CONTROL SYSTEMS

Time Response Analysis

Dr. S. SUMATHI

Associate Professor

RNSIT

Bengaluru

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Time Response Analysis The time response of a system is the output response of the

system as a function of time.

The time response of a control system is usually

divided into two parts:

Transient response

Steady state response

( ) ( ) ( )t ssc t c t c t

c(t) denote the time response of a continuous data system

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Transient response

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The transient response is defined as that part of the

time response that goes to zero as time tends to

infinity.

Steady state response

The steady state response is the part of the total

response that remains after the transient has died out.

( ) 0ttLt c t

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First step is to obtain a mathematical model of the system

Transfer function is an Important mathematical model for a

LTI system

For any specific input signal, a complete time response

can then be obtained - Laplace transform inversion of c(s)

If the input signal is such that ,we can not find Laplace

transformable, then time response is obtained through

convolution Integral.

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Steady state behavior of the system can be obtained from

c(t) expression, with time tending to infinity.

In case of simple deterministic signals steady state

response can be obtained by the use of Final value

theorem

Time response analysis is normally carried out after

stability analysis through indirect tests

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INPUT SIGNALS Inputs to the system are not known ahead of time.

Input signals to some of the systems are random in nature

The actual signals which severely stain the a control

system are

a sudden shock

a sudden change

a constant velocity

a constant acceleration

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Hence the system dynamic behavior is analyzed under

the application of standard test signals

Standard test signals

Impulse signal (sudden shock)

Step signal (sudden change)

Ramp signal (constant velocity)

Parabolic signal (constant acceleration)

Nature of transient response depends on the system

poles and not on the type of input.

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STEP SIGNAL

The step is a signal whose value changes from one level

(usually zero) to another level A in zero time.

The mathematical representation of the step function is

r(t) = Au(t)

where u(t) = 1; t>0

= 0; t<0

u(t) is called the unit step function.

In the Laplace transform, U(s)=A/s

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RAMP SIGNAL

The ramp is a signal which starts at a value of zero and

increases linearly with time.

Mathematically,

r(t) = At; t>0

= 0; t<0

In the Laplace transform form,

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PARABOLIC SIGNAL

The parabolic function represents a signal that is represented as

In Laplace transform form

; t > 0

= 0 ; t<0

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IMPULSE SIGNAL

A unit impulse is defined as a signal which has zero value

everywhere except at t=0 , where its magnitude is infinite.

It is generally called the δ function and has the following

property.

( ) 0; 0t t

( ) 1t dt

( ) 1s

Where

tends to zero.

In the Laplace transform

( ) 1s

0 t

δ(t)

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IMPULSE RESPONSE

Impulse signal is derivative of step signal.

The impulse response of a system with transfer function

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

( )G s

1( ) ( ) ( )c t L G s g t

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(As R(s)=1)

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Thus, the impulse response of a system, indicated by

( )g t

which is the inverse Laplace transform of its transfer

function.

This is also referred to as weighting function of the

system.

The weighting function of a system can be used to find the

system’s responses to any input r(t) by means of

convolution integral

0

( ) ( ) ( )

t

c t g t r d 5/31/2019 13

dt

• Impulse ẟ (t)

• Step u(t)

• Ramp r(t)

• Parabolic p(t)

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For linear systems transient response is obtained for

one of the standard test signals,

normally for a step input .

As the system response depends on the system poles

and not upon the type of input

Steady state response is then examined for step signal

as well as other test signals.

For frequency response analysis of systems a

sinusoidal signal with variable frequency is used.

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TIME RESPONSE OF FIRST ORDER SYSTEMS

The transfer function is given by

( ) 1

( ) 1

C s

R s Ts

+

_

R(s) C(s)

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UNIT STEP RESPONSE OF FIRST ORDER SYSTEMS

( ) 1/R s s

1 1( ) .

1C s

s Ts

( ) 1

( ) 1

C s

R s Ts

Therefore output response is given by

By applying partial fraction expansion to C(s) we have

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UNIT STEP RESPONSE OF FIRST ORDER SYSTEMS cont..

Ttetutc /)()(

By taking Laplace inverse C(s) we have

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T is called the TIME CONSTANT

The initial slope of the output curve is given by

Time constant is indicative of how fast the system tends to

reach the final value.

A large time constant corresponds to a sluggish system

A small time constant corresponds to a fast response.

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The error response of the system is given by

( ) ( ) ( )tTe t r t c t e

The steady state error is given by

( ) 0sst

e Lt e t

Thus the first order system tracks the unit step

input with zero steady state error.

ERROR RESPONSE

20

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21

RAMP RESPONSE OF FIRST ORDER SYSTEMS

The transfer function of a first order system is given by

( ) 1

( ) 1

C s

R s Ts

For a ramp input

Therefore

By applying partial fraction expansion to C(s) we have

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22

By taking Laplace inverse C(s) we have

Let us differentiate the ramp response

This is same as step response

g (t) ie impulse response

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TIME RESPONSE OF SECOND ORDER SYSTEMS

A general second-order

system is shown below

It is a Type-1 second

order system

n

Un-damped natural frequency of oscillations

Damping factor

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TIME RESPONSE OF SECOND ORDER SYSTEMS

Its closed loop transfer function is given by

22

2

2 nn

n

sssR

sC

)(

)(

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• The time response of any system is characterized by the

roots of the denominator polynomial q(s) .

• The denominator polynomial q(s) is therefore called the

characteristic polynomial

• q(s)=0 is called the characteristic equation

• Roots of the characteristic equation are same as the poles

of the system

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Hence the characteristic equation of the system

under consideration is given by

2 22 0n ns s

Two roots of the characteristic equation or poles of

the system are

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According the value of ξ , a second-order system can be

classified into one of the four categories

Overdamped ( ξ >1) - the system poles are real and distinct .

-a -b -c σ

jω

The response rises slowly and reaches the final value

without any oscillations

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Critically damped (ξ = 1) - the system has two real but

equal poles

The two poles are

-ωn -b -c

σ

jω

The response rises slowly and reaches the final value

without any oscillations

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Underdamped (0 < ξ <1) - the system has a pair of

complex conjugate poles

-ξωn σ

jω

The transient response is oscillatory.

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Undamped (ξ = 0) -the system has two

imaginary poles.

-a -b -c

σ

jω

The system response is oscillatory.

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Most control systems are designed as under damped systems

to have fast response

Step Response of an under damped system 0 1

22

2

2 nn

n

sssR

sC

)(

)(( ) 1/R s s

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2 2

21

( 2 )

n

n n

s

s s s

Solving for A, B and C we can find that A=1 B= -1 and

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222222 2

21

nnnn

n

ss

s

ssC

)(

2221

21

nn

n

s

s

ssC )(

22

21

dn

n

s

s

ssC

)(

• Where , is the frequency of transient

oscillations and is called damped natural frequency.

21d n

2222

1

dn

n

dn

n

ss

s

ssC

)(

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2 2 2 2

1.

( ) ( )

n n d

n d d n d

s

s s s

2 2 2 22

1.

( ) ( )1

n d

n d n d

s

s s s

Taking the inverse Laplace transform of the above equation

2( ) 1 cos sin

1

n

n

tt

d d

ec t e t t

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2

21 1 cos sin

1

nt

d d

et t

2

1 sin cos cos sin1

nt

d d

et t

21 sin( )

1

nt

d

et

22 1

2

11 sin( 1 tan )

1

nt

n

et

ξ

ϴ

1

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0 time t 0

1

c(t)

Step Response of an under damped system C(t)

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Step Response of an un damped system C(t)

The step response of un damped system can be obtained

from that of Under damped system by substituting ξ =0

22 1

2

11 sin( 1 tan )

1

nt

n

et

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Step Response of an critically damped system

For step input R(s)=1/s, therefore

By applying partial fraction expansion

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The step response is characterized by following

performance indices

i) Delay time

ii) Rise time

iii) Peak time

iv) Maximum overshoot

v) Settling time

Time Response Specifications

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i) Delay time ii) Rise time iii) Peak time

ii) Maximum overshoot v) Settling time

Time Response Specifications

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DELAY TIME The delay time is the time

required for the response to reach

50% of the final value for the very

first time.

rt

RISE TIME The rise time is the time

required for the response to rise

from 0 to 100% of the final value

for under damped system and

from 10% to 90% of the final

value for over damped systems.

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MAXIMUM OVERSHOOT

The peak or maximum overshoot is the maximum peak value

of the response curve measured from unity.

If the final steady state value of the response differs from unity,

then it is common to use the maximum percent overshoot.

Maximum percent overshoot = ( ) ( )

100%( )

pc t cX

c

PEAK TIME

The peak time is the time required for the response to reach

the first peak of the overshoot.

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SETTLING TIME

The settling time is the time required for the response

curve to reach and stay within a range about the final

value of size specified by absolute percentage of the

final value (usually 2% or 5%).

STEADY STATE ERROR

It indicates the error between the actual output and the

desired output as t tends to infinity.

[ ( ) ( )]sst

e Lt r t c t

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i) Delay time ii) Rise time iii) Peak time

ii) Maximum overshoot v) Settling time

Time Response Specifications

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EXPRESSION FOR TIME RESPONSE SPECIFICATIONS

RISE TIME

The output of a second order under damped system excited

by a unit step input is given by

2( ) 1 sin( )

1

nt

d

ec t t

Rise time is defined as the time taken by the output to rise

from 0 to 100% of the final value

rt t ( ) 1rc t Therefore at

,

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21 1 sin( )

1

n rt

d r

et

2

sin( ) 01

n rt

d r

et

20

1

n rte

sin( ) 0 sind rt

d rt d rt

21

2

1tan

1r

d n

t

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PEAK TIME

The output of a second order under-damped system

excited by a unit step input is given by

2( ) 1 sin( )

1

nt

d

ec t t

Peak time is defined as the time at which the

maximum value of magnitude occurs

Therefore, at pt t slope of c(t) must be zero

2 2

( )| cos( ). sin( ) ( ) | 0

1 1

n n

p p

t t

t t d d d n t t

dc t e et t

dt

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2sin( ) 1 cos( ) 0n d p n d pt t

2sin( ) 1 cos( ) 0d p d pt t

cos sin( ) sin cos( ) 0d p d pt t

sin( ) 0 sind pt d pt

21p

d n

t

ξ

ϴ

1

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PEAK OVERSHOOT

2( ) 1 sin( )

1

nt

d

ec t t

The peak overshoot is the difference between the peak value

and the reference input. Therefore,

2( ) 1 1 sin( ) 1

1

nt

p p d p

eM c t t

2sin( )

1

n pt

d p

et

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21

2sin

1

n

n

d

d

e

21

2sin

1

e

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SETTLING TIME

2( ) 1 sin( )

1

nt

d

ec t t

4 44s

n

t T

Assuming ξ to be small

3 33s

n

t T

2% criterion

5% criterion

The equations for obtaining are valid only for the standard second order system

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STEADY STATE ERROR

2( ) 1 sin( )

1

nt

d

ec t t

2( ) [1 ( )] sin( ) 0

1

nt

ss dt t t

ee Lt e t Lt c t Lt t

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Numericals 1)The closed loop transfer function of certain second order unity

feedback system is given below. Determine the type of damping in the system

2

( ) 8

( ) 3 8

C s

R s s s

Solution Comparing the given transfer functions with the standard form of the

transfer function of a second order system

2

2 2 2

( ) 8

( ) 3 8 2

n

n n

C s

R s s s s s

2 8n 8 2.82n 2 3n 3 3

0.532 2 2.82n

1 Hence it is an under damped system

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Numericals 2)The closed loop transfer function of certain second order unity

feedback system is given below. Determine the type of damping in the system

Solution Comparing the given transfer functions with the standard form of the

transfer function of a second order system

2

( ) 2

( ) 4

C s

R s s

2

2 2 2

( ) 2

( ) 4 2

n

n n

C s

R s s s s

2 0n 2 4n 2n

0

0

Hence it is an undamped system

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Numericals 3) Measurements conducted on a servomechanism show the

system response to be

when subjected to step input of magnitude 2. Determine the undamped

natural frequency and damping ratio.

Solution

Taking Laplace transform of the above equation we get

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For step input

Therefore

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Comparing with standard second order system equation

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4) Obtain the response of a unity feedback system whose open

loop transfer function is for a unit step input

3( )

( 4)G s

s s

Solution

( ) ( )

( ) 1 ( )

C s G s

R s G s

3

( 4)

31

( 4)

s s

s s

2

3 3

4 3 ( 1)( 3)s s s s

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For a unit-step input, r(t)=1. Therefore 1

( )R ss

3( )

( 1)( 3)C s

s s s

3 11 2 2

1 3s s s

Taking the inverse Laplace transform, the response is

33 1( ) 1

2 2

t tc t e e

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5) What is the response of the system for a unit-step input with

Solution

The closed loop transfer function of the system is

2

10

( ) 10( 3)

10( ) 4 101 (0.1 1)

( 3)

C s s s

R s s ss

s s

For a unit step input 1

( )R ss

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2

1 10( )

4 10C s

s s s

2 4 10

A Bs C

s s s

2

1 4

4 10

s

s s s

2 2 22

1 2 2 6.

6 ( 2) ( 6)( 2) ( 6)

s

s ss

Taking the inverse Laplace transform, the response is

2 22( ) 1 cos 6 sin 6

6

t tc t e t e t

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6)The open loop transfer function of a unity feedback system is 4

( )( 1)

G ss s

Determine the nature of response of the closed loop system for a unit step input.

Also determine the rise time, peak time, peak overshoot and settling time

Solution

The closed loop transfer function is 2

4

( ) 4( 1)

4( ) 41

( 1)

C s s s

R s s s

s s

Comparing it with the standard form of the closed loop transfer function

of a second order system

2

2 2 2

( ) 4

( ) 4 2

n

n n

C s

R s s s s s

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2 4n 2n 2 1n 1 1

0.252 2 2n

21d n 22 1 0.25 1.936 rad/s

2 21 11 1 0.25

tan tan 1.3100.25

The rise time 3.141 1.310

0.9451.936

r

d

t s

rad/s

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The peak time in sec 3.141

1.6221.936

p

d

t s

21

pM e

The peak overshoot = 0.4326

Therefore, percentage of peak overshoot is 100% 43.26%pM

The settling time for 5% error is 3 3

60.25 2

s

n

t

sec

The settling time for 2% error is

4 48

0.25 2s

n

t

sec

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7) A unity feedback system is characterized by an open-loop transfer

function

( )( 10)

KG s

s s

Determine the gain K so that the system will have a damping ratio of 0.5.

For this value of K, determine the settling time, peak overshoot and time

to peak overshoot for a unit step input

Solution The closed loop transfer function of the given feedback system is

2

( ) ( ) ( 10)

( ) 1 ( ) 101

( 10)

K

C s G s Ks s

KR s G s s s K

s s

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Comparing it with the standard form of the transfer function

of the second order system, we have

2

2 2 2

( )

( ) 10 2

n

n n

C s K

R s s s K s s

2

n K n K 2 10n

2 0.5 10n 10n

2 210 100nK

So the gain K=100 so that the system will have a damping ratio of 0.5

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The settling time for 5% criterion is

4 40.8

0.5 10s

n

t s

The settling time for 2% criterion is

3 30.6

0.5 10s

n

t s

The peak overshoot is

% 100 0.163 100 16.3%p pM M

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The peak time

21p

d n

t

2

3.140.363

10 1 0.5s

The time to peak overshoot is 0.363pt s

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8) A second order system is represented by a

transfer function given below

0

2

( ) 1

( )

Q s

T s Js fs K

where

0 ( )Q s the proportional to output and T is is the torque

input. A step input 10 N-m is applied to the system and the

test results are given below

Peak overshoot

pM

pM =6% Peak time pt =1s

The steady state output of the system is 0.5 radian

Determine the values of J, K and f.

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Given, the input torque T is a step of 10 N-m. Therefore 10

( )T ss

0 2 2

( ) 10( )

( )

T sQ s

Js fs K s Js fs K

00

( )sLt sQ s

The steady state value of output =

20

100.5 .

( )sLt s

s Js fs K

100.5

K 20K

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Given the peak overshoot 6% 0.06pM

210.06 e

0.667

Given that peak time 1pt s

21

1d n

4.21n rad/s

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Comparing the given equation with the standard form of

characteristic equation of a second-order system

2 2 22 0n n

f Ks s s s

J J

2

n

K

J 2

2 2

201.128

4.21n

KJ kg m

2 n

f

J

2 2 0.667 4.21 1.128 6.34 / /nf J N m rad s

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Steady State Error of Unity Feedback Systems

The closed-loop transfer function is

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For Unity feedback system

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Static Position Error Constant (Kp)

The steady-state error of the system for a unit-step input is

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The static position error constant Kp is defined by

The steady state error for step input is

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Static Velocity Error Constant (K v)

• The steady-state error of the system for a unit-ramp

input is

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• The static velocity error constant K v is defined by

• Thus, the steady-state error for a ramp input is

given by

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Static Acceleration Error Constant (Ka)

• The steady-state error of the system for parabolic

input is

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• The static acceleration error constant Ka is defined

by

• Thus, the steady-state error for parabolic input is

given by

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For N=0 there is no pole at origin and the system is

referred as TYPE 0 system.

If N=1 the system is referred as TYPE 1 system etc.

The highest degree of the characteristic polynomial gives the order of the system

TYPE and ORDER of a system

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• For a Type 0 system

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• For a Type 0 system

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• For a Type 1 system

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• For a Type 1 system

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• For a Type 2 system

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For a Type 2 system

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STEADY STATE ERROR

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1) For a closed loop system whose open loop transfer function

Find the steady state error when the input is

Solution

For

Steady state error Ess=

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For

=5

Steady state error Ess =

For

= 0

Steady state error Ess =

Total steady state error = 0 + 0.4 + =

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2)The block diagram of a unity feedback control system

with

inner output derivative feedback is shown below.

i)Calculate the steady state error for unit ramp input.

ii) choose the value of K such that the unit step response of

the system has no over shoot and yet it is as fast as

possible

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Solution

The forward transfer function with inner loop

Steady state error for unit ramp input 1/Kv

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Closed loop transfer function is

For no overshoot

For fast response

K=0.2

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Controllers

Control systems are designed to meet three time

response specifications

Steady state accuracy

Peak overshoot to step input (damping factor)

Settling time

It can be proved that to meet these specifications a

second order system needs to modified.

This modification is termed as compensation

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In the proportional control algorithm, the controller output is

proportional to the error signal, which is the difference between

the reference signal and the feedback signal

If the input error variable, e(t) the output of the controller p(t) is

p(t) = Kp*e(t)

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Therefore it increases the forward gain of the system

Hence it increases the natural undamped frequency

but decreases damping ratio

Therefore the

steady state accuracy improves, but the

transient response becomes more oscillatory

ξ

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Proportional plus Derivative Control

The controller output is equal to proportional plus

derivative of the error signal

Kd*de(t)/dt Kp*e(t) +

Taking Laplace transform of above equation gives

The effect of Increasing the coefficient of s term in

equation, which increases the damping of the system

Hence the transient performance improves

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The effect of Increasing the coefficient of s term in equation,

which increases the damping of the system

Hence the transient performance improves

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Ki*∫e(t)dt

Proportional plus Integral controller Control

The controller output is equal to proportional plus

integral of the error signal Kp*e(t) +

Taking Laplace transform of above equation gives

Integral error compensation increases the order of the system

.

If the system forward path has type-1 T.F , the Integral

compensator changes it to type 2 system.

Hence Improves the steady state accuracy

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PID Controller

In General the effects of dynamic performance caused by

PID controller is not obvious

If G(s) is second order system Introduction of controller

converts the characteristic equation to third order.

Certain values of Kp, Kd,Ki may cause Instability in the

system

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In General the effects of increasing parameters is:

Parameter: Rise Time Overshoot Settling Time S.S.Error

Kp Decrease Increase Small Change Decrease

Ki Decrease Increase Increase Eliminate

Kd Small Change Decrease Decrease None