s 2 + 3s + K = 0 1+ 2 s(s + 3)(s + s + 1)
\ s4
1
4
K The Rouths array is,
s3
4
3
0
s2
3.25
K
0
s1
9.75 - 4K 3.25
0
s0
K
For marginal value of K, row of s 1 must be row of zeros. \ 9.75 – 4 Kmar = 0 i.e. Kmar = 2.4375 For this value of K, the auxiliary equation is, A(s) = 3.25 s 2 + Kmar = 0 i.e. s 2 = \s = ± j 0.866 TM
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- Kmar -2.4375 3.25 3.25
Control System Engineering
10 - 7
The Stability of Linear Feedback Systems
Thus the frequency of oscillations in 0.866 rad/sec. Example 10.9.6 Solution :
The characteristic equation is,
s(s 2 + 25 +2) (s 2 + 65 +10) + K = 0
s5 + 8s 4 + 24s 3 + 32s 2 + 20s + K = 0
i.e.
Routh's array is, For marginal K1 row of s 1 must be row of zeros. s5
1
24
20
s4
8
32
K
s3
20
160 - K 8
0
480 + K 20
s2
\
For this value,
\
\( K + 3541.6845) (K - 21.6845) = 0 … Selecting + ve value 480 + K mar ö 2 A(s) = æç ÷ s + Kmar = 0 20 ø è s2 = -
\
\76800 + 160K - 480 K - K2 - 3200 k = 0 i.e. - K2 - 3520K + 76800 = 0
K
Kmar = + 21 .6845
480 + K ö æ 160 - K ö \æç ÷ç ÷ - 20K = 0 è 20 ø è 8 ø
21.6845 = 0.8644 (480 + 21.6845) 20
s = ± j 0.9297 = ± jw
Hence the frequency of oscillations is 0.9297 rad/sec. Example 10.9.7 Solution : The characteristic equation of the system is 1 + G(s) H(s) = 0 1+
K(s + 5) = 0 s(1 + Ts) (1 + 2s)
\
s(1 + Ts) (1 + 2s) + K(s + 5) = 0
\
s(2Ts 2 + (T + 2)s + 1) + K(s + 5) = 0
\
2Ts 3 + (T+2) s 2 + s(K + 1) + 5K = 0
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\
The Stability of Linear Feedback Systems
Routh’s array is s3
2T
K+1
s2
(T + 2)
5K
s1
(T + 2) (K + 1) - 2T ´ 5K (T + 2)
0
s0
5K
From last row 5K > 0 K must be positive. From row of s 1 (T + 2) (K + 1) – 10KT > 0 KT + 2K + T + 2 – 10KT > 0 \
2K + T + 2 – 9KT > 0
Limiting value is 2K + T + 2 – 9KT > 0 i.e.
2+ T 9T - 2
K <
\ Region in which a closed loop system is stable is T
Unstable
K=
Unstable
Stable
Unstable
0
0.1
Considering positive values of K and T
2+T 9T – 2
0.2
0.3
0.4
0.5
0.6
0.7
K
Example 10.9.8 Solution : (i) Loop T.F. has pole at s = 0, s = –1, s = –3 and zero at s = –5, gain K = 10 10(s + 5) G(s) H(s) = s(s + 1) (s + 3) Characteristic equation. 1 + G(s) H(s) = 0 1+
10(s + 5) = 0 s(s + 1) (s + 3) TM
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The Stability of Linear Feedback Systems
s(s 2 + 4s + 3) + 10s + 50 = 0 s 3 + 4s 2 + 13s + 50 = 0 s3
1
13
s2
4
50
s1
0.5
0
s0
50
No sign change in the first column hence system is absolutely stable in nature. ii) Type 1, Kv = 10 sec -1 , Poles at s = –3, s = – 6 K(1 + T1 s)... Type 1 G(s) H(s) = s(1 + T2 s) (1 + T3 s)... G(s) H(s) =
Now
Now \
Kv =
K/ 18 s s s æç 1 + ö÷ æç 1 + ö÷ 3 6 è øè ø 5 ´ K/ 18 K = 18 s s s æç 1 + ö÷ æç 1 + ö÷ 3 øè 6ø è
lim lim sG(s) H(s) = s®0 s®0
K = 10 \ K = 180 18 180 G(s) H(s) = s(s + 3) (s + 6)
Charcteristic equation : 1 + G(s) H(s) = 0 s(s 2 + 9s + 18) + 180 = 0 s 3 + 9s 2 + 18s + 180 = 0 s3
1
18
s2
9
180
s1
–2
0
s0
180
There are two sign changes so two roots are located in R.H.S. of s-plane Hence system is Unstable. Example 10.9.9 Solution : The open loop T.F., K G(s)H(s) = s (1 + sT1 ) (1 + sT2 ) TM
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The Stability of Linear Feedback Systems
The characteristic equation is, 1 + G(s)H(s) = 0 1+
\
K = 0 s (1 + sT1 ) (1 + sT2 )
T1 T2 s 3 + (T1 + T2 ) s 2 + s + K = 0
\
The Routh's array is, s3
T1T2
1
s2
(T1 + T2 )
K
s1
(T1 + T2 ) - T1T2 K (T1 + T2 )
0
s0
K
K>0 For stability, and
(T1 + T2 ) - T1 T2 K > 0
\
T1 T2 K < T1 + T2
\
1 ö æ 1 K 82.26
system is unstable.
Example 11.7.23 Solution : Step 1 : P = 4, Z = 0, N = 4, All branches approaching to ¥. Starting points –3 ± 9 – 12 i.e. 0, –3, –1.5 ± j 0.866. s = 0, –3 and 2 Step 2 :
Pole-zero plot is as shown.
Minimum one breakaway point exists between 0 and – 3. Step 3 and 4 : changed.
Real parts of poles are not
j 0.866 NRL
NRL –3 –1.5
0 –j 0.866
Centroid –1.5 and angles q 1 = 45°, q 2 = 135°, q 3 = 225°, q 4 = 315°. TM
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Step 5 :
11 - 25
The Root Locus Method
Breakaway point 1 + G(s)H(s) = 0
1+
K s (s + 3) (s 2 + 3 s + 3)
s 4 + 6s 3 + 12s 2 + 9s + K
=
0
=
0
\ K = – s 4 – 6s 3 – 12s 2 – 9s \
dK = – 4s 3 – 18s 2 – 24s – 9 = 0 ds
... (1) i.e.
4s 3 + 18s 2 + 24s + 9 = 0
Solving, s = – 1.5, – 0.633, – 2.366 \ Breakaway points are = –1.5, – 0.633, –2.366 and all are valid. For s = –1.5, K = 1.6875
... Using equation (1)
s = - 0.633 , K= 2.25 ü ý These two occur simultaneously. s = - 2. 366 , K= 2.25 þ
... Using equation (1)
Step 6 :
Intersection with imaginary axis.
Characteristic equation : s 4 + 6s 3 + 12s 2 + 9s + K = 0 Routh’s array, s4
1
12
K
s3
6
9
0
s2
10.5
K
0
s1 s0
Step 7 :
94.5 – 6 K 10.5
\
94.5 – 6K = 0 Kmar = 15.75 A(s) = 10.5s 2 + K = 0
\10.5s 2 + 15.75 = 0 s2 =
0 \
K
s =
– 1.5 ± j 1.224
Angle of departure.
fd at –1.5 + j 0.866 is – 90º. While fd at –1.5 – j 0.866 is + 90º. Step 8 : In this problem s = –1.5 will occur as a breakaway point first as K = 1.6875 for this breakaway point. But branches from complex poles will reach at s = –1.5 rather than branches from real open loop poles. This is because the complex poles are more closer to s = – 1.5 than the real poles. The remaining breakaway points will occur later simultaneously. These breakaway points will exist after the complex poles branches break into real branches at s = –1.5. The root locus is shown in the Fig. 11.12.
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11 - 26
The Root Locus Method jw
K = 2.25 Breakaway point = –0.633
q2
¥
q1
¥
j 1.224 K = 2.25 Breakaway point = –2.366
j0.866
–3
Kmar = 15.75
s
0 –j 0.866
–j 1.224 ¥
Centroid & B.P. at s = –1.5 K = 1.6875
q3
¥ q4
Fig. 11.12
Step 9 :
For 0 < K < 15.75 system is stable. At K = 15.75 system is marginally stable. For K > 15.75 system is unstable.
Example 11.7.24 Solution : G(s)H(s) = Step 1 :
K(s + 4) (s + 2) (s + 3)
2 breakaway points exist
P = 2, Z = 1, N = P = 2
NRL X –4 –3
P – Z = 1 branch approach to ¥ Starting points : s = –2, –3
NR L X –2
Terminating points : s = – 4, ¥ Step 2 :
Sections of real axis.
Step 3 :
Angle of asymptote (2q + 1)180° ,q=0 q = P-Z
\
q1 = 180°
Step 4 :
Centroid not required as only one asymptote exists.
Step 5 : Breakaway point K(s + 4) = 0 i.e. 1+ (s + 2)(s + 3) \ \ \
s 2 + 5s + 6 + K(s + 4) = 0
i.e. K =
- s 2 - 5s - 6 (s + 4)
(s + 4)( -2s - 5) - ( - s 2 - 5s - 6)(1) dK = =0 ds (s + 4) 2 i.e. -2s 2 - 8s - 5s - 20 + s 2 + 5s + 6 = 0 s = – 2.5857, – 5.4142
s 2 + 8s + 14 = 0 TM
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…(1)
Control System Engineering
11 - 27
The Root Locus Method
Both are valid breakaway points. At s = – 2.5857, K = + 0.1715
…From equation (1)
At s = – 5.4142, K = + 5.8284 Step 6 :
…From equation (1)
Intersection with imaginary axis i.e. s 2 + 5s + 6 + K(s + 4) = 0
1 + G(s)H(s) = 0
\ s 2 + s(5 + K) + (6 + 4K) = 0 s2
1
6 + 4K
From row of s 1 , for Kmar,
s1
5+K
0
5+K=0
s0
6 + 4K
i.e. Kmar = – 5 As it is negative, root locus does not intersect imaginary axis and lies completely in left half of s-plane.
Step 7 : Angle of departure not required as no complex poles. Step 8 : The root locus is as shown in the Fig. 11.13.
Imj Scale : On both axes 2 Units = 1 Unit
Breakin point –5.41
Breakaway point –2.58
q1 = 180º ¥
–4
X –3
X –2
0
Real
Fig. 11.13
The system is absolutely stable as for any value of K > 0, all the roots lie in the left half of s-plane. TM
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Control System Engineering
11 - 28
The Root Locus Method
Example 11.13.4 Solution : Without derivative rate feedback controller, compare denominator with s 2 + 2xwn s + wn2 , w2n = 16 i.e. wn = 4
and
2xwn = 1.6 i.e. x = 0.2
wd = wn 1 - x 2 = 3.9192 rad/sec, \
Tr =
ö ÷ = 1.3694 rad ÷÷ ø
p-q p = 0.4521 sec, Tp = = 0.8016 sec wd wd
% Mp = e - px / For
æ 1 - x2 q = tan -1 çç x ç è
1- x 2
´ 100 = 52.662 %
C(s) 16 16 = , H(s) = 1 , G(s) = 2 R(s) s(s + 1.6) s + 1.6s + 16
\
A 1 s ´ 16 = 0.1 = = 10, e ss = K 10 s(s + 1.6) s ®0 v
Kv = Lim sG(s)H(s) = Lim s ®0
Now use rate feedback controller, C(s) G(s) 16 = = \ R(s) 1 + G(s)H(s) s 2 + s(1.6 + 16 K ) + 16 t Comparing denominator with s 2 + 2xwn s + wn2 , w2n = 16
\
i.e. wn = 4
2xwn = 1.6 + 16 Kt
… Remains same
i.e.
x=
1.6 + 16 Kt = 0.8 (given) 2´ 4
\
Kt = 0.3
\
wd = wn 1 - x 2 = 4 ´ 1 -( 0.8) 2 = 2.4 rad/sec R
… For x = 0.8
16 s(s + 1.6)
1 –
–
C
s Kt Rate feedback controller
Fig. 11.14 (a)
16 G(s) = s(s + 1.6 + 16 K ) t H(s) = 1
R
–
16 s(s + 1.6 + 16 K t)
Fig. 11.14 (b) TM
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C
Control System Engineering
11 - 29
é 1 - x2 q = tan -1 ê ê x ë
\
\
Tr =
ù ú = 0.6435 rad ú û
p-q = 1.04 sec, wd
% Mp = e - px /
1- x 2
Tp =
p = 1.308 sec. wd
´ 100 = 1.516 %
Kv = Lim sG(s)H(s) = Lim s ´ s®0
\
e ss
The Root Locus Method
s®0
16 = 2.5 s(s + 1.6+ 16 ´ 0.3)
A 1 = 0.4 = = Kv 2.5
The steady state error increases but overshoot decreases drastically. Example 11.13.5 Solution :
i) With K0 = 0, the system is, KA C(s) s(s + 2) = R(s) KA 1+ s(s +2) =
R(s)
KA –
1 s(s + 2)
C(s)
KA S2
+ 2s + KA
Comparing denominater with s 2 + 2x wn s + w2n , w2n = KA
i.e. wn = KA = 10 = 3.16 rad / sec
2xw2n = 2 G(s)H(s) =
i.e.
2 2 ´ 10
= 0.3162
KA 10 = s(s + 2) s(s + 2)
\
Kv = Lim sG(s)H(s) =
\
e ss
s®0
=
x =
10 =5 2
A 1 = = 0.2 Kv 5
... For unit ramp A = 1
ii) With K0 , the system becomes , 10 \
\
C(s) = R(s)
2
s + s(2 + K0 ) 10 = 2 10 s + s (2 + K0 ) + 10 1+ s 2 + s (2 + K0 )
w2n = 10
i.e. wn = 10 rad/sec TM
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11 - 30
The Root Locus Method
1
s2 + s(2 +K0) Minor loop
–
–
C(s)
1 s(s + 2)
KA = 10
»
R(s)
sK0
2xwn = 2 + K0
and But given \ \
i.e
x = 0.6 hence,
x=
R(s)
KA = 10 –
2 + K0 2 10 2 + K0
0.6 =
2 10
K0 = 1.7947 G(s)H(s) =
10 s2
+ s(2 + 1.7949)
\
Kv = Lim sG(s)H(s) =
\
e ss =
s®0
=
10 s(s + 3.7949)
10 = 2.6351 3.7949
A 1 = = 0.3795 Kv 2.6351
Example 11.13.6 Solution : Without controller, 100 H(s) = 1 G(s) = s (s + 12) \ \
C(s) 100 = 2 R(s) s + 12s + 100 w2n = 100
\ wn = 1 0
2x wn = 12 \ \
wd = wn % Mp = e Ts =
\
x = 0.6
1 - x 2 = 10 ´ 0.8 = 8 rad/sec
- p x
1 – x2
= 9.47 %
4 = 0.666 sec x wn
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1 s(s + 2) sK0 1+ s(s + 2)
C(s)
Control System Engineering
11 - 31
The Root Locus Method
With controller we get, R(s)
100 s (s + 12)
+
– s
C(s)
1 30
æ 1 + s ö 100 ç (s + 30) ´ 3.33 30 ÷ø G(s) = è , H(s) = 1 = s(s + 12) s (s + 12) C(s) = R(s)
= \
(s + 30) 3.33 3.33(s + 30) s(s + 12) = 2 (s + 30) 3.33 s + 12s + 3.33s + 100 1+ s (s + 12) 3.33 (s + 30) s2
+ 15.33s + 100
w2n = 100
\ wn = 10
2x wn = 15.33 \
x =
15.33 = 0.7665 2 ´ 10
\ x is improved, wd = wn
1 - x 2 = 10 1 - (0.7665) 2
= 6.4224 rad/sec % Mp = e - p x
1 - x2
´ 100 = 2.353 %
Overshoot decreased to 2.3 % from 9.47 %. 4 = 0.5218 sec Ts = x wn
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11 - 32
The Root Locus Method
Comparison : Following figure shows comparison between system with controller and system without controller. c(t)
9.47%
c(t)
1
2.353%
1
0.66 sec Without controller
time
0.5218 sec With controller
time
qqq
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Frequency Response Methods (Bode Plot Method)
12
Solutions of Selected Examples for Practice Example 12.6.4 Solution : The open loop transfer function is K and H(s) = 1 G(s) = s(1 + sT)
\
K C(s) G(s) s(1 + sT) K K/ T = = = = 2 R(s) 1 + G(s)H(s) K 2 Ts + s + K s + 1 s + K 1+ T T s(1 + sT)
Comparing this with standard form, wn = x = Now Solving, \ \ \
K T
... (1)
1 2 KT
... (2)
Mp = 20 %
i.e.
x = 0.455
and
0.2 = e - p x /
1- x 2
wr = 6 rad/sec
... Given
wr = wn 1 - 2 x 2 6 = wn 1 - 2 ´ (0.455) 2 wn = 7.8382 rad/sec
Using equation (1),
7.8382 =
K T
i.e.
61.437 =
K T
Using equation (2),
0.455 =
1 2 KT
i.e.
0.8281 =
1 KT
Substituting from equation (1) into equation (2), 61.437 = \
1 0.8281T 2
i.e.
T 2 = 0.0196
i.e.
T = 0.14
K = 8.6133 (12 - 1) TM
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Control System Engineering
Mr =
12 - 2
1
=
2 x 1 - x2
Frequency Response Mehtods (Bode Plot Method)
1 2 ´ 0.455 1 - (0.455) 2
= 1.234
Example 12.6.5 Solution : For the second order system, Mp = 10 % -px 1 - x2
= ln ( 0.1)
px
i.e.
1 - x2
´ 100
= 2.3025
x = 0.5911
Solving,
Tr = \
1- x2
10 = e - p x
i.e.
0.1 =
p-q wd
where
p - 0.9383 wn 1 - x 2
q = tan -1
i.e. wn =
1 - x2 = 0.9383 rad x 2.2032
0.1 ´ 1 - ( 0.5911) 2
\ wn = 27.3158 rad/sec From these values, Mr and B.W. can be calculated as, 1 1 = Mr = 2 2 x 1 -x 2 ´ 0.5911 ´ 1 - ( 0.5911) 2
= 1.0487
B.W. = wn ´ 1 - 2 x 2 + 2 - 4 x 2 + 4 x 4 = 27.3158 ´ 1 - 2 ´ ( 0.5911) 2 + 2 - 4 ´ ( 0.5911) 2 + 4 ( 0.5911) 4 = 31.686 Example 12.6.6 K , H(s) = 1 s(1 + st) K K C(s) s(1 + st) K t = = = 2 +s+K R(s) K 2 +1 + K ts s s 1+ t t s(1 + st)
Solution : G(s) =
\
Comparing denominator with s 2 + 2 x wn s + w2n \ and
w2n = 2xwn =
K i.e. wn = t
K t
...(1)
1 1 i.e. x = t 2 Kt
...(2)
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Control System Engineering
Mr =
12 - 3
1 2x 1 – x2
Frequency Response Mehtods (Bode Plot Method)
= 1.06
\
x 1- x 2 = 0.4716
i.e.
x 2 (1 - x 2 ) = 0.2225
\
x 4 - x 2 + 0.2225 = 0
i.e.
x 2 = 0.6658, 0. 3341
\
x = 0.8159, 0.578
\
x = 0.578
Using equation (1),
K t
wn =
Using equation (2), 0.578 = \
t
but x cannot be more than 0.707.
i.e.
1 2 144t ´ t
K t
i.e.
0.578 =
1 2 ´ 12 ´ t
12 =
i.e.
= 0.072, K = 10.3806
B.W. = wn 1 - 2 x 2 + 2 - 4 x 2 + 4 x 4 = 12 1 - 2 ´ ( 0 . 578) 2 + 2 - 4 ´ ( 0 . 578) 2 + 4 ´ ( 0 . 578) 4 = 14.1246 rad/sec Example 12.6.7 Solution : From the table Mp = 0.12 i.e. But,
Mp =
2 100 e - p x / 1- x
12 % and Tp = 0.2 sec. 1- x 2
i.e. 0.12 = e - p x /
Solving, x = 0.5594 Tp =
p = wd
p wn 1 - x 2
i.e. 0.2 =
p wn 1 - ( 0.5594) 2
wn = 18.9504 rad/sec. \
Mr =
1 2x 1-x
2
=
1 2 ´ 0.5594 ´
1 - (0.5594) 2
wr = wn
1 - 2 x 2 = 11.5914 rad/sec
wb = wn
1 - 2 x2 +
= 1.078
2 - 4 x 2 + 4 x 4 = 22.1162
Example 12.6.8 Solution : For the given pole-zero diagram, T(s) =
K 104 = (s + 2 - j10) (s + 2 + j10) s 2 + 4 s + 104
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K = 144 t
Control System Engineering
12 - 4
Frequency Response Mehtods (Bode Plot Method)
Comparing denominater with s 2 + 2 x wn s + w2n , w2n = 104 2 x wn = 4 Mr =
i.e. 10.198 rad/sec. i.e. 1
2x
wr = wn
1 - x2
x = 0.1961 =
1 1 - (0.1961) 2
2 ´ 0.1961 ´
1 – 2 x 2 = 10.198 ´
wb = B.W. = wn 1 - 2 x 2 +
= 2.6
1 - 2 ´ (0.1961) 2 = 9.4136 rad/sec. 2 - 4 x 2 + 4 x 4 = 15.4066
Example 12.6.9 Solution : Mp = 50 %,
Td = Time period = 0.2 sec.
Mp = 100 e - p x
Now,
ln (0.5) =
-px 1 - x2
1 - x2
i.e.
0.5 = e - p x
i.e. 0.2206 =
1 - x2
x 1 - x2
x = 0.2154
Solving,
fd =
1 = 5 Hz Td
And
Td = 0.2 sec.
\ \
wd = 2 p fd = 10 p rad / sec = wn 1 - x 2 10 p = 32.1711 rad/sec wn = 1 - (0. 2154) 2
i)
Mr =
1 2x
ii)
wr = wn
iii)
B.W. = wn
1 - x2
i.e.
= 2.377
1 - 2 x 2 = 30.6421 rad/sec. 1 - 2 x2 + 2 - 4 x2 4 x4
= 48.336
Example 12.15.17 Solution : Step 1 :
G(s) =
4 (1 + 0.5 s)
… Time constant form
æ s 2 ö÷ s (1 + 2s) ç 1 + 0.05 s + ç 64 ÷ ø è
Step 2 : Factors i) K = 4 ii)
i.e. 20 Log 4 = 12 dB
1 i.e. one pole at the origin. So magnitude plot is straight line of slope s – 20 dB/dec, passing through intersection point of w = 1 and 0 dB. TM
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Control System Engineering
iii)
12 - 5
1 i.e. simple pole i.e. T1 = 2 1 + 2s
Frequency Response Mehtods (Bode Plot Method)
i.e. wC1 =
1 = 0.5 T1
So straight line of 0 dB till w = 0.5 and then slope of – 20 dB/dec. i.e. T2 = 0.5
iv) 1 + 0.5 s i.e. simple zero
i.e. wC2 =
1 =2 T2
So straight line of 0 dB till w = 2 and then slope of + 20 dB/dec. v)
1 æ ö ç 1 + 0.05 s + ÷ ç 64 ÷ è ø s2
w2n
, quadratic pole, comparing with
1 2x s2 1+ s+ wn w2
n
= 64
i.e.
wC
3
= wn = 8 rad / s
Straight line of slope – 40 dB/dec for w ³ 8. The resultant slope table is, Range of w
0 < w< 0.5
0.5 £ w < 2
2 £w < 8
8 £w < ¥
Resultant slope in dB/dec
–20
– 20 –20 = – 40
–40 + 20 = – 20
–20 – 40 = – 60
Step 3 : Phase angle plot G (jw) =
4(1 + 0.5 jw) éæ ù w2 ö÷ + 0.05 jwú jw(1 + 2 jw)êç 1 – ç 64 ÷ êëè úû ø
w
1 jw
– tan –1 2 w
+ tan –1 0.5 w
é ù ê 0.05w ú – tan –1 ê ú w2 ú ê êë1 – 64 úû
fR
0.2
–90º
– 21.8º
+ 5.71º
– 0.573º
– 106.67 º
4
–90º
– 82.87º
+ 63.43º
– 14.93º
– 124.37 º
7
–90º
– 85.91º
+ 74.05º
– 56.19º
– 158.05 º
8
–90º
– 86.42º
+ 75.96º
– 90º
– 190.45 º
¥
–90º
– 90º
+ 90º
– 180º
– 270 º
Step 4 :
The Bode plot is shown in the Fig. 12.1.
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Fig. 12.1
–270º
–240º
–210º
–180º
–150º
0.1
2
4
1
2
wgc 1.3
– 40 dB/dec
P.M. +64º
5 6 7 891
– 20 dB/dec
3 3 4
wpc = 7.4
2
10
G.M. = +22 dB
–20 dB/dec
5 6 7 891 3
100
5 6 7 891
– 60 dB/dec
4 2
3 5 6 7 891 2
Log w
System stable
(1/s)
K=4
3 4
5 6 7 891
Semi-log paper (5 cycles ´ 1/10)
Scale On Y axis : 1 unit = 30º, 20 dB
4
12 - 6
–120º
–90º
fR in deg
–20
0 dB
– 40 dB/dec +20 – 20 dB/dec
+40
+60 – 60 dB/dec
MR in dB
1
Control System Engineering Frequency Response Mehtods (Bode Plot Method)
Control System Engineering
12 - 7
Frequency Response Mehtods (Bode Plot Method)
Example 12.15.18 Solution : Step 1 : G(s)H(s) in time constant form. Step 2 : In this example, K is unknown. So draw the magnitude plot without K. The effect of K is to shift the magnitude plot upwards or downwards by 20 Log K dB. Thus shift the magnitude plot so as to match the given specifications. Equating this shift required to 20 Log K, the required value of K can be determined. The various factors other than K are, 1 1) , pole at the origin, straight line of slope –20 dB/dec passing through intersection s point of w = 1 and 0 dB. 2)
1 1 , simple pole, T1 = 0.5, wC = =2 1 1 + 0.5 s T1
Straight line of slope –20 dB/dec for w ³ 2. 3)
1 1 , simple pole, T2 = 0.2, wC = =5 2 1 + 0.2 s T2
Straight line of slope –20 dB/dec for w ³ 5. Range of w Resultant slope in dB/dec
0
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