1. Relative stability from margins If there is one wgc, and multiple wpc’s And if system is minimum phase (all zeros in left half plane) And if gain plot is generally decreasing –PM>0, all GM>0: closed-loop system is stable –PM>0, and at wpc right to wgc GM>0: closed- loop system is stable –PM<0, and at wpc right to wgc GM<0: closed- loop system is unstable

3. ans = 1.0e+002 * -1.7082 -0.2888 -0.0310 0.0040 + 0.0341i  0.0040 - 0.0341i  Two right half plane poles,  unstable

4. Conditionally stable systems Closed-loop stability depends on the overall gain of the system For some gains, the system becomes unstable Be very careful in designing such systems Type 2, or sometimes even type 1, systems with lag control can lead to such Need to make sure for highest gains and lowest gains, the system is stable

5. Relative stability from margins If there are multiple wgc’s –Gain plot cannot be generally decreasing –There may be 0, or 1 or multiple wpc’s –If all PM>0: closed-loop system is stable –If one PM<0: closed-loop system is unstable

6. poles = -25.3788 -4.4559 -0.2653 stable

7. Relative stability from margins If there are multiple wgc’s –Gain plot cannot be generally decreasing –There may be 0, or 1 or multiple wpc’s –If all PM>0: closed-loop system is stable –If one PM<0: closed-loop system is unstable

8. poles = 4.7095 +11.5300i 4.7095 -11.5300i -1.1956 -0.3235 Unstable

9. poles = 4.8503 + 7.1833i 4.8503 - 7.1833i 0.3993 -0.1000 Unstable

10. Poles = 28.9627 -4.4026 + 4.5640i -4.4026 - 4.5640i -0.2576 Unstable

11. Limitations of margins Margins can be come very complicated For complicated situations, sign of margins is no longer a reliable indicator of stability In these cases, compute closed loop poles to determine stability If transfer function is not available, use Nyquist plot to determine stability

12. Stability from Nyquist plot The complete Nyquist plot: –Plot G(jω) for ω = 0 + 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

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

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

19. That is: G(jω) needs to encircle the “–1” point counter clock wise P times. If open loop is stable to begin with, G(jω) cannot encircle the “–1” point for closed-loop stability In previous example: 1.No encirclement, N = 0. 2.Open-loop stable, P = 0 3.Z = P + N = 0, no unstable poles in closed-loop, stable

20. Example:

21. 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

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

23. Example: 1.Get G(jω) for ω = 0 + to +∞ 2.Use conjugate to get G(jω) for ω = –∞ to 0 – 3.How to go from ω = 0 – to ω = 0 + ? At ω ≈ 0 :

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

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

29. Solution: 1.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? ________

30. Note that the total loop T.F. is KG(s). If K is not = 1, Nyquist plot of KG(s) is a 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? ______

31. 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 ∞?

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

33. Incorrect Correct

34. Example: G(s) stable, P = 0 G(jω) for ω > 0 as given. 1.Get G(jω) for ω < 0 by conjugate 2.Connect ω = 0 – to ω = 0 +. But how?

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

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

37. 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 ω = 0, G(s) makes a full circle near ∞. choice a) is impossible, but choice b) is possible.

38. Incorrect

39. Example:G(s) stable,P = 0 1.Get conjugate for ω < 0 2.Connect ω = 0 – to ω = 0 +. Needs to go one full circle with radius ∞. Two choices.

40. Choice a) : N = 0 Z = P + N = 0 closed-loop stable Incorrect!

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

42. Which way is correct? For stable & non-minimum phase systems,

43. Example: G(s) has one unstable pole P = 1, no unstable zeros 1.Get conjugate 2.Connect ω = 0 – to ω = 0 +. How? One unstable pole/zero If connect in c.c.w.

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

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

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

47. 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)

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