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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
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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
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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)
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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.
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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.
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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.
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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
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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
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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
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Stability Margins
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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
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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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
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.
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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
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