1. Margins on Bode plots
G(s) + -

2. Margins on Nyquist plot
Suppose:
Draw Nyquist plot G(jω) & unit circle
They intersect at point A
Nyquist plot cross neg. real axis at –k

3. Stability from Nyquist plot
G(s) +
The complete Nyquist plot:
Plot G(jω) for ω = to +∞
Get complex conjugate of plot, that’s G(jω) for ω = 0– to –∞
If G(s) has pole on jω-axis, treat separately
Mark direction of ω increasing
Locate point: –1 -

5. Encirclement of the -1 point
As you follow along the G(jω) curve for one complete cycle, you may “encircle” the –1 point
Going around in c.w. once is +1 encirclement
c.c.w. once is –1 encirclement

9. Nyquist Criterion Theorem
# (unstable poles of closed-loop) Z = # (unstable poles of open-loop) P + # encirclement N or: Z = P + N To have closed-loop stable: need Z = 0, i.e. N = –P

10. That is: G(jω) needs to encircle the “–1” point c.c.w. P times.
If open loop is stable to begin with, G(jω) cannot encircle the “–1” point for closed-loop stability
In previous example:
No encirclement, N = 0.
Open-loop stable, P = 0
Z = P + N = 0, no unstable poles in closed-loop, stable

11. function mynyquist(n,d,w)
%mynyquist-- plots a nonlinearly compressed nyquist plot.
% Near the origin (distance less than 2), it preserves the
% original Nyquist plot. Far away from the origin (distance
% more than 2), the distance is compressed by taking the log. %
%Syntax: mynyquist(n,d), or mynyquist(n,d,w)
% where n and d are the numerator and denominator of the transfer
% function, w is the desired frequency points.
%The program does not handle jw-axis poles.
if nargin <3
[m,p,w]=bode(n,d);
d1=log10(min(w));
d2=log10(max(w));
w=logspace(d1, d2, 200);
end;

12. G=polyval(n, j*w)./polyval(d,j*w); %or G=evalfr(tf(n,d),j*w);
GA=abs(G);
GP=angle(G);
GAA=(GA<2).*GA+(GA>=2).*(2+2*log(GA/2)); %if GA<2, do nothing; if>=2, do log compression
plot(GAA.*cos(GP),GAA.*sin(GP),'b');
hold;
plot(GAA.*cos(GP),-GAA.*sin(GP),'r'); %this plots the negative freq part
axis equal;
%axis([ ]); %this sets x, y limits for the graph window
grid;
a=0:2*pi/200:2*pi;
plot(cos(a),sin(a),':'); %this plots the unit circle for phase margin
%you can also use the rectangle command

13. Matlab nyquist plot for: 23(s+0.25)(s2+0.7s+1)
G(s) =
(0.15s+1)(s3+10s2)(s2+0.5s+0.25)
nyquist(23*conv([1 .25],[ ]),conv([0.15 1],conv([ ],[ ])))
Hard to see any useful information.
Essentially useless.

14. Without the “axis” command line.
mynyquist(23*conv([1 .25],[ ]),conv([0.15 1],conv([ ],[ ])))
Without the “axis” command line.

15. With the “axis” command line.
You can change the x, y limits.

16. Example:

17. As you move around
from ω = –∞ to 0–,
to 0+, to +∞, you go
around “–1” c.c.w.
once.
# encirclement N = – 1.
# unstable pole P = 1

18. Check: c.l. pole at s = –3, stable.
i.e. # unstable poles of closed-loop = 0
closed-loop system is stable.
Check:
c.l. pole at s = –3, stable.

19. Example:
Get G(jω) for ω = 0+ to +∞
Use conjugate to get G(jω) for ω = –∞ to 0–
How to go from ω = 0– to ω = 0+? At ω ≈ 0 :

22. # encirclement N = _____
# open-loop unstable poles P = _____
Z = P + N = ________
= # closed-loop unstable poles.
closed-loop stability: _______

23. Q: 1. Find stability margins
Example:
Given:
G(s) is stable
With K = 1, performed open-loop sinusoidal tests, and G(jω) is on next page
Q: 1. Find stability margins
2. Use Nyquist criterion to determine closed-loop stability

25. Solution:
Where does G(jω) cross the unit circle? ________ Phase margin ≈ ________ Where does G(jω) cross the negative real axis? ________ Gain margin ≈ ________ Is closed-loop system stable with K = 1? ________

26. Note that the open loop T.F. is KG(s).
If K is not = 1, Nyquist plot of KG(s) is a radial scaling of G(jω).
e.g. If K = 2, scale G(jω) by a factor of 2 in all directions.
Q: How much can K increase before GM becomes lost? ________
How much can K decrease? ______

27. Some people say the gain margin is 0 to 5 in this example
Q: As K is increased from 1 to 5, GM is lost, what happens to PM?
What’s the max PM as K is reduced to 0 and GM becomes ∞?

28. To use Nyquist criterion, need complete Nyquist plot.
Get complex conjugate
Connect ω = 0– to ω = 0+ through an infinite circle
Count # encirclement N
Apply: Z = P + N
o.l. stable, P = _______
Z = _______
c.l. stability: _______

29. Correct
Incorrect

30. Example:
G(s) stable, P = 0
G(jω) for ω > 0 as given.
Get G(jω) for ω < 0 by conjugate
Connect ω = 0– to ω = 0+. But how?

31. Incorrect Choice a) : Where’s “–1” ? # encirclement N = _______
Z = P + N = _______
Make sense? _______
Incorrect

32. Correct Choice b) : Where is “–1” ? # encir. N = _____
Z = P + N = _______
closed-loop stability _______
Correct

33. Note: If G(jω) is along –Re axis to ∞ as ω→0+, it means G(s) has in it.
when s makes a half circle near s = 0 and sweeps in CCW direction 180o, G(s) makes a full circle near ∞ and sweeps in CW direction 360o.
choice a) is impossible, but choice b) is possible.

34. Incorrect

35. Example: G(s) stable, P = 0
Get conjugate for ω < 0
Connect ω = 0– to ω = 0+. Needs to go one full circle with radius ∞. Two choices.

36. G(s) sweeps in CW direction!
Choice a) :
N = 0
Z = P + N = 0
closed-loop stable
Incorrect!
G(s) sweeps in CW direction!

37. Correct! Choice b) : N = 2 Z = P + N = 2
Closed loop has two unstable poles
Correct!

38. Which way is correct?

39. Example: G(s) has one unstable pole P = 1, no unstable zeros
Get conjugate
Connect ω = 0– to ω = 0+. How? If connect in c.c.w.,
we get =>
This is wrong,
0- to 0+ should be in CW dir!

40. # encirclement N = ?
If “–1” is to the left of A
i.e. A > –1
then N = 0
Z = P + N = = 1
but if gain is increased, “–1” could be inside, N = –2
Z = P + N = –1
c.c.w. is impossible

41. If connect c.w.:
For A > –1 N = ______
Z = P + N
= ______
For A < –1 N = ______
Z = ______
No contradiction This is the correct way.

42. Example: G(s) stable, minimum phase
G(jω) as given:
get conjugate.
Connect ω = 0–
to ω = 0+,

43. If A < –1 < 0 :. N = ______. Z = P + N = ______. stability of c
If A < –1 < 0 : N = ______ Z = P + N = ______ stability of c.l. : ______
If B < –1 < A : A=-0.2, B=-4, C=-20 N = ______ Z = P + N = ______ closed-loop stability: ______
Gain margin: gain can be varied between 1/0.2 and 1/4,
or can be less than 1/20

44. If C < –1 < B :. N = ______. Z = P + N = ______
If C < –1 < B : N = ______ Z = P + N = ______ closed-loop stability: ______
If –1 < C : N = ______ Z = P + N = ______ closed-loop stability: ______