# 3320100XZ8L_Control System Engineering_Solution Manual

March 25, 2018 | Author: Santhosh Kumar H O | Category: Control System, Systems Engineering, Cybernetics, Psychology & Cognitive Science, Cognitive Science

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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)

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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

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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Mathematical Models of Control Systems

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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Block Diagram Models of Control Systems

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)

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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Block Diagram Models of Control Systems

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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Y

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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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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

C

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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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5 - 28

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)

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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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5 - 33

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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Signal Flow Graph Models of Control Systems

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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6-2

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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Signal Flow Graph Models of Control Systems

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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6-5

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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Y4

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Signal Flow Graph Models of Control Systems

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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6-8

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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6-9

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

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6 - 10

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

Control System Engineering

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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6 - 13

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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6 - 14

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

TM

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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

TM

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7

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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7-2

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

TM

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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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7-5

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

TM

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7-6

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)

Control System Engineering

\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

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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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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)

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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

Control System Engineering

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)

=

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.

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.

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

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

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.

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

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

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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10 - 5

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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10 - 8

\

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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10 - 9

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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10 - 10

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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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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Control System Engineering

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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Control System Engineering

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

... 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

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 =

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

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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TM

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