b_lecture12 frequency response and plotautomatic control system

Frequency Domain Analysis

Introduction

In practice, the performance of a control system is

more realistically measured by its time-domain

characteristics. The reason is that the performance of

most control systems is judged based on the time

responses due to certain test signals.

This is in contrast to the analysis and design of

communication systems for which the frequency

response is of more importance, since most of the

signals to be processed are either sinusoidal or

composed of sinusoidal components.

Introduction

We learned in time-domain that the time response of a control system is usually more difficult to determine analytically, especially for high-order systems.

In the frequency domain, there is a wealth of graphical methods available that are not limited to low-order systems.

There are correlating relations between the frequency-domain and the time-domain performances in a linear system, so the time-domain properties of the system can be predicted based on the frequency-domain characteristics.

Frequency Response

r

0

0

t

t

tAt 1sin

tAt 2sin

t

t0

0

c

)sin( 222 tAc

)sin( 111 tAc

tr tcsystem

1

1s

Ts

sinr ru A t 2 2r

r

AU s

s

2 21

1

rc r

AU s s U s

Ts s

1

2 2 2 2sin( arctan )

1 1

r rTc

A T Au t e t T

T T

2 2

lim sin( arctan )1

rc

t

Au t t T

T

2 2 ( )

1

rc

AA arctg T

T

Frequency Response

First-order system

Input

Transient response Steady state response

If the system is stable, then the transient response0 when t

The steady state response

22)(

s

AsR

n

i

ips

sm

sn

sms

1

)(

)(

)(

)()( tAtr sin)(

js

a

js

a

ps

k

ps

k

ps

k

s

A

ps

smsC

n

n

n

i

i

21

2

2

1

1

22

1

)(

)()(

R(s) C(s) )(sGeneral system

Where pi is assumed to be distinct poles (i=1,2,3n). Then, in partial fraction form, we have

Frequency Response

js

a

js

a

ps

k

ps

k

ps

ksC

n

n

21

2

2

1

1)(

Taking the inverse Laplace transform yields

For the stable system all poles (pi) have negative real parts, then the transient response

the steady state response:

0)(1

n

i

tp

itiektc

js

a

js

aLtcs

211)(

js

a

js

aLekekektc

tp

n

tptp n 211

2121)(

Transient response )(tct Steady state response )(tcs

Frequency Response

)90)((

12

)(

2)( )(

))(()(

ojj

js

ejA

j

Ajjs

jsjs

Asa

)90)((*

122

)( ojjejA

aa

js

a

js

aLtcs

211)(

))(sin()(

2)(

)(

)90)(()90)((

21

jtjA

eejA

eaeatc

oo jtjjtj

tjtj

s

Frequency Response

sincos

formula

je

Euler

j

j

ee

ee

jj

jj

2sin

2cos

The amplitude ratio of the steady-state output cs(t) versus sinusoid input r(t):

istic charactermagnitude jR

jC(j

A

jA

)(

)()

)(

The phase difference between the steady-state output and sinusoid input:

acteristicphase charjRjCjtjt )()( )()]([

Then we have :

jss

jR

jCj

)(

)(

)()(

Compare with the sinusoid input tAtr sin)( , we have:

)sin()()( 211

tjA

js

a

js

aLtcs

where )( j

Frequency Response

Definition : frequency response (or characteristic) the ratio of the complex vector of the steady-state output versus sinusoid

input for a linear system, that is:

Here: input sinusoid theoftion representactor complex ve the)( jR

output the of tionrepresentavector complex the )( jC

stic)characterir response(o frequency )( j

jss

jR

jCj

)(

)(

)()(

)] (sin[)()( output state-steady the

)sin()(

jtAjtc

tAtrwhen

rrs

rr

Frequency Response

Thus the steady-state response depends only on the magnitude

and phase of , at a specific frequency . )( j

A unity feedback control system, the open-loop transfer function:

ssG

5.0

1)(

Determine the steady-state response of the system.

)6028.6sin(10)( : ottrIf

Solution:

The closed-loop transfer function : 15.0

1

5.011

5.01

)(1

)(

)(

)()(

ss

s

sG

sG

sR

sCs

5.0 10 28.6 : TAknown r

2 2

1( ) ( )

1arctg T

T

2

0

1 6.28 ( ) 0.3

(0.5 6.28) 1

( ) (0.5 6.28) 72.4

when

arctg

)12.4-3sin(6.28t

)4.726028.6sin(103.0)] (sin[)()(

0

00

tjtAjtc rrs

Example

Graphic expression of the frequency response

1. Rectangular coordinates plot

2. Polar plot (Nyquist curve)

3. Bode diagram(logarithmic plot)

Frequency Response Plot

1. Rectangular coordinates plot

Example )2()2(1

10

12

10)(

12

10)(

1

2

tgj

jGs

sG

o

o

o

o

o

o

o

. .

jGjG

29849950 5

875.8224.14

538.8064.13

964.754.22

435.6347.41

4507.75.0

0100

)()(

-90o

0.5 1 2 3 4 50

)( jG

1

5

10

)( jG


2. Polar plot (Nyquist curve )

The magnitude and phase response:

Calculate and for different . )(G

The polar plot is easily useful for investigating system stability.


)}(Im{)}(Re{)()( GjGsGjG js

)()()()( )( jGjGejGjG jj

It is done in polar coordinates as varies from 0 to .

)( jGr )( jG

)(G

0

1

10

)( 1jG

Im

0

or

3. Bode diagram(logarithmic plot) Plot the frequency characteristic in a semilog coordinate:

Magnitude response Y-coordinate in decibels: )(log20 10 jG

X-coordinate in logarithm of : 10log

Phase response Y-coordinate in radian or degree: )( jG

X-coordinate in logarithm of : 10log


Note:

The logarithm of the magnitude is normally expressed in terms

of the logarithm to the base 10, multiplying by 20,where the units are

decibels (dB).

A unit change in in the rectangular coordinates is

equivalent to one decades of variation in ,that is from 1 to 10, 10 to 100, ,and so on.

)(log20 10 jG

10log 10log


Note:

The logarithm of the magnitude is normally expressed in terms

of the logarithm to the base 10, multiplying by 20,where the units are

decibels (dB).

A unit change in in the rectangular coordinates is

equivalent to one decades of variation in ,that is from 1 to 10, 10 to 100, ,and so on.

)(log20 10 jG

10log 10log

)dB( )(L

)/( srad

20

40

0 2 4 6 10 1 8 20 40 60

60

2

2

3

Bode Plot

)(lg20)( jGL

)()( jG

80 100

1. Inertial element

Frequency Response of The Typical Elements

1

1

TssG

1

1

TjjG

)(1

1

22

TarctgGT

Gjs

bandwidth- 3

3

1

b

sb

s

b t

TtT

TG

G 1

4

707.0

Characteristic point

1

707.0

0

4

2

b

T1

G

G

Rectangular coordinates plot

Polar plot (Nyquist curve )


1. Inertial element

0

T/1

1 045

Im

0

GjtgTj eGeT

jG

arg

22 1

1

1

1

TssG

1

1

TjjG

4db 3707.0lg20 20lg

, 1

arctgTG

G

T

arctgTG

TG

TG

1lg20lg20

1

1

22

22


1. Inertial element

Bode diagram(logarithmic plot)

T/1 Break frequency or Corner frequency

1

1

TssG

1

1

TjjG

Characteristic point

Asymptotic plot 1lg20lg2022 TG

1 with compare when neglected T)( because 0lg20 1 1 2 isGTT

.a

.b lg20lg20lg20 1 1 22 TTGTT

dBnTT

dBTT

dBTT

dBTT

nn 2010lg20lg2010

4010lg20lg2010

2010lg20lg2010

01lg20lg201

22

Blue curve: exact plot

Red curve: asymptotic plot(corner plot)


1. Inertial element

[-20]

22

2

2 nn

n

sssG

22

2

2 nn

n

jjjG

2

22

21

1

nn

jG

2

1

2

n

narctgjG


2. Oscillating element

eakresonant pG

requencyresonant f

Gd

jGd

m

m

m

nm

12

1

21

02

2

2

2

1

G

G

n

bandwidth-

707.0|2

1

707.0

b

707.0

nb

nG


Rectangular coordinates plot

Polar plot


[-40]

22

2

21lg20lg20

nn

G

2

1

2

n

narctgG


Bode diagram


dB0||lg20 1or Gn

n

)lg(40)lg(20lg20 1or 2

nnn

n |G|

n Break frequency or Corner frequency

Asymptotic plot


Transfer function: KsG )(

Frequency response:

ojG

KjGLKjGKjG

0)()(

lg20)(lg20)()()(

3. Proportional element

Re

Im

K

0dB, 0o

100 10 1 0.1 )(lg

)( ),( L

dB log20)( KL o0)(

Polar plot Bode diagram

0

)(L

4. Integrating element

Transfer function: s

sG1

)(

Frequency response:

ojG

jGLjG

jjG

90)()(

lg20)(lg20)(1

)(1)(

Polar plot

0

Bode diagram

decdB /20

o90)(


Re

Im

0dB, 0o

10 1 0.1 0.01 )(lg

)( ),( L

0

5. Differentiating element

Transfer function

aldifferentiordersecondss

aldifferentiorderfirsts

aldifferentis

sG

nn 1)/(2)/(

1)(2

Polar plot

Re

Im

Re

Im

1

Re

Im

1

differential first-order differential second-order differential


Because of the transfer functions of the differentiating elements are

the reciprocal of the transfer functions of Integrating element, Inertial

element and Oscillating element respectively

]1)/(2)//[(11)/(2)/(

111

1

22

nn

inverse

nn

inverse

inverse

ssss

Tss

ss

Bode curves of the differentiating elements are symmetrical to the lg-axis with the Bode curves of the Integrating element, Inertial element and

Oscillating element respectively.

Then we have the Bode diagrams of the differentiating elements:


5. Differentiating element

0dB, 0o

100 10 1 0.1

)(lg

)( ),( L

decdB /20 o90)(

differential

)( ),( L

1th-order differential

0dB, 0o

100 10 1 0.1 )(lg

decdB /20o45

o90


0dB, 0o

100 10 1 0.1 )(lg

)( ),( L

decdB /40

o180

o90

n

2th-order differential

6. Delay element

Transfer function: sesG )(

)()(

0)(lg20)(1)()(

jG

jGLjGejG j

Polar plot

Re

Im R=1

0dB, 0o 0

)(lg

)(L

Bode diagram

)(


0

)(L

b_lecture12 frequency response and plotautomatic control system

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