time domain specifications of second order system

TIME DOMAIN SPECIFICATIONS OF SECOND ORDER SYSTEM

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1SYED HASAN SAEED

SYED HASAN SAEED 2

BOOKS

1. AUTOMATIC CONTROL SYSTEM KUO &GOLNARAGHI

2. CONTROL SYSTEM ANAND KUMAR

3. AUTOMATIC CONTROL SYSTEM S.HASAN SAEED

SYED HASAN SAEED 3

Consider a second order system with unit step input andall initial conditions are zero. The response is shown in fig.

1. DELAY TIME (td): The delay time is the time requiredfor the response to reach 50% of the final value infirst time.

2. RISE TIME (tr): It is time required for the response torise from 10% to 90% of its final value for over-damped systems and 0 to 100% for under-dampedsystems.

We know that:

SYED HASAN SAEED 4

21

2

2

1tan

1sin1

1)(

te

tc n

tn

Where,

Let response reaches 100% of desired value. Put c(t)=1

SYED HASAN SAEED 5

01sin1

1sin1

11

2

2

2

2

te

te

n

t

n

t

n

n

0 tneSince,

)sin())1sin((

0))1sin((

2

2

nt

t

n

n

Or,

Put n=1

SYED HASAN SAEED 6

2

2

1

)1(

n

r

rn

t

t

2

21

1

1tan

n

rt

Or,

Or,

3. PEAK TIME (tp): The peak time is the time requiredfor the response to reach the first peak of the timeresponse or first peak overshoot.

For maximum

SYED HASAN SAEED 7

te

tc n

tn

2

21sin

11)(Since

)1(01

1sin

11cos1

)(

0)(

2

2

22

2

tnn

nn

t

n

n

et

te

dt

tdc

dt

tdc

Since,

Equation can be written as

Equation (2) becomes

SYED HASAN SAEED 8

0 tne

sin1

1sin11cos

2

222

tt nn

Put cosand

cos1sinsin1cos 22 tt nn

cos

sin

))1cos((

))1sin((

2

2

t

t

n

n

SYED HASAN SAEED 9

nt

nt

pn

n

)1(

))1tan((

2

2

The time to various peak

Where n=1,2,3,…….

Peak time to first overshoot, put n=1

21

n

pt

First minimum (undershoot) occurs at n=2

2min

1

2

n

t

4. MAXIMUM OVERSHOOT (MP):

Maximum overshoot occur at peak time, t=tp

in above equation

SYED HASAN SAEED 10

te

tc n

tn

2

21sin

11)(

21

n

ptPut,

2

2

2

1

1.1sin

11)(

2

n

n

n

n

etc

SYED HASAN SAEED 11

2

2

1

2

2

1

2

1

11

1)(

1sin

sin1

1)(

)sin(1

1)(

2

2

2

etc

etc

etc

Put,

sin)sin(

SYED HASAN SAEED 12

2

2

2

1

1

1

11

1)(

1)(

eM

eM

tcM

etc

p

p

p

100*%21

eM p

5. SETTLING TIME (ts):

The settling time is defined as the time required for thetransient response to reach and stay within theprescribed percentage error.

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Time consumed in exponential decay up to 98% of the input. The settling time for a second order system is approximately four times the time constant ‘T’.

6. STEADY STATE ERROR (ess): It is difference betweenactual output and desired output as time ‘t’ tends toinfinity.

n

s Tt

44

)()( tctrLimitet

ss

EXAMPLE 1: The open loop transfer function of a servosystem with unity feedback is given by

Determine the damping ratio, undamped natural frequencyof oscillation. What is the percentage overshoot of theresponse to a unit step input.

SOLUTION: Given that

Characteristic equation

SYED HASAN SAEED 15

)5)(2(

10)(

sssG

1)(

)5)(2(

10)(

sH

sssG

0)()(1 sHsG

SYED HASAN SAEED 16

0207

0)5)(2(

101

2

ss

ss

Compare with 02 22 nnss We get

%92.1100*

7826.0

7472.4**2

sec/472.420

72

20

22 )7826.0(1

7826.0*

1

2

eeM

rad

p

n

n

n

%92.1

7826.0

sec/472.4

p

n

M

rad

EXAMPLE 2: A feedback system is described by thefollowing transfer function

The damping factor of the system is 0.8. determine theovershoot of the system and value of ‘K’.

SOLUTION: We know that

SYED HASAN SAEED 17

KssH

sssG

)(

164

12)(

2

016)164(

16)164(

16

)(

)(

)()(1

)(

)(

)(

2

2

sKs

sKssR

sC

sHsG

sG

sR

sC

is the characteristic eqn.

Compare with

SYED HASAN SAEED 18

K

ss

n

n

nn

1642

16

02

2

22

.sec/4radn

K1644*8.0*2 15.0K

%5.1

100*100*22 )8.0(1

8.0

1

p

p

M

eeM

EXAMPLE 3: The open loop transfer function of a unityfeedback control system is given by

By what factor the amplifier gain ‘K’ should be multiplied sothat the damping ratio is increased from 0.3 to 0.9.

SOLUTION:

SYED HASAN SAEED 19

)1()(

sTs

KsG

0

/

)(

)(

1.)1(

1

)1(

)()(1

)(

)(

)(

2

2

T

K

T

ss

T

K

T

ss

TK

sR

sC

sTs

K

sTs

K

sHsG

sG

sR

sC

Characteristic Eq.

Compare the characteristic eq. with

Given that:

SYED HASAN SAEED 20

02 22 nnss

T

K

T

n

n

2

12

We get

TT

K 12

T

Kn

KT2

1Or,

9.0

3.0

2

1

TK

TK

2

2

1

1

2

1

2

1

SYED HASAN SAEED 21

21

2

1

2

1

2

2

1

9

9

1

9.0

3.0

KK

K

K

K

K

Hence, the gain K1 at which 3.0 Should be multiplied

By 1/9 to increase the damping ratio from 0.3 to 0.9