Chapter 09 Frequency Response Method[1]

9/17/2001

1

EEE 352 Automatic Control Systems

Prof. Dr. Ahmet Uçar

© Dr. Ahmet Uçar EEE 352 Chapter 9 1

G(j) R(j) Y(j)

H(j)

E (j)

Chapter 09: Frequency Response Analysis


)()()( tytyty sst

By the term frequency response, we mean the steady-state response of a system to asinusoidal input r(t)=Xsint where “” is measured in rad/sec.

)(tyss

the transient response

(t0 t tf).

the steady state response

(tf < t ).

)(ty

0

)(tyt

t0 ts=tf t

To obtain and study frequency-response of the system G(j), the input signalfrequency is varied over a certain range.The information obtained from such analysis is different from root-locus analysis.In fact, the frequency response and root-locus approaches complement each other.The frequency-response methods are most powerful in conventional (classical)control theory.

Frequency Response Analysis

G(s)R(s) Y(s)

y(t)r(t) = Xsint

)()(

)(,),(

)(

)(

jG

jR

jYjssG

sR

sY

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2


G(s)R(s) Y(s)

H(s)

E(s) )()()( tytyty sst

)(tyss


(t0 t tf).


(tf < t ).

)(ty

0

)(tyt

t0 ts=tf t

Frequency-response methods were developed in 1930s and 1940s by Nyquist,Bode, Nichols, and many others.The Nyquist stability criterion enables us to investigate both the absolute andrelative stabilities of linear closed-loop systems from a knowledge of their open-loop frequency response characteristics, G(j)H(j).The introduction of logarithmic plots, often called Bode plots, simplifies thedetermination of the graphical portrayal of the frequency response.


)()(1

)()(

)(

)(

,),()(

)(

jHjG

jGjT

jR

jY

jssTsR

sY


)()()( tytyty sst

The logarithmic plots are called Bode plots in honor of H. W. Bode, who usedthem extensively in his studies of feedback amplifiers.

)(tyss


(t0 t tf).


(tf < t ).

)(ty

0

)(tyt

t0 ts=tf t


)()(1

)()(

)(

)(

,),()(

)(

jHjG

jGjT

jR

jY

jssTsR

sY

G(s)R(s) Y(s)

H(s)

E(s)

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The steady state outputs to sinusoidal input:The steady-state response of system G(s) can be obtained directly from thesinusoidal input (r(t)=Xsint) transfer function G(j ), that is, the transferfunction in which s=j is replaced by j, where is frequency.

G(s)R(s) Y(s)

)()()(

)(

jGsGsR

sY

js

)(

)()()(

jG

jGjGjG

s=j

j

0

s=+j plane


1

jImG(j)=jX()

0

G(j) planePolar Plane

1

)( 1jG

ReG(j)=jR()

)( 1jG = 0


The steady state outputs to sinusoidal Input:Suppose that the transfer function G(s)

))...()((

)(

)(

)()(

)(

)(

21 nssssss

sZ

sP

sZsG

sR

sY

G(s)

R(s) Y(s)

polessystem

n

n

Input

ss

b

ss

b

ss

b

js

a

js

a

s

XsGsRsGsY

2

2

1

122

)()()()(

The steady-state response of a stable, linear, time-invariant system to a sinusoidalinput does not depend on the initial conditions.

where a and the b, (where i = 1,2, . . . , n) are constants and ā is the complexconjugate of a. The inverse Laplace transform of system output Y(s) gives

polessystem

tsn

tsts

Input

jj nebebebeaaety 21

21)(

For a stable system, the terms contain e-st approach zero as t approaches infinity.


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The steady state outputs to sinusoidal Input:For sinusoidal input, r(t)=Xsint, the output of transfer function G(j ) is yss(t)

G(s)R(s) Y(s)

polessystemty

tsn

tsts

inputty

jj

t

n

ss

ebebebeaaety

;0)(

21

);(

21)(

jjss eaaety )(

Thus, regardless of whether the system is of the distinct-pole type, the steady state response becomes

where the constant a and ā are

j

jXGjs

s

XsGa

j

jXGjs

s

XsGa

jsjs 2

)()()(,

2

)()()(

2222

Since G(j) is a complex quantity, it can be written in the following form:

jejGjG )()(

where |G(j)| represents the magnitude and represents the angle of G(j); that is,

)(Re

)(Imtan)( 1

jG

jGjG The angle may be negative, positive, or zero.



The steady state outputs to sinusoidal Input:Similarly, we obtain the following expression for G(-j):

Then,

j

j

ejG

ejGjG

)(

)()(

j

ejGXa

j

ejGXa

jj

2

|)(|,

2

|)(|

The steady state response is

)sin()sin(|)(|

2|)(|)(

)()(

tYtjGX

j

eejGXeaaety

tjtjjj

ss

G(s)R(s) Y(s)

y(t)r(t) = Xsint

ts t

x(t)

yss(t)

The steady state response

Results: A stable, linear, time-invariant system subjected to a sinusoidal input will,at steady state, have a sinusoidal output of the same frequency as the input. But theamplitude and phase of the output will, in general, be different from those of theinput.

)()(

)(

jG

jR

jY

The function G(j) is called the sinusoidal transfer function.


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Example 9.1: Consider the following RC system where the transfer function G(s)is

For sinusoidal input, r(t)=Xsint, the steady state response of the system yss(t) is obtained by letting s= j;

G(s)R(s) Y(s)

y(t)r(t) = Xsint

)(tan1)(

1

|)(|1

1)(

)(

)(

1

2

jGj

jGjR

jY

RCs

sGsR

sY

,1

1)(

)(

)(

))(tansin(1)(

)sin(|)(|)( 1

2

t

XtjGXtyss


r(t) vc=y(t)R

CI

s=j

j

0

s=+j plane

1

jImG(j)=jX()

0


1

)( 1jG

ReG(j)=jR()

)( 1jG = 0


Example 9.1: The steady state response of an RC circuits, r(t)=Xsint:

0

1

111

452

1

)1(tan2

1

1,|)(|

)(

)(

1

RC

jGjR

jY

RC


j

0

s=+j plane

)45sin(2

)sin(|)(|)( 0111 t

XtjGXtyss

jImG(j)

0


1=1/RC

2

1)( 1 jG

ReG(j)

45 = 0

1

X, Mag. (0)

0.01 1 -0.5729

0.1 0.995 -5.7106

1 0.707 -45.0000

2 0.4472 -63.4349

3 0.3162 -71.5651

4 0.2425 -75.9638

5 0.1961 -78.6901

6 0.1644 -80.5377

7 0.1414 -81.8699

8 0.124 -82.8750

9 0.11 -83.6598

10 0.099 -84.2894

100 0.01 -89.4271

RC

1

11

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Example 9.1: The steady state response of an RC circuits, r(t)=Xsint:

0

1

111

452

1

)1(tan2

1

1,|)(|

)(

)(

1

RC

jGjR

jY

RC

From yss(t), it can be seen that for 0, the amplitude of the steady-stateoutput yss(t) is almost equal to X. The phase shift of the output is small for small.The magnitude of output is X|G(j1)|=X/2 and phase (1)=450 at=1=1/=1/RC.For , the amplitude of the output is small and almost inverselyproportional to . The phase shift approaches -900 and yss(t)=0 as . Thisis a phase-lag network.


j

0s=+j plane

)45sin(2

)sin(|)(|)( 0111 t

XtjGXtyss

jImG(j)

0


1=1/RC

2

1)( 1 jG

ReG(j)

45 = 0

1


Example 9.1: The steady state response of an RC circuits:

0

1

111

452

1

)1(tan2

1

1,|)(|

)(

)(

1

RC

jGjR

jY

RC

Results: The magnitude of the transfer function |G(j)| is almost equal to onefor the range of frequency of 0 1=1/. However it rapidly degreased tozero for 1=1/ < ∞. Thus the frequency 1=1/ is called corner frequencyfort the simple poles/zeros.


j

0s=+j plane

jImG(j)

0


1=1/RC

2

1)( 1 jG

ReG(j)

45 = 0

1

G(s)R(s) Y(s)

y(t)r(t) = Xsint

r(t) vc=y(t)R

CI

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Presenting Frequency-Response Characteristics in Graphical Forms:The transfer function G(j) is characterized by its magnitude and phase angle,with frequency as the parameter.

)(

)()()()(

)(

jG

jGjGjGjR

jY

G(j)R(j) Y(j)

j

0s=+j plane

0

There are three common methods are used study the frequency response of thesystems:1. Bode diagram or logarithmic plot2. Nyquist diagram or polar plot3. Log-magnitude-versus-phase plot (Nichols plots)In this lecture Bode diagram of feedback systems will be subject t be studied.



A Bode diagram consists of two graphs:One is a plot of the logarithm of the magnitude in decibels (dB);

)()()()(

)(

)(jGjGjGjG

jR

jY

G(j)R(j) Y(j)

Bode Diagram: Logarithmic Plot

G

(j

)(d

eg)

-1800

|G(j

)(d

B)

0

10-2 10-1 100 101 102 103

10-2 10-1 100 101 102 103

dBjGa ,)(log20) 10

deg),() jGb

Both are plotted against the frequency on a logarithmic scale where the base ofthe logarithm is 10; log10

Bode Diagrams or Logarithmic Plots:

the other is a plot of thephase angle in degrees(deg);

log10

log10

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Bode Diagram: Logarithmic PlotExample 9.2: Bode diagram of an RC circuits:

r(t) vc=y(t)R

CI

G

(j

)(d

eg)

|G(j

)(d

B)

110

010

10

210

110

1

10

210

20100

-10-20-30-40

00

300

600

900

1200

1500

1800

a) Mag. (dB) b) Angle (0)

0.01 -0.0004 -0.5729

0.1 -0.0432 -5.7106

1 -3.0103 -45.0000

2 -6.9897 -63.4349

4 -12.3045 -75.9638

5 -14.1497 -78.6901

6 -15.6820 -80.5377

8 -18.1291 -82.8750

10 -20.0432 -84.2894

100 -40.0004 -89.4271

1/

deg),(tan)()

,1)(

1log20)(log20)

1

1)(

)(

)(

1

21010

jGb

dBjGa

jjG

jR

jY


Bode Diagram: Logarithmic PlotExample 9.2:

|G(j

)(d

B)

110

1

10

210

20100

-10-20-30-40

For small frequencies - that is, <<1/ - the logarithmic gain is

/1,,01)(log2001)(log201log201)(

1log20 2

102

1010210

dB

For large frequencies - that is, >>1/ - the logarithmic gain is

/1,),(log201)(log200)(log20 102

1010 dBjG

and at =1/ , the logarithmic gain is

/1,01.32log20)(log20 1010 dBjG

The frequency =1/ is often called the break frequency or corner frequency.

deg),(tan)()

,1)(

1log20)(log20)

1

1)(

)(

)(

1

21010

jGb

dBjGa

jjG

jR

jY

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Bode Diagram: Logarithmic PlotExample 9.2:

Results: An interval of two frequencies with a ratio equal to 10 is called a decade, so that the range of frequencies from 1 to 2, where 2 = 101, is called a decade. The magnitude curve has two asymptotes one has 0 dB/decade gain for <<1/ the other has -20 dB/decade gain for >> 1/.

|G(j

)(d

B)

110

1

10

210

20100

-10-20-30-40

a decade

0 dB/dec-20 dB/dec

break frequency or corner frequency


Bode Diagram of Feedback Control Systems

)()(1)(, jHjGjPjs

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

)(

)(

)(sHsGsP

sHsG

sG

sR

sY

G(s)

R(s) Y(s)

H(s)

deg180)()()

1)()()

1)()(0)(

0

jHjGb

jHjGa

jHjGjP

Bode diagram of feedback systems is obtained from the knowledge of their looptransfer (open-loop) function frequency response characteristics, G(j)H(j) toenables us to investigate both their stability and frequency performances.

Frequency response characteristics of the characteristic equation;

A Bode diagram gives the frequency performance of the closed loop system bystudying the loop transfer function G(j)H(j) only.

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A Bode diagram consists of two graphs:a) Magnitudes graph; A plot of the logarithm of the magnitude of a sinusoidalloop transfer function G(j)H(j) in the decibel, usually abbreviated dB;20log10|G(j)H(j)| = 0 (dB)b) Angle graph; A plot of the phase angle;G(j)H(j) = -1800

dB01log20 10

deg180)()()

0)()(log20)

0

10

jHjGb

dBjHjGa


)()(1

)(

)(

)(

jHjG

jG

jR

jY

G(s)R(s) Y(s)

H(s)

deg180)()(

1)()(1)()(0)(

0

jHjG

jHjGjHjGjP


deg180)()()

0)()(log20)

0

10

jHjGb

dBjHjGa


G

(j

)H(j

) (deg

)

-1800

|G(j

)H(j

)| (dB

)

0

10-2 10-1 100 101 102 103

10-2 10-1 100 101 102 103

The magnitude and the phase diagram are plotted against the frequency on alogarithmic scale where the base of the logarithm is 10; log10

dBjHjGa 0)()(log20) 10

deg180)()() 0 jHjGb

)()(1

)(

)(

)(

jHjG

jG

jR

jY

G(s)R(s) Y(s)

H(s)

1)()( jHjG

0

)()(1)(

jHjGjP

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Basic Factors of G(j)H(j). The basic factors that very frequently occur in anarbitrary loop transfer function G(j)H(j) are:1. Gain K2. Integral and derivative factors (j) 1

3. First-order factors (1+jT) 1

4. Quadratic factors (1+ j 2 (j/n) + (j/n)2 ) 1

Once the logarithmic plots of these basic factors are well understood, it ispossible to utilize them in constructing a composite logarithmic plot for anygeneral form of loop transfer function G(j)H(j) by sketching the curves foreach factor and adding individual curves graphically, because adding thelogarithms of the gains corresponds to multiplying them together.

Plotting Bode Diagram

G(s)R(s) Y(s)

H(s)

deg180)()()

0)()(log20)

1)()(

0)(

)()(1)(

0

10

jHjGb

dBjHjGa

jHjG

jP

jHjGjP

)()(1

)(

)(

)(

jHjG

jG

jR

jY


1) Gain G(j)H(j) = K, where K > 0

dBKjGjG 1010 log20|)()(|log20

K Mag. (dB)

0.01 -40

0.1 -20

1 0

2 6.02

4 12.04

5 13.97

6 15.56

8 18.06

10 20

100 40

dBnKK n 20log20)10(log20 1010

00)()( jHjG

nKjHjG 10)()(

Magnitudes graph;

Angle graph;


deg180)()()

0)()(log20)

1)()(

0

10

jHjGb

dBjHjGa

jHjGG(s)

R(s) Y(s)

H(s)

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Magnitudes graph;

Angle graph;

KsHsG )()( 00)()( KjHjG

00)()( jHjG

10-1 100 101 102

40

20

0

-20

-40Magnitu

de (

dB

)

(rad/sec)


G(s)R(s) Y(s)

H(s)

1) Gain G(j)H(j) = K, where K > 0

K Mag. (dB)

0.01 -40

0.1 -20

1 0

2 6.02

4 12.04

5 13.97

6 15.56

8 18.06

10 20

100 40

octave

decade

K=10

K=0

K=0.1

K=0.01

0

-45

-90

-135

-180

KjHjG )()(

(rad/sec)

Phase

(deg) octave

decade

10-1 100 101 102


2. Integral factors G(j)H(j) = (j) -1

Magnitudes graph;

Angle graph;

ssHsG

1)()( 0

290

11)()(

j

jjHjG

,log201log201

log201

log20)()(log20 1010101010

j

jHjG dB01log20 10

1010 log20)()(log20 jHjG

decadedB /20)log20(log

10

10

The slope of the line is

090)()( jHjG


deg180)()()

0)()(log20)

1)()(

0

10

jHjGb

dBjHjGa

jHjGG(s)

R(s) Y(s)

H(s)

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Magnitudes graph;

Angle graph;

1010 log20)()(log20 jHjG

090)()( jHjG

Mag. (dB) Angle (deg)

0.01 40 -900

0.1 20 “

1 0 “

2 -6.02 “

4 -12.04 “

5 -13.97 “

6 -15.56 “

8 -18.06 “

10 -20 “

100 -40 “

10-2

10-1

100

101

102

-40

-20

0

20

40

decade

octave

Magnitu

de (

dB

)

(rad/sec)


2. Integral factors G(j)H(j) = (j) -1


10

10

-20 dB/dec

10-2

10-1

100

101

102

-180

-135

-90

-45

0

decade

octave

(rad/sec)

Phase

(deg)


2. n Integral factors (j)-n

Magnitudes graph;

Angle graph;

0901

)()(

jHjG


nssHsG

1)()(

dBnnjHjG 10101010 log20log201log20)()(log20

10-1

100

101

-60

-40

-20

0

20

40

60

Magnitude(d

B)

Frequency (rad/sec)

n=3

n=2

n=1

decade

octave

decadedBnjHjG

/20log

|))()((|

10

090)()( njHjG

The phase angle of (j)-n is equal to -n900 over the entire frequency range.

The magnitude curves will pass through the point(0 dB, = 1).


G(s)R(s) Y(s)

H(s)

-60 dB/dec

-40 dB/dec

-20 dB/dec

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2. Derivative factors (j)n

Magnitudes graph;

Angle graph;

090)()( jHjG


ssHsGn )()(;1

101010 log20log20)()(log20 jjHjG


10

10

10-2

10-1

100

101

102

-40

-20

0

20

40

Magnitude(d

B)

Frequency (rad/sec)

decade

octave

090)()(;1 jHjGn

Mag (dB)

0.01 -40

0.1 -20

1 0

2 6.02

4 12.04

5 13.97

6 15.56

8 18.06

10 20

100 40 The phase angle of (j) is equal to 900 over the entire frequency range.


G(s)R(s) Y(s)

H(s)

20 dB/dec


Magnitudes graph;

Angle graph;

090)()( nnjHjG


nssHsGn )()(;

dBnjHjG 1010 log20)()(log20

10-1

100

101

-60

-40

-20

0

20

40

60

Magnitude(d

B)

Frequency (rad/sec)

n=3

n=2

n=1

decade

octave

decadedBnjHjG

/20log

|))()((|

10

090)()( njHjG

The phase angle of (j)n is equal to n900 over the entire frequency range.

The magnitude curves will pass through the point(0 dB, = 1).

2. Derivative factors (j)n


G(s)R(s) Y(s)

H(s)

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G(s)R(s) Y(s)

H(s)Figure 1:

3. First-order factors (1+j)-n , Simple pole with negative real axes. (Phase Lag)

Magnitudes graph;

For low frequencies, such that << 1/ (corner frequency), the log magnitude may be approximated by

1

1)()(,1,1

ssHsGn

)(tan1)(

1()()()()()( 1

2

jHjGjHjGjHjG

dBjHjG 2101010 1log201log20)()(log20

dBdBjHjG 01log20)(1log20)()( 210

For high frequencies, such that >> 1/ (corner frequency),

dBdBjHjG )log(20)(1log20)()( 210

The slope is 0 dB/decade

The slope is -20 dB/decade

Results: At = 1/ , the log magnitude equals 0 dB; at = 10/, the logmagnitude is -20 dB. Thus, the value of -20 log dB decreases by 20 dB forevery decade of . For >>1/, the log-magnitude curve is thus a straight linewith a slope of -20 dB/decade (or -6 dB/octave).



Magnitudes graph;

Angle graph;

3. First-order factors (1+j)-1

1

1)()(,1,1

jjHjGn


)(tan)()( 1 jHjG

Mag. (dB) Angle (0)

0.01 -0.0004 -0.5729

0.1 -0.0432 -5.7106

1 -3.0103 -45.0000

2 -6.9897 -63.4349

4 -12.3045 -75.9638

5 -14.1497 -78.6901

6 -15.6820 -80.5377

8 -18.1291 -82.8750

10 -20.0432 -84.2894

100 -40.0004 -89.4271

1/

10-1

100

101

-20

-15

-10

-5

0

Magnitude(d

B)

10-1

100

101

-80

-60

-40

-20

0

Pha

se(d

eg)

Frequency (rad/sec)

<<1/ >>1/

-20 dB/decade

1/0 dB/dacade

The maximum error occurs at the corner frequency 1/ and is approximately equal to -3 dB.


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Magnitudes graph;

Angle graph;

3. First-order factors (1+j)-1

1

1)()(,1

jjHjGn

dBjHjG 2101010 )(1log201log20)()(log20

)(tan)()( 1 jHjG



G(s)R(s) Y(s)

H(s)Figure 1:

3. First-order factors (1+j)n , Simple pole with negative real axes. (Phase Lead)

Magnitudes graph;

For low frequencies, such that << 1/ (corner frequency), the log magnitude may be approximated by

1)()(,1,1 jjHjGn

dBdBjHjG 01log20)(1log20)()(log20 21010

For high frequencies, such that >> 1/ (corner frequency),

dBdBjHjG )log(20)(1log20)()(log20 21010



Results: At = 1/ , the log magnitude equals 0 dB; at = 10/ , the logmagnitude is 20 dB. Thus, the value of 20 log dB increases by 20 dB forevery decade of . For >>1/, the log-magnitude curve is thus a straight linewith a slope of 20 dB/decade (or 6 dB/octave).

)(tan1)(

)()()()()()(

12

jHjGjHjGjHjG

21010 1log20)()(log20 jHjG


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Magnitudes graph;

Angle graph;

3. First-order factors (1+j)n


)(tan)()( 1 jHjG

1/

Mag. (dB) Angle (0)

0.01 0.0004 0.5729

0.1 0.0432 5.7106

1 3.0103 45.0000

2 6.9897 63.4349

4 12.3045 75.9638

5 14.1497 78.6901

6 15.6820 80.5377

8 18.1291 82.8750

10 20.0432 84.2894

100 40.0004 89.4271

10-1

100

101

0

10

20

Magnitude(d

B)

10-1

100

101

0

20

40

60

80

Pha

se(d

eg)

Frequency (rad/sec)

20 dB/dacade

>>1/ <<1/1/

0 dB/dacade

dBjHjG 311log20)()( 10

1)()(,1,1 jjHjGn



G(s)R(s) Y(s)

H(s)Figure 1:

4. Quadratic factors (1+ j 2 (j/n) + (j/n)2 ) -1, Complex conjugate poles with

negative real axes. (Phase Lag)

Magnitudes graph;

For high frequencies, such that >> (corner frequency),


22

2

2)()(

nn

n

sssHsG

1)(

2)(

1

)(2)()()(

2

222

2

nn

nn

n

jjjjjHjG

0

001)()(lim

jHjG 01800)()(lim

jHjGand

22

2

2

1010 )2()1(log201log20)()(nn

jHjG

For low frequencies, such that << n(natural frequency), the log magnitude may be approximated by

01log20)()( 10 jHjGn

nn

n jHjG

102

2

10 log40log20)()( The slope is -40 dB/decade


9/17/2001

18


4. Quadratic factors (1+ j 2 (j/n) + (j/n)2 )-1, (Phase Lag)

Magnitudes graph;

For >> n(natural frequency), asymptote is a straight line having the slop -40 dB/decade

22

2

2

1010 )2()1(log201log20)()(nn

jHjG

For << n(natural frequency), the low-frequency asymptote is thus a horizontal line at 0 dB.

01log20)()( 10 jHjGn

nn

n jHjG

102

2

10 log40log20)()(

The high-frequency asymptote intersects the low-frequency one at = n (natural frequency), since at this frequency

dBjHjGn

n 01log40log40)()( 1010

1)(

2)(

1)()(

2

2

nn

jjjHjG

0

001)()(lim

jHjG 01800)()(lim

jHjGand

G(s)R(s) Y(s)

H(s)Figure 1:

Angle graph;

)1/(2tan)()(

2

21

nn

jHjG



4. Quadratic factors (1+ j 2 (j/n) + (j/n)2 ) -1, (Phase Lag)

Magnitudes graph; 22

2

2

1010 )2()1(log20)()(log20nn

jHjG

Angle graph;

)1/(2tan)()(

2

21

nn

jHjG

(=0.5) (n=1)

(rad/s) |G(j)| (dB) G(j)

0.01 0.1 12456810100

0.00040.04321.1094 -11.14-23.82-27.78-31.01-36.06-39.95-79.99

-0.57-5.76-33.69 -146.30-165.06-168.23-170.27-172.76-174.23-179.42

10-1

100

101

-40

-30

-20

-10

0

Ma

gnitude

(dB

)

10-1

100

101

-150

-100

-50

0

Pha

se

(de

g)

Frequency (rad/sec)

= n

<< n

>> n

-40 dB/d

0 dB/d

Natural frequencyn=1


9/17/2001

19


4. Quadratic factors (1+ j 2 (j/n) + (j/n)2 ) -1, (Phase Lag) for a range of 0 < < 1.

Note that the optimum value of = 0.707.

10-1

100

101

-40

-30

-20

-10

0

10

Magnitude(d

B)

10-1

100

101

-150

-100

-50

0

Pha

se(d

eg)

Frequency (rad/sec)

( = 0.1)

( = 0.2)

( = 1)

(n=

1)

)1(

2

tan

)()(

2

2

1

n

n

jHjG

22

2

2

10 )2()1(log20

)()(

nn

jHjG



4. Quadratic factors (1+ j 2 (j/n) + (j/n)2 ) -1, (Phase Lag)

Magnitudes graph; 22

2

2

1010 )2()1(log20)()(log20nn

jHjG

)2/(1)()( jjHjG nn and

At the natural frequency = n

)2/(1|)()(| jjHjG nn

The two asymptotes just derived are independent of the value of damping ratio .Near the frequency =n, a resonant peak occurs. The damping ratio determines the magnitude of this resonant peak.Errors obviously exist in the approximation by straight-line asymptotes. Themagnitude of the error depends on the value of . It is large for small values of .

Angle graph;

)1/(2tan)()(

2

21

nn

jHjG


9/17/2001

20


4. Quadratic factors (1+ j 2 (j/n) + (j/n)2 )-1, (Phase Lag) for a range of 0 < < 1.

Note that the optimum value of = 0.707.

Figure shows exact log-magnitude curves, together with the straight-lineasymptotes and the exact phase-angle curves for the quadratic factor withseveral values of .

10-1

100

101

-40

-30

-20

-10

0

10

Magnitude

(dB

)

( = 0.1)

( = 0.2)

( = 1)

(n= 1)

22

2

2

10 )2()1(log20)()(nn

jHjG

Frequency (rad/sec)



4. Quadratic factors (1+ j 2 (j/n) + (j/n)2 ) -1, (Phase Lag) for a range of 0 < < 1.

The phase angle is a function of both o and . At = 0, the phase angle equals 00. At the corner frequency = n, the phase angle is -90" regardless of , since

At = , the phase angle becomes -1800. The phase-angle curve is skew symmetricabout the inflection point-the point where = -90°. There are no simple ways tosketch such phase curves. We need to refer to the phase-angle curves shown inabove Figure.

( = 0.1)

( = 0.2)

10-1

100

101

-150

-100

-50

0

Pha

se(d

eg)

Frequency (rad/sec)

(n= 1)

( = 1)

)1(

2

tan)()(

2

2

1

n

njHjG

011 90tan0

2tan)()(

jHjG


9/17/2001

21


G(s)R(s) Y(s)

Plant

H(s)

E(s)

Sensors

deg180)()()2(

0)()(log20)1(

0

10

jHjG

dBjHjG

1) First rewrite the sinusoidal transfer function G(j)H(j) as a product of basicfactors discussed above.

2) Then identify the corner frequencies for first order factors of zeros/poles andnatural frequencies for quadratic factors of zeros/poles.

3) Finally, draw the asymptotic log-magnitude curves, |G(j)H(j)| with proper slopes between the corner frequencies and natural frequencies with the damping demping ratio. The exact curve, which lies close to the asymptotic curve, can be obtained by adding proper corrections.

4) The phase-angle curve of G(j)H(j) can be drawn by adding the phase-angle curves of individual factors.

Characteristic equation;

Angle graph;

Magnitudes graph;

1)()(0)( jHjGjP

General Procedure for Plotting Bode Diagrams


G(s)R(s) Y(s)

Plant

H(s)

E(s)

Sensors

The use of Bode diagrams employing asymptotic approximations requires muchless time than other methods that may be used for computing the frequencyresponse of a transfer function.The ease of plotting the frequency-response curves for a given transfer functionand the ease of modification of the frequency-response curve as compensation isadded are the main reasons why Bode diagrams are very frequently used inpractice.

deg180)()()2(

0)()(log20)1(

0

10

jHjG

dBjHjG


Angle graph;

Magnitudes graph;

1)()(

0)()(1

0)(

jHjG

jHjG

jP

General Procedure for Plotting Bode Diagrams

9/17/2001

22


Example 9.3: Plot the Bode diagram of following system.

Solution 9.3: The system transfer function;

There is no zeros of Gc(j)G(j)H(j) The corner frequency of the simple poles of Gc(j)G(j)H(j) are; = 1 ve = 2 rad/sec

G(s)R(s) Y(s)

H(s)

E(s)Gc(s)

Controller

U(s)

ssssG

sHsGc

23

2)(

1)(,5)(

23

))2

(tantan90(0)()()( 110 jHjGjGc

)2)(log201log20log20(

10log20)()()(log20

2210

221010

1010

jHjGjGc

)()()(1

)()(

)(

)(

sHsGsG

sGsG

sR

sY

c

c

The characteristic equation; )()()(1)( sHsGsGsP c

The loop transfer function;)2)(1(

10)()()(

ssssGsHsGc

Magnitudes graph;

Angle graph;



))2

(tantan90(0

)()(

110

1

jHjG

)2)(log201log20log20(10log20)()()(log20 2210

2210101010 jHjGjGc

Magnitudes graph;

Angle graph;

Solution 9.3:

Mag. (dB) Angle (0)

10-3 73 -90

10-2 53 -90.8

10-1 33.5 -98.57

10-0 10 -161.56

2 -2.04 -198.4

101 -40.2 -252.9

102 -100 -268.28

-150

-100

-50

0

50

100

Magnitude

(dB

)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Pha

se

(deg)

Bode Diagram

(rad/sec)

53

10

=2 =1-2

-20 dB/dec

-40 dB/dec

-60 dB/dec

-90.8

-161.5

-198.4


9/17/2001

23


Example 9.4: Plot the Bode diagram of the following system.

Solution 9.4; The loop transfer function;

The zero of G(j) is =20 rad/secThe corner frequency of the simple pole of G(s) is = 2 rad/sec andthe natural frequency of the second order pole is n 13 rad/sec

Magnitudes graph;

Angle graph;

)200)(2(

)20(10)(

2

ssss

ssGG(s)

R(s) Y(s)

)200)(2(

)20(10

)200))((2(

)20(10)(

22

jjj

j

jjjj

jjG

j

s-düz.

n13 13.14d

5.0 n

= 87038.087cos

0

dB

jG

])200(log202log20log20[

20log2010log20)(log20

22210

221010

22101010

)200

tan2

tan90(20

tan0)(2

1101

jG



Magnitudes graph;

Angle graph;

Solution 9.4:

Mag. (dB) Angle (0)

0.1 13.96 -92.60

2 -14.83 -129.87

13 -27.6 -161

20 -49.11 -123.57

200 -117.97 -274.84

dB

jG

])200(log202log20log20[

20log2010log20)(log20

22210

221010

22101010

)200

tan2

tan90(20

tan0)(2

1101

jG

)200)(2(

)20(10)(

2

ssss

ssG G(s)

R(s) Y(s)


9/17/2001

24


10-1

100

101

102

103

-200

-150

-100

-50

0

50

Magnitude(d

B)

10-1

100

101

102

103

-400

-300

-200

-100

0

Pha

se(d

eg)

Frequency (rad/sec)

-180

G(s)R(s) Y(s)

=2

=13

=20

-20 dB/dec

-40 dB/dec

-60 dB/dec

)200)(2(

)20(10)(

2

ssss

ssG

-80 dB/dec


Solution 9.4:


G(s)R(s)

H(s)

E(s) Y(s)

Homework 9.1: Consider the system shown in Figure where system blockstransfer function are;

a) Plot Bode diagram for K=5.b) Plot Bode diagram for K=50.c) Compare the results of (a) and (b).

1)(,)20)(1(

)(

sHsss

KsG

G(s)R(s)

H(s)

E(s) Y(s)


a) Plot Bode diagram for K=5.b) Plot Bode diagram for K=50.c) Compare the results of (a) and (b).

Plotting Bode Diagram: Homework

1)(,)20)(1(

)5()(

sH

sss

sKsG

9/17/2001

25


bode(sys) draws the Bode plot of the LTI model SYS (created with either TF, ZPK, SS, or FRD).

G(s)R(s) Y(s)

Plant

H(s)

E(s)

Sensors

deg180)()()2(

0)()(log20)1(

1)()(

0

10

jHjG

dBjHjG

jHjG

Magnitudes graph;

Angle graph;

Plotting Bode Diagram: MATLAB

Transfer function modelsys = tf(num,den)

bode(sys )

bode(sys )

State-space modelsys = ss(A,B,C,D)

)()( jHjG

DuCxy

BuAxx


0

)()(1)(

jHjGjP


bode(sys,{wmin,wmax}) draws the Bode plot for frequencies between WMIN and WMAX (in radians/second).

bode(sys1,sys2,...,w) graphs the Bode response of multiple LTI models SYS1,SYS2,... on a single plot.

bode(sys1,'r',sys2,'y--',sys3,'gx')

[mag,phase] = bode(sys,w) and [mag,phase,w] = bode(sys) return the response magnitudes and phases in degrees (along with the frequency vector W if unspecified). No plot is drawn on the screen. mag(:,:,k) and phase(:,:,k) determine the response at the frequency w(k).

To get the magnitudes in dB,

type magdb = 20*log10(mag).

G(s)R(s) Y(s)

Plant

H(s)

E(s)

Sensorsdeg180)()()2(

0)()(log20)1(

1)()(

0

10

jHjG

dBjHjG

jHjG

Magnitudes graph;

Angle graph;


9/17/2001

26


Plotting Bode Diagram: MATLABExample 9.5: The block diagram of a disk drive head position control, including effect of flexure head mount where the blocks dynamics are given in equations (1). Use Matlab to plot Bode diagram of the system for the control gain K=400.

PD Control

Gc(s)

R(s) Y(s)E(s) Motor coil

G1(s)Arm

G2(s)Flexure and head

G3(s)

Sensor

H(s)

U(s)

(1)

1)(

)1885011310(

18850

)2()(

)20(

1

)105.0(

05.0)(

)1000(

5000

)110(

5)(

),1()(

22

2

22

2

3

2

31

sH

sssssG

sssssG

sssG

sKsG

nn

n

c


deg180)()()()()2(

0)()()()(log20)1(

0321

32110

jGjGjGjG

dBjGjGjGjG

c

c

)1885011310)(1000)(20(

)1(*18850*5000*400)()()()()(

22

2

321

sssss

ssHsGsGsGsGc

0

)()()()(1)( 321

jGjGjGjGjP c

Plotting Bode Diagram: MATLABSolution 9.5:

PD Control

Gc(s)


G1(s)Arm


G3(s)

Sensor

H(s)

U(s)

The characteristic equation of the feedback system and the loop transfer function for the control gain K=400 are;

9/17/2001

27


)1885011310)(1000)(20(

)1(*18850*5000*400)()()()()(

22

2

321

sssss

ssHsGsGsGsGc

Plotting Bode Diagram: MATLABSolution 9.5:

M-file to plot Bode diagramclear allnum=400*5000*18850^2*[1 1]; den=conv(conv([1 0],[1 20]),conv([1 1000],[1 11310 18850^2]));sys=tf(num,den); bode(sys)grid

PD Control

Gc(s)


G1(s)Arm


G3(s)

Sensor

H(s)

U(s)



Example 9.5:

-100

-50

0

50

Ma

gnitu

de

(dB

)

10-1

100

101

102

103

104

105

-360

-270

-180

-90

0

Pha

se

(de

g)

Frequency (rad/sec)

PD Control

Gc(s)


G1(s)Arm


G3(s)U(s)

1)(

)1885011310(

18850)(

)20(

1)(

)1000(

5000)(

),1()(

22

2

3

2

1

sH

sssG

sssG

ssG

sKsGc

1 2

3

4

1

2

3

4

1 = 1 2 = 20

3 = 1000

n = 18850

-20 dB/d0 dB/d

-20 dB/d

-40 dB/d

-80 dB/d

9/17/2001

28


Disk drive head position control, including effect of flexure head mount, K=400.

© 2001 by Prentice Hall, Upper Saddle River, NJ.

)1885011310)(20)(100(

18850*)1(*500*400)()()()(

22

2

321

sssss

ssGsGsGsG c


Example 9.5:


num=[1 20]; den=conv([1 1],[1 200]);sys=tf(num,den); w=logspace(-1,2);[mag,phase]=bode(sys,w);magd=20*log10(mag);subplot(2,1,1),semilogx(w,magd(1,:)),gridylabel('Magnitude(dB)')subplot(2,1,2),semilogx(w,phase(1,:)),gridylabel('Phase(deg)'),xlabel('Frequency (rad/sec)')

)200)(1(

)20()()(

ss

ssHsG

10-1

100

101

102

103

-80

-60

-40

-20

10-1

100

101

102

103

-80

-60

-40

-20

0

Pha

se(d

eg)

Frequency (rad/sec)

0 dB/decade

-20 dB/decadeMagnitude(d

B)

0 d

B/d

eca

de

-20

dB

/dec

ade

G(s)R(s) Y(s)

Plant

H(s)

E(s)

Sensors

Plotting Bode Diagram: MATLABExample 9.6: Use Matlab to plot Bode diagram of the following system.

9/17/2001

29


The main advantage of using the Bode diagram is that multiplication factors innumerator of G(j)H(j) are converted to addition and in denominator ofG(j)H(j) are converted to subtraction factors.

)10)(1(

)5(5)()(,,

)10)(1(

)5(5)()(

jjj

jjHjGjs

sss

ssHsG

Magnitudes graph;

2210

21010

22101010

10log201log20log20

5log205log20|)()(|log20

jHjG

Angle graph;

)10/(tan)(tan90)5/(tan0)()( 1101 jHjG

Advantages of Using Bode Diagram

G(s)R(s) Y(s)

H(s)deg180)()()

0)()(log20)

1)()(

0

10

jHjGb

dBjHjGa

jHjG


A simple method based on asymptotic approximations for sketching anapproximate magnitudes graph (log-magnitude) curve is available. Suchapproximation by straight line asymptotes is sufficient if only rough information onthe frequency-response characteristics is needed. Should the exact curve be desired,corrections can be made easily to these basic asymptotic plots.

Expanding the low frequency range by use of a logarithmic scale for thefrequency is highly advantageous since characteristics at low frequencies are mostimportant in practical systems. Although it is not possible to plot the curves rightdown to zero frequency because of the logarithmic frequency (log 0 = -), thisdoes not create a serious problem.

The experimental determination of a transfer function can be made simple iffrequency response data are presented in the form of a Bode diagram.

Advantages of Using Bode Diagram

9/17/2001

30


Minimum-Phase Systems and Nonminimum-Phase Systems.

Transfer functions having neither poles nor zeros in the right-half s= jcomplex plane are minimum-phase transfer functions, whereas those having polesand/or zeros in the right-half s plane are nonminimum-phase transfer functions.

Systems with minimum-phase transfer functions are called minimum-phase systems,whereas those with nonminimum-phase transfer functions are called nonminirnum-phase systems.

For systems with the same magnitude characteristic, the range in phase angle of the

minimum-phase transfer function is minimum among all such systems, while therange in phase angle of any nonminimum-phase transfer function is greater than thisminimum.

It is noted that for a minimum-phase system, the transfer function can be uniquelydetermined from the magnitude curve alone. For a nonminimum-phase system, thisis not the case.

Multiplying any transfer function by all-pass filters does not alter the magnitudecurve, but the phase curve is changed.

Bode Diagram of Minimum and Nonminimum Phase Systems


Minimum-Phase Systems and Nonminimum-Phase Systems.Nonminimum-phase situations may arise in two different ways. One is simplywhen a system includes a nonminimum-phase element or elements. The othersituation may arise in the case where a minor loop is unstable.

For a minimum-phase system, the phase angle at = becomes -90°(n- m), wherem and n are the degrees of the numerator and denominator polynomials of thetransfer function, respectively.

Minimum-Phase Systems

The magnitude graph satisfies 20log10G(j) = -20(n-m) dB/decade for

The angle graph satisfies G(j) = -900(n-m) for

Bode Diagram of Minimum and Nonminimum Phase Systems

9/17/2001

31


Example 9.7; The following system has minimum phase loop transfer function,G(j)H(j);

G(s)R(s) Y(s)

H(s)

n

jj

m

iijHjG

ddBmnjHjG

11

10

)()(

/)(20)()(log20lim

G(j)H(j); has n =2 poles and no zeros

)1(

1)()(

)1(

1)()(

jjjHjG

sssHsG

0

0

0

10

180

)02(90

)(90)()(lim

/40

/)02(20)()(log20lim

)1(

1)()(

mnjHjG

ddB

ddBjHjG

jjjHjG

-40

-20

0

20

40

60

80

Magnitu

de (

dB

)

10-1

100

101

102

-180

-135

-90

Phase

(deg)

Bode Diagram

Frequency (rad/sec)

-40 dB/d

-1800

Bode Diagrams of Minimum-Phase Systems

0

50

100

Magnitu

de (

dB

)

10-1

100

101

102

90

135

180

Phase (

deg)

Bode Diagram

Frequency (rad/sec)



G(s)R(s) Y(s)

H(s)

n

jj

m

iijHjG

ddBmnjHjG

11

10

)()(

/)(20)()(log20lim

G(j)H(j); has n = 0 poles and m =2 zeros

)1()()(

)1()()(

jjjHjG

sssHsG

40 dB/d

1800

0

0

0

10

180

)20(90

)(90)()(lim

/40

/)20(20)()(log20lim

)1()()(

mnjHjG

ddB

ddBjHjG

jjjHjG

Bode Diagrams of Minimum-Phase Systems

9/17/2001

32


,)10)(5)(1(

)2()()(

)10)(5)(1(

)2(50)()(

jjjj

jjHjG

ssss

ssHsG


G(s)R(s) Y(s)

H(s)

n

jj

m

iijHjG

ddBmnjHjG

11

10

)()(

/)(20)()(log20lim

G(j)H(j); has n =4 poles and m = 1 a zero

0

00

10

270

)14(90)(90)(lim

/60

/)14(20)(20)(log20lim

mnjG

ddB

ddBmnjG

Bode Diagram of Minimum Phase Systems


)10)(5)(1(

)2(50)()(

ssss

ssHsG


G(s)R(s) Y(s)

H(s)

-100

-50

0

50

Ma

gn

itud

e (

dB

)

10-1

100

101

102

-270

-225

-180

-135

-90

Ph

as

e (

deg

)

Bode Diagram

Frequency (rad/sec)

0

0

10

270

)(90)(lim

/60

/)(20)(log20lim

mnjG

ddB

ddBmnjG


9/17/2001

33


Nonminimum Phase Systems: Nonminimum-phase situations may arise in twodifferent ways. One is simply when a system includes a nonminimum-phaseelement or elements. The other situation may arise in the case where a minor loopis unstable.

For a nonminimum-phase system, the slope of the log-magnitude curve at = isequal to -20(n–m) dB/d.But the phase angle at = differs from -90°(n-m).

Nonminimum Phase Systems

The magnitude graph satisfies 20log10G(j) = -20(n-m) dB/decade for

The angle graph does not satisfy G(j) = -900(n-m) for

Results: To detect whether the system is minimum phase is necessary to examineboth the slope of the high-frequency asymptote of the log-magnitude curve and thephase angle at = .If the slope of the log-magnitude curve as approaches infinity is -20(n - m) dB/dand the phase angle at = is equal to -90°(n - m), then the system is minimumphase.



Example 9.10; Consider the feedback system given in the Figure. The loop transfer function is given in (1) where 0<T<T1.

Results: The system is nonminimumphase system.

G(s)R(s) Y(s)

H(s))1()1(

)1()()(

1

sT

TssHsG

j

0

1

1

T

T

1

1

1

1

s-plane

1111

10

)()(

/)(20)()(log20lim

n

jj

m

iijHjG

ddBmnjHjG

10-1

100

101

102

103

-40

-30

-20

-10

0

Magnitu

de(d

B)

10-1

100

101

102

103

-200

-150

-100

-50

0

Phase(d

eg)

Frequency (rad/sec)

501

T1

1

1

T

0 dB/d

0 dB/d

-20 dB/d

Bode Diagram of Nonminimum Phase Systems

9/17/2001

34


Example 9.11; Consider the feedback system given in the Figure. The loop transfer function is given in (1) where 0<T<T1.

Results: The system is minimum phase system.

G(s)R(s) Y(s)

H(s))1()1(

)1()()(

1

sT

TssHsG

0)(90)()(lim

,)()(

/)(20)()(log20lim

11

10

mnjHjG

jHjG

ddBmnjHjG

j

0

s-plane1

1

T

T

1

1

11

10-1

100

101

102

103

-40

-30

-20

-10

0

Magnitude(d

B)

10-1

100

101

102

103

-80

-60

-40

-20

0

Phase(d

eg)

Frequency (rad/sec)

501

T1

1

1

T

0 dB/d

0 dB/d

-20 dB/d


-60

-40

-20

0

20

40

Ma

gn

itud

e (

dB

)

10-1

100

101

102

-90

-45

0

Ph

ase

(d

eg

)

Bode Diagram

Frequency (rad/sec)© Dr. Ahmet Uçar EEE 352 Chapter 9 68

Example 9.11; Consider the feedback system given in the Figure. The loop transfer function is given in (1).

Results: The system is nonminimumphase system.

G(s)R(s) Y(s)

H(s)

)2(

20)()(

sssHsG

j

0

1

1

s-düzlemi

2

0

0

10

180

)02(90)(lim

/40

/)02(20)(log20lim

jG

ddB

ddBjG


9/17/2001

35


In designing a control system, we require that the system be stable. Furthermore, itis necessary that the system have adequate relative stability.One of the important problems in analyzing a control system is to find all closed-loop poles or at least those closest to the j axis (or the dominant pair of closed-loop poles).If the open-loop frequency-response characteristics of a system are known, it maybe possible to estimate the closed-loop poles closest to j axis.Although the stability of closed loop systems are usually studied with Nyquiststability criterion, however, Bode diagram can effectively be used.

G(s)R(s) Y(s)

H(s)0)()(1)(

,)()(1

)(

)(

)(

sHsGsP

sHsG

sG

sR

sY

dBjHjG

jHjG

gc

gc

0|)()(|log20

1|)()(|)1(

10

0180|)()()2( pc

jHjG

Bode Diagrams: Relative Stability


Phase and Gain Margins.In general, the closer the Magnitude and angle graphs of the loop transfer functionG(j)H(j) to 1 (0 dB) and -1800 respectively, the more oscillatory is the systemresponse. The closeness of the Magnitude graph of the loop transfer function,|G(j)H(j)| to the reference point 1 (0 dB) and the angle graph =G(j)H(j) tothe reference point -1800 can be used as a measure of the margin of the stability.

G(s)R(s) Y(s)

H(s)

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

)(

)(

)(sHsGsP

sHsG

sG

sR

sY

dBjHjG

jHjG

pc

pc

0|)()(|

1|)()(|)1(

0180|)()()2( gcjHjG

log

log

0

|G(j

)H(j

)|[d

B]

-1800

gc

pc

GM

PM


9/17/2001

36


The gain crossover frequency, gc, is the frequency at which |G(j)H(j)|, themagnitude of the open loop transfer function, is unity.The phase margin PM is 1800 plus the phase angle = G(j)H(j) of the open-loop transfer function at the gain crossover frequency, gc, or

G(s)R(s) Y(s)

H(s)

log

log

0

|G(j

)H(j

)|[d

B]

-1800

gc

pc

GM

PM

1)()(0)()(1)( jHjGjHjGjP

dBjHjGpc

0|)()(|)1(

0180|)()()2( gcjHjG



G(s)R(s) Y(s)

H(s)

)()(180 gcgc jHjGPM log

log

0

|G(j

)H(j

)|[d

B]

-1800

gc

pc

GM

PM


dBjHjGpc

0|)()(|)1( 0180|)()()2( gc

jHjG

Phase Margin: The phase margin is defined as the change in open loop phase shift required to make a closed loop system unstable. The phase margin is that amount of additional phase lag at the gain crossover frequency required to bring the system to the verge of instability.The phase margin is the difference in phase between the phase curve and -180 deg at the point corresponding to the frequency that gives us a gain of 0dB (the gain cross over frequency, gc).


9/17/2001

37


G(s)R(s) Y(s)

H(s)

)()(180 gcgc jHjGPM log

log

0

|G(j

)H(j

)|[d

B]

-1800

gc

pc

GM

PM


dBjHjGpc

0|)()(|)1( 0180|)()()2( gc

jHjG

Phase Margin:Likewise, the gain margin is the difference between the magnitude curve and 0dBat the point corresponding to the frequency that gives us a phase of -180 deg (the phase cross over frequency, gc).



Figures illustrate the phase margin of both a stable system and an unstable systemin Bode diagrams.The phase margin is positive for PM >0 and negative for PM<0. For a minimumphase system to be stable, the phase margin must be positive.In the Bode diagram, the critical point, j axis, in the complex plane correspondsto the 0 dB and -180" lines.

log

log

0

|G(j

)H(j

)|[d

B]

-1800

gc

pc

GM

PM

Figure 1: The stable system

log

log

0

|G(j

)H(j

)|[d

B]

-1800

gc

pc

-GM

PM

Figure 2: The unstable system

)()(180 gcgc jHjGPM


9/17/2001

38


Gain Margin: The gain margin is defined as the change in open loop gainrequired to make the system unstable. Systems with greater gain margins canwithstand greater changes in system parameters before becoming unstable inclosed loop.Keep in mind that unity gain in magnitude is equal to a gain of zero in dB.

G(s)R(s) Y(s)

H(s)

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

)(

)(

)(sHsGsP

sHsG

sG

sR

sY

0180|)()()2( gcjHjG

log

log

0

|G(j

)H(j

)|[d

B]

-1800

gc

pc

GM

PM

dBjHjG

jHjG

pc

pc

0|)()(|

1|)()(|)1(



Gain margin: The gain margin is the reciprocal of the magnitude |G(j)H(j)| at thefrequency, pc, at which the phase angle is -1800. Defining the phase crossover frequency pc, to be the frequency at which the phase angle of the open-loop transfer function equals -1800 gives the gain margin Kg:

G(s)R(s) Y(s)

H(s)

|)()(|

1

pcpc

gjHjG

K

log

log

0

|G(j

)H(j

)|[d

B]

-1800

gc

pc

GM

PM


dBjHjGpc

0|)()(|)1( 0180|)()()2( gc

jHjG

In terms of decibels,

|)()(|20

log20

pcpc

g

jHjG

KGM


9/17/2001

39


The gain margin expressed in decibels is positive if Kg is greater than unity andnegative if Kg is smaller than unity. Thus, a positive gain margin (in decibels)PM>0 means that the system is stable, and a negative gain margin (in decibels)PM<0 means that the system is unstable. The gain margin is shown in Figuresillustrate the phase margin of both a stable system and an unstable system in Bodediagrams.

log

log

0

|G(j

)H(j

)|[d

B]

-1800

gc

pc

GM

PM

Figure 1: The stable system, gc < pc

log

log

0

|G(j

)H(j

)|[d

B]

-1800

gc

pc

-GM

PM

Figure 1: The unstable system gc > pc

|)()(|20log20 pcpcg jHjGKGM



For a stable minimum-phase system, the gain margin indicates how much the gaincan be increased before the system becomes unstable.For an unstable system, the gain margin is indicative of how much the gain mustbe decreased to make the system stable.The gain margin of a first or second order system is infinite since the Bodediagrams for such systems do not cross the reference points 0 dB and -1800

degrees.Thus, theoretically, first- or second order systems cannot be unstable.

Comments:

The phase and gain margins of a control system are a measure of the closeness ofthe Bode diagrams to the reference point 0 dB and -1800.

Therefore, these margins may be used as design criteria. It should be noted thateither the gain margin alone or the phase margin alone does not give a sufficientindication of the relative stability. Both should be given in the determination ofrelative stability.


9/17/2001

40


Comments:

For a minimum-phase system, both the phase and gain margins must be positivefor the system to be stable. Negative margins indicate instability.

Proper phase and gain margins ensure us against variations in the systemcomponents and are specified for definite positive values. The two values boundthe behavior of the closed-loop system near the resonant frequency.

For satisfactory performance, the phase margin should be between 300 and 60°,300<PM< 60°, and the gain margin should be greater than 6 dB, GM>6 dB.

With these values, a minimum-phase system has guaranteed stability, even if theopen loop gain and time constants of the components vary to a certain extent.



Transient response performance:

Although the phase and gain margins give only rough estimates of the effectivedamping ratio of the closed-loop system, they do offer a convenient means fordesigning control systems or adjusting the gain constants of systems.

For minimum-phase systems, the magnitude and phase characteristics of the openloop transfer function are definitely related.

The requirement that the phase margin be 300<PM< 60° means that in a Bodediagram the slope of the log magnitude curve at the gain crossover frequency, gc

should be more gradual than -40 dB/decade.

In most practical cases, a slope of -20 dB/decade is desirable at the gain crossoverfrequency, gc, for stability. If it is -40 dB/decade the systems could be eitherstable or unstable. (Even if the system is stable, however, the phase margin issmall.) If the slope at the gain crossover frequency is -60 or steeper, the system ismost likely unstable.


9/17/2001

41

© Dr. Ahmet Uçar EEE 352 Chapter 9 81© Dr. Ahmet UÇAR 81

The system is stable

log

log

0

|G(j)H(j)|[dB]

G(j)H(j)

-1800

gc

pc

GM

PM

pcgc

][|)()(|log200 10 dBjHjGGM pcpc


deg)()(180 gcgc jHjGPM

G(s)R(s) Y(s)

H(s)


0180|)()()2(

0|)()(|)1(

gc

pc

jHjG

dBjHjG




G(s)R(s) Y(s)

H(s)


0

GMPM

pcgc

log

log

0

|G(j)H(j)|[dB]

G(j)H(j)-1800

gc

pc


The system is critical

0180|)()()2(

0|)()(|)1(

gc

pc

jHjG

dBjHjG

9/17/2001

42




G(s)R(s) Y(s)

H(s)

1)()(

0)()(1)(

jHjG

jHjGjP

0180|)()()2(

0|)()(|)1(

gc

pc

jHjG

dBjHjG


log

log

0

|G(j)H(j)|[dB]

G(j)H(j)

-1800

gc

pc

-GM

-PM

pcgc The system is unstable


Nonminimum phase system:For nonminimum phase systems, the correct interpretation of stability marginsrequires careful study. The best way to determine the stability of nonminimumphase systems is to use the Nyquist diagram approach rather than Bodediagram approach.

Bode DiagramsRelative Stability of Nonminimum phase system

9/17/2001

43


G(s)R(s) Y(s)

H(s)

Example 9.12; a) Obtain phase crossover frequency pc and gain margin GM ofthe given system given in (1).

b) Obtain gain crossover frequency gc and phase margin PM of the given system.c) Draw the Bode diagram of the system and show pc, GM, gc, and PM on theBode diagram.

)1()50)(5(

2500)()(

ssssHsG

Solution 9.12; The system transfer function and characteristic equation are;

0)()(1)(,)()(1

)(

)(

)(

sHsGsP

sHsG

sG

sR

sY

dBjHjG

jHjG

gc

gc

0|)()(|log20

1|)()(|)1(

10

0180|)()()2( pcjHjG

0)()(1)( jHjGjP



Solution 9.12; a) The system characteristic equation is

2322

3

2322

2

3223

)250()55(

)250(2500

)250()55(

)55(25001

)250(55

25001

25055

25001

)50)(5(

25001)(

j

jjjjjjjP

Phase crossover frequency pc:

0)250()55(

)250(25000)]()(Im[)]()(1Im[

2322

3

jHjGjHjG

sradpc /82.15250

00)250(0)250(2500

22

123

2

log -1800

pc


9/17/2001

44


2322

3

2322

2

)250()55(

)250(2500

)250()55(

)55(25001)(

jjP

The gain margin GM is the margin at the phase crossover frequency pc at which the phase angle is -1800.

Solution 9.12; a) The system characteristic equation is

dB

jHjG pcpcpcpc

81.14

)39.3439.2498.23(95.67

))50(log205log20log20(2500log20)()(log20 2210

2210101010

dBGM 14.8114.8167)(0

][|)()(|log200 10 dBjHjGGM pcpc log

log

0

|G(j

)H(j

)|[d

B]

G

(j

)H(j

)

-1800

pc =15.82

GM = 14.82



Method 1: i) The gain crossover frequency gc can approximately be readfromthe Bode diagram directly.ii) The gain crossover frequency gc can approximately be obtained from theslope log magnitude graph of Bode diagram.First, the slope between two corner frequencies f, and l is determined, wherethe positive sign of log magnitude equation at f, changes to negative at l.

Solution 9.12; b) The gain crossover frequency gc and phase margin PM:The gain crossover frequency gc can be obtained approximately from method1,based on slope of the log magnitude at the gain crossover frequency gc, or frommethod 2, based on analytically solution of the log magnitude equation;

dBjHjGjHjGgcgc

0|)()(|log201|)()(|)1( 10


2210101010 jHjG

Mag .(dB) Phase

0.1 39.99 -91.26

5 3 -140.71

50 -37.03 -219.28

As seen from the table that the positive log magnitudeequation at f = 5 rad/s changed to negative at l = 50rad/s and;


9/17/2001

45


Solution 9.12; b) The gain crossover frequency gc and phase margin PM:Method 1: ii) gc can approximately be obtained as follows;


2210101010 jHjG

Genlik (dB)

Açı

0.1 39.99 -91.26

5 3 -140.71

50 -37.03 -219.28

gc

f

dB

dB

22

11

,0

5,3

40

5log

3

5loglog

30

5loglog

/40log

12

gcgcgc

dBdB

decdBdB

sradgc

gc/95.510510

540

3

40

3

The log magnitude equation is positive at f = 5 rad/sand changed to negative at l = 50 rad/s.

log

|G(j

)H(j

)|[d

B]

1 = 5

-40 dB/decade

3 dB

2 =gc

0 dB

Actual gainApproximated

gain

Comment: The correctness of the gain crossover frequency gc depends on thedistance of the both frequencies.



Solution 9.12; b) The gain crossover frequency gc and phase margin PM:Method 2:The analytical solution of the log magnitude equation is;

20log10|G(jpc)H(jpc)| = 0 dB or |G(jpc)H(jpc)| =1

)50)(5(25001505

25001|)()(| 222222

2222

jHjG

sec/22.67.38

7.3802500625002525

,02500625002525

02500)50)(5(

223

22246

222222

rad

uuuu

u

log 0|G(j

)H(j

)|[d

B]

gc

Comment: Although this gives the correct result for the gain crossover frequencygc but it may need high order polynomial to be solved.


9/17/2001

46


Solution 9.12; b) Fore gain crossover frequency gc =5.96 rad/s (Method 1) andphase margin PM is;

0

96.58.146)()(

gc

jHjG

PM = 180-146.8 = 33.20

][|)()(|log200 10 dBjHjGGM pcpc


log

log

0

|G(j

)H(j

)|[d

B]

G

(j

)H(j

)-1800

gc=5.96

pc =15.82

GM = 14.82

PM=33.20

1

2

The system is stable

Angle equation;

The phase margin PM:



Solution 9.12; c) Draw the Bode diagram and pc, GM, gc, and PM of the system;

]50log20

5log20log20[

2500log20)()(

2210

221010

10

jHjG

]50

tan5

tan90[0

)()(

110

jHjG

Angle graph;

sradpc /82.15

sradgc /95.5

PM = 180-146.8 = 33.20

dBGM 14.8167

Obtained from Method 1;

-150

-100

-50

0

50

Magnitude (dB

)

10-1 100 101 102 103-270

-225

-180

-135

-90

Pha

se (deg)

Gm = 14.8 dB (at 15.8 rad/sec) , Pm = 31.7 deg (at 6.22 rad/sec)

Frequency (rad/sec)

PM

GM

gc= 6.22

pc=15.8

Log magnitude graph;,)50)(5(

2500)()(

jjjjHjG


9/17/2001

47


G(s)R(s) Y(s)

H(s)

Example 9.13; Consider the system given in (1)

a) Draw the Bode diagram of the system for K=10 and determine its stability.b) Draw the Bode diagram of the system for K=100 and determine its stability.

)1()5)(1(

)()(

sss

KsHsG

Solution 8.13; a) For K=10 phase crossover frequency, pc and gain margin GM;

0)()(1)(,)()(1

)(

)(

)(

sHsGsP

sHsG

sG

sR

sY

0)()(1)( jHjGjP

2322

3

2322

2

23

)5()6(

)5(10

)5()6(

)6(101

56

101

)5)(1(

101)()(1)(

j

jjjjjjHjGjP



Solution 9.13; a) For K=10 phase crossover frequency, pc and gain margin GM;

2322

3

2322

2

)5()6(

)5(10

)5()6(

)6(101)()(1)(

jjHjGjP

Phase crossover frequency, pc;

sradjP

pg /23.25,0

0)5(0)5(100

)5()6(

)5(10)](Im[

21

2

3

2322

3

2

Gain margin GM;

dB

jHjGpc

-9.49

)23.2)5(log2023.21log2023.2log20(10log20)()(log20 2210

2210101023.210

dBGM 9.499.49)(0 log

log

0

|G(j

)H(j

)|[d

B]

G

(j

)H(j

)

-1800

pc =2.23

GM = 9.49.82

The gain at pc= 2.23 rad/sec


9/17/2001

48


Solution 9.13; a) For K=10 gain crossover frequency, gc and phase margin PM;

Gain crossover frequency, gc;

Method 1: ii) gc can approximately be obtained from log magnitude as follows;

)5(log201log20log20(10log20)()(log20 2210

2210101010 jHjG

Gain (dB) Phase

0.1 25.97 -96.8

1 2.84 -146.3

5 -25.1 -213.6

gcdB

dB

22

11

,0

1,284

log 1=1

-40 dB/decade

2.84 dB

gc

0 dB

Actual gain

Approximatedgain

40log

84.2

1loglog

/40log

12

gcgc

dBdB

decdBdB

sradgcgc /1.11040

1



Solution 9.13; a) The gain crossover frequency gc and phase margin PM:Method 2:The analytical solution of the log magnitude equation is;

20log10|G(jpc)H(jpc)| = 0 dB or |G(jpc)H(jpc)| =1

)10)(1(101101

101|)()(|

0)()(1)(

22222

222

jHjG

jHjGjP

22246

22222

,010100101

010)10)(1(

u

sec/7844.06153.0

6153.0

010100101 223

radu

u

uuu

gc

log 0|G(j

)H(j

)|[d

B]

gc=0.78


9/17/2001

49


The phase margin PM;

Solution 9.13; a) For K=10 gain crossover frequency, gc and phase margin PM;

)5

1.1tan

1

1.1tan90(0)()( 110

gc

jHjG

0

1.1150)()(

gc

jHjG

PM = 180-150 =30 0

log

log

0|G

(j

)H(j

)|[d

B]

-1800

gc=1.1

pc =2.23

GM=9.49

PM=300

Since gc =1.1 < pc =2.23, the system is stable.

pcgc

The angle at gc =1.1 rad/s



Solution 9.13; b) For K=100, By using the same methods

The gain crossover frequency, gc, phase margin PM,

The phase crossover frequency, pc and gain margin GM:

sradgc /5.4 PM = 180-210= -300

sradpc /1.2

dBGM 12log

log

0

|G(j)H(j)|[dB]

G(j)H(j)

-1800

gc

pc

-GM

-PM1.25.4 pcgc

pcgc

Since

the system is unstable.


9/17/2001

50

|G(j

)H(j

)|[d

B]

|G(j

)H(j

)|[d

B]


Solution 9.13:Bode Diagrams: Relative Stability



1)(,)50)(10)(1(

1)(

)(

sHsss

sG

KsG

p

cc

a) Draw the Bode diagram of the loop transfer function,b) Determine the gain Kc such that the phase margin is PM=50°.What is the gain margin PM in this case?

Gp(s)R(s)

H(s)

E(s) Y(s)Gc(s)


1)(,)2(

1)(,)(

2

sH

sssGKsG pcc

a) Plot the frequency response for this system when Kc = 4.b) Calculate the phase and magnitude at = 0.5,1,2, 4, and .

Gp(s)R(s)

H(s)

E(s) Y(s)Gc(s)

Bode Diagrams: Homework

9/17/2001

51


Homework 9.5: Consider the system represented in state space form

ux

xy

ux

x

x

x

]0[11

5

0

32

10

2

1

2

1

2

1

a) Determine the transfer function representation of the system,b) Sketch the Bode plot.,



In addition to the phase margin PM, gain margin GM, resonant peak Mr, and resonantfrequency r, there are other frequency-domain quantities commonly used in performancespecifications. They are the cutoff frequency, bandwidth BW, and the cutoff rate.

Cutoff Frequency and Bandwidth: Both cutoff Frequency and Bandwidth are related to the closed loop system Bode diagram such that the frequency b, at which the magnitude of the closed-loop frequency response is 3 dB below its zero-frequency value is called the cutoff frequency. Thus

G(s)R(s) Y(s)

H(s)

bfordBjR

jY

jHjG

jG

jR

jY

,3

)0(

)0(

)()(1

)(

)(

)(

For systems in which |Y(j0)/R(j0)|= 0 dB, bfordBjHjG

jG

jR

jY

,3

)()(1

)(

)(

)(

The closed-loop system filters out the signal components whose frequencies are greaterthan the cutoff frequency and transmits those signal components with frequencies lowerthan the cutoff frequency.

Closed Loop Frequency Performances: Cutoff Frequency and Bandwidth

9/17/2001

52


Bandwidth: The frequency range 0≤≤ b in which the magnitude of the closed loop doesnot drop -3 dB is called the bandwidth of the system. The bandwidth indicates the frequencywhere the gain starts to fall off from its low-frequency value. Thus, the bandwidth indicateshow well the system will track an input sinusoid. Note that for a given natural fequency n

the rise time increases with increasing damping ratio . On the other hand, the bandwidthdecreases with the increase in . Therefore, the rise time and the bandwidth are inverselyproportional to each other.

G(s)R(s) Y(s)

H(s)

bfordBjR

jY

jHjG

jG

jR

jY

,3

)0(

)0(

)()(1

)(

)(

)(

For systems in which |Y(j0)/R(j0)|= 0 dB,

bfordBjHjG

jG

jR

jY

,3

)()(1

)(

)(

)(

dB

-3

Bandwidth

b in log scale

dB

-3

Bandwidth

b in log scale

0

Closed Loop Frequency Performances: Bandwidth


Bandwidth: The specification of the bandwidth may be determined by the following factors:1. The ability to reproduce the input signal. A large bandwidth corresponds to a small rise time, or fast response. Roughly speaking, we can say that the bandwidth is proportional to the speed of response. 2. The necessary filtering characteristics for high-frequency noise.For the system to follow arbitrary inputs accurately, it must have a large bandwidth. Fromthe viewpoint of noise, however, the bandwidth should not be too 1arge.Thus there areconflicting requirements on the bandwidth, and a compromise is usually necessary for gooddesign. Note that a system with large bandwidth requires high-performance components, sothe cost of components usually increases with the bandwidth.

G(s)R(s) Y(s)

H(s)

bfordBjR

jY

jHjG

jG

jR

jY

,3

)0(

)0(

)()(1

)(

)(

)(

dB

-3

Bandwidth

b in log scale

Closed Loop Frequency Performances: Bandwidth

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53


Cutoff Rate. The cutoff rate is the slope of the log-magnitude curve near the cutofffrequency. The cutoff rate indicates the ability of a system to distinguish the signal fromnoise.

G(s)R(s) Y(s)

H(s)

bfordBjR

jY

jR

jY

jHjG

jG

jR

jY

,3)0(

)0(

)(

)(

)()(1

)(

)(

)(

It is noted that a closed-loop frequency response curve with a steep cutoff characteristicmay have a large resonant peak magnitude, which implies that the system has a relativelysmall stability margin.

dB

-3

Bandwidth

b in log scale

Cutoff Rate.

Closed Loop Frequency Performances: Cutoff Rate


The resonant peak is the value of the maximum magnitude (in decibels) of the closed-loop frequency response. The resonant frequency is the frequency that yields the maximum magnitude.

G(s)R(s) Y(s)

H(s)

bfordBjR

jY

den

num

jHjG

jG

jR

jY

,3

)0(

)0(

)()(1

)(

)(

)(

dB

-3

Bandwidth

b in log scale

% M-File 1 for obtaining the resonant peak and resonant frequency [mag,phase,w] = bode(num,den,w); or [mag,phase,w]= bode(sys,w);[Mp,k]= max(mag);resonant-peak = 20*log10(Mp);resonant-frequency = w(k)% M-File 2 The bandwidth can be obtained by entering the following lines in the program:n = l ;while 20*log10(mag(n)) > = -3; n = n + 1;endbandwidth = w(n)

Cutoff Rate.

Closed Loop Frequency Performances: Cutoff Frequency and Bandwidth

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54


Example 9.14; Consider the system shown in Figure . Using MATLAB, obtain a Bode diagram for the closed loop transfer function. Obtain also the resonant peak, resonant frequency, and bandwidth.

G(s)R(s) Y(s)

H(s)

1)(,)1)(15.0(

1)(

sH

ssssG

Solution 9.14;

15.15.0

1

1)1)(15.0(

1

)()(1

)(

)(

)(23

sssssssHsG

sG

sR

sY

bfordBjR

jY

jR

jY

den

num

jR

jY

jHjG

jG

jR

jY

,3)0(

)0(

)(

)(

)(

)(

)()(1

)(

)(

)( dB

-3

Bandwidth

b in log scale

Cutoff Rate.

The closed loop system transfer function is

Resonant Peak, Resonant Frequency, and Bandwidth: MATLAB


G(s)R(s) Y(s)

H(s)

Solution 9.14;

M-file produces a Bode diagram for the closed-loop system as well as the resonant peak, resonant frequency, and bandwidth.

15.15.0

1

)()(1

)(

)(

)(23

ssssHsG

sG

sR

sY

% M-file for bandwidth. clear all; numcl=[1]; dencl=[0.5 1.5 1 1]; syscl=tf(numcl,dencl);w = logspace(-1,1);bode(syscl,w); grid;[mag,phase,w] = bode(numcl,dencl,w); [Mp,k] = max(mag);ResonantPeak = 20*log10(Mp)ResonantFrequency = w(k)% M-file for bandwidth. n = 1;while 20*log10(mag(n)) > -3; n = n +1;endBandwidth = w(n)

RestultsResonantPeak = 5.2388ResonantFrequency = 0.7906Bandwidth = 1.2649


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55


Solution 9.14; The resulting Bode diagram is shown in

)85.024.0)(5.2(

1

15.15.0

1

)()(1

)(

)(

)(23 jssssssHsG

sG

sR

sY

The resonant peak is obtained as 5.2388 dB.

The resonant frequency is 0.7906 rad/sec.

The bandwidth is 1.2649 rad/sec.

These values can be verified from the Bode diagram.

)()(1

)(

)(

)(

jHjG

jG

jR

jY

Cutoff Rate=-40 dB/d

-60 dB/dr= 0.7906 rad/sec.

Mr= 5.2388 dB.

b =1.2649 rad/sec.

-60

-40

-20

0

20

Ma

gnitud

e (

dB

)

10-1

100

101

Bode Diagram

Frequency (rad/sec)



Remark 1; The frequency performances such Phase Margin PM, Gain margin GM andstability are related to the loop (oplen loop) frequency responce. That is open loopsystem Bode diagram.

G(s)R(s) Y(s)

H(s)

0)()(1)(,)()(1

)(

)(

)(

sHsGsP

sHsG

sG

sR

sY

0

10

180|)()()2(

0|)()(|log20

1|)()(|)1(

pc

gc

gc

jHjG

dBjHjG

jHjG

Remark 2; Resonant Peak, Resonant Frequency, Cutoff Frequency and Bandwidth arerelated to the closed loop system Bode diagram.

bfordBjR

jY

jR

jY

jHjG

jG

jR

jY

,3)0(

)0(

)(

)(

)()(1

)(

)(

)(

G(s)R(s) Y(s)

H(s)

Frequency Performances of Closed Loop Systems

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56


The static position, velocity, and acceleration error constants describe the low-frequency behaviour of type 0, type 1, and type 2 systems, respectively.

0)()(1)(,)()(1

)(

)(

)(

sHsGsP

sHsG

sG

sR

sY

dBjHjG

jHjG

pc

pc

0|)()(|log20

1|)()(|)1(

10

0180|)()()2( gcjHjG

Consider the following feedback control system.

log 0

|G(j

)H(j

)|[d

B]

gc

Region I

Region II

Region III

-20 dB/d

G(s)R(s) Y(s)

H(s)

Bode Diagrams: The steady state performance


log 0

|G(j

)H(j

)|[d

B]

gc

Region ISteady state performance

Region IITransient

performance

Region IIIShow

system’s complexity

-20 dB/d

The static position, velocity, and acceleration error constants describe the low-frequency behaviour (region I) of type 0, type 1, and type 2 systems, respectively.For a given system, only one of the static error constants is finite and significant.(The larger the value of the finite static error constant, the higher the loop gain is as approaches zero.)The type of the system determines the slope of the log-magnitude curve at lowfrequencies (region I). Thus, information concerning the existence and magnitudeof the steady state error of a control system to a given input can be determined fromthe observation of the low-frequency region (region I) of the log-magnitude curve.


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57


Figure 2 shows an example ofthe log-magnitude plot of a type0 system. In such a system, themagnitude of G(j)H(j) equalsKp at low frequencies, or

Determination of static position error constants Kp from Bode diagram.Consider the feedback control system shown in Figure. Assume that the loop(open-loop) transfer function is given by

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

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

21

sTsTsTs

sTsTsTKsHsG

pN

mba G(s)R(s) Y(s)

H(s)

log 0

|G(j

)H(j

)|[d

B]

gc

-20 dB/d20 log10Kp

-40 dB/d

0 dB/decade

Figure 2:

pKjHjG

)()(lim0

From Bode diagram20log10|G(j)H(j)|= 20log10Kp pKjHjG

|)()(|lim

0

KeKsHsGK ss

sp

1

1,)()(lim

0The steady state error, ess;



Determination of static velocity error constants Kv from Bode diagram.Consider the feedback control system shown in Figure. Assume that the loop(open-loop) transfer function is given by

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

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

21

sTsTsTs

sTsTsTKsHsG

pN

mbaG(s)

R(s) Y(s)

H(s)

log 0

|G(j

)H(j

)|[d

B]

gc

-20 dB/d

-40 dB/dFigure 2:2 =1

vK10log20

1

Figure 2 shows an example of the log magnitude plot of a type 1 system. Theintersection of the initial -20 dB/d segment (or its extension) with the line 1 = 1has the magnitude 20 log Kv.This may be seen as follows: In a type 1 system

1,)()(

forj

KjHjG v

vv K

j

Klog20log20

Thus


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58


Determination of static velocity error constants Kv from Bode diagram.Consider the feedback control system shown in Figure. Assume that the loop(open-loop) transfer function is given by

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

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

21

sTsTsTs

sTsTsTKsHsG

pN

mbaG(s)

R(s) Y(s)

H(s)

.,1|||)()(| 1

1

11

vv Kor

j

KjHjG

The intersection of the initial -20dB/dsegment (or its extension) with the 0dB line has a frequency numericallyequal to Kv. To see this, define thefrequency at this intersection to be 1;then

v

sss

vK

esHssGLimK1

)(),()(0

The steady state error, ess;



Determination of static acceleration error constants Ka from Bode diagram.Consider the feedback control system shown in Figure. Assume that the loop(open-loop) transfer function is given by

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

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

21

sTsTsTs

sTsTsTKsHsG

pN

mbaG(s)

R(s) Y(s)

H(s)

Figure 2 shows an example of the log-magnitude plot of a type 2 system. The intersection of the initial -40-dB/decade segment (or its extension) with the = 1line has the magnitude of 20 log Ka. Since at low frequencies

1,)(

)()(2

forj

KjHjG a

aa K

j

Klog20|

)(|log20 12

it follows that

log 0

|G(j

)H(j

)|[d

B]

-40 dB/d

-20 dB/d

Figure 2: =1

aK10log20

1

-60 dB/d

aa K


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59


Determination of static acceleration error constants Ka from Bode diagram.For the given feedback control system the loop (open-loop) transfer function is;

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

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

21

sTsTsTs

sTsTsTKsHsG

pN

mbaG(s)

R(s) Y(s)

H(s)

The frequency a at the intersection of the initial -40 dB/d segment (or itsextension) with the 0 dB line gives the square root of Ka numerically. This can beseen from the following:

2aaK

log 0|G

(j

)H(j

)|[d

B]

-40 dB/d

-20 dB/d

Figure 2: =1

aK10log20

-60 dB/d

aa Kwhich yields

1)(

0)(

log20|)()(|log20

22

210

a

a

a

a

a

a

K

j

K

j

KjHjG

a



Determination of static acceleration error constants Ka from Bode diagram.For the given feedback control system the loop (open-loop) transfer function is;

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

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

21

sTsTsTs

sTsTsTKsHsG

pN

mba G(s)R(s) Y(s)

H(s)

2

22

210

1)(

0)(

log20

aa

a

a

a

a

a

a

K

K

j

K

j

K

a

ss

sa

Ke

sHsGsLimK

1)(

)()(2

0

The steady state error, ess;


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60


G(s)R(s) Y(s)

H(s)

Example 9.15; Consider the system given in (1). Determine the finite andsignificant static error constants for a given system. Draw the magnitude graph ofthe Bode diagram of the system. Obtain the finite static error constants of thesystem from the driven bode diagram.

Solution 9.15; Since the system is type 1, it has finite static velocity error constant, Kv. The system Bode diagram is;

)1()50)(5(

2500)()(

ssssHsG

)50log20

5log20log20(

2500log20)()(log20

2210

221010

1010

jHjG

10-1

100

101

102

-60

-40

-20

0

20

40

Magnitude(d

B)

Frequency (rad/sec)

1

i) From the magnitude graph of the Bode diagram

11 10 sKv



G(s)R(s) Y(s)

H(s)

Solution 9.15;)1(

)50)(5(

2500)()(

ssssHsG

)50log205log20log20(2500log20)()(log20 2210

2210101010 jHjG

10-1

100

101

102

-60

-40

-20

0

20

40

Magnitude(d

B)

Frequency (rad/sec)

1

ii) From Magnitude equation of the Bode diagram:

1

1

10

10

10

10

1log

20log20

s

K

K

K

v

v

v

iii) Without using the Bode diagram:

1

0010

)50)(5(

2500)()(

s

ssssLimsHssGLimK

ssv


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a) Plot the Bode Diagram the system.b) Obtain Phase Margin, PM, and Gain Margin, PM. c) Find the finite static error constant.

Gp(s)R(s)

H(s)

E(s) Y(s)Gc(s)

Figure 1: Satellite repair.

Homework 9.6: The space shuttle has been used to repair satellites and the Hubble telescope. Figure 1 illustrates how a crew member, with his feet strapped to the platform on the end of the shuttle's robotic arm, used his arms to stop the satellite's spin.

Figure 2: Satellite repair.

The control system of the robotic arm has a closed-loop transfer function;


1)(,)12(

1)(,60)(

sH

sssGsG pc



10

10)(

14.1

1)(,)(

2

ssH

sssGKsG pcc

a) Sketch the Bode plot of the loop transfer function for Kc=2.b) Obtain Phase Margin, PM, and Gain Margin, PM. c) Find the finite static error constant.d) Obtain the closed loop system response y(t) to a unit step input, R(s) = 1/s.e) Determine the bandwidth of the system.

Gp(s)R(s)

H(s)

E(s) Y(s)Gc(s)


9/17/2001

62

Electrical & Electronics Engineering

Chapter 09: Control Systems Design by the Root Locus Method

Remarks and Questions?


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Chapter 09 Frequency Response Method[1]