Systems Analysis and Control

Matthew M. PeetIllinois Institute of Technology

Lecture 23: Drawing The Nyquist Plot

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

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)

M. Peet Lecture 23: Control Systems 3 / 30

ReviewThe Nyquist Contour

Definition 1.

The Nyquist Contour, CN is 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 =∞.

• At r =∞, loops around clockwise.

• Returns to the origin along imaginary axis.

We want to know if

1

k+G(s)

has any zeros in the Nyquist Contour

r = ∞

M. Peet Lecture 23: Control Systems 4 / 30

ReviewContour Mapping Principle

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

∠s∗ = ∠G(s)

• We measure θ, not phase.

To measure the 360◦ resets in ∠G(s)

• We count the number of +360◦ resets in θ!

• We count the number of times CG encircles the origin Clockwise.

The number of clockwise encirclementsof 0 is

• The #poles −#zeros in the RHPs

s*= G(s)

θ = < G(s)

M. Peet Lecture 23: Control Systems 5 / 30

The Nyquist ContourClosed Loop

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

• N is the number of clockwiseencirclements of −1k .

• 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?

M. Peet Lecture 23: Control Systems 6 / 30

Plotting the Nyquist DiagramExample

How are we to plot the Nyquist diagram for

G(s) =1

(τ1s+ 1)(τ2s+ 1)

• τ1 = 1• τ2 = 1

10

First lets take a look at the root locus.

Obviously stable for any k > 0.

−12 −10 −8 −6 −4 −2 0 2−6

−4

−2

0

2

4

6

Root Locus

Real Axis

Imag

inar

y A

xis

M. Peet Lecture 23: Control Systems 7 / 30

The Nyquist Plot

Bode Plot: Lets look at the Frequency Response.

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

Point ω ∠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

Ma

gn

itu

de

(d

B)

10-2

10-1

100

101

102

103

-180

-135

-90

-45

0

Ph

ase

(d

eg

)

Bode Diagram

Frequency (rad/sec)

A BC

D

E

A

B

C

D

E

M. Peet Lecture 23: Control Systems 8 / 30

The Nyquist Plot

Plot the points from the Bode Diagram.

Point ω ∠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

Ima

gin

ary

Axi

s

A

B

C

D

E

M. Peet Lecture 23: Control Systems 9 / 30

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

Ima

gin

ary

Axi

s

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

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 0.5−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Root Locus

Real Axis

Imag

inar

y A

xis

M. Peet Lecture 23: Control Systems 11 / 30

The Nyquist Plot

Bode Plot: Lets look at the Frequency Response.

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

Point ω ∠G |G|A .1 0◦ 1B .28 −45◦ .95C 1 −135◦ .35D 1.8 −180◦ .1E 10 −260◦ .001

-70

-60

-50

-40

-30

-20

-10

0

10

Ma

gn

itu

de

(d

B)

10-2

10-1

100

101

-270

-225

-180

-135

-90

-45

0

Ph

ase

(d

eg

)

Bode Diagram

Frequency (rad/sec)

A B

C

D

E

AB

C

D

E

M. Peet Lecture 23: Control Systems 12 / 30

The Nyquist Plot

Plot the points from the Bode Diagram.

Point ω ∠G |G|A .1 0◦ 1B .28 −45◦ .95C 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

Ima

gin

ary

Axi

s

A

B

C

D E

M. Peet Lecture 23: Control Systems 13 / 30

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

Ima

gin

ary

Axi

s

A

B

C

D E

Point D: Two CW encirclements when − 1k < −.1 (N=2).

• Instability for − 1k < −.1

• Stable for k < 10.

• Could have used Routh Table.

M. Peet Lecture 23: Control Systems 14 / 30

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 ∞ doesn’t 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

M. Peet Lecture 23: Control Systems 15 / 30

The Nyquist Plot

Case 2: Not Strictly Proper.

• Angle to all poles and zeros is the same.

• Degree of n(s) and d(s) the same.I 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 2.2−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

Nyquist Diagram

Real Axis

Imag

inar

y A

xis

M. Peet Lecture 23: Control Systems 16 / 30

The Nyquist Plot

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

G(s) =s+ 1

s2 +√

gl

Magnitude goes to ∞ at ω =√

gl .

Question How do we plot the NyquistDiagram?

−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

Imag

inar

y A

xis

Matlab can’t help us this time.

−40

−20

0

20

40

60

80

100

120

140

160

Mag

nitu

de (

dB)

10−2

10−1

100

101

102

−135

−90

−45

0

45

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

M. Peet Lecture 23: Control Systems 17 / 30

The Nyquist Plot

Problem: The Nyquist Contour passes through a pole.

Because of the pole, the argument principle is invalid.

What to do? r = ∞

We Modify 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

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.

M. Peet Lecture 23: Control Systems 19 / 30

The Nyquist Plot

Look at the following example:

G(s) =s+ 2

s2

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

I ∠G(0) = −180◦I |G(0)| =∞

• 2 poles means −360◦ loop at ω = 0M. Peet Lecture 23: Control Systems 20 / 30

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

Ima

gin

ary

Axi

s

___ω=√(g/l)

Now we can figure out what goes on at ∞.

• There is a −180◦ loop at each ω =√

gl .

M. Peet Lecture 23: Control Systems 21 / 30

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

Ima

gin

ary

Axi

s

___ω=√(g/l)

For 0 < −1k < 1, we have N = 1

−400 −200 0 200 400 600 800 1000 1200 1400−1500

−1000

−500

0

500

1000

1500

Nyquist Diagram

Real Axis

Imag

inar

y A

xis

M. Peet Lecture 23: Control Systems 22 / 30

Stability Margins

Recall the definitions of Gain Margin.

Definition 2.

The Gain Margin, Km = 1/|G(ıω)|when ∠G(ıω) = 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 −1Kmwhich

destabilizes

M. Peet Lecture 23: Control Systems 23 / 30

Stability MarginsExample

Recall

G(s) =1

(s+ 1)3

Stability: Stable for k < 10.

Km = 10 or 20dB.

-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

Ima

gin

ary

Axi

s

A

B

C

D E

M. Peet Lecture 23: Control Systems 24 / 30

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

Imag

inar

y A

xis

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

Ima

gin

ary

Axi

s

}9.5

M. Peet Lecture 23: Control Systems 25 / 30

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

Imag

inary

Axi

s

-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

inary

Axi

s

M. Peet Lecture 23: Control Systems 26 / 30

Stability Margins

Recall the definitions of Phase Margin.

Definition 3.

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

−3 −2 −1 0 1 2 3−3

−2

−1

0

1

2

3

Nyquist Diagram

Real Axis

Ima

gin

ary

Axi

s ΦM

−3 −2 −1 0 1 2 3−3

−2

−1

0

1

2

3

Nyquist Diagram

Real Axis

Ima

gin

ary

Axi

s

M. Peet Lecture 23: Control Systems 27 / 30

Stability MarginsExample

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

Ima

gin

ary

Axi

s

ΦM

Looking at the intersection with the circle:• Phase Margin: ΦM

∼= 40◦

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

Stability MarginsExample

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

Ima

gin

ary

Axi

s

ΦM

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

• Phase Margin: ΦM∼= 35◦

Gain Margin is technically undefined because open loop is unstable.

• There is a minimum gain, not a maximum.

M. Peet Lecture 23: Control Systems 29 / 30

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

M. Peet Lecture 23: Control Systems 30 / 30