443 Lecture 23

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Systems Analysis and Control

Matthew M. Peet

Illinois Institute of Technology

Lecture 23: Drawing The Nyquist Plot

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Overview

In this Lecture, you will learn:

Review of Nyquist

Drawing the Nyquist Plot Using the Bode Plot

What happens at r=

Poles on the imaginary axis

Phase Margin and Gain Margin

Reading Stability Margins off the Nyquist Plot

M. Peet Lecture 23: Control Systems 2 / 30

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ReviewSystems in Feedback

The closed loop iskG(s)

1 +kG(s)

We want to know when1 +kG(s) = 0

Question: Does 1k

+G(s) have any zeros in the RHP?

G(s)k+

-

y(s)u(s)


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ReviewThe Nyquist Contour

Definition 1.The Nyquist Contour, CNis a contour which contains the imaginary axis andencloses the right half-place. The Nyquist contour is clockwise.

A Clockwise Curve

Starts at the origin. Travels along imaginary axis till r= .

Atr= , loops around clockwise.

Returns to the origin along imaginary axis.

We want to know if1

k+G(s)

has any zeros in the Nyquist Contour

r =


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ReviewContour Mapping Principle

Key Point: For a point on the mapped contour, s =G(s),

s = G(s)

We measure , not phase.

To measure the 360 resets in G(s)

We count the number of+360

resets in ! We count the number of times CG encircles the origin Clockwise.

The number of clockwise encirclementsof0 is

The #poles #zeros in the RHPs

s*= G(s)

= < G(s)


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The Nyquist ContourClosed Loop

The number of unstable closed-loop poles isN+P, where

N is the number of clockwiseencirclements of 1

k .

P is the number of unstable open-looppoles.

If we get our data from Bode, typically P = 0

-1/k

How to Plot the Nyquist Curve?


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Plotting the Nyquist DiagramExample

How are we to plot the Nyquist diagram for

G(s) = 1(1s+ 1)(2s+ 1)

1 = 1 2 =

1

10

First lets take a look at the root locus.

Obviously stable for any k >0.

12 10 8 6 4 2 0 26

4

2

0

2

4

6

Root Locus

Real Axis

ImaginaryAxis


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The Nyquist Plot

Bode Plot: Lets look at the Frequency Response.

The Bode plot can give us informationon |G| at different frequencies.

Point G |G|

A .1 0

1B 1 45 .7C 3 90 .3D 10 135 .07E 100 175 .001

The last two columns give us points onthe Nyquist diagram.

-120

-100

-80

-60

-40

-20

0

20

Mag

nitude(dB)

10-2

10-1

100

101

102

103

-180

-135

-90

-45

0

Phase

(deg)

Bode Diagram

Frequency (rad/sec)

A BC

D

E

A

B

C

D

E


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The Nyquist Plot

Plot the points from the Bode Diagram.

Point G |G|

A .1 0 1B 1 45 .7C 3 90 .3D 10 135 .07E 100 175 .001

We get

the upper half of the Nyquist diagramfrom symmetry.

-0.2 0 0.2 0.4 0.6 0.8 1 1.2-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Nyquist Diagram

Real Axis

ImaginaryAxis

A

B

C

D

E


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The Nyquist Plot

-0.2 0 0.2 0.4 0.6 0.8 1 1.2-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Nyquist Diagram

Real Axis

Imag

inaryAxis

A

B

C

D

E

There are no encirclements of 1k

.

Stable for all k >0.

We already knew that from Root Locus.M. Peet Lecture 23: Control Systems 10 / 30

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The Nyquist PlotExample 2

G(s) = 1

(s+ 1)3

First lets take a look at the root locus.

We expect instability for large k.

2.5 2 1.5 1 0.5 0 0.52

1.5

1

0.5

0

0.5

1

1.5

2

Root Locus

Real Axis

ImaginaryAxis


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The Nyquist Plot

Bode Plot: Lets look at the Frequency Response.

The Bode plot can give us informationon |G| at different frequencies.

Point G |G|

A .1 0 1

B .28 45 .95C 1 135 .35D 1.8 180 .1E 10 260 .001

-70

-60

-50

-40

-30

-20

-10

0

10

Magnitude(dB)

10-2

10-1

100

101

-270

-225

-180

-135

-90

-45

0

Phase(deg)

Bode Diagram

Frequency (rad/sec)

A B

C

D

E

A B

C

D

E


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The Nyquist Plot

Plot the points from the Bode Diagram.

Point G |G|

A .1 0 1B .28 45 .95

C 1 135 .35D 1.8 180 .1E 10 260 .001

Point D is especially important.

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

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Nyquist Diagram

Real Axis

ImaginaryAxis

A

B

C

D E


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The Nyquist Plot

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

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Nyquist Diagram

Real Axis

ImaginaryAxis

A

B

C

D E

Point D: Two CW encirclements when 1k

< .1 (N=2).

Instability for 1k

< .1

Stable for k

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The Nyquist Plot

Conclusion: We can use the Bode Plot to map theimaginary axis onto the Nyquist Diagram.

Question: What about the other part of the Nyquistcontour at r= ?

r =

Case 1: Strictly Proper.

lims

|G(s)| = 0

What happens at doesnt matter.

Case 2: Not Strictly Proper.

lims

|G(s)| =c

Constant Magnitude at .

Im(s)

Re(s)

< s-p1

< s-z = < s-p1= < s-p

2= < s-p

3


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The Nyquist Plot

Case 2: Not Strictly Proper.

Angle to all poles and zeros is the same.

Degree ofn(s) and d(s) the same. Number of Poles and Zeros the same.

The total angle is

G(s) =

n

i=1

(s zi)

n

i=1

(s pi)

= 0

The contour map at has

Constant magnitude. Zero angle.

The infinite loop is mapped to a single point!

Either (0, 0) or(c, 0).

Im(s)

Re(s)

< s-p1

< s-z = < s-p1= < s-p

2= < s-p

3

0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.20.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.8

Nyquist Diagram

Real Axis

ImaginaryAxis


Th N i Pl

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The Nyquist Plot

Another Problem: Recall the non-inverted pendulumwith PD feedback.

G(s) = s+ 1s2 +

gl

Magnitude goes to at =

gl.

Question How do we plot the Nyquist

Diagram?

2 1.5 1 0.5 0 0.5 1 1.5 2 2.5

6

4

2

0

2

4

6

Nyquist Diagram

Real Axis

ImaginaryAxis

40

20

0

20

40

60

80

100

120

140

160

Magnitude(dB)

102

101

100

101

102

135

90

45

0

45

Phase(deg)

Bode Diagram

Frequency (rad/sec)


Th N i Pl

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The Nyquist Plot

Problem: The Nyquist Contour passes through a pole.

Because of the pole, the argument principle is invalid.

What to do? r =

WeModify the Nyquist Contour.

We detour around the poles. Can detour to the right or left.

If we detour to the left, then the poles count as unstable open loop poles.

P=2

Assume we detour to the right.

P=0M. Peet Lecture 23: Control Systems 18 / 30

Th N i Pl

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The Nyquist Plot

Look at the detours at small radius.

Obviously, magnitude

Before the Detour, the phase from the pole is

(s p) = 90

In the middle of the Detour, the phase from thepole is

(s p) = 0

At the end of the Detour, the phase from the poleis

(s p) = 90

< (s - p) = 80o

< (s - p) = -5o

The total phase change through the detour is 180.

Corresponds to a CW loop at large radius.

If there are two or more poles, there is a -180 loop for each pole.


Th N i t Pl t

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The Nyquist Plot

Look at the following example:

G(s) =s+ 2

s2

There are 2 poles at the origin. At = 0,

G(0) =180

|G(0)|=

2 poles means 360

loop at = 0M. Peet Lecture 23: Control Systems 20 / 30

Th N i t Pl t

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The Nyquist Plot

Lets re-examine the pendulum problem with derivative feedback.

10 5 0 5 10

10

5

0

5

10

Nyquist Diagram

Real Axis

Imaginary

Axis

___=(g/l)

Now we can figure out what goes on at .

There is a 180 loop at each =

gl.


The Nyquist Plot

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The Nyquist Plot

Conclusion: The loops connect in a non-obvious way!

10 5 0 5 10

10

5

0

5

10

Nyquist Diagram

Real Axis

ImaginaryAxis

___=(g/l)

For 0< 1k

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Stability Margins

Recall the definitions of Gain Margin.

Definition 2.

The Gain Margin, Km = 1/|G()|when G() = 180

Let Km is the maximum stable gain inclosed loop.

KmG(s) is unstable in closed loop

Sometimes expressed in dB

It is easy to find the maximum stablegain from the Nyquist Plot.

Find the point 1Km

whichdestabilizes


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Stability Margins

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Stability MarginsExample

Suspension System with integral feedback

There is a pole at the origin.

CW loop at .

20 0 20 40 60 80 100

50

40

30

20

10

0

10

20

30

40

50

Nyquist Diagram

Real Axis

ImaginaryAxis

Conclusion: Stable for 1k

> 9.5.

Stable for k < .105

Km =.105 or 19.5dB

-14 -12 -10 -8 -6 -4 -2 0 2-8

-6

-4

-2

0

2

4

6

8

Nyquist Diagram

Real Axis

ImaginaryAxis

}

9.5


Stability Margins

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Stability Margins

Question: What is the effect of a phase change on the Nyquist Diagram.

A shift in phase changes the angle of all points.

A Rotation about the origin. Will we rotate into instability?

-0.2 0 0.2 0.4 0.6 0.8 1 1.2-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Nyquist Diagram

Real Axis

ImaginaryAxis

-0.2 0 0.2 0.4 0.6 0.8 1 1.2-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

Nyquist Diagram

Real Axis

ImaginaryAxis


Stability Margins

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Stability Margins

Recall the definitions of Phase Margin.

Definition 3.

The Phase Margin, M is the uniformphase change required to destabilize thesystem under unitary feedback.

3 2 1 0 1 2 33

2

1

0

1

2

3

Nyquist Diagram

RealAxis

ImaginaryAxis

M

3 2 1 0 1 2 33

2

1

0

1

2

3

Nyquist Diagram

RealAxis

ImaginaryAxis


Stability Margins

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The Suspension Problem

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Nyquist Diagram

Real Axis

ImaginaryA

xis

M

Looking at the intersection with the circle:

Phase Margin: M=40

Gain Margin is infinite.M. Peet Lecture 23: Control Systems 28 / 30

Stability Margins

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The Inverted Pendulum Problem

-2 -1.5 -1 -0.5 0 0.5 1

-1.5

-1

-0.5

0

0.5

1

1.5

Nyquist Diagram

Real Axis

Imag

inaryAxis

M

Even though open-loop is unstable, we can still find the phase margin:

Phase Margin: M=35

Gain Margin is technically undefined because open loop is unstable.

There is a minimum gain, not a maximum.


Summary

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Summary

What have we learned today?

Review of Nyquist

Drawing the Nyquist Plot

Using the Bode Plot

What happens at r=

Poles on the imaginary axis

Phase Margin and Gain Margin

Reading Stability Margins off the Nyquist Plot

Next Lecture: Controller Design in the Frequency Domain


443 Lecture 23

Documents

Transcript of 443 Lecture 23