System Responses

7/27/2019 System Responses

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System Responses


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Two parts:

Steady-state Response Transient response

System Responses

u

bK

Mx

Transient response

Steady State Response

Input:

Output:


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First Order System

The first order system has only one pole.

Where K is the D.C gain and T is the time constantof the system.

Time constant is a measure of how quickly a 1storder system responds to a unit step input.

D.C Gain of the system is ratio between the inputsignal and the steady state value of output.

1TsK

sRsC)()(


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Introduction

The first order system given below.

1310

s

sG )(

5

3

ssG )( 151

53

s/

/

D.C gain is 10 and time constant is 3 seconds.

And for following system

D.C Gain of the system is 3/5 and time constant is 1/5

seconds.


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Impulse Response of 1st Order System

Consider the following 1st order system

1Ts

K)(sC)(sR

0t

(t)

1

1 )()( ssR

1Ts

KsC )(


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Re-arrange following equation as

1 Ts

KsC )(

TsTKsC/

/)(1

TteT

Ktc /)(

In order represent the response of the system in time domain

we need to compute inverse Laplace transform of the above

equation.

atCe

as

CL

1


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Tt

eT

K

tc

/)( IfK=3 and T=2s then

0 2 4 6 8 10

0

0.5

1

1.5

Time

c(t)

K/T*exp(-t/T)


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Step Response of 1st Order System


1Ts

K)(sC)(sR

ssUsR 1 )()(

1

Tss

KsC )(

1 Ts

KT

s

KsC )(

In order to find out the inverse Laplace of the above equation, we

need to break it into partial fraction expansion

Forced Response Natural Response


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Taking Inverse Laplace of above equation

1

1

TsT

sKsC )(

TtetuKtc /)()(

Where u(t)=1Tt

eKtc

/)( 1

KeKtc 63201 1 .)(

When t=T


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IfK=10 and T=1.5s thenTteKtc /)( 1

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6

7

8

9

10

11

Time

c(t)

K*(1-exp(-t/T))

Unit Step Input

Step Response

1

10

Input

outputstatesteadyKGainCD

.

%63


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IfK=10 and T=1, 3, 5, 7TteKtc /)( 1

0 5 10 150

1

2

3

4

5

6

78

9

10

11

Time

c(t)

K*(1-exp(-t/T))

T=3s

T=5s

T=7s

T=1s


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Step Response of 1st order System

System takes five time constants to reach its

final value.


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IfK=1, 3, 5, 10 and T=1TteKtc /)( 1

0 5 10 1

0

1

2

34

5

6

7

8

9

10

11

Time

c(t)

K*(1-exp(-t/T))

K=1

K=3

K=5

K=10


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Relation Between Step and impulse

response

The step response of the first order system is

Differentiating c(t) with respect to t yields

TtTt KeKeKtc //)( 1

Tt

KeKdt

d

dt

tdc /)(

TteT

K

dt

tdc /)(


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Ramp Response of 1st Order System


1Ts

K)(sC)(sR

2

1

ssR )(

12 Tss

KsC )(

The ramp response is given as

TtTeTtKtc /)(


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0 5 10 15

0

2

4

6

8

10

Time

c(t)

Unit Ramp Response

Unit Ramp

Ramp Response


IfK=1 and T=1Tt

TeTtKtc/

)(

error


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0 5 10 150

2

4

6

8

10

Time

c(t)

Unit Ramp Response

Unit Ramp

Ramp Response


IfK=1 and T=3Tt

TeTtKtc

/

)(

error


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Parabolic Response of 1st Order System


1Ts

K)(sC)(sR

3

1

ssR )(

13 Tss

KsC )(

Do it yourself

Therefore,


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1st Order System with a Zero

Zero of the system lie at -1/and pole at -1/T.

1

1

Ts

sK

sR

sC )(

)(

)(

11

Tss

sKsC

)()(

Step response of the system would be:

1Ts

TK

s

KsC

)()(

TteTT

KKtc /)()( Tt

eKtc

/

)(

1


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1st Order System with & W/O Zero

1

1

TssK

sRsC )()()(

TteT

T

KKtc /)()( TteKtc /)( 1

1TsK

sRsC)()(

IfT> the response will be same

Tt

enT

K

Ktc/

)()(

TteT

KnKtc /)( 1


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IfT> the response of the system would look like

0 5 10 156.5

7

7.5

8

8.5

9

9.5

10

Time

c(t)

Unit Step Response

13

2110

s

s

sR

sC )(

)(

)(

332

3

1010

/)()( tetc


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IfT


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1st Order System with a Zero

0 5 10 15

6

7

8

9

10

11

12

13

14

Time

U

nitStepResp

onse

Unit Step Response of 1st Order Systems with Zeros

T

T

t /


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0 2 4 6 8 100

2

4

6

8

10

12

14

Time

UnitStepResp

onse

Unit Step Response of 1st Order Systems with Zeros

T

T

1st Order System Without Zero


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Second Order Systems Have discussed the affect of location of poles and zeros on the

transient response of 1st order systems.

Compared to the simplicity of a first-order system, a second-ordersystem exhibits a wide range of responses that must be analyzedand described.

Varying a first-order system's parameter (T, K) simply changes thespeed and offset of the response

Whereas, changes in the parameters of a second-order system can

change theform ofthe response.

A second-order system can display characteristics much like a first-order system or, depending on component values, display dampedorpure oscillations for its transient response.


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Second Order Spring Mass Damper Example

ODE :

Transfer function :

k b

y(t)

F(t)=ku(t)

m


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General Form

A general second-order system is characterized

by the following transfer function.

22

2

2 nn

n

sssR

sC

)(

)(

un-damped natural frequency of the second order system,

which is the frequency of oscillation of the system withoutdamping.

n

damping ratioof the second order system, which is a measure

of the degree of resistance to change in the system output.


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Polar vs. Cartesian representations.Cartesian representation :

Imaginary part (frequency)

Real part (rate of decay)


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Polar representation :

damping ratio

natural frequency


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Polar representation :

damping ratio

natural frequency


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Polar vs. Cartesian representations.

Overdamped Critically damped

Underdamped

Undamped

Significance of the damping ratio :

System transfer function :


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Polar vs. Cartesian representations.System transfer function :



Underdamped

Undamped

P l C t i t ti


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Polar vs. Cartesian representations.

System transfer function :



Underdamped

Undamped

All 4 cases Unless overdamped


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Poles of Second Order System

Pole Locations22

21,2

2 0

( 1)

n n

n n

s s

p

Over damped

(two real distinct roots = two 1st

order systems with real poles)Critically damped

(a single pole of multiplicity two,

highly unlikely, requires exact

matching)

Underdamped

(complex conjugate pair of poles,

oscillatory behavior, most

common)

Real

Im.

n

n

1. If 0 <


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Pole position vs time response


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Pole variation vs time response

TRANSIENT

CHARACTERISITCS AREAFFECTED BY POLE

LOCATIONS


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Definitions of Transient Response Specifications

1. Delay time, : The delay time is the time required for the

response to reach half the final value.

2. Rise time, : The rise time is the time required for the response

to rise from 10% to 90%, 5% to 95%, or 0% to 100% of its final

value. For underdamped second-order systems, the 0% to 100%rise time is normally used. For overdamped systems, the 10% to

90% rise time is commonly used.

3. Peak time, : The peak time is the time required for the

response to reach the first peak of the overshoot.

dt

rt

pt


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Definitions of Transient Response Specifications

4. Maximum (percent) overshoot, : The 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.

It is defined by

Maximum percent overshoot =

5. 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%).

%100)(

)()(

c

ctc p

st

pM


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The system is assumed to be underdamped.

Rise time: Referring to Eq. (5-3), we obtain the rise time by letting

.

Then, we have

rdrd

t

r

ttetc rn

sin

1cos11)(

2

rt

1)( rtc

d

n

nrdt

2211

tan

d

d

drt

1tan

1

Rise Time


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Peak time: Referring to Eq. (5-3), we may obtain the peak time by

differentiating with respect to time and letting this derivative equal to

zero. Since

Then, we have

The last equation yields the following equation:

or

Since the peak time corresponds to the first peak overshoot, Hence

te

ttettedt

dc

d

tn

dd

dd

t

dd

t

n

n

nn

sin

1

cos1

sinsin1

cos

2

22

0sin1 2

pd

tn

tt

tedt

dcpn

p

0sin pdt

Peak Time

,3,2,,0 pdt

dpt


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Maximum overshoot: The maximum overshoot occurs at the peak timeor . It is possible to obtain

The maximum percent overshoot is .

)1/(

2

)/(

2

sin1

cos1)(

e

etcM dnpp

%100)1/( 2

e

Maximum Overshoot

dptt /


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Settling time: For convenience in comparing the responses ofsystems, the settling time is defined to be

or

)criterion%2(44

4n

s Tt

)criterion%5(33

3n

s Tt

Pair of envelope curves forhe unit-step response curve

Settling Time


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Max Overshoot vs curve

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