3320100XZ8L_Control System Engineering_Solution Manual

Solutions of Examples for Practice

Example 3.3.7

Solution : Taking Laplace transform of the network,

Divide the network into two parts, Part 1 : E_i(s) = I(s) 1 s I(s) 1 1 R C + = I(s) R 1 sC 1 1 + é ëê ù ûú ... (1) but, V₁(s) = 1 sC₁ I(s) I(s) = V₁(s) (sC₁) \ E_i(s) = V (s) (s C ) R 1 sC 1 1 1 1 + é ëê ù ûú \

( )

( )

V s E s 1 i =

(

R C s 1_{1 1}1 +

)

Part 2 : V (s) = æçR + 1 ö÷ I(s)

3

Transfer Function and Impulse Response

1 sC₁ sC1₂ R₁ R₂ E (t)_i E (s)_o Fig. 3.1 (a) 1 sC₁ R₁ E (s)_{i I(s) V (s)} 1 Fig. 3.1 (b) 1 sC₂ R₂ E (s)_o I(s) V (s)₂ Port 2

E (s)_o = I(s) 1 sC₂ E (s) V (s) o 2 = 1 sC R 1 sC 2 2 2 + æ è ç ö ø ÷ = 1 sC R 1 sC 1 R C s 1 2 2 2 2 2 + æ è ç ö ø ÷ = ₊

Amplifier gain = 10 hence V₂(s) = 10 V₁(s)

E_o(s) (1+R₂C₂s) = 10E (s) 1 R C s i 1 1 + i.e. E (s) E (s) o i =

(

)(

)

10 1 sR C+ _{1 1} 1 sR C+ _{2 2} 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 - + = sCI(s) I(s) R V (s)i 0 \ V (s)_i = I(s) 1 sC+ R é ëê ùûú \ V (s) I (s) i ₌ 1 sC+ R é ëê ùûú = 1 s s + CR C

And, V (s)_o = I(s) R i.e. V (s)

I(s) R

o ₌

Example 3.3.9

Solution : Applying we get the equations as, E_i = iR + Ldi dt+ 1 C

ò

idt ... (1) Input = E_i; Output = E_o Laplace transform of

ò

f(t) dt = F(s)

s , … Neglecting initial conditions

and Laplace transform of df(t)

dt =sF(s) … Neglecting initial conditions

Take Laplace transform,

\ E_i(s) = I(s) R + sL + 1 sC é ëê ù ûú i.e. I(s) E s)_i( = 1 R + sL + 1 sC é ëê ù ûú ... (2)

Control System Engineering 3 - 2 Transfer Function and Impulse Response

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R V (s)_o V (s)_i I(s) 1 sC + – + – + – Fig. 3.2 (a)

Now E_o = 1

C

ò

idt i.e. E (s)o = 1

sCI(s) i.e. I(s) = sCE (s)o ... (3) Substituting value of I(s) in equation (2),

\ sCE (s) E (s) o i = 1 R + sL + 1 sC é ëê ùûú \ E (s) E (s) o i = 1 sC R + sL + 1 sC = 1 RsC + s LC + 12 é ëê ù ûú

So we can represent the system as in the Fig. 3.3 (a).

Example 3.3.10

Solution : The s-domian network is shown in the Fig. 3.4 (a).

\ T.F. = E (s)_{E (s) Z}Z_Z R 1 s R C R 1 sR C R o 1 2 1 2 2 2 2 1 1 1 = ₊ = + + +1 sR C+ 22 2 \ T.F. = R (1 s R C ) (R R ) + sR R (C C ) 2 1 1 1 2 1 2 1 2 + + +

Control System Engineering 3 - 3 Transfer Function and Impulse Response

E (s)_i R1 R₂ 1 ––– sC₁ 1 ––– sC2 E (s)o R || 11 –––_sC 1 =–––––––R 1+sR C 1 1 1 R || 12 –––_sC 2 =–––––––R 1+sR C2_{2 2} E (s)_i Z (s)₁ Z (s)₂ E (s)_o (a) (b) Fig. 3.4 E (s)_i 1 E (s)_o s LC + sRC + 12

Example 3.4.2 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.

\ T(s) = K(s + 3) (s + 1) (s + 2 + j) (s + 2 j)- = K(s + 3) (s + 1) (s + 2)[ 2 -(j)2] = K(s + 3) (s + 1) s + 4s + 5[ 2 ] Now d.c. gain is value of T(s) at s = 0 which is given as 10.

\ d.c. gain = T(s)|_{at s= 0} i.e. 10 = K 3_´´ 1 5 i.e. K = 50 3 = 16.667 \ T(s) = _(s+16.667 (s + 3)_{1) (s}2 _+4s+5) ... Transfer function 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

Control System Engineering 3 - 4 Transfer Function and Impulse Response

Solutions of Examples for Practice

Example 4.6.1

Solution : As discussed earlier work done by the gears remains same.

\ T T 1 2 = N N = 1 2 2 1 q q Now T₂ is supplying torque to load.

\ T (t)₂ = J d dt + B d dt 2 2 2 2 2 2 q q ... (1) \ As T₂ = N N T 2 1 1 , Substituting in equation (1) \ T₁ = N N J d dt + N N B d dt 1 2 2 2 2 2 1 2 2 2 q q ... (2) Now q₂ = N N 1 2 1 q , Substituting in equation (2) T₁ = N N J N N d dt + N N B N N d dt 1 2 2 1 2 2 1 2 1 2 2 1 2 1 æ è ç ö ø ÷ æ è ç ö ø ÷ q q ... (3) Hence and J N N 2 1 2 2 é ëê ù

ûú = Equivalent M.I. referred to primary = Jeq

B N N 2 1 2 2 é ëê ù

ûú = Equivalent friction referred to primary = Beq \ Taking Laplace of equation (3) and assuming initial conditions zero.

T (s)₁ = J_eq s2 q₁(s) + B_eq sq₁(s)= q₁(s) s J[ 2 _eq + s B_eq] \ q1 1 (s) T (s) = 1 s s J[ _eq + B_eq]

Example 4.9.10

Solution : Two displacements : No element under x (t)₁ 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 x₂ and as B and K_{2 2} are connected to fixed supports both are under x₂(t) only as shown in the equivalent system.

At node 1, f(t) = K (x_{1 1} -x )₂ ... (1) At node 2, 0 = K (x x ) + K x M d x dt B dx dt 1 2 1 2 2 2 2 2 2 2 2 - + + ... (2) i) F-V analogy : M ® L, B ® R, K ® 1/C, x ® i dt

ò

, dx dt ® i, d x dt di dt 2 2 ® v(t) = 1 C₁

ò

(i1-i )2 dt ... (3) Loop (1) 0 = 1 C (i i ) dt + L di dt + 1 C i dt + R i 1 2 1 2 2 2 2 2 2

-ò

ò

... (4) Loop (2)

The F-V analogous network is shown in the Fig. 4.1.

ii) F - I analogy : M® C, B ® 1/R, K ® 1/L, x ® v dt

ò

, dx dt ® v, d x dt dv dt 2 2 ® \ i(t) = 1 L₁

ò

(v1-v )2 dt ... (5) Node (1) 0 = 1 L (v v ) dt + C dv dt + v R + 1 L v dt 1 2 1 2 2 2 2 2 2

-ò

ò

... (6) Node (2)

Control System Engineering 4 - 2 Mathematical Models of Control Systems

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM F x₁ x₂ B₂ K₁ K₂ M₂ Equivalent system V(s) L₂ R₂ C₁ C₂ I₁ I₂ (1/K )₁ (1/K )₂

Same displacement same current.

The F-I analogous system is shown in the Fig. 4.1 (a).

Example 4.9.11

Solution : There are three displacements x , x_{i o} and x_y. The input is x_i and output is x_o. So transfer function of the mechanical system is X (s)_o /X (s)_i . The equilibrium equations are, B d(x x ) dt + K (x x ) + B d(x x ) dt 1 o i 1 o i 2 o y - _- -= 0. ... (1) B d(x x ) dt + K x 2 y o 2 y -= 0 ... (2)

Taking Laplace transform of both the equations, neglecting initial conditions, B sX (s) B sX (s) + K X (s) K X (s) + B sX (s) B sX (s_{1 o} - _{1 i 1 o} - _{1 i 2 o} - _{2 y} ) = 0 \ X (s)[sB + K + sB ] X (s)[sB + K ] B sX (s)o 1 1 2 - i 1 1 - 2 y = 0 ... (3) B sX (s) B sX (s) + K X (s)_{2 y} - _{2 o 2 y} = 0 i.e. X (s)_y = B s sB + K X (s) 2 2 2 o ... (4) Substituting in equation (3), X (s)[sB + K + sB ] X (s)[sB + K ] sB sB X (s) sB + K o 1 1 2 i 1 1 2 2 o 2 2 - - = 0 Solving, X (s) X (s) o i = (sB + K ) (sB + K ) s B B + sB K + sK B + K K + s B + sK B 1 1 2 2 2 1 2 1 2 1 2 1 2 2 22 2 2-s B2 22 \ X (s)_{X (s)}o i = 1 sB K 1 s B K 1 sB K 1 s B K 1 1 2 2 1 1 2 2 + æ è ç ö ø ÷æ + è ç ö ø ÷ + æ è ç ö ø ÷æ + è ç ö ø ÷ + sB_K2 1

... Required transfer function

Control System Engineering 4 - 3 Mathematical Models of Control Systems

I(s) v₁ v₂ C₂ L (1/K )_{1 1} (1/K )₂ (1/B )₂ R₂ L₂ 2 1

Same displacement same voltage.

The s domain network for the given electric network is, Now I(s) = E (s) Z(s) R 1 sC i 1 1 + + and E (s)o = I(s) R1 _sC1 1 + é ëê ù ûú with Z(s) = R || 1 sC 2 2 \ E (s)_o = E (s) R 1 sC Z(s) R 1 sC i 1 1 1 1 + é ëê ù ûú + + = E (s)(1+ sR C ) sC Z(s) sC R 1 i 1 1 1 + 1 1+ Substituting Z(s) = R 1 sC R 1 sC 2 2 2 2 ´ + = R 1 sR C 2 2 2 + \ E (s) E (s) o i = (1 sR C ) sC R (1 sR C ) (1 sR C ) 1 1 1 2 2 2 1 1 + + + + = (1 sR C ) (1 sR C ) (1 sR C ) (1 sR C ) sC R 1 1 2 2 1 1 2 2 1 2 + + + + + As K 1 C

® , B®R the two transfer functions are identical hence the two systems are analogous in nature.

Example 4.9.12

Solution : The displacements areq₁ and q as shown in the Fig. 4.3 (a). The J₁ is underq₁. The displacement changes toq due to K. While J₂ is underq.

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,

T = J dt K 1 ₂ d2 ₁ 1 q q q + ( - ) …(1)

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0 = K J d dt 2 2 2 (q q- ₁)+ q …(2)

Taking Laplace transform of equation (1) and equation (2), \ T(s) = J s₁ 2q₁(s) K+ q₁( )s -K sq( ) …(3) \ 0 = Kq(s) K- q₁(s) + J s₂ 2q(s) …(4) From equation (4), q₁(s) = K+ J s2 (s) 2 K é ë ê ê ù û ú úq …(5)

Using equation (5) in equation (1),

T(s) = q₁(s) [ J s + K] K (s)₁ 2 - q = K+ J s2 ( J s + K) (s) K (s) 2 1 2 K é ë ê ê ù û ú ú q - q \ T(s) = (K J s ) (J s K) K K (s) 2 2 1 2 + + -ì í ï îï ü ý ï þï q \ q(s) T(s) = K (K s J ) (K s J ) K+ 2 ₂ + 2 ₁ - 2 = K s J J4 _{1 2} +Ks J2 ₂ +Ks J2 ₁ \ q(s) T(s) = K s [s J J2 2 _{1 2} +K( J₁+J )]₂ Example 4.9.13

Solution : The mass M₁, K₁ and

friction f_v1 are under the

displacement x₁(t) while mass M₂, K₃ and friction f_v2 are under the displacement x₂(t). The spring K₂ and the friction f_v3 are between x₁(t) and x₂(t). The equivalent mechanical

Control System Engineering 4 - 5 Mathematical Models of Control Systems

J2 J1 T q (Same) (Same) K q q1 q1 K1 M1 K3 K₂ x (t)1 f(t) M2 f_v1 f_v2 fv3 (a) (b) Fig. 4.3 T _{J1 J2} q₁ K q Fig. 4.3 (c) M₁ K₁ K₃ K₂ x₁(t) f(t) M₂ f_v1 f_v2 f_v3

f(t) = M₁ d x dt 2 1 2 + K1 x1 + fv1 dx dt 1 _{+ K} 2(x1 – x2) + fv3 d (x x ) dt 1- 2 _…(1) 0 = K₂(x₂ – x₁) + f_v3 d (x x ) dt 2 - 1 _{+ M} 2 d x dt 2 2 2 + fv2 dx dt 2 _{+ K} 3 x2 …(2)

Taking Laplace transform of both the equations,

F(s) = X₁(s) [M₁s2 + s(f_v1 + f_v3) + (K₁ + K₂)] – X₂(s) [sf_v3+ K₂] …(3) 0 = – X₁(s) [K₂ + sf_v3] + X₂(s) [M₂s2 + s (f_v2+ f_v3) + (K₂+ K₃)] …(4) From equation (4), X₁(s) = M s s(f f ) (K K ) K sf 2 2 v2 v3 2 3 2 v3 + + + + + é ë ê ê ù û ú úX2(s) Using in equation (3), F(s) = X (s)₂ [M s1 s(f f ) (K K )][M s s(f f ) (K K )] 2 v1 v3 1 2 2 2 v2 v3 2 3 + + + + + + + + (sf_v3+ -(sfv3+ ì í î ü ý þ K₂₎ K2) \ X (s) F(s) 2 ₌ (sf K ) [M s s(f f ) (K K )] [M s s(f f ) v3 2 1 2 v1 v3 1 2 2 2 v2 v3 + + + + + + + +(K₂+K )] [sf₃ - _v3+K ]₂ 2 Example 4.9.14

Solution : Write the equilibrium equation

At 1, F(t) = K(x₁-x₂) ... (1) At 2, 0 = K(x x ) + Md x dt + B d x dt 2 1 2 2 2 2 - ... (2)

Taking Laplace of both, neglecting initial conditions

F(s) = KX (s)₁ -KX (s)₂ ... (3) 0 = KX (s)₂ -KX (s) + Ms X (s) + BsX (s)₁ 2 _{2 2} ... (4) from equation (3) X (s)₁ = 1 K[F(s) + KX (s)]2 Substituting in (4) 0 = KX (s) K 1 K[F(s) + KX (s)] Ms X (s) + Bs X (s) 2 - ìí_î 2 üý_þ 2 2 2 \ 0 = KX (s)₂ -F(s) KX (s) + Ms X (s) + Bs X (s)₂ 2 _{2 2} \ F(s) = sX (s) [Ms + B]₂ \ X (s) F(s) 2 ₌ 1 s(Ms B)+

Control System Engineering 4 - 6 Mathematical Models of Control Systems

Example 4.9.15

Solution : The equilibrium equation at node X_o is, B d (x x ) dt K (x x ) B dx dt K x 1 o i 1 o i 2 o 2 o -+ - + + = 0

Taking Laplace transform and neglecting initial conditions, B s X (s) B s X (s) K X (s) K X (s) B sX (s) K_{1 o} - _{1 i} + _{1 o} - _{1 i} + _{2 o} + ₂ X (s)_o = 0 \ X (s) [s B_{o 1}+K₁+s B₂+K ]₂ = [K₁+s B ] X (s)_{1 i} \ X (s) X (s) o i = s B K s (B B ) (K K ) 1 1 1 2 1 2 + + + +

For force-current analogy, B® 1 R , K ® 1 L , x® f, x · = V, x =

ò

V dt \ 1 R [V V ] 1 L (V V ) dt 1 o- i + 1

ò

o- i + 1 +

_ò

R V 1 L V dt 2 o 2 o = 0

The force-current equivalent analogous electrical network is as shown in the 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 = K_T i(t) e_b = K d (t) dt b q

Now torque produced has to overcome spring torque and M. I. J of system.

\ T = K (t) + Jd (t) dt s 2 2 q q

Now analysing coil circuit,

e(t) = i(t) R + e (t)_b i.e. e(t) = i(t) R + K d (t) dt b

q Taking Laplace of all the equations,

T(s) = K_T I(s) ... (1)

Control System Engineering 4 - 7 Mathematical Models of Control Systems

V_i V_o R (_{1 B}1) 1 L (_{1 K}1) 1 R (1/B )2₂ L (1/K )_{2 2} V_i V_o Fig. 4.5 (a)

\ Equating (1) and (2), K_T I(s) = Ks q(s) + s J (s)2 q \ I(s) = q é ë ê ê ù û ú ú (s) K + s J K s 2 T \ Substituting in (3), e(s) = q(s) R ( K + s J ) q K + K s (s) s 2 T b = q é ë ê ê ù û ú ú (s) R (K + s J) + K s K K s 2 b T T \ q (s) e(s) = K J R s + K K s + K R T 2 b T s

is the required transfer function Example 4.9.17

Solution : Due to force applied, mass M₁ will be displaced by x (t)₁ . Now K₃ will consume same energy hence one end of B₁ will get displaced by displacement x (t)₂ which is less than x (t)₃ . Now due to B₁ and K₂, the force on M₂ will be different and M₂ will be displaced by x (t)₁ which is other than x (t)₁ and x (t)₂ . So mass M₁ and K₁ are under influence of x (t)₁ .

Spring K₃ under influence of x₁-x₂ Spring K₂ under influence of x₁-x₃ Friction B₁ under influence of x₂ -x₃.

Hence equivalent mechanical system will be as shown in above figure. The equations of equilibrium are,

At x₁, F(t) = M d x dt K x K (x x ) K (x x ) 1 2 1 2 + 1 1+ 3 1- 2 + 2 1- 3 ... (1) At x₂, 0 = K (x x ) B d(x x ) dt 3 2- 1 + 1 2 3 -... (2) At x₃, 0 = B d(x x ) dt K (x x ) M d x dt 1 3 2 2 3 1 2 2 3 2 -+ - + ... (3)

Use F-V analogy and the replacements,

M® ,L B® ,R K 1 C ® , x® ,q dx dt ® ,i d x dt di dt 2 2 ® and x® ®q

ò

i dt

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B₁ F(t) x₁ x₂ x₃ K₁ K₂ K₃ M₁ M₂

V(t) = L di dt 1 C i dt 1 C (i i )dt 1 C (i i )dt 1 1 1 1 3 1 2 2 1 3 +

_ò

+

_ò

- +

_ò

- ... (1) 0 = 1 C₃

ò

(i2-i )dt R (i1 + 1 2 -i )3 ... (2) 0 = R (i i ) 1 C (i i )dt L di dt 1 3 2 2 3 1 2 3 - +

_ò

- + ... (3)

Simulating using loop basis,

Currents through various elements,

L , G1 1 i1alone

C₃ i₁-i₂

R₁ i₂-i₃

C₂ i₁-i₃

L_{2 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 M K₁, ₁ in parallel ® L , C_{1 1} in series K , B_{3 1} in series ® C , R_{3 1} in parallel

(K , B_{3 1} series) parallel with (K₂) ® (C , R_{3 1} in parallel) series with (C₂)

Control System Engineering 4 - 9 Mathematical Models of Control Systems

V C (1/K )_{1 1} R₁ L₂ L₁ C (1/K )_{2 2} C (1/K )_{3 3} i₁ i₂ i₃ +_

Example 4.9.18

Solution : Due to applied force, M₁ will be displaced by x₁. Spring K₁ will be under same displacement.

Due to friction B₁, K₃ will get displaced by x₂ as friction causes change in displacement. Due to spring K₂, friction B₂ will be under influence of x₃ which is different than x₂. This is due to storing of energy by K₂.

Due to B₂ and K₃, mass M₂ will again under different displacement x₄.

Due to friction B₃, again there will be change in force applied to K₄ hence K₄ will be under the influence of displacement x₅.

So mass M₁andK₁ are under x₁. B₁ between x₁ and x .₂ K₂ between x₁and x .₃ K₃ between x₂ and x .₄ B₂ between x₃ and x .₄ B₃ between x₄ and x .₅ K₄ under x₅ alone.

Hence equivalent system is as shown,

The equations of equilibrium are, At x₁, F(t) = M d x dt K x B d(x x ) dt K (x x ) 1 2 1 2 1 1 1 1 2 2 1 3 + + - + - ... (1) At x₂, 0 = B d(x x ) dt K (x x ) 1 2 1 3 2 4 -+ - ... (2) At x₃, 0 = K (x x ) B d(x x ) dt 2 3- 1 + 2 3 4 -... (3)

Control System Engineering 4 - 10 Mathematical Models of Control Systems

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM B₁ B₂ B₃ F(t) x₁ x₂ x₃ x₄ x₅ K₁ K₂ K₃ K₄ M₁ M₂

At x₄, 0 = K (x x ) B d(x x ) dt B d(x x ) dt M d x dt 3 4 2 2 4 3 3 4 5 2 2 4 2 - + - + - + ... (4) At x₅, 0 = K x B d(x x ) dt 4 5+ 3 5 4 -... (5) Using F-V analogy, the replacements are,

M® L, B® R K® 1 C, x ® q, dx dt ® ,i d x dt di dt 2 2 ® and x® ®q

ò

i dt as i dq dt = V(t) = L di dt 1 C i dt R (i i ) 1 C (i i )dt 1 1 1 1 1 1 2 2 1 3 +

_ò

+ - +

_ò

- ... (6) 0 = R (i i ) 1 C (i i )dt 1 2 1 3 2 4 - +

_ò

- ... (7) 0 = 1 C₂

ò

(i3-i )dt R (i1 + 2 3-i )4 ... (8) 0 = 1 C (i i )dt R (i i ) R (i i ) L di dt 3 1 2 2 4 3 3 4 5 2 4 - + - + - +

ò

... (9) 0 = 1 C₄

ò

i dt R (i5 + 3 5 -i )4 ... (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

B , K_{1 3}series R , C_{1 3} parallel K , B_{2 2} series C , R_{2 2} parallel B , K_{1 3}series in parallel with K , B2 2 R C_{1 3} parallel in series with C ,R2 2parallel

Thus the F-V analogous network based on loop basis is,

Check the elements and currents as per the table made above. Example 4.9.19

Solution : For the mass M₁, the displacement is x₁ due to force f(t), against the friction force, \ f (t) = M d x dt 1 2 1 2 + B dx dt 1 1 K (1)

Let force transmitted to mass M₂ be f (t)₂ which will cause the displacement of x₂ of mass M₂ against springK₂ and friction B₂.

\ f (t)₂ = M d x dt 2 2 2 2 +K x1 2 +B dx dt 2 2 K (2)

Now taking moments about fulcrum, f (t)´ L₁ = f₂ (t)´ L₂ \ f₂ (t) = _LL1

2 ´ f (t) ... (3)

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The equivalent mechanical system is,

For F-V analogy, M ® L , B ® R, K ® 1/C, x ® q, dx/dt ® I Hence equations (1) and (2) get modified as,

V(t) = L₁ dI_dt1 +I R_{1 1} K (4) V (t)₂ = L dI dt 2 2₂ + I R2 2 + _C1 I dt 2

ò

2 K (5) and V (t)₂ = _LL1 2. V(t) K (6)

The equation (6) indicates that it acts as a transformer with turns ratio L

L1₂. Hence F-V analogous network is,

For obtaining the transfer function X (s) /X (s)_{2 1} , take laplace transform of the equations (1) and (2),

F (s) = M s₁ 2´X (s) B s X (s)₁ + _{1 1} K (7)

F (s)₂ = M s X (s) B s X (s) + K X (s)₂ 2 ₂ + _{2 2 2 2} ... (8) and F (s)₂ = _LL1

2 F (s) ... (9)

Control System Engineering 4 - 13 Mathematical Models of Control Systems

f(t) _{f (t)2} M₁ M₂ K₂ x₁ x2 B₁ L / L_{1 2} B₂ v(t) L₁ R₁ L₂ R₂ C₂(1/K )₂ I₁ I₂ (L₁/ L₂) V₂

Substituting (7) and (8) in (9) we get, M s X (s) + B s X (s) + K X₂ 2 _{2 2 2 2 2}(s) = _LL1

[

M s X (s) B s X (s)

]

2 1 2 1 + 1 1 \ X (s) M s₂

[

₂ 2+B s + K_{2 2}

]

= _{L X}L1

[

M s B s

]

2 1 1 2 1 ( )s + X s X s 2 1 ( ) ( ) =

(

)

(

L s M s B

)

L M s B s K 1 1 1 2 2 2 2 2 + + +

This is the required transfer function.

qqq

Control System Engineering 4 - 14 Mathematical Models of Control Systems

Solutions for Examples for Practice

Example 5.3.11 Solution :

The blocks G₁ and G₂ are in series.

5

Models of Control Systems

Block Diagram

G₆ R(s) ₊ – C(s) G G 1 + G G H1 2_{1 2 1} G + G + G3 4 5 Series H₂ G₆ R(s) ₊ – C(s) G G (G + G + G ) 1 + G G H 1 2 3 4 5 1 2 1 Minor loop G G_{1 2} H₁ G5 G₄ G₆ G₃ R(s) ₊ – H₂ C(s) Minor loop 3 blocks in parallel + + +

Example 5.3.12

Solution : Shifting the take off point to the right as shown we get,

The dotted portion shown has a parallel combination of block of 1 and G / G_{3 2}, so they add up. Then shift summing point to the left of G₁.

Control System Engineering 5 - 2 Block Diagram Models of Control Systems

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM G₆ R(s) C(s) G G (G + G + G ) 1 + G G H 1 2 3 4 5 1 2 1 1 + G G (G + G + G )H 1 + G G H 1 2 3 4 5 2 1 2 1 R(s) G G G (G + G + G ) C(s) 1 + G G H + G G H G + G G H G + G G G H 1 2 6 3 4 5 1 2 1 1 2 2 3 1 2 2 4 1 2 5 2 Fig. 5.1 – R(s) G1 G2 1 H1 – C(s) G G32 + + Parallel – R(s) _G 1 H₁ G₂ – + C(s) G G3₂ 1 + – R(s) G₁ G₂ H₁ – C(s) G +G G 2 3 2 1 G₁

Interchanging the positions of summing points using Associative law, ... Rule 1

Combining two blocks in forward path we get canonical form from which we can write, C(s) R(s) = G G (G G ) G (1 G G H ) 1 G G (G G ) G (1 G G H ) 1 1 2 2 3 2 1 2 1 1 2 2 3 2 1 2 1 + + + ₊ + × G₁ = G (G G ) 1 G G H G G 1 2 3 1 2 1 2 3 + + + + ... Ans Example 5.3.13

Solution : Combine the parallel blocks G₂ and G₃ to give G₂+G₃ and minor loop of G and H_{1 1}.

Reduce the minor feedback loop of G₁ and H₁ to give G 1+ G H

1 1 1

.

Control System Engineering 5 - 3 Block Diagram Models of Control Systems

– – C(s) R(s) _G 1 G2 G + G––––––2_G 3 2 G G –––––––––_{1+G G H}1 2 1 2 1 H₁ 1 –– G₁

Minor feedback loop

G + G_{2 3} G –––––––––– 1 + G H 1 1 1 H₂ G₄ + – C(s) R(s) Minor feedback loop G₄ + _C(s) R(s) G (G + G )_{1 2 3}

Reducing inner minor loop we get,

The two blocks are in parallel, so adding them we get the single equivalent block as

Example 5.3.14

Solution : The blocks G₂ and G₃ in parallel, use rule 3.

Control System Engineering 5 - 4 Block Diagram Models of Control Systems

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R(s) G G + G G1 2 1 3 C(s) 1 + G H + G G H + G G H_{1 1 1 2 2 1 3 2} G +₄ G + G G H + G G G H +G G G H + G G + G G_{4 4 1 1 4 1 2 2 4 1 3 2 1 2 1 3} 1 + G H + G G H + G G H_{1 1 1 2 2 1 3 2} … Ans. C(s) R(s) = G +₂ G₃ G₅ G₆ R(s) C(s) G₄ G₄ G₇ G₁ – – + + – + Minor loop of 1 and G₆ G +₂ G₃ G₅ R(s) + _{1 + G}G6 C(s) 6 G +₂ G₃ G₁ G₄ G₇ + + – –

Minor loop with G = 1, H = G (G + G )_{1 2 3}

\ \ G_eq = G G G G G G G G (1 G G G G ) (1 G ) 2 5 6 3 5 6 4 6 1 2 1 3 6 + + + + + H_eq = G₇ Example 5.3.15

Solution : Rearranging the block diagram we get,

Control System Engineering 5 - 5 Block Diagram Models of Control Systems

R(s) + _{1 + G}G6 C(s) 6 1 1 + G (G + G )_{1 2 3} G (G + G )5 2 3 G₄ G₇ G + G (G + G )_{4 5 2 3} Parallel blocks, Rule 3 – + + R(s) + C(s) G₇ – 1 1 + G (G + G )_{1 2 3} G + G (G + G )4 5 2 3 _{1 + G}G6 6 G 1 + G H eq eq eq R(s) C(s) – + R _G 3 G8 G₆ G₅ G₂ G₁ + – + + G₇ G₄ C Separate paths – + R _G 3 G8 G₆ G₅ G₂ G₁ + – + + G₇ G₄ C

Separate the feedback paths through G₄ and G₇, combined at summing point before G₆. So G₆ will appear separately in each path of G₄ and G₇, as shown below.

Reducing inner minor feedback loop,

Hint : Combine inner blocks, reduce minor feedback loop. Again combine blocks in forward path to get simple form.

Using G 1 GH+ we get, \ C(s)_R(s) = G G G G G G G G G G G G G G G G G G 1 2 3 5 8 6 7 8 2 4 5 6 8 1 2 3 5 8 1+ + + ... Ans.

Control System Engineering 5 - 6 Block Diagram Models of Control Systems

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM – R _G 3 G8 G₆ G₆ G₄ G₇ G G_{2 5} G₁ – – C Minor loop – R _G 3 G G2 5 G G_{4 6} G₁ – C G 1 + G G G_{6 7 8}8

Example 5.3.16

Solution : Shifting take off point before G₂. ... Rule 6

Shifting take off point before summing point using critical rule No. 10,

Separating the paths in the feedback path shown dotted,

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 H₂, carry the negative sign and then H G_{1 2} to get – H H G_{1 2 2}.

Control System Engineering 5 - 7 Block Diagram Models of Control Systems

– + + – C(s) R(s) G₃ G₁ G₂ H₂ H₁ G₂ Critical rule – + + – C(s) R(s) G₃ G₁ G₂ H₂ H G_{1 2}

For this negative sign, include block of '–1' while separating the path.

– – + – C(s) R(s) G₃ G₁ G₂ H₂ H G_{1 2} H G_{1 2} –1 H₂

Shifting summing point as shown and then interchanging the two summing points using

associative Law we get, ... Rule 4 and 1

Rule 8 gives, C(s) R(s) = G 1 G H eq eq eq + = G (G + G ) 1 + G H + H G G G G G H H 1 2 3 2 2 1 1 2 - 1 2 3 1 2

Control System Engineering 5 - 8 Block Diagram Models of Control Systems

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM – + – C(s) R(s) G₃ G₂ G₁ 1 + H G G_{1 1 2} G2 H₂ – H₁H G_{2 2} 1 Parallel Minor loop – C(s) R(s) G₃ G₂ G₁ 1 + H G G_{1 1 2} G₂ 1 + G H_{2 2} – H₁H G_{2 2} 1 + Rule 3 Rule 8 – C(s) R(s) G₁(G + G ) (1 + G H ) (1 + H G G ) 2 3 2 2 1 1 2 – H₁H G_{2 2} G_eq H_eq Minor loop

Example 5.3.17

Solution : Shift summing point before G₁ and take off point of H₂ after the block G₄.

Control System Engineering 5 - 9 Block Diagram Models of Control Systems

R Y – G₁ G₂ G₃ G₄ 1/G₁ H₂ 1/G₄ H₃ H₁ – – Minor loop G G ––––––––_{1+G G H}3 4 3 4 3 Minor loop = G G G G –––––––––– 1+G G H ––––––––––––––––– 1+––––––––––G G G G 1+G G H H –––– G G 1 2 3 4 3 4 3 1 2 3 4 3 4 3 2 1 4 ´ R Y H –––– G G_{1 4}2 G G_{1 2} ––––––––G G 1+G G H3 4_{3 4 3} H₁ – – H₁ Y G G G G –––––––––––––––– 1+G G H +G G H_{3 4 3}1 2 3 4_{2 3 2} R R – Y G G G G –––––––––––––––– 1+G G H +G G H –––––––––––––––––– = 1 2 3 4 3 4 3 2 3 2 G G G G 1+ –––––––––––––––––– H 1+G G H +G G H 1 2 3 4 1 3 4 3 2 3 2 ´ G G G G ––––––––––––––––––––––––––––– 1+G G H +G G H +G G G G H 1 2 3 4 3 4 3 2 3 2 1 2 3 4 1

Example 5.3.18

Solution : Shift the take off point of V₃(s) after G₂.

Control System Engineering 5 - 10 Block Diagram Models of Control Systems

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G₁ G₂ H₂ 1/G₂ G₃ H₃ H₁ + – + + – – R(s) C(s) Minor loop Parallel G ––––––_{1+G H}2 2 2 1+––1 G = 1+G –––– G 2 2 2 G –––––– 1+G H3_{3 3} Minor loop H₁ G₁ ––––––_{1+G H}G2 2 2 G –––––– 1+G H3_{3 3} 1+G –––––_G2 2 R(s) – C(s) Minor loop –––––––G G 1+G H –––––––––– 1+–––––––G G H 1+G H =––––––––––––––G G 1+G H +G G H 1 2 2 2 1 2 1 2 2 1 2 2 2 1 2 1 1+G –––––_G2 2 G –––––– 1+G H3_{3 3} G G –––––––––––––– 1+G H +G G H 1 2 2 2 1 2 1 R(s) C(s) Series R(s) G G (1+G ) C(s) –––––––––––––––––––––––––––––––––––––––– 1+G H +G G H +G H +G G H H +G G G H H 1 3 2 2 2 1 2 1 3 3 2 3 2 3 1 2 3 1 3 Fig. 5.2

Example 5.3.19

Solution : Shifting take off point before G₂.

Shifting take off point before summing point using critical rule No. 10,

Separating the paths in the feedback path shown dotted,

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 H₂, carry the negative sign and then H G_{1 2} to get – H H G_{1 2 2}.

Shifting summing point as shown and then interchanging the two summing points using Associative Law we get,

Control System Engineering 5 - 11 Block Diagram Models of Control Systems

– + + – C(s) R(s) G₃ G₁ G₂ H₂ H₁ G₂ – + + – C(s) R(s) G₃ G₁ G₂ H₂ H G_{1 2} – – – + – C(s) R(s) G₃ G₁ G₂ H₂ H G_{1 2} – H H G_{1 2 2} Minor feedback loop

C(s) R(s) = G (G + G ) (1 + G H ) (1 + H G G ) G (G + G ) ( H H G ) (1 + 1 2 3 2 2 1 1 2 1 2 3 1 2 2 1+ -G H ) (1 + H -G -G )_{2 2 1 1 2} \ C(s) R(s) = G (G + G ) 1 + G H + H G G G G G H H 1 2 3 2 2 1 1 2 - 1 2 3 1 2 Example 5.3.20

Solution : Separating the feedback paths,

Control System Engineering 5 - 12 Block Diagram Models of Control Systems

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM – + – C(s) R(s) G G 3 2 G 1 + H G G 1 1 1 2 G2 H₂ – H H G_{1 2 2} Blocks in parallel

Minor feedback loop

– C(s) R(s) G G 3 2 G 1 + H G G 1 1 1 2 G 1 + G H 2 2 2 – H H G_{1 2 2} 1 + G + G G 2 3 2 – C(s) R(s) G (G + G ) (1 + G H ) (1 + H G G ) 1 2 3 2 2 1 1 2 – H H G_{1 2 2}

Separating the dotted paths from the summing points 'S'.

Separating the paths from T to S_{1 1}, T to S_{1 2}, T to S_{2 1} and T to S_{2 2}.

Control System Engineering 5 - 13 Block Diagram Models of Control Systems

G₁ G₂ G₃ H₁ H₂ S R(s) – + – C(s) – – G₁ G₂ G₃ R(s) T₁ S1 S2 T₂ H₂ H₁ C(s) 2 + – 1 4 – – – H₁ 3 G₁ H₁ H₁ G₂ G₃ H₁ – H₂ H₁ –H₂ 1 C(s) R(s) – – – – Minor feedback loop Parallel – 2 3 4 G₃ 1 + G (–H H )_{3 1 2} Shift the summing point C(s) H₁ G – H_{1 1} G₂ – H H_{1 2} R(s) – – _–

\ C(s) R(s) =

[

G G H (1 G ) G

]

1 G H H G G H H 1 G G 1 2 1 2 3 3 1 2 2 3 1 2 1 - + - -+

[

2 1 2

]

3 3 1 2 2 3 1 2 H (1 G ) G 1 G H H G G H H - + -

-Control System Engineering 5 - 14 Block Diagram Models of Control Systems

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G –H_{1 1} G₂ G3 1 – G H H_{3 1 2} C(s) Interchange the summing points H₁ Interchange 1 G₂ – –H H_{1 2} – R(s) – G –H_{1 1} G G2 3 1 – G H H_{3 1 2} –H H_{1 2} C(s) – R(s) + H₁ G₂

Parallel Minor loop

– – G₁– H₁– H_G1 2 G G_{2 3} 1 – G H H_{3 1 2} 1 + G G2 3 1 – G H H_{3 1 2}× –H H1 2 C(s) G G_{2 3} 1 – G H H –G G H H_{3 1 2 2 3 1 2} G G –H G –H_{1 2 1 2 1} G₂ R(s) – C(s) G G – H_{1 2 1}(1 + G )₂ G₂ R(s) 1 – G₃H H_{1 2}–G G H H_{2 3 1 2} G₂G₃ × –

\ C(s) R(s) = G G G H G H G G 1 G H H G G H H G G 1 2 3 1 3 1 2 3 3 1 2 2 3 1 2 1 2 - -- - + G₃-H G_{1 3}-H G G_{1 2 3} Example 5.3.21

Solution : Separating two feedback from second takeoff point which is after block having transfer function G₂ as shown, we get,

Shifting summing point behind the block having transfer function ‘G₁’ as shown we get,

Control System Engineering 5 - 15 Block Diagram Models of Control Systems

+ + G₂ G₃ C(s) R(s) H₂ G₁ Minor feedback loop H₁ H₁ G₄ – – + + G 1 + G H 2 2 1 G3 C(s) R(s) H₂ G₁ H₁ G₄ – + + G 1 + G H 2 2 1 G3 G₁ C(s) R(s) H₂ 1 / G₁ H₁ –

Use Associative law for the two summing points and interchange their positions, we get,

G₄ and other block are in parallel hence,

\ C(s) R(s) = G + G G H – G G G H + G G G H + G G G 1 + G H – G G H + G 4 4 2 1 4 1 2 1 2 3 4 2 1 2 3 2 1 1 2 1 2 G3 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).

Control System Engineering 5 - 16 Block Diagram Models of Control Systems

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM + G G G 1 + G H – G G H 1 2 3 2 1 1 2 1 C(s) R(s) H G 2 1 G₄ – Minor loop + G G G 1 + G H – G G H 1 2 3 2 1 1 2 1 G G G 1 + G H – G G H + G G H 1 2 3 2 1 1 2 1 2 3 2 C(s) R(s) G₄ H G 2 1 G G G 1 + G H – G G H 1 2 3 2 1 1 2 1 1 + + + G 1 + G H 2 2 1 G3 G₁ C(s) R(s) H G 2 1 H₁ Series G₄ – Minor loop G G 1 + G H 1 2 2 1 G G 1 + G H 1 2 2 1 1 – × H₁ =_{1 + G H – G G H}G G1 2 2 1 1 2 1 C(s) R(s) G G G 1 + G H – G G H + G G H 1 2 3 2 1 1 2 1 2 3 2 G +₄

Interchange the summing points and solve minor feedback loops. Loop 1 = 1 sC R 1 1 sC R 1 1 1 1 + = 1 1 sC R+ _{1 1} Loop 2 = 1 sC R 1 1 sC R 2 2 2 2 + = 1 1 sC R+ _{2 2} \

( )

( )

C s R s =

(

) (

)

1 1 sC R 1 1 sC R 1 1 1 sC R 1 1 sC R sC 1 1 2 2 1 1 2 2 1 + æ è ç ö ø ÷ æ ₊ è ç ö ø ÷ + + + ´ R2

( )

( )

C s R s =

(

)

1 s C C R R2 _{1 2 1 2}+ s R C + R C_{1 1 2 2}+R C_{2 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 G₁ as shown, we get,

Control System Engineering 5 - 17 Block Diagram Models of Control Systems

1 sC₁ R1_{1 –} sC1₂ R1₂ + sC₁ R₂ + – – R(s) C(s) Fig. 5.3 (a) 1 sC R_{1 1 –} + – – R(s) _C(s) sC R_{1 2} 1 sC R_{2 2} Loop 1 Loop 2 Fig. 5.3 (b) 1 1 + sC₁R₁ – R(s) _C(s) sC R_{1 2} 1 1 + sC₂R₂ Fig. 5.3 (c)

Use Associative law for the summing points, we get,

Control System Engineering 5 - 18 Block Diagram Models of Control Systems

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– – + G₂ C(s) R(s) G₃ 1 / G₁ G₁ H₁ H₂ – – + G₂ C(s) R(s) G G 3 1 G₁ H₁ H₂ G 1 + —— G 3 1 Parallel Minor loop – G₂ C(s) R(s) G 1 + ——_G3 1 G 1 + G H 1 1 1 H₂ G + G G 1 3 1 _Series – – C(s) C(s) R(s) G + G R(s) G 1 3 1 (G + G ) G 1 3 1 G G 1 + G H 1 2 1 1 G G 1 + G H 1 2 1 1 H₂ H₂ – C(s) C(s) R(s) G (G + G ) R(s) 1 + G H 2 1 3 1 1 G (G + G ) 1 + G H 2 1 3 1 1 H₂ G (G + G ) H 1 + G H 2 1 3 2 1 1 1 +

\ C(s) R(s) = G G + G G 1 + G H + G G H + G H G 1 2 2 3 1 1 1 2 2 2 2 3 Example 5.3.24

Solution : Separate all the paths linked by take-off point and summing point in the feedback path.

Control System Engineering 5 - 19 Block Diagram Models of Control Systems

G₁ G₂ G₃ R(s) + – – + + + + C(s)

Minor loop Parallel of ‘1’ and G3

Separate the dotted paths H₁ G_{1 –––––––}G 1 + G 2 2 R(s) + – – C(s) Minor loop H₁ H₁ 1 + G₃ G₁ R(s) + – C(s) H₁ 1 + G₃ G ––––––– 1 + G 2 2 G H –––––––_{1 + G}2 1 2 1 + G –––––––––––– 1 + + 2 2 2 1 G G H

\ C(s) R(s) = G G (1 G ) 1 G 2G H G G H 1 2 3 2 2 1 2 3 1 + + + + 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 R₁ acting, R₂ = 0 and C₂ not considered R₂ = 0 and C₂ is suppressed (not considered). C₂ suppressed does not mean that C₂ = 0. Only it is not the focus of interest while C₁ is considered. As R₂= 0, summing point at R₂ can be removed but block of '–1' must be introduced in series with the signal which is shown negative at that summing point.

C R 1 1 = G 1 [G ] [ G G G ] 1 1 2 3 4 + - = G 1 G G G G 1 1 2 3 4 -For C R 2 1 , assume C₁ suppressed.

Control System Engineering 5 - 20 Block Diagram Models of Control Systems

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM G₁ R(s) + – C(s) H₁ G (1 + G ) ––––––––––– 1 + G + G H 2 3 2 2 1 G₁ R(s) _{1 + G + G H}–––––––––––G (1 + G )2 3 _C(s) 2 2 1 G (1 + G )H ––––––––––– 1 + G + G H 2 3 1 2 2 1 1 + Minor loop – R₁ R₁ C₁ C₁ G₁ G₂ –1 G₄ G₃ – G₁ – G G G_{2 3 4} » Forward path Feedback path Minor loop – R₁ R₁ C₂ C₂ G₁ G₂ –1 G₄ G₃ – – G G G_{1 3 2} G₄

»

Feedback path Forward path Minor loop

\ C R 2 1 = _{+ -}-G G G 1 [ G G G ] [G ] 1 3 2 1 3 2 4 = _--G G G 1 G G G G 1 2 3 1 2 3 4 For C R 1 2 , R₁ = 0 and C₂ is suppressed. \ C R 1 2 = + -G -G -G 1 [ G G G ] [G ] 1 2 4 1 2 4 3 = -G -G -G 1 G G G G 1 2 4 1 2 3 4 For C R 2 2 , R₁ = 0 and C₁ is suppressed. C R 2 2 = G 1 [G ] [ G G G ] 2 2 1 3 4 + - = G 1 G G G G 2 1 2 3 4 -Example 5.4.4

Solution : i) With N(s) = 0 block diagram becomes

Control System Engineering 5 - 21 Block Diagram Models of Control Systems

– R₂ R₂ C₁ C₁ G₁ G₂ –1 G₄ G₃ – – G G G_{1 2 4} G₃

»

Feedback path Forward path Minor loop – R₂ R₂ C₂ C₂ G₁ G₂ –1 G₄ G₃ – – G G G_{1 3 4} G₂

»

Forward path Feedback path Minor loop

Minor feedback loop = 10 s (s + 1) 1 + 10 s (s + 1)0.5s = 10 s + s + 5s2 = 10 s + 6s2

Assume output of second summing points as X(s),

Hence E(s) = R(s) – C(s) …(i) and C(s) = X(s)10 (s + 4)

s + 6s2 ... (ii)

While X(s) = E(s) + 3

s + 4R(s) ... (iii)

Substituting value of X(s) and R(s) from (i) and (ii) in (iii) we get, s + 6s 10 (s + 4) 2 C(s) = E(s) + 3 s + 4E(s) + 3 s + 4C(s) Solving, C(s) E(s) = 10 (s + 7) s + 6s2 -30 when N(s) = 0 ii) To find C(s)

R(s), we have to reduce block diagram solving minor feedback loop and shifting summing point to the left as shown earlier in (i).

Control System Engineering 5 - 22 Block Diagram Models of Control Systems

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– – + _C(s) R(s) E(s) 3 s + 4 _{s(s + 1)}10 0.5 s Minor loop – + C(s) R(s) E(s) 3 10 s + 6s2 s + 4 – + _C(s) R(s) E(s) 3 s + 4 10 s + 6s2 X(s) s + 4

So referring to block diagram after these two steps i.e.

Exchanging two summing points using associative law,

\ C(s)_R(s) = s + 7 s + 4 10 (s + 4) s + 16s + 402 æ è ç ö ø ÷ ´ æ è ç ç ö ø ÷ ÷= 10(s + 7) s + 16s + 402

iii) With R(s) = 0 block diagram becomes,

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.

Control System Engineering 5 - 23 Block Diagram Models of Control Systems

– + _C(s) R(s) E(s) 3 s + 4 10 s + 6s2 s + 4 – + _C(s) R(s) 3 s + 4 10 (s + 4) s + 6s2 Parallel of '1' and = 1 + 3 –––– s + 4 3 –––– s + 4 _{Minor loop} G –––––– 1 + GH 10(s+4) ––––––––––– s +16s+402 = = 10 (s + 4) –––––––– s + 6s2 10 (s + 4) –––––––– s + 6s2 1 + = s + 7––––_{s + 4} – – + + + C(s) N(s) s + 4 10 s(s + 1) 0.5 s

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.

Removing summing point, as sign is positive no need of adding a block.

\ C(s) N(s) = 1 1 10 (1.5s + 4) s(s + 1) -é -ëê ù ûú = 1 1 +15s + 40 s(s + 1) = s(s + 1) s + 16s + 402 Example 5.4.5

Solution : Consider R₁ alone R , R , R_{2 3 4} are zero.

Note Whenever R₁ is zero, as the sign of feedback at R₁ is negative, while removing summing point at R₁, do not forget to insert a block of '–1' to consider effect of negative sign.

Control System Engineering 5 - 24 Block Diagram Models of Control Systems

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM – + ₊ + _C(s) N(s) s + 4 – 1 10 s(s + 1) 0.5 s + + + C(s) N(s) 10 s(s + 1) – (s + 4) – 0.5 s After simplification ( – 1.5 s – 4) – + + + C(s) N(s) 10 s(s + 1) – (s + 4) 0.5 s Parallel + + C(s) N(s) 10 s(s + 1) – (1.5 s + 4) + + C(s) N(s) – 10 (1.5 s + 4) s(s + 1)

\ C R₁ = G G 1 G H 1 G G H 1 G H 1 2 2 2 1 2 1 2 2 + + + = G G 1 G H G G H 1 2 2 2 1 2 1 + + i.e. C = G G R 1 G H G G H 1 2 1 2 2 1 2 1 + +

Consider R₂ alone, with R = R = R = 0_{3 1 4}

\ C R₂ = G 1 G H 1 G 1 G H ( G H ) 2 2 2 2 2 2 1 1 + - +æ_èç ö ø ÷ -i.e. C = G R 1 G H G G H 2 2 2 2 1 2 1 + + Consider R₃ alone, R = R = R = 0_{1 2 4}

Control System Engineering 5 - 25 Block Diagram Models of Control Systems

G₁ R₁ G₂ – – C H₁ H₂ Minor loop G₁ R₁ G 1 + G H 2 2 2 – C H₁ G₁ R₂ G₂ + + – C H₁ H₂ –1 Minor loop Note this _R 2 G 1 + G H 2 2 2 + + C –G H_{1 1} G₁ R₃ G₂ + – – C H₁ H₂ –1 Note this +

Combining two summing points we get, \ C R₃ - = G 1 G H 1 G 1 G H ( G H ) 2 2 2 2 2 2 1 1 + - +æ_èç ö ø ÷ -= G 1 G H G G H 2 2 2 1 2 1 + + \ _{C =} -+ + R G 1 G H G G H 3 2 2 2 1 2 1

Consider R₁ alone, with R = R = R = 0_{1 2 3}

\ C R₄ = -+ - - æ ₊ è ç ö ø ÷ G G H 1 G H 1 ( G H ) G 1 G H 1 2 1 2 2 1 1 2 2 2 = -+ + G G H 1 G H G G H 1 2 1 2 2 1 2 1 \ _{C =} -+ + G G H R 1 G H G G H 1 2 1 4 2 2 1 2 1

Combining all the values of C, we get

C = G G R G (R R ) G G H R 1 G H G G H 1 2 1 2 2 3 1 2 1 4 2 2 1 2 1 + - -+ +

Control System Engineering 5 - 26 Block Diagram Models of Control Systems

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R₃ G 1 + G H 2 2 2 + C –G H_{1 1} – R₃ G₂ + C –G H_{1 1} – – H₂ Minor loop R₄ G 1 + G H 2 2 2 + C –G H_{1 1} G₁ G₂ + + + – C H₁ H₂ R₄ –1 Note this

Example 5.4.6

Solution : i) Input R at station I

Control System Engineering 5 - 27 Block Diagram Models of Control Systems

G₁ R(s) G₂ + – C(s) H₁ G₃ H₂ H₃ – + + – Shift the takeoff point G₁ R(s) G₂ + – C(s) H₁ G₃ H₂ H₃ – + + – 1 G₃ Minor loop G₁ R(s) G₂ + – C(s) H₁ – + H G 3 3 Minor loop G 1 + G H 3 3 2 – C(s) R(s) G₁ G G 1 + G H 2 3 3 2 G G 1 + G H 2 3 3 2 H₃ 1 + G G 1 + G H + G H 2 3 3 2 2 3 + ×_G 3

\ C(s)_R(s) = G G G 1 G H G H 1 G G G 1 G H G H H 1 2 3 3 2 2 3 1 2 3 3 2 2 3 + + + + + ´ 1 = G G G 1 G H G H G G G H 1 2 3 3 2 2 3 1 2 3 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 H₁ as it is. Thus add a block of '– 1' as shown in the system.

Shift the summing point to the left of G₂ and interchange the two summing points.

Control System Engineering 5 - 28 Block Diagram Models of Control Systems

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G₁ + G₂ C(s) H₁ –1 Note this H₃ – + – G3 H₂ R(s) + G₁ + H₁ – – –1 G₂ G₃ H₃ H_{2 G}1 2 + + R(s) C(s) Minor loop + – G H_{1 1} – G₃ H G2₂ + R(s) C(s) G 1+G H2_{2 3 +} Blocks in parallel

\ C(s) R(s) = G G ( G G H H ) 1 G H 3 3 1 2 1 2 2 3 1 – - ₊ -\ C(s) R(s) = G (1 G H ) 1 G H G H G G G H 3 2 3 2 3 3 2 1 2 3 1 + + + + Example 5.4.7

Solution : Assume R s

( )

= but to consider negative0

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).

( )

( )

Y s W s = 4 1 1 1 14 s 1 ´ - ´ -+éëê ùûú é ë ê ê ê ê ù û ú ú ú ú = 4 s 1

(

)

s 15 + + Example 5.4.8

Solution : Solving the minor feedback loop of ‘G₄’ and ‘1’.

Control System Engineering 5 - 29 Block Diagram Models of Control Systems

G₃ + R(s) C(s) G 1+G H22 3 + – G G G 1 2 2 H – H_{1 2} H₂ G₂ – G H_{1 1}– G₃ + R(s) C(s) + – G G 1 + G H1 2_{2 3} H – H_{1 2} 1 s + 1 4 W(s) 2 –1 7 Y(s) Note this + (a) W(s) 4 – 14 s + 1 + Y(s) (b) Fig. 5.5

As C₂ is not the focus of interest, G₆ becomes meaningless in the block diagram. Shifting take off point to the right of G₂.

Control System Engineering 5 - 30 Block Diagram Models of Control Systems

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM + + + – R₁ C₁ G₁ G₂ G₃ G 1 + G 4 4 G5 H₂ H₁ + + – R₁ C₁ G₁ G₂ G₃ G G 1 + G 4 5 4 H₂ 1 G₂ H₁ Minor loop Shift + R₁ C₁ G G 1 + G G 1 2 1 2 G₃ G G 1 + G 4 5 4 H_{2 G}1 2 1 G₃ H₁ +

Combining all the blocks in series. \ C R 1 1 = G G G G G H (1 + G G (1 + G ) 1 G G G G G H (1 + G G ) (1 1 2 3 4 5 1 1 2 4 1 2 3 4 5 1 1 2 ) -+ G ) H G G 4 2 2 3 × \ C R 1 1 = G G G G G H 1 + G G + G + G G G G G G H H 1 2 3 4 5 1 1 2 4 1 2 4 - 1 4 5 1 2 Example 5.5.2

Solution : The s-domain network, currents and voltages are shown in the Fig. 5.6

I (s)₁ = V (s) V (s) 1 sC i 1 1 -= sC [V (s) V (s)]_{1 i} - ₁ …(1) V (s)₁ = R [I (s) I (s)]_{1 1} - ₂ …(2) I (s)₂ = sC [V (s) V (s)]_{2 1} - _o …(3) V (s)_o = R I (s)_{2 2} …(4)

The block diagrams for equations are,

Control System Engineering 5 - 31 Block Diagram Models of Control Systems

+ C₁ R₁ G G G G G H (1 + G G ) (1 + G ) 1 2 3 4 5 1 1 2 4 H G G 2 2 3 V_o(s) R₁ R2 1 sC₁ V_i(s) I₁(s) I₂(s) V (s)₁ sC12 Fig. 5.6 I₂(s) R₂ V (s)o I₁(s) V (s)_i R1 V (s)1 V (s)i R₁ I2(s) sC₂ I2(s) sC₂ V (s)o I₁(s) V (s)₁ – – – sC₁ sC₁

The complete block diagram is shown in the Fig. 5.7.

Control System Engineering 5 - 32 Block Diagram Models of Control Systems

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sC₁ I1 R₁ V1 sC₂ R₁ R₂ V (s)o I₂ V (s)_i sC₁ sC₂ Shift Shift Fig. 5.7 sC₁ R₁ 1 –– R₁ sC₂ R₂ R₁ ––R1₂ sC₂ V (s)_o – sC₁ – – V (s)_i Interchange Interchange sC₁ R₁ sC₂ sC₁ R –––––– 1+s R C11 1 R ––––––––– 1+s R C22 2 R₂ sC₂ 1 –– R1x R1x 1––R2= 1––R2 V (s)_o V (s)_i Minor loops – – sC_{1 1+sC R}–––––––R1 1 1 sC2 R ––––––– 1+sR C22 2 V (s)o Minor loop Series 1 –– R₂ V (s)_i – sC₁ V (s)_i (1 + sC R )(1 + sC R )s R R C sR R C _{x 1} R1 (1 + sC R )(1 + sC R ) 1 2 2 1 1 2 2 1 2 2 1 1 2 2 V (s)_o 1 +

\ V (s) V (s) o i = sC sR R C (1 sC R ) (1 sC R ) sR C 1 1 2 2 1 1 2 2 1 2 ´ + + + =

[

]

s R R C C s R R C C s C R C R R C 1 2 1 2 1 2 2 1 2 1 2+ 1 1+ 2 2+ 1 2 +

qqq

Solutions of Examples for Practice

Example 6.5.14

Solution : T₁ = G G G G_{1 2 3 4} ... Only one forward path No combination of nontouching loops.

\ D = 1– [L₁+L₂+L ]₃ = 1 G H+ _{2 1}+G H_{3 2}+G G G G H_{1 2 3 4 3}

D₁ = 1 ... All loops touching to T₁

\ C(s) R(s) = T_{1 1}D D = G G G G 1 G H G H G G G G H 1 2 3 4 2 1 3 2 1 2 3 4 3 + + + 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.

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6

Models of Control Systems

Signal Flow Graph

G₂ G₃ – H₂ –H₁ G₁ G₂ G₃ G₄ – H₃ L = –G H_{1 2 1} L = –G H_{2 3 2} L = –G G G G H_{3 1 2 3 4 3} – – – C(s) R(s) _G 1 s₁ t₁ t₂ + t₃ s₂ s₃ s₄ G₂ H₁ H2 G₄ G₃ (6 - 1)

Number of forward paths are K = 2 Forward path gains are

T₁ = G G G_{1 2 3} T₂ = G₄

Individual feedback loops are

There are no combination of nontouching loops \ D = 1-

[

L + L + L + L + L_{1 2 3 4 5}

]

All the loops are touching to both the forward paths

\ D₁ = D₂ = 1

\ Using Mason’s gain formula C(s)

R(s) =

T₁ D₁+ T₂ D₂ D

Control System Engineering 6 - 2 Signal Flow Graph Models of Control Systems

G₁ s₁ s₂ t₁ s₃ G₂ t₂ s₄ t₃ –H₁ –1 –H₂ G₃ G₄ 1 1 1 1 1 R(s) C(s) –H₂ G₂ G_{3 1} –H₁ G₂ G₁ 1 –H₁ –H₂ G₄ G₂ 1 1 L = –G G H_{1 1 2 1} L = –G G H_{2 2 3 2} L = +G G H_{3 2 4 1}H₂ –1 G₂ G₁ G_{3 1} 1 –1 G₄ 1 1 1 L = –G_{5 4} L = –G_{4 1}G G_{1 3}

\ C(s)_R(s) = G G G + G 1 + G G H + G G H + G G G +

1 2 3 4

1 2 1 2 3 2 1 2 3 G4-G G H H2 4 1 2 Example 6.5.16

Solution : Name the various summing and take-off points to draw the signal flow graph as shown,

The corresponding signal flow graph is,

Forward path gains are,

T = G G_{1 1 2} T = G G_{2 3 2} The various loops are,

L₁ =– G G H_{1 2 2} L₂ = -G G_{2 3} H₂ L₃ = –G H_{2 1} No combinations of non toucting loops.

\D = 1 [L + L + L ]– _{1 2 3} = 1 + G G H_{1 2 2} +G G_{2 3} H + G H_{2 2 1} For T₁, All loops are touching, \ D =₁ 1

For T₂, All loops are touching, \ D =₂ 1

Control System Engineering 6 - 3 Signal Flow Graph Models of Control Systems

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R(s) C(s) + + + G1 G2 H₂ H₁ G₃ – – Input Output s₁ s₂ s₃ t₁ t2 +G₃ G₁ G₂ 1 1 1 t₁ t₂ –H₁ –H₂ Input Output s₁ s₂ s₃ 1 G₁ G₂ 1 –H₂ +G₃ G₂ 1 –H₂ G₂ 1 _–H 1 1

According to Mason's gain formula, C(s) R(s) = T₁ D₁+ T₂ D₂ D C(s) R(s) = G G G G 1 + G G H + G G H + G H 1 2 2 3 1 2 2 2 3 2 2 1 + 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.

Forward paths : T₁= G₁G₂G₃G₄, T₂ = G₆G₂G₃G₄, T₃ = G₁G₂G₅, T₄ = G₆G₂G₅

Individual feedback loops : L₁= – G₂G₃H₁, L₂ = – G₁G₂G₃G₄H₂, L₃ = – G₁G₂G₅H₂ No combination of two non touching loops.

\ D = 1–

[

L₁+L₂+L₃

]

= 1 + G₂G₃H₁+ G₁G₂G₃G₄H₂ + G₁G₂G₅H₂ All loops are touching to T₁, T₂, T₃, T₄hence D₁=D₂ =D₃ =D₄ =1

\

( )

( )

C s R s = T_{1 1}D T_{2 2}D T₃D₃ T_{4 4}D D + + + = G G G G G G G G G G G G G G 1 G G H G G G G H 1 2 3 4 6 2 3 4 1 2 5 6 2 5 2 3 1 1 2 3 4 2 + + + + + + G G G H_{1 2 5 2} Example 6.5.18

Solution : The various forward paths are,

The various individual loop gains are,

The combinations of two non-touching loops are,

Control System Engineering 6 - 4 Signal Flow Graph Models of Control Systems

R(s) 1 1 G1 G2 G3 G4 1 1 C(s) G₆ G₅ –H₁ –H₂ Fig. 6.1 G₁ T₁= G G_{1 2} T₂= G G_{3 4} T₃= G₇ T₄= G G G_{3 5 2} T₅= G G G_{1 6 4} G₂ G3 G4 G₇ G₃ G₅ G₂ G₄ G₆ G₁ Fig. 6.2 H₂ H₁ G₅ G₆

\ D = 1 – [L₁ + L₂+ L₃] + [L₁L₂] All loops are touching to T₄ and T₅.

\ D₁ = 1 – L₁ , D₂ = 1 – L₂ , D₃ = 1 – [L₁+ L₂ + L₃] + [L₁ L₂] , D₄ = D₅ = 1 \ C(s) R(s) = T₁D₁ T₂ D₂ T₃D₃ T₄D₄ T₅D₅ D + + + + \ C(s) R(s) = G G (1 H ) + G G (1 H ) G [1 H_{1 2} - _{2 3 4} - ₁ + ₇ - ₁-H₂-G₅ G H H ] G G G G G G 1 6 1 2 3 5 2 1 6 4 + + + -H₁-H₂-G G + H H_{5 6 1 2} Example 6.5.19

Solution : System node variables are Y , Y , Y , Y_{1 2 3 4}.

Consider equation 1 : This indicates Y₂ depends on Y and Y_{1 3} Consider equation 2 : This indicates Y₃ depends on Y , Y_{1 2} and Y₃

Consider equation 3 : This indicates Y₄ depends on, Y and Y_{3 2}

Control System Engineering 6 - 5 Signal Flow Graph Models of Control Systems

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM H₂ H₁ G₁ G₂ G₄ G₃ G₇ L non-touching to T_{1 1} L non-touching to T_{2 2} H₂ G₆ G₅ H₁ All loops are non-touching to T₃ Fig. 6.2 (b) G₁ Y₂ Y₃ G₃ S.F.G. for equation (1) G₅ Y₃ Y₂ Y₁ G₄ G₂ S.F.G. for equation (2) Y₄ Y₃ Y₂ G₆ G₇ S.F.G. for equation (3) G₅ Y₃ Y₂ Y₁ Y₄ G₄ G₆ G₂ G₃ G₇ G₁ Fig. 6.3

Combining all three we get, complete S.F.G. as shown in Fig. 6.3, No. of forward paths = K = 4

\ T.F. = TK K K D D =

å

1 4

= T1D +1 T2 D +2_D T3 D +3 T4 4D ... Mason's gain formula T = G G G_{1 1 2 7},T₂ =G G_{4 7}, T₃ =G G_{1 6}, T₄ =G G G_{4 3 6}

Individual loops are,

\ D = 1 -[L + L ] = 1 G G_{1 2} - _{2 3} -G₅ No nontouching loop combinations.

For T , T_{1 2} and T₄, all loops are touching. \ D₁ =D₂ =D₄ =1 And for T₃, `G₅' self loop is nontouching, \ D = -₃ 1 G₅ Y Y 4 1 = T1D1+ T2 D2 + T_D 3 D +3 T4 D4 = G G G1 2 7×1 + G G4 7×1 + G G (1 G ) + G G G_D1 6 - 5 4 3 6×1 \ Y_Y4 1 = G G G + G G + G G (1 G ) + G G G 1 G G 1 2 7 4 7 1 6 5 4 3 6 2 3 -- - G₅ Example 6.5.20

Solution : There are two inputs and two outputs hence there are four gains possible as C R 1 1 , C R 2 2 , C R 2 1 and C R 2 2

. Let us find the gain C R 1 1 by Mason's gain formula. The input R₂= 0 and C₂ is suppressed as shown in the Fig. 6.4 (a). The forward paths are, T₁ = G₁ (only one forward path).

The individual feedback loops are,

Control System Engineering 6 - 6 Signal Flow Graph Models of Control Systems

G₅ G₂ G₃ SELF LOOP L₁ L₂ L = G G_{1 2 3} L = G_{2 5} G₅ G₆ G₁ T₃ L₂ L nontouching to T_{2 3} R₁ 1 G1 1 C₁ G₂ H₁ H₂ G₃ G₄ Fig. 6.4 (a)