FR – What is Frequency Response (RF)? y(t y t ss (€¦ · · 2009-02-14• Steady state response yss (t) (after transient dies out) ... GGZ Week 12-13: Frequency Response Page

2009 Spring ME451 - GGZ Page 1Week 12-13: Frequency Response

• We would like to analyze a system property by applying

a test sinusoidal input u(t) and observing a response y(t).

• Steady state response yss(t) (after transient dies out) of a

system to sinusoidal inputs is called frequency

response.

SystemSystem

FR FR –– What is Frequency Response (RF)?What is Frequency Response (RF)?

)(ty)(ty

ss

tAtu ωsin)( =


• RC circuit

• Input a sinusoidal voltage u(t)

• What is the output voltage y(t)?

RR

CC

FR FR –– A Simple Example A Simple Example (1)(1)

)(1

)( sIsC

sY =

)(tu

)()1

()( sIsC

RsU +=

)(ty

1

1)(

+=

RCssG


• TF (R=C=1)

• Let

0 5 10 15 20 25 30 35 40 45 50-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

r

y

At steadyAt steady--state, state, uu((tt)) and and yy((tt)) has same frequency, has same frequency,

but different amplitude and phase!but different amplitude and phase!


1

1)(

+=

ssG

ttu sin)( =

uy


• Derivation of y(t)

• Inverse Laplace

0 as t goes to infinity.0 as t goes to infinity.

Partial fraction expansionPartial fraction expansion

(Derivation for general (Derivation for general GG((ss)) is given at the end of lecture slide.)is given at the end of lecture slide.)


)1

1

1

1(

2

1

1

1

1

1)()()(

22 +

+−+

+=

+⋅

+==

s

s

ssssUsGsY

( )ttety sincos2

1)(

1 +−= −

( ) )45sin(2

1sincos

2

1)(

o

sstttty −=+−=


• How is the steady state output of a linear system when

the input is sinusoidal?

• Steady state output:

– Frequency is same as the input frequency

– Amplitude is that of input (A) multiplied by

– Phase shifts

GainGain

FR FR –– System Response to Sinusoidal InputSystem Response to Sinusoidal Input

)(ty )(tyss

tAtu ωsin)( =)(sG

))(sin()()( ωωω jLGtjGAtyss

+=

ω)( ωjG

)( ωjG∠


• For a stable system G(s), G(jw) (ω is real and positive) is called frequency response function (FRF).

• FRF is a complex number, and thus, has an amplitude and

a phase.

• First order example

ReRe

ImIm

FR FR –– Frequency Response FunctionFrequency Response Function

1

1)(

+=

ssG

−=+∠−∠=∠+

=

− ωωωω

ω

1

2

tan)1()1()(1

1)(

jjG

jG

1

1)(

+=

ωω

jjG


• Second order system

ReRe

ImIm

23

2)(

2 ++=

sssG

ωωωωω

32

2

2)(3)(

2)(

22 jjjjG

+−=

++=

FR FR –– Another FRF ExampleAnother FRF Example

−−=

+−∠−∠=∠

+−=

−

2

1

2

222

2

3tan

)32()2()(

9)2(

2)(

ω

ω

ωωω

ωωω

jjG

jG

ω3j

22 ω−


• FRF

• Two graphs representing FRF

– Bode diagram (Bode plot)

– Nyquist diagram (Nyquist plot)

FR FR –– First Order Example RevisitedFirst Order Example Revisited

1

1)(

+=

ωω

jjG


• Bode diagram consists of gain plot & phase plot

LogLog--scalescale

FR FR –– Body Diagram (Plot) of Body Diagram (Plot) of GG(j(jωω))

(deg) )( ωjG∠

(dB) )(log2010

ωjG


• TF

10-2

10-1

100

101

102

-50

-40

-30

-20

-10

0

10-2

10-1

100

101

102

-100

-80

-60

-40

-20

0

Corner frequencyCorner frequency

FR FR –– Body Diagram of a 1Body Diagram of a 1stst Order SystemOrder System

1

1)(

+=

TssG

<<

>>≈

+=

T

T

TjjG

Tjω

ω

ωω

ω1 if

1 if1

1

1)(

1

T

1

45−

90−

dB/decade20−

1

1)(

+=

ssG


• First order system

FR FR –– Sketching Body DiagramSketching Body Diagram

1

1)(

+=

ssG

11.0

1)(

+=

ssG

110

1)(

+=

ssG


• Bode diagram shows amplification and phase shift of a

system output for sinusoidal inputs with various frequencies.

• It is very useful and important in analysis and design of

control systems.

• The shape of Bode plot contains information of stability, time responses, and much more!

• It can also be used for system identification. (Given FRF

experimental data, obtain a transfer function that matches

the data.)

FR FR –– Remarks on Body DiagramRemarks on Body Diagram


• Sweep frequencies of sinusoidal signals and obtain FRF

data (i.e., gain and phase).

• Select G(s) so that G(jω) fits the FRF data.

Agilent Technologies: FFT Dynamic Signal Analyzer

UnknownUnknown

systemsystem

Generate sin signalsGenerate sin signals

Sweep frequenciesSweep frequencies

Collect FRF dataCollect FRF data

Select Select G(sG(s))

FR FR –– Body Diagram: System IdentificationBody Diagram: System Identification


• Frequency response is a steady state response of

systems to a sinusoidal input.

• For a linear system, sinusoidal input generates

sinusoidal output with same frequency but different amplitude and phase.

• Bode plot is a graphical representation of frequency

response function. (“bode.m”)

• Next, Bode diagram of simple transfer functions

FR FR –– Body Diagram SummaryBody Diagram Summary


Term having denominator of Term having denominator of G(sG(s))

0 as t goes to infinity.0 as t goes to infinity.

FR FR –– Derivation of Frequency ResponseDerivation of Frequency Response

)()()()()( 21

22sC

js

k

js

k

s

AsGsUsGsY

g+

−+

+=

+==

ωωω

ω

==+

−=

−=

−−=

++=

−→

−→

j

jAG

j

AjG

s

AsGjsk

j

jAG

j

AjG

s

AsGjsk

js

js

2

)(

2)()()(lim

2

)(

2)()()(lim

222

221

ω

ω

ωω

ω

ωω

ω

ω

ωω

ω

ωω

ω

ω

4444 34444 21))(sin(

))(())((

2)()(

ωω

ωωωω

ω

jGt

jGtjjGtj

ssj

eejGAty

∠+

∠+−∠+ −=

)}({)( 1

21sCLekekty

g

tjtj −− ++= ωω


• Basic functions

– Constant gain

– Differentiator and integrator

– Double integrator

– First order system and its inverse

– Second order system

– Time delay

• Product of basic functions

1. Sketch Bode plot of each factor, and

2. Add the Bode plots graphically.

Main advantage of Bode plot!Main advantage of Bode plot!

FR FR –– Sketching Body DiagramSketching Body Diagram


10-2

10-1

100

101

102

19

19.5

20

20.5

21

10-2

10-1

100

101

102

-1

-0.5

0

0.5

1

• TF

FR FR –– Body Diagram (Constant Gain)Body Diagram (Constant Gain)

ωωω ∀=∠=⇒= ,0)( ,)()( ojGKjGKsG

10=K


10-2

10-1

100

101

102

-40

-20

0

20

40

10-2

10-1

100

101

102

89

89.5

90

90.5

91

• TF

FR FR –– Body Diagram (Differentiator)Body Diagram (Differentiator)

ωωωω ∀=∠=⇒= ,90)( ,)()( ojGjGssG

dB/decade20+


• TF

10-2

10-1

100

101

102

-40

-20

0

20

40

10-2

10-1

100

101

102

-91

-90.5

-90

-89.5

-89

Mirror image of Mirror image of

the bode plot of the bode plot of

1/s with respect 1/s with respect

to to ωω--axis.axis.

ωωω

ω ∀−=∠=⇒= ,90)( ,1

)(1

)( ojGjG

ssG

e20dB/decad-

FR FR –– Body Diagram (Integrator)Body Diagram (Integrator)


10-2

10-1

100

101

102

-100

-50

0

50

100

10-2

10-1

100

101

102

-181

-180.5

-180

-179.5

-179

• TF

FR FR –– Body Diagram (Double Integrator)Body Diagram (Double Integrator)

ωωω

ω ∀−=∠=⇒= ,180)( ,1

)(1

)(22

ojGjG

ssG

e40dB/decad-


• TF

10-2

10-1

100

101

102

-50

-40

-30

-20

-10

0

10-2

10-1

100

101

102

-100

-80

-60

-40

-20

0

Corner frequencyCorner frequency

Straight lineStraight line

approximationapproximation

e20dB/decad-

FR FR –– Body Diagram (1Body Diagram (1stst Order System)Order System)

1

1)(

+=

TssG

<<

>>≈

+=

T

T

TjjG

Tjω

ω

ωω

ω1 if

1 if1

1

1)(

145−

90−

1

1)(

+=

ssG

T

1


• TF

10-2

10-1

100

101

102

0

10

20

30

40

50

10-2

10-1

100

101

102

0

20

40

60

80

100

Mirror image of the Mirror image of the

original bode plot original bode plot

with respect to with respect to ωω--

axis.axis.

FR FR –– Body Diagram (Inversed 1Body Diagram (Inversed 1stst Order System)Order System)

1

1

11)(

−

+=+=

TsTssG

e20dB/decad+


• TF

• Resonant freq

• Peak gain

10-1

100

101

-60

-40

-20

0

20

10-1

100

101

-200

-150

-100

-50

0

resonanceresonance

FR FR –– Body Diagram (2Body Diagram (2ndnd Order System)Order System)

22

2

2)(

nn

n

sssG

ωςω

ω

++=

221: ςωω −=nr

1 if

2

1

12

1

2

<<

≈−

ς

ςςς

90−

180−

1.0=ς

1=ς

1=ς

dB/decade40−


1 0-2

1 0-1

1 00

1 01

1 02

-1

-0 .5

0

0 .5

1

1 0-2

1 0-1

1 00

1 01

1 02

-6 0 0 0

-4 0 0 0

-2 0 0 0

0

• TF

Huge phase lag!Huge phase lag!

As can be explained with As can be explained with NyquistNyquist stability criterion, stability criterion,

this phase lag causes instability of the closedthis phase lag causes instability of the closed--loop system,loop system,

and hence, the difficulty in control.and hence, the difficulty in control.

FR FR –– Body Diagram (Time Delay System)Body Diagram (Time Delay System)

(rad) )( , ,1)()( TjGjGesG Ts ωωωω −=∠∀=⇒= −


• Use Matlab “bode.m” to obtain precise shape.

• ALWAYS check the correctness of

– Low frequency gain (DC gain)

– High frequency gain

• Example

FR FR –– Body Diagram RemarkBody Diagram Remark

)0(G

)(∞G

dB62)0( ⇒=G

dB2010)( ⇒=∞G

5

)1(10)(

+

+=

s

ssG


• Sketch bode plot.

FR FR –– Body Diagram ExercisesBody Diagram Exercises

110

1)(

+=

ssG

1)( =sG 1.0)( =sG 10)( −=sG

2)( ssG =3)( ssG =

3

1)(

ssG =

110

10)(

+=

ssG 110)( += ssG

12)( += ssG 15

)( +=s

sG42

4)(

2 ++=

sssG


• Bode plot of various simple transfer functions.

– Constant gain

– Differentiator, integrator

– 1st order and 2nd order systems

– Time delay

• Sketching Bode plot is just ….

– to get a rough idea of the characteristic of a system.

– to interpret the result obtained from computer.

– to detect erroneous result from computer.

FR FR –– Body Diagram SummaryBody Diagram Summary


10-2

10-1

100

101

102

19

19.5

20

20.5

21

10-2

10-1

100

101

102

-1

-0.5

0

0.5

1

10-2

10-1

100

101

102

-40

-20

0

20

40

10-2

10-1

100

101

102

89

89.5

90

90.5

91

FR FR –– Body Diagram of Basic Functions Body Diagram of Basic Functions (Review (Review -- 1)1)

KsG =)(

K10

log20

ssG =)(

dB/decade20+


10-2

10-1

100

101

102

-100

-50

0

50

100

10-2

10-1

100

101

102

-181

-180.5

-180

-179.5

-179

10-2

10-1

100

101

102

-40

-20

0

20

40

10-2

10-1

100

101

102

-91

-90.5

-90

-89.5

-89


ssG

1)( =

2

1)(

ssG =

dB/decade20+dB/decade20−


10-2

10-1

100

101

102

-50

-40

-30

-20

-10

0

10-2

10-1

100

101

102

-100

-80

-60

-40

-20

0

10-2

10-1

100

101

102

0

10

20

30

40

50

10-2

10-1

100

101

102

0

20

40

60

80

100


1

1)(

+=

TssG

1)( += TssG

dB/decade20+

dB/decade20−

T

1

T

1


10-1

100

101

-60

-40

-20

0

20

10-1

100

101

-200

-150

-100

-50

0

Bode plot of a 2nd order system


resonanceresonance

22

2

2)(

nn

n

sssG

ωςω

ω

++=

221: ςωω −=nr

1 if

2

1

12

1

2

<<

≈−

ς

ςςς

90−

180−

1.0=ς

1=ς

1=ς

dB/decade40−

Resonant frequency

Peak Gain


Bode plot of a time delay


1 0-2

1 0-1

1 00

1 01

1 02

-1

-0 .5

0

0 .5

1

1 0-2

1 0-1

1 00

1 01

1 02

-6 0 0 0

-4 0 0 0

-2 0 0 0

0







• Bode plot of a series connection G1(s)G2(s) is the addition of

each Bode plot of G1 and G2.

– Gain

– Phase

• We use this property to design C(s) so that G(s)C(s) has a

“desired” shape of Bode plot.

FR FR –– Advantages of Body DiagramAdvantages of Body Diagram

)(log20)(log20)()(log202101102110

ωωωω jGjGjGjG +=

)()()()(1121

ωωωω jGjGjGjG ∠+∠=∠


• Use polar representation

Then,

Therefore,

FR FR –– Short Proofs of Gain and Phase FormulaShort Proofs of Gain and Phase Formula

)()(

212121)()()()(

ωωωωωω jGjjGjejGjGjGjG

∠+∠=

)(log20)(log20

))()((log20)()(log20

210110

21102110

ωω

ωωωω

jGjG

jGjGjGjG

+=

=

)(

222)()(

ωωω jGjejGjG

∠=)(

111)()(

ωωω jGjejGjG

∠=

)()()()(21

)()(

2121 ωωωω ωω

jGjGejGjGjGjjGj

∠+∠=∠=∠∠+∠


• Sketch the Bode plot of a transfer function

1. Decompose G(s) into a product form:

2. Sketch a Bode plot for each component on the same

graph.

3. Add them all on both gain and phase plots.

FR FR –– Example 1 Example 1 (1)(1)

ssG

110)( ⋅=

ssG

10)( =


dBdB

degdeg

--2020

FR FR –– Example 1 Example 1 (2)(2)

ssG

10)( =

ssG

1)(

2=

10)(1 =sG


dBdB

degdeg

--2020

FR FR –– Example 2Example 2

ssG

1)(

2=

1.0)(1 =sG

ssG

1.0)( =


dBdB

degdeg

--2020

--4040ssG

1)(

2=

)12(

1)(

+=

sssG


12

1)(1

+=

ssG


dBdB

degdeg

+20+20

--2020

+45+45 --4545

--4545


)10(

)1(10)(

+

+=

s

ssG

1)(1

+= ssG

11.0

1)(

2+

=s

sG


dBdB

degdeg

+20+20

--4040

+45+45 --4545

--9090

--2020


2)(1

=sG

1)(2

+= ssG

22)11.0(

1)(

+=

ssG

22)10(

)1(200)(

+

+=

s

ssG


• Sketch the Bode plot of:

• Find a transfer function having the gain plot:

0dB0dB

20dB20dB

22 2020

--4040+20+20

Ans.Ans.

FR FR –– ExercisesExercises

2)1(

20)(

+=

sssG

2)1(

8)(

+=

s

ssG

2

2)(

s

ssG

+=

)2(

2)(

2 +=

sssG

)1()1(

5)(

212

201 ++

=ss

ssG


• Consider the feedback system

• Fundamental questions

– If G, C, and H are stable, is closed-loop system always

stable?

– If G, C, and H are unstable, is closed-loop system always

unstable?

FR FR –– Stability of Feedback SystemsStability of Feedback Systems

)(sG)(sC

)(sH

)(sR )(sY


• Closed-loop stability can be determined by the roots of

the characteristic equation

• CL system is stable if the Characteristic Equation has all

roots in the open left half plane.

• How to check the stability?

– Compute all the roots.

– Routh-Hurwitz stability criterion

– Nyquist stability criterion

FR FR –– ClosedClosed--Loop Stability CriterionLoop Stability Criterion

0)(1 =+ sL )()()(:)( sHsCsGsL =


• Nyquist path

(very large)(very large)

• Nyquist plot

ss

L(sL(s) when s ) when s

moves on moves on

NyquistNyquist path path

ReRe

ImIm

ReRe

ImIm

jjωωωωωωωω

FR FR –– NyquistNyquist PlotPlot


• First order system

10-2

10-1

100

101

102

-40

-20

0

10-2

10-1

100

101

102

-80

-60

-40

-20

0

Bode plotBode plot NyquistNyquist plotplot

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

FR FR –– Example of Body and Example of Body and NyquistNyquist Plot Plot (1)(1)

1

1)(

+=

ssL


• Second order system

Bode plotBode plot

10-2

10-1

100

101

102

-100

-50

0

10-2

10-1

100

101

102

-150

-100

-50

0

NyquistNyquist plotplot

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1


2)1(

1)(

+=

ssL


• Third order system

Bode plotBode plot

10-2

10-1

100

101

102

-150

-100

-50

0

10-2

10-1

100

101

102

-200

-100

0

NyquistNyquist plotplot

-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1


3)1(

1)(

+=

ssL


• Z: # of CL poles in open RHP

• P: # of OL poles in open RHP (given)

• N: # of clockwise encirclement around -1

by Nyquist plot of OL transfer function L(s)

(counted by using Nyquist plot of L(s))

Remark: N = -1: a counter-clockwise encirclement

FR FR –– NyquistNyquist Stability CriterionStability Criterion

0: stable is system CL =+=⇔ NPZ


• Clockwise

ReRe

ImIm

• Counter-clockwise

ReRe

ImIm

FR FR –– Encirclement on Encirclement on NyquistNyquist PlotPlot


• If Nyquist plot passes the point -1, it means that the

closed-loop system has a pole on the imaginary axis

(and thus, not stable).

ReRe

ImIm

FR FR –– Encirclement on Encirclement on NyquistNyquist Plot RemarkPlot Remark

00 somefor 1)( ωω −=jL

00 somefor 0)(1 ωω =+ jL

0at pole a has system CL ωj


-1.5 -1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

-1.5 -1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

For For L(sL(s) of 2) of 2ndnd order and constant numerator, order and constant numerator,

gain increase never lead to unstable CL system!gain increase never lead to unstable CL system!

Gain increaseGain increase

FR FR –– Example for 2Example for 2ndnd Order Order L(sL(s))

2)1(

1)(

+=

ssL

2)1(

10)(

+=

ssL

stable CL0,0 ⇒== NP stable CL0,0 ⇒== NP


-5 0 5 10 15-15

-10

-5

0

5

10

15

-2 -1.5 -1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

Gain increaseGain increase

For For L(sL(s) with relative degree 3, ) with relative degree 3,

gain increase eventually lead to unstable CL system!gain increase eventually lead to unstable CL system!

FR FR –– Example for 3Example for 3rdrd Order Order L(sL(s))

stable CL0,0 ⇒== NP unstable CL1,0 ⇒== NP

3)1(

1)(

+=

ssL

3)1(

15)(

+=

ssL


BB

• An ray is drawn from -1 point in any convenient direction. Then,

ReRe

ImIm

AA

Count N for cases:Count N for cases:

�� A = A = --11

�� B = B = --11

FR FR –– How to Count # of EncirclementsHow to Count # of Encirclements

direction) wisecountclockin plot Nyquist by ray of crossing of (# -

direction) clockwiseplot Nyquist by ray of Crossing of #(=N


• Nyquist stability criterion allows us to determine the stability

of CL system from a knowledge of the G(jω) of OL system.

• If an OL system is stable, it requires only frequency

response data of OL system (TF model L(s) is not

necessary).

• It can deal with time delay, which Routh-Hurwitz criterion

cannot.

• We often draw only half of Nyquist plot. (The other half is

mirror image w.r.t. real axis.)

FR FR –– Notes on Notes on NyquistNyquist Stability CriterionStability Criterion


• Nyquist plot (Matlab command: “nyquist.m”)

• Nyquist stability criterion for feedback stability

• Exercises: For (half of) Nyquist plot below, count N for

each case.

•• A = A = --1 1 (Ans. N=2)(Ans. N=2)

•• B = B = --1 1 (Ans. N=0)(Ans. N=0)

•• C = C = --1 1 (Ans. N=2)(Ans. N=2)

•• D = D = --1 1 (Ans. N=0)(Ans. N=0)

•• E = E = --1 1 (Ans. N=2)(Ans. N=2)

•• F = F = --1 1 (Ans. N=0)(Ans. N=0)

AA

BB

CC

DD

EE

FF

FR FR –– NyquistNyquist Summary and Summary and ExcerciseExcercise


• Consider the stability of the feedback system

• To analyze the stability of the feedback system, we can use Nyquist stability criterion.

• Nyquist stability criterion utilizes the Nyquist plot, a

representation for G(jω).

FR FR –– Stability of Feedback SystemsStability of Feedback Systems

)(sG)(sC

)(sH

)(sR )(sY


• Nyquist path


• Nyquist plot

ss

L(sL(s) when s ) when s

moves on moves on

NyquistNyquist path path

ReRe

ImIm

ReRe

ImIm

jjωωωωωωωω

FR FR –– NyquistNyquist Plot (Review)Plot (Review)


• Z: # of CL poles in open RHP

• P: # of OL poles in open RHP (given)

• N: # of clockwise encirclement around -1

by Nyquist plot of OL transfer function L(s)

(counted by using Nyquist plot of L(s))

Remark: N = -1: a counter-clockwise encirclement

FR FR –– NyquistNyquist Stability Criterion (Review)Stability Criterion (Review)

0: stable is system CL =+=⇔ NPZ


• Unstable L(s)

• Stable L(s)

• L(s) with an integrator

• L(s) with a double integrator

• L(s) with a time-delay

FR FR –– NyquistNyquist Stability (More Examples)Stability (More Examples)


-1.2 -1 -0.8 -0.6 -0.4 -0.2 0-0.1

-0.05

0

0.05

0.1

FR FR –– NyquistNyquist Stability Example Stability Example (Unstable (Unstable L(sL(s) (1))) (1))

stable is system CL0)1(1: ⇒=−+=+= NPZ

)3)(2)(1(

8)(

++−=

ssssL


-2 -1.5 -1 -0.5 0

-0.1

-0.05

0

0.05

0.1

0.15

-1 -0.8 -0.6 -0.4 -0.2 0

-0.06

-0.04

-0.02

0

0.02

0.04

0.06Gain increaseGain increase Gain decreaseGain decrease


)3)(2)(1(

11)(

++−=

ssssL

)3)(2)(1(

5)(

++−=

ssssL

unstable is system CL

2)1(1:

⇒

=++=+= NPZ

unstable is system CL

1)0(1:

⇒

=++=+= NPZ


Interpretation by root locus

11--22--33

ReRe

ImIm

Open loop gain increase will first stabilize, Open loop gain increase will first stabilize,

and then, destabilize the closedand then, destabilize the closed--loop system.loop system.Stabilizing!Stabilizing!

Destabilizing!Destabilizing!


)3)(2)(1(

1)(

++−=

ssssL


• IF P=0 (i.e., if L(s) has no pole in open RHP or stable)

This fact is very important since openThis fact is very important since open--loop systems loop systems

in many practical problems have no pole in open RHP!in many practical problems have no pole in open RHP!

FR FR –– NyquistNyquist Stability Criterion: A Special CaseStability Criterion: A Special Case

0stable is system CL =+=⇔ NPZ:

0stable is system CL =⇔ N


-1 0 1 2 3-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

-5 0 5 10 15 20-15

-10

-5

0

CL stableCL stable CL unstableCL unstable

FR FR –– NyquistNyquist Stability Example Stability Example (stable (stable L(sL(s) (1))) (1))

)3)(2)(1(

20)(

+++=

ssssL

)3)(2)(1(

100)(

+++=

ssssL


Interpretation by root locus

--11--22--33

ReRe

ImIm

Open loop gain increase will Open loop gain increase will

destabilize the closeddestabilize the closed--loop system.loop system.

FR FR –– NyquistNyquist Stability Example Stability Example (stable (stable L(sL(s) (2))) (2))

)3)(2)(1(

1)(

+++=

ssssL


• We modify Nyquist path:


ss

ReRe

ImIm

(very small)(very small)

FR FR –– NyquistNyquist Example Example ((L(sL(s) with an integrator (1))) with an integrator (1))


-1 -0.8 -0.6 -0.4 -0.2 0

-15

-10

-5

0

5

10

15

FR FR –– NyquistNyquist Example Example ((L(sL(s) with an integrator (2))) with an integrator (2))

)1(

1)(

+=

sssL

Stable CL0,0 ⇒== NP


• For small |s|,

• When s moves as

L(s) moves as

ReRe

ImIm

(very small)(very small)

ss

L(sL(s))

Note: If Note: If L(sL(s) has no open RHP pole,) has no open RHP pole,

we are interested in (and draw) we are interested in (and draw) NyquistNyquist plot plot

ONLY around the critical point ONLY around the critical point --1.1.

FR FR –– NyquistNyquist Example Example ((L(sL(s) for modified Path)) for modified Path)

ssssL

1

)1(

1)( ≈

+=

ojes 900, →== θε θ

ojesL 900,

1)( 1 →=≈ − θ

ε

θ


-5 -4 -3 -2 -1 0-5

0

5

FR FR –– NyquistNyquist Example Example ((L(sL(s) with an double integrator)) with an double integrator)

)1(

1)(

2+

=ss

sL

unStable CL2,0 ⇒== NP


-1 0 1 2 3-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

RouthRouth--Hurwitz is NOT applicable!Hurwitz is NOT applicable!

FR FR –– NyquistNyquist Example Example ((L(sL(s) with a time delay (1))) with a time delay (1))

)3)(2)(1(

20)(

+++=

ssssL

unStable CL1,0 ⇒== NP

Stable CL0,0 ⇒== NP

)3)(2)(1(

20)(

7.0

+++=

−

sss

esL

s


1 0-2

1 0-1

1 00

1 01

1 02

-1

-0 .5

0

0 .5

1

1 0-2

1 0-1

1 00

1 01

1 02

-6 0 0 0

-4 0 0 0

-2 0 0 0

0

• TF






FR FR –– NyquistNyquist Example Example ((L(sL(s) with a time delay (2))) with a time delay (2))


• Examples for Nyquist stability criterion

• Next,

– PID control

– Relative stability

• Exercise: Suppose L(s) is stable and has Nyquist plot below. Find the range of OL gain K>0 for which CL

system is stable.

--0.50.5--22--3.33.3

(Ans. 0 < K< 1/3.3, (Ans. 0 < K< 1/3.3, ½½ < K <2)< K <2)

--11

FR FR –– NyquistNyquist SummarySummary

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