Closed-Loop Control System - IES Academy

Control System 

Contents 

Chapter Topic Page 

Chapter-1 

Chapter-2 

Chapter-3 

Chapter-4 

Introduction 

Theory at a glance 

Previous Years GATE Questions 

Previous Years IES Questions 

Previous Years GATE Answer 

Previous Years IES Answer 

Transfer Function, Block Diagrams 

and Signal Flow Graphs 






Mathematical Modeling Control 

System 



35Previous Years IES Answer 

Time Response Analysis of 







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India’s No 1 

IES Academy 


Contents 

Chapter-5 

Chapter-6 

Chapter-7 

Chapter-8 

Chapter-9 

Frequency Analysis 




Stability Analysis of Control 

System 






Root — Locus Technique 






Compensators 






Industrial Controllers 






Chapter-10 Introduction to State Space 

Variable 






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IES Academy Chapter 1 

Contents for this chapter 

Introduction 

1. Introduction 

2. Open–Loop System 

3. Mathematical Model for Open-loop Control System 

4. Closed-Loop Control System 

5. Mathematical Model for Closed-Loop System 

6. Comparison of Open Loop and Closed Loop 

7. Laplace Transform 

8. Basic Laplace Transform Theorem 

9. Summary 

1. Introduction 

Theory at a Glance 

(For IES, GATE, PSU & JTO) 

Control system is a combination of elements arranged in a planned manner. Where each 

element causes an effect to produce a desire output. 

Example of control systems 

1. System for the control of position. 

2. System for the control of velocity. 

2. Open–Loop System 

1. No feedback in open loop system is used. 

2. Control system (open-loop) depends only on the accuracy of input calibration. 

Example of open-loop control system 

1. Traffic signal light 

2. Electric lift 

3. Automatic washing machine 

3. Mathematical Model for Open-loop Control System 

C =G 

R 


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Where, 

G = gain of system 

C = o/p of system 

R = input 

Points: 

1. Feedback system is not used for improving stability. 

2. An open-loop system may become unstable when we used negative feedback. 

4. Closed-Loop Control System 

In a closed loop control system the output has an effect on control action through a feedback. 

Example of closed-loop system: 

1. D.C. Motor speed control 

2. Radar tracking system 

3. Auto pilot system 

5. Mathematical Model for Closed-Loop System 

C 

R 

G 

= 

1+GH 

1* Here feedback is negative 

2. This form is also called control canonical form 

From figure 

C(S) = G(S) = 

E(S) 

As a Forward path transfer function 

B(S) =H(S)= 

C(S) 

As a feedback transfer function 

The o/p of summing point 

E(S) 

[ R(S) – B(S) ] 

= ; 

C(S) =R(S)–B(S) ; 

G(S) 

C(S) =R(S)–C(S) H(S) ; 

G(S) 

C(S) = R(S) G(S) – G(S) C(S) H(S) ; 

C(S) [ 1+G(S) H(S) ] = R(S) G(S) 


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C(S) G(S) 

= 

R(S) 1+G(S) H(S) 

6. Comparison of Open Loop and Closed Loop 

Open Loop System 

1. The accuracy of an open loop system 

depends on the calibration of the i/p. 

Closed Loop System 

1. As the error between the reference input 

and the output is continuously measuredthrough 

feedback. 

2. The open loop system is more stable. 2. The closed loop system is less stable. 

3. It is less accurate. 3. It is more accurate. 

4. It is cheap and less complex. 4. It is expensive and more complex circuit. 

5. Effect of Noise and disturbance is more 

in open loop control system. 

7. Laplace Transform 

Laplace transformation is very great tool in control system. 

The mathematical expression for laplace transforms 

LF(t) 

= 

∞ 

–st 

F(S) = ∫ F(t) e dt 

0 

5. Effect of Noise and disturbance is less in 

closed loop control system. 

F(S) 

The term “laplace transform of F(t)” is used for the letter LF(t). 

8. Basic Laplace Transform Theorem 

Basic theorems of laplace transform are given below — 

Theorem 1: Multiplication by a constant 

Let k be a constant and F(S) be the laplace transform of F(t), then 

[ Kf(t) ] 

Theorem 2: Sum and difference 

= 

KF(S) 

Let F1(S) and F2(S) be the laplace transform of f ( t ) f ( t) 

( ) ( ) 

1 2 

± , then 

⎡⎣f1 t ± f2 t ⎤⎦ 

= F(S)± 

1 

F 

2(S) 


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Theorem 3: Differentiation 

dF(t) 

i. L = SF(S)-F(0) 

dt 

[ ] 

2 

dF(t) 2 1 

ii. L = ⎡S F(S)-F(0)-f (0) 

2 

dt 

1 dF(0) 

where, F (0) = dt 

In general, for higher order derivatives or F(t) 

n 

⎡d F() 

t ⎤ 

L ⎢ ⎥ = sFS 

n 

− s f O − s f − f 

⎣ dt ⎦ 

⎣ 

n n−1 n−2 (1) ( n−1) 

( ) ( ) (0) (0) 

Where, F 1 (0) denotes the i th order derivative of f(t) with respect to t1, 

Theorem 4: Integration 

⎡ 

i. L∫ 

F(t) = ⎢ + 

⎣ S 

1 

F(S) F (0) 

⎡ 

ii. L∫∫ 

F(t) = ⎢ + + 

⎣ S S S 

S 

⎤ 

⎥ 

⎦ 

F(S) 

1 

F (0) 

2 

F (0) 

2 2 

Theorem 5: Shift in time 

The laplace transform of F(t) delayed by time T is equal to the laplace transform F(t) 

multiplied by e –ST that is 

-ST 

L [ F(t – T)u 

s(t – T) ] = e F(S) 

Where US(t–T) denotes the unit step function that is shifted in time to the right by T. 

Theorem 6: Complex shifting 

The laplace transform of F(t) multiplied by 

transform F(S), with S replaced by ( S ± α ) 

L ⎡ 

⎣e 

∓αt 

Theorem 7: Initial-value theorem 

If the laplace transform of F(t) is F(S), then 

t→0 

αt 

e ∓ 

⎤ 

⎦ 

⎤ 

⎥ 

⎦ 

, where α is a constant is equal to the laplace 

that is 

F(t) ⎤ 

⎦=F(S±α) 

lim F( t) = lim SF( S ) 

S→∞ 


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Theorem 8: Final value theorem 

lim F( t) = lim SLF( t) 

t→∞ 

S→0 

lim F( t) = lim SF( S) 

t→∞ 

S→0 

Point to be Remember 

If the denominator of SF(S) has any root having real part as zero or positive, then final value 

theorem is not valid. [GATE 2007] 

USEFUL TRANSFORM (LAPLACE) PAIR 

1 F(t) F(S) = LF(t) 

2 δ(t) unit impulse 1 

3 U(t) 

1 

S 

4 U(t–T) 

1 −st 

e 

S 

1 

5 t 

2 

S 

6 

7 

8 

9 

10 

11 

12 

2 

t 

2 

1 

S 

n 

t n 1 

s + 

1 

at 

e − 

at 

e 

−at 

te 

at 

te 

3 

n 

s+ 

a 

1 

s− 

a 

1 

( s+ 

a) 2 

1 

( s− 

a) 2 

13 

h –αt 

n+ 

te ∠ n/ 

( s+ 

a ) 1 


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ω 

14 sinωt 

2 2 

s +ω 

−αt 

15 e cosωt 

−αt 

16 e sinωt 

17 sin hα 

t 

18 cos hα 

t 

s + α 

( ) 2 2 

s + α + ω 

ω 

( ) 2 2 

s + α + ω 

α 

2 2 

s −α 

s 

2 2 

s −α 

9. Summary 

1. Open loop control system no feedback used. 

2. In closed loop control system we used feedback. 

3. Open loop system is more stable. 

4. Closed loop system is more accurate. 

5. Final value theorem can not used if denominators of SF(S) have real part as a zero or 

positive. 


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ASKED OBJECTIVE QUESTIONS (GATE, IES) 


Basic Laplace Transform Theorem 

GATE-1. 

If the Laplace Transform of a signal y(t) is 

1 

Y() s = , 

ss ( −1) 

then its final 

value is: 

[GATE-2007] 

(a) -1 (b) 0 (c) 0 (d) Unbounded 

−t 

GATE-2. The unit impulse response of a system is f () t = e , t ≥ 0 [GATE-2006] 

For this system, the steady-state value of the output for unit step input 

is equal to 

(a) -1 (b) 0 (c) 1 (d) ∞ 


Closed-Loop Control System 

IES-1. 

When a human being tries to approach an object, his brain acts as 

(a) An error measuring device (b) A controller [IES-1999] 

(c) An actuator 

(d) An amplifier 


Assertion (A): Feedback control systems offer more accurate control 

over open-loop systems. 

[IES-2000] 

Reason (R): The feedback path establishes a link for input and output 

comparison and subsequent error correction. 

(a) Both A and R are true and R is the correct explanation of A 

(b) Both A and R are true but R is NOT the correct explanation of A 

(c) A is true but R is false 

(d) A is false but R is true 

IES-3. Consider the following statements: [IES-2000] 

1. The effect of feedback is to reduce the system error 

2. Feedback increases the gain of the system in one frequency range 

but decreases in another 

3. Feedback can cause a system that is originally stable to become 

unstable 

Which of these statements are correct 

(a) 1, 2 and 3 (b) 1 and 2 (c) 2 and 3 (d) 1 and 3 

IES-4. Consider the following statements which respect to feedback control 

systems: 



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1. Accuracy cannot be obtained by adjusting loop gain. 

2. Feedback decreases overall gain. 

3. Introduction of noise due to sensor reduces overall accuracy. 

4. Introduction of feedback may lead to the possibility of instability of 

closed loop system. 

Which of the statements given above are correct 

(a) 1, 2, 3 and 4 (b) Only 1, 2 and 4 

(c) Only 1 and 3 (d) Only 2, 3 and 4 

IES-5. A negative-feedback closed-loop system is supplied to an input of 5V. 

The system has a forward gain of 1 and a feedback gain of a 1. What is 

the output voltage 


(a) 1.0 V (b) 1.5 V (c) 2.0 V (d) 2.5 V 

Basic Laplace Transform Theorem 

ω 

IES-6. Consider the function F(s) = where F(s) is the Laplace transform 

2 2 

s + ω 

of f(t). What is the steady-state value of f(t) 


(a) Zero (b) One (c) Two (d) A value between -1 and +1 


The transfer function of a linear-time-invariant system is given as 

1 

. What is the steady-state value of the unit-impulse response 

( s + 1) 


(a) Zero (b) One (c) Two (d) Infinite 


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Answers with Explanation 

(Objective) 


GATE-1. Ans. (d) 

1 

Y() 

s = 

SS ( + 1) 

Final value of Y(s) 

-1 -1 ⎡ 1 ⎤ -1 ⎡ 1 1 ⎤ 

LT ( Y( s)) = LT ⎢ = LT − + 

( + 1) 

⎥ ⎢ 

⎣ −1⎥ 

⎣SS ⎦ S S ⎦ 

Yt () = e 

t 

−ut 

() 

final value = ∞ 

t →∞ 

GATE-2. Ans. (c) Unit impulse response of a system is 

f() t = e 

−t 

t≥0 

1 

f() 

s = S +1 

O/P for unit step I/P 

= 1 1 

S + 1 

× S 

1 

= 

SS ( + 1) 

1 

( Cs ( )) lim S = 1 

t =∞ = × s→0 

SS ( + 1) 


IES-1. Ans. (b) 

IES-2. Ans. (a) 

IES-3. Ans. (d) Feedback is applied to reduce 

the system error. Consider the 

example. 

Cs ( ) Gs ( ) 

= 

Rs 1 − GsHs 

( ) 

( ) ( ) 

1 

s 1 1 

= 

+ 

= 

1 s − 1 

1 − s+ 

1 

Thus, we see that the closed loop system is unstable while the open loop system is 

stable. 

IES-4. Ans. (d) 


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IES-5. Ans. (d) Output voltage = 

Vin 

1 

A 

AB 

1 

= 5x = 2.5V 

+ 1+ 

( 1x1x ) 

IES-6. Ans. (d) This is the Laplace transform of sin t. 

So, f(t) = sin t 

Steady-state value of f(t) is undetermined because poles of F(s) are not in LHS of 

s-plane. Therefore, steady-state value will vary between - 1 and + 1. 

1 

IES-7. Ans. (a) Steady state value = lims 1 = 0 

s→ 0 s+ 

1 

( ) 


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Closed-Loop Control System - IES Academy

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