Control Systems

Basics of Control System

ECE - 2008

1. A signal flow graph of a system is given below

The set of equations that correspond to this signal flow graph is

(A) d

dt(

x1

x2

x3

) = [β −γ 0γ α 0

−α −β 0] (

x1

x2

x3

) + [1 00 00 1

] (u1

u2)

(B) d

dt(

x1

x2

x3

) = [

0 α γ0 −α −γ0 β −β

] (

x1

x2

x3

) + [0 00 11 0

] (u1

u2)

(C) d

dt(

x1

x2

x3

) = [−α β 0−β −γ 0α γ 0

] (

x1

x2

x3

) + [1 00 10 0

] (u1

u2)

(D) d

dt[

x1

x2

x3

] = [−γ 0 βγ 0 α

−β 0 −α] (

x1

x2

x3

) + [0 10 01 0

] (u1

u2)

ECE - 2010

2. The transfer function Y(s)/R(s) of the system shown is

(A) 0

(B) 1

s+1

(C) 2

s+1

(D) 2

s+3

ECE - 2014

3. For the following system,

when X1(s) = 0,

+

−

s

s + 1

X2(s)

1

𝑠

Y(s) +

+

X1(s)

1

s + 1

1

s + 1

∑

Y(s)

R(s) +

−

− +

∑

−γ

1/s

1/s

1/s

-𝛼

α

γ

β −β

1

1

u1

u2

1

1

Control Systems

the transfer functionY(s)

X2(s) is

(A) s + 1

s2

(B) 1

s + 1

(C) s + 2

s(s + 1)

(D) s + 1

s(s + 2)

4. Consider the following block diagram in the figure.

The transfer functionC(s)

R(s) is

(A) G1G2

1+G1G2

(B) G1G2 + G1 + 1

(C) G1G2 + G2 + 1

(D) G1

1+G1G2

EE - 2007

1. The system shown in figure below.

Can be reduced to the form

With

(A) X =c0s + c1,

Y= 1 (s2 + a0s + a1), Z = b0s + b1⁄

(B) X = 1, Y = (c0s + c1)

(s2 + a0s + a1),⁄

Z = b0s + b1

(C) X = c1s + c0 ,

Y =(b1s + b0)

(s2 + a1s + a0), Z = 1⁄

(D) X = c1s + c0 , Y = 1(s2 + a1s + a0), Z = b1s + b0

⁄

EE - 2008

Z

Σ X Y P +

+

X Y P Σ +

+

Σ

b0 c0 b1 C1

1/s p 1/s

Σ

Σ

a

1 a

0

Σ

Σ Σ

+

+

+ + + +

− a0 a1

C(s)

+

+ G2 G1

+

+ R(s)

Control Systems

2. A function y(t) satisfies the following differential equation, dy(t)

dt + y(t) = δ(t), where δ (t) is

the delta function. Assuming zero initial condition and denoting the unit step function by

u(t), y(t) can be of the form

(A) et

(B) e−t

(C) etu(t)

(D) e−tu(t)

EE - 2010

3. As shown in the figure, a negative feedback system has an amplifier of gain 100 with ±10%

tolerance in the forward path, and an attenuator of value 9/100 in the feedback path. The

overall system gain is approximately:

(A) 10±1%

(B) 10±2%

(C) 10±5%

(D) 10±10%

EE - 2014

4. The closed-loop transfer function of a system is T(s) =4

(s2+0.4s+4). The steady state error

due to unit step input is________

5. The signal flow graph of a system is shown below. U(s) is the input and C(s)is the output.

Assuming h1 = b1 and hO = b0 − b1a1, the input-output transfer function G(s) =C(s)

U(s) of the

system is given by

(A) G(s) =b0s + b1

s2 + a0s + a1

(B) G(s) =a1s + a0

s2 + b1s + b0

(C) G(s) =b1s + b0

s2 + a1s + a0

(D) G(s) =a0s + a1

s2 + b0s + b1

6. The block diagram of a system is shown in the figure

R(s) 1

s

G(s) S −

− − + + C(s)

C(s)

1

1

s

1

s

1 1 h0

−a1

−a0

h1

U(s)

100 ± 10%

9100⁄

+

−

Control Systems

If the desired transfer function of the system is C(s)

R(s)=

s

s2+s+1 then G(s) is

(A) 1

(B) s

(C) 1/s

(D) −s

s3+s2−s−2

IN - 2006

1. The signal flow graph representation of a control system is shown below. The transfer

function Y(s)

R(s)is computed as

(A) s

1

(B) 2)s(s

1s2

2

(C) 2s

1)s(s2

2

(D) 1 −1

s

IN - 2007

2. A feedback control system with high K, is shown in the figure below:

Then the closed loop transfer function is.

(A) sensitive to perturbations in G(s) and H(s)

(B) sensitive to perturbations in G(s) and but not to perturbations H(s)

(C) sensitive to perturbations in H(s) and but not to perturbations G(s)

(D) insensitive to perturbations in G(s) and H(s)

IN - 2009

3. A filter is represented by the signal flow graph shown in the figure. Its input x(t) and

output is y(t).

The transfer function of the filter is

(A) −(1+KS)

S+K (C)

−(1−KS)

S+K

X(s)

Y(s)

1

1 1

1

k −k

−1/S

R(s) C(s) G(s)

H(s)

K +

_

R(s) Y(s)

1

s

1/s 1/s 1/s

−1 −1

1 1

Control Systems

(B) (1+KS)

S−K (D)

(1−KS)

S+K

IN - 2011

4. The signal flow graph of a system is given below.

The transfer function (C/R) of the system is

(A) (G1G2+G1G3)

(1+G1G2H2)

(B) (G1G2+G1G3)

(1−G1H1+G1G2H2)

(C) (G1G2+G1G3)

(1−G1H1+G1G2H2+G1G3H2)

(D) (G1G2+G1G3)

(1−G1H1+G1G2H2+G1G3H2+G1G2G3H1)

IN - 2012

5. The transfer function of a Zero-order-Hold system with sampling interval T is

(A) 1

s(1 − e−Ts)

(B) 1

s(1 − e−Ts)2

(C) 1

se−Ts

(D) 1

s2 e−Ts

IN - 2013

6. The complex function tanh(s) is analytic over a region of the imaginary axis of the complex

s- plane if the following is TRUE everywhere in the region for all integers n

(A) Re(s) =0

(B) Im (s) ≠ nπ

(C) Im(s) ≠nπ

3

(D) Im(s)≠(2n+1)π

2

7. The signal flow graph for a system is given below. The transfer function Y(s)

U(s) for this system

is given as

(A) s+1

5s2+6s+2

(B) s+1

s2+6s+2

(C) s+1

s2+4s+2

(D) 1

5s2+6s+2

IN - 2014

8. A plant has an open-loop transfer function,

Gp(s) =20

(s + 0.1)(s + 2)(s + 100)

Y(s) U(s) 1 s−1

1

−4

−2

1 s−1

G2

R

G3

G1

H1

−H2

C 1 1 1

Control Systems

The approximate model obtained by retaining only one of the above poles, which is

closest to the frequency response of the original transfer function at low frequency is

(A) 0.1

s+0.1

(B) 2

s+2

(C) 100

s+100

(D) 20

s+0.1

Answer Keys and Explanations

ECE

1. [Ans. D]

x1

′ = −αx1 − βx2 + u1

x2′ = β1x1 − γx2 + u2

x3′ = α x1 + γx2

[

x1

x2

x3

] = [

−α −β 0β −γ 0α γ 0

] [

x1

x2

x3

] + [1 00 10 0

] [u1

u2]

One can denote any state by any name

So, that answer is

[

x1

x2

x3

] = [−γ 0 βγ 0 α

−β 0 −α] (

x1

x2

x3

) + [0 10 01 0

] (u1

u2)

2. [Ans. B]

Q(s) = P(s) [1

s + 1−

1

s + 1] = 0

So, P(s) = R(s) − 0 = R(s)

Y(s) =P(s)

s + 1=

R(s)

s + 1

Therefore,Y(s)

R(s)=

1

s + 1

3. [Ans. D]

Q(s)

1

s + 1

1

s + 1 Y(s)

P(s) R(s)

∑

∑

−

+

+ −

x3

−γ

u1

u2

x1′

β

x2′ 1

1

−β

x2

x1

α

γ

1

S 𝐱𝟑

′

1

S

1

S

−α

Control Systems

When x1(s) = 0

∴ y(s)

x2(s)=

1

s

1 +1

s.

s

s+1

=s + 1

s(s + 2)

4. [Ans. C]

C(s)

R(s)= 1 + G2 + G1G2

EE

1. [Ans. D]

Signal flow graph of the block – diagram

c0

b0

1/s P

a0

Σ Σ

c1 Σ b1

a1

+

+ ③ ⑤ 1/s

④

+ + ②

①

⑥

C/s R(s)

1

G1 G2 + + R(s) C(s)

G2 1 + G1 R(s) C(s)

+

R(s) C(s) G2 + G1G2

⇓

G2 G1 +

R(s) C(s)

⇓

X2(s) +

−

1

s

Y(s)

s

s + 1

Control Systems

There are forward paths

P1 = 1 × C0 ×1

S×

1

S× P =

C0P

S2

P2 = 1 × C1 × 1 ×1

S× P =

C1P

S

These are four individual loops

L1 = −a1 ×1

s= −

a1

s

L2 =1

s×

1

s× −a0 = −

a0

s2

L3 =1

s×

1

s× P × b0 =

b0P

s2

L4 = b1 × P ×1

s=

b1P

s

All the loops touch forward paths

Δ1 = Δ2 = 1

Δ = 1 − (L1 + L2 + L3 + L4)

= 1 +a0

s2+

a1

s−

b0P

s2−

b1P

s

Using Mason’s gain formula

C(s)

R(s)=

P1Δ1 + P2Δ2

s

=

c0P

s2 +c1P

s

1 +a1

s+

a0

s2 −b0P

s2 −b1P

s

C(s)

R(s)=

c0P + c1Ps

s2 + (a0 − b1)s + (a0 − b0P)

C(s)

R(s)=

P(c0 + c1s)/(s2 + a1s + a0)

1 − (b0 + b1s)P/(s2 + a1s + a0). . (i)

C(s)

R(s)=

xyP

1 − yzP

Comparing eg. (I) and (ii), we get

xy =c0 + c1s

s2 + a1s + a0

yz =b0 + b1s

s2 + a1s + a0

Hence option (D) is correct

Σ

+

+

X Y P

Z

C(s)

−𝑎0 b0

−𝑎1

①

③ 1/s ④ c0

1

1/s ⑤ P 1

b1 c1

C(s) R(s)

②

Control Systems

2. [Ans. D]

Take LT on both sides,

s.Y(s)+Y(s) =1 Y(s) = 1

(S+1)

y(t) = e−t. u(t)

3. [Ans. A]

T.F = G

1+GH=

100

1+9

100.100

=10

Also, SGT =

1

1+GH=

1

10%

overall change in T.F for G = 10 is 1

10. 10

= 1%

∴ T = 10 ± 1%

4. [Ans. 0]

T(s) =4

s2 + 0.4s + 4

R(s) =1

s (for unit step)

Y(s) =4

s(s2 + 0.4s + 4)

y(t) = lims→0

sy(s) =4s

s(s2 + 0.4s + 4)= 1

Steady state error = 1 − 1 = 0

5. [Ans. C]

Using Mason’s gain formula

Paths:

P1 = h0 ×1

s×

1

s

P2 = h1 × (1

s) × (1)

Loops:

L1 = 1 ×1

s× (−a1)

L2 =1

s×

1

s× (−a0)

C(s)

U(s)= ∑ =

n

i=1

ΔiPi

Δ

n = number of paths

∑ΔiPi

Δ

2

i=1

=Δ1P1 + Δ2P2

Δ

Δ = 1 − (L1 + L2) = 1 +a1

s+

a0

s2

Δ1 = 1; Δ2 = 1 − L1 = 1 +a1

s

C(s)

U(s)=

P1 + P2(1 − L1)

Δ

Control Systems

h0

s2 +h1

s(1 +

a1

s)

1 +a1

s+

a0

s2

=

b0−a1b1

s2 +b1(s+a1)

s2

1 +a1

s+

a0

s2

=b0 − a1b1 + b1s + b1a1

s2 + a1s + a0

=b0 + b1s

s2 + a1s + a0

6. [Ans. B]

Convert the given block diagram to signal flow graph and use Mason’s rule

𝑃1 = 𝐺(𝑠),

𝛥 = 1 − [−𝐺(𝑠) + −𝐺(𝑠)

𝑠− 𝑠𝐺(𝑠)]

𝐶(𝑠)

𝑅(𝑠)=

𝑆𝐺(𝑠)

𝑠2𝐺(𝑠) + 𝑠(𝐺(𝑠) + 1) + 𝐺(𝑠)

So put G(s) = s

𝐶(𝑠)

𝑅(𝑠)=

𝑠

𝑠2 + 𝑠 + 2

IN

1. [Ans. A]

M1 = 1 ×1

s=

1

s, M2 = 1 ×

1

s×

1

s×

1

s× 1

=1

s3

Δ1 = 1 , Δ2 = 1

There individual loops with gains:

(1

s) (

1

s) (−1), (

1

s) (

1

s) (−1),

1

s(−1) (

1

s) (−1)

Δ = 1 − [−1

s2 −1

s2 +1

s2] = 1 +1

s2 =s2+1

s2

Y(s)

R(s)=

M1Δ1 + M2Δ2

Δ=

1

s+

1

s3

s2+1

s2

=

1

s(

s2+1

s2 )

(s2+1

s2 )=

1

s

2. [Ans. C]

In this case, open loop gain sensitivity is measured with respect to KG. Because here open

loop gain = KG

R(s) 1 1/s G(s) s

−1

C(s)

−1

1

−1

R(s) 1

s G(s) S

− C(s)

+

−

−

−

Control Systems

T =KG

1 + KGH

SKGT = (

∂T

T∂(KG)

KG

) = (∂T

∂KG) (

KG

T)

= (∂T

∂G) (

G

T) (K is count)

=K + K2GH − K2GH

(1 + KGH)2(

G

KG) (1 + KGH)

=1

1 + KGH→ 0(for large k)

sKGT =

1

(1 + KGH)2=

1

K2G2H2 KGH ≫ 1

sHT =

−KG. KG

(1 + KGH)2.

H

KG(1 + KGH)

=−KGH

1 + KGH≃ −1

So close loop transfer function is sensitive to perturbation in H(s) but not to perturbation

in G(s) for large k values.

[Ans. A]

P1 =−1

S; P2 = −K; ∆1= 1; ∆2= 1;

∆= 1 − (−k

S); L1 =

−k

S

Y(s)

U(s)=

−1

S −k

1+k

S

= −(kS+1)

S(S+k)

S

=−(kS+1)

(S+k)

3. [Ans. C]

P1 = G1G2; P2 = G1G3

L1 = G1H1; L2 = −G1G2H2;

L3 = −G1G3H2

∆1= 1; ∆2= 1

∆ = 1−(L1 + L2 + L3)

= 1 − G1H1 + G1G2H2 + G1G3H2

∴C

R=

(P1∆1 + P2∆2)

∆

=(G1G2 + G1G3)

(1 − G1H1 + G1G2H2 + G1G3H2)

4. [Ans. A]

The transfer function of a Zero-order-Hold system having a sampling interval T is 1

s(1 − e−Ts)

5. [Ans. D]

tan h(s) =sin h(s)

cos h(s)=

es − e−s

es + e−s=

ejω − e−jω

ejω + e−jω

For analytical region there should be no poles

⇒ ejω + e−jω ≠ 0

Control Systems

⇒ cos ω + jsin ω + cos ω − jsinω ≠ 0

⇒ 2cos ω ≠ 0

⇒ ω ≠(2n + 1)π

2

6. [Ans. A]

forward path =

P1 = S−1S−1 = S−2

P2 = S−1 |

Δ1 = 1Δ2 = 1

Loop =

L1 = −4S−1

L2 = −2S−1S−1 = −2S−2

L3 = −2S−1

L4 = −4

T(s) =S−2 + S−1

1 + 4 + 2S−1 + 2S−2 + 4S−1

=S−1 + S−2

S + 6S−1 + 2S−2=

S + 1

5S2 + 6S + 2

7. [Ans. A]

Writing the given open loop transfer in form of 1

1+jω

ωc

We get 1

(1 +s

0.1) (1 +

s

2) (1 +

s

100)

So, at low frequencies, the approximate model equivalent to original transfer function will

be (1

1 +S

0.1

)

y(s) u(s) 1 s−1

1

−4

−2

1 s−1