Chapter 6

Lecture 8

Feedback Control Systems

President University Erwin Sitompul FCS 8/1

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com

2 0 1 3


Chapter 6

The Frequency-Response Design Method

Feedback Control Systems


Frequency Response A linear system’s response to sinusoidal inputs is called the

system’s frequency response. Frequency response can be obtained from knowledge of the

system’s pole and zero locations.

( )( ),

( )

Y sG s

U s

0( ) sin( ) 1( ),u t A t t

02 2

0

( ) .( )A

U s u ts

L

02 2

0

( ) ( ) .A

Y s G ss

Consider a system described by

where

With zero initial conditions, the output is given by

Chapter 6 The Frequency-Response Design Method


Frequency Response

*0 01 2

1 2 0 0

( ) .... ,n

n

Y ss p s p s p s j s j

Assuming all the poles of G(s) are distinct, a partial fraction expansion of the previous equation will result in

1 21 2 0 0( ) .... 2 sin( ), 0,np tp t p t

ny t e e e t t

01

0

Im( )tan .

Re( )

where

Transient response Steady-state response

1( ) ,

1G s

s

( ) sin(10 )u t t



( )( )

( )

Y sG s

U s

Frequency ResponseExamine

00( ) ,( )

s jG j G s

0 0 0( ) ( ) ( ),G j G j G j

2 2

0 0 0( ) Re ( ) Im ( )G j G j G j

010

0

Im ( )( ) tan

Re ( )

G jG j

G j

0( ) 0.U j A

0 0 0( ) ( ) ( ),Y j G j U j

,M

.

But the input is

Therefore

0( ) ,Y j AM 0( ) sin( ).y t AM t • Meaning?

Magnitude

Phase



Frequency Response A stable linear time-invariant system with transfer function

G(s), excited by a sinusoid with unit amplitude and frequency ω0, will, after the response has reached steady-state, exhibit a sinusoidal output with a magnitude M(ω0) and a phase Φ(ω0) at the frequency ω0.

The magnitude M is given by |G(jω)| and the phase Φ is given by G(jω), which are the magnitude and the angle of the complex quantity G(s) evaluated with s taking the values along the imaginary axis (s = jω).

The frequency response of a system consists of the frequency functions |G(jω)| and G(jω), which describe how a system will respond to a sinusoidal input of any frequency.



Frequency Response Frequency response analysis is interesting not only because it

will help us to understand how a system responds to a sinusoidal input, but also because evaluating G(s) with s taking on values along jω axis will prove to be very useful in determining the stability of a closed-loop system.

As we know, jω axis is the boundary between stability and instability. Therefore, evaluating G(jω) along the frequency band will provide information that allows us to determine closed-loop stability from the open-loop G(s).



Frequency Response

1( ) , 1.

1

TsD s K

Ts

Given the transfer function of a lead compensation

(a) Analytically determine its frequency response characteristics and discuss what you would expect from the result.

1( ) ,

1

TjD j K

Tj

2

1 1

2

1 ( )tan ( ) tan ( ).

1 ( )

TT TK

T

M = DΦ

• At low frequency, ω→0, M→K, Φ→0.• At high frequency, ω→∞, M→K/α, Φ→0.• At intermediate frequency, Φ > 0.



Frequency Response(b) Use MATLAB to plot D(jω) with K = 1, T = 1, and α = 0.1 for

0.1 ≤ ω ≤ 100 and verify the prediction from (a).

1( ) .

0.1 1

sD s

s

Using bode(num,den), in this case bode([1 1],[0.1 1]), MATLAB produces the frequency response of the lead compensation.



y t

nt

2

2 2( ) ,

2n

n n

G ss s

Frequency Response For second order system having the transfer function

we already plotted the step response for various values of ζ.

• The damping and rise time of a system can be determined from the transient-response curve.



2

1( ) ,

( ) 2 ( ) 1n n

G ss s

Frequency Response The corresponding frequency response of the system can be

found by replacing s = jω

This G(jω) can be plotted along the frequency axis, for various values of ζ.

2

1( ) .

( ) 2 ( ) 1n n

G jj j



Frequency Response

• The damping of a system can be determined from the peak in the magnitude of the frequency response curve.

• The rise time can be estimated from the bandwidth, which is approximately equal to ωn.

The transient-response curve and the frequency-response curve contain the same information.

?



Frequency ResponseBandwidth is defined as the maximum frequency at which the

output of system will track an input sinusoid in a satisfactory manner.

By convention, the bandwidth is the frequency at which the output is attenuated to a factor of 0.707 times the input.

The maximum value of the frequency-response magnitude is referred to as the resonant peak Mr.

10.707

2



Bode Plot TechniquesAdvantages of working with frequency response in terms of

Bode plots:1. Dynamic compensator design can be based entirely on

Bode plots.2. Bode plots can be determined experimentally.3. Bode plots of systems in series can be simply added,

which is quite convenient.4. The use of a logarithmic scale permits a much wider

range of frequencies to be displayed on a single plot compared with the use of linear scales.

Bode plot of a system is made of two curves, 1. The logarithm of magnitude vs. the logarithm of

frequency, log M vs. log ω, or also Mdb vs. log ω.2. The phase versus the logarithm of frequency

Φ vs. log ω.



Bode Plot Techniques

1 2

1 2

( )( ) ( )( ) .

( )( ) ( )m

n

s z s z s zKG s K

s p s p s p

For root locus design method, the open-loop transfer function is written in the form

For frequency-response design method, s is replaced with jω to write the transfer function in the Bode form

1 20

( 1)( 1) ( 1)( ) .

( 1)( 1) ( 1)m

a b n

j j jKG j K

j j j




12

1

( )( ) ,

( )

s zKG s K

s s p

To draw the Bode plot of a transfer function, it must be rewritten in magnitude equation and phase equation, i.e.,

10 2

( 1)( ) .

( ) ( 1)a

jKG j K

j j

0 12

( ) ( 1) ( ) ( 1),a

KG j K jj j

Then

0 1

2

( 1),( )

( 1)( ) a

K jKG j

jj

0 12

log log log ( 1)( )

log log ( 1) ,( ) a

K jKG j

jj

0 1db2

20log 20log ( 1)( )

20 log 20log ( 1) .( ) a

K jKG j

jj

Phase Equation

Magnitude Equation

(log)

(db)

Magnitude Equation

Magnitude Equation



Bode Plot TechniquesExamining all the transfer functions we have dealt with so far,

all of them are the combinations of the following four terms:

( )nj

12( ) 2 ( ) 1n nj j

1( 1)j

0K • Gain

• Pole or zero at the origin

• Simple pole or zero

• Quadratic poles or zeros

Once we understand how to plot each term, it will be easy to draw the composite plot, since log M and Φ are the additive combination of the magnitude logarithms and the phases of all terms.



Bode Plot Techniques0K

Magnitude

Phase

Gain



Bode Plot Techniques( )nj

Magnitude

Phase

90n

log log( )n n jj

Pole or zero at the origin



Bode Plot TechniquesAs example, Bode plot of a zero (jω) at origin will be as

follows:



Approximation Magnitude Phase

Bode Plot Techniques1( 1)j Simple pole or zero

1

1

1

1( 1)j

2( 1) 1j j

( 1)j

1 0

( 1) 45j

( ) 90j

( 1) 1j

( 1) 1j j

( 1)j j

• The point where ωτ = 1 or ω = 1/τ is called the break point.

( 1)j



Bode Plot TechniquesAs example, Bode magnitude plot of a simple zero (jωτ+1) is

given below, with τ = 10.• The break point lies at ω = 1/τ = 0.1.

db

10( 1)j

db

1

20log( 2) 3 db( 1)j

db

120log( )( 1)j

• Correction of Asymptote



Bode Plot Techniques The corresponding Bode phase plot of a simple zero (jωτ+1)

is given as:

1 1 0

1( 1) 45j

1( ) 90j

• Corrections of Asymptotes by 11°,at ω = 0.02 and ω = 0.5.

• Corresponds to 1/5ωbreak and 5ωbreak



Bode Plot TechniquesQuadratic poles or zeros

12( ) 2 ( ) 1n nj j

• Asymptotes can be used for rough sketch.• Afterwards, correction must be made

according to the value of damping factor ζ.



Bode Plot TechniquesChapter 6 The Frequency-Response Design Method


5

Bode Plot TechniquesChapter 6 The Frequency-Response Design Method


Bode Plot: ExamplePlot the Bode magnitude and phase for the system with the transfer function

2000( 0.5)( ) .

( 10)( 50)

sKG s

s s s

0.5

10 50

2000 0.5( 1)( )

10( 1) 50( 1)

j

j jKG jj

Convert the function to the Bode form,

0.5

10 50

2( 1).

( 1)( 1)

j

j jj

• ωb1 = 0.5, ωb2 = 10, and ωb3 = 50

• One pole at the origin

5 terms will be drawn separately and finally composited1 1 1

0.5 10 502, ( ) , ( 1), ( 1) , ( 1) .j j jj




40

60

20

0

–20

–40

db2

1( )j

0.5( 1)j

1

10( 1)j

1

50( 1)j

ωb1 = 0.5 ωb2 = 10 ωb3 = 50

: Rough composite

–20 db/dec

–20 db/dec

–40 db/dec

0 db/dec




40

60

20

0

–20

–40

db

ωb1 = 0.5 ωb2 = 10 ωb3 = 50

: Rough composite

: 3-db-corrected composite

+3 db–3 db

–3 db

Final Result




2

1( )j

0.5( 1)j

1

10( 1)j

1

50( 1)j

ωb1 = 0.5 ωb2 = 10 ωb3 = 50

0.1

2.52

5010

250

+11°

–11° –11°

+11° –11°

+11°



Bode Plot Techniquesωb1 = 0.5 ωb2 = 10 ωb3 = 50

Final Result



Homework 8No.1, FPE (6th Ed.), 6.4.

Hint: Draw the Bode plot in logarithmic and semi-logarithmic scale accordingly.

No.2.

(a) Derive the transfer function of the electrical system given above.

(b) If R1 = 10 kΩ , R2 = 5 kΩ and C = 0.1 μF, draw the Bode plot of the system.


Due: Thursday, 21 November 2013.

Chapter 6

Documents

systems frequency response

frequency responsefor

frequency responseexamine

low frequency

frequency band

high frequency

intermediate frequency

frequency functions