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Margins on Bode plot

Margins on Bode plot. Margins on Nyquist plot. Suppose: Draw Nyquist plot G ( j ω ) & unit circle They intersect at point A Nyquist plot cross neg. real axis at – k. Relative stability from margins. One of the most widely used methods in determine “how stable the system is”

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Margins on Bode plot

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  1. Margins on Bode plot

  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. Relative stability from margins • One of the most widely used methods in determine “how stable the system is” • Margins on based on open-loop transfer function’s frequency response • Basic rule: • PM>0 and GM>0: closed-loop system stable • PM + Mp 70 • As PM or GM  0: oscillates more • PM=0 and GM=0: sustained oscillation • PM<0: unstable

  4. If no wgc, gain never crosses 0dB or 1: • Gain > 1: Closed loop system is unstable. • Gain < 1: Closed loop system is stable G(s) unstable

  5. If no wgc, gain never crosses 0dB or 1: • Gain > 1: Closed loop system is unstable. • Gain < 1: Closed loop system is stable stable G(s)

  6. Relative stability from margins • If there is one wgc and multiple wpc’s all > wgc • PM>0, all GM>0, and closed-loop system is stable • If there is one wgc but > one wpc’s • Closed-loop system is stable if margins >0 • PM and GM reduce simultaneously • PM and GM becomes 0 simultaneously, at which case the closed loop system will have sustained oscillation at wgc=wpc

  7. 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 wgcGM<0: closed-loop system is unstable

  8. ans = 1.0e+002 * -1.7071 -0.2928 -0.0168 -0.0017 + 0.0083i -0.0017 - 0.0083i All poles negative (in left half plane)  Closed loop system is stable

  9. 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 wgcGM<0: closed-loop system is unstable

  10. ans = 1.0e+002 * -1.7435 -0.0247 + 0.1925i -0.0247 - 0.1925i -0.1748 -0.0522 Closed loop system poles are all negative  System is stable

  11. 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 wgcGM<0: closed-loop system is unstable

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

  13. 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

  14. 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

  15. poles = -25.3788 -4.4559 -0.2653 stable

  16. 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

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

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

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

  20. 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

  21. Stability from Nyquist plot The completeNyquist 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

  22. 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 clock wise direction once is +1 encirclement • Counter clock wise direction once is –1 encirclement

  23. Nyquist Criterion Theorem # (unstable poles of closed-loop) Z= # (unstable poles of open-loop) P + # encirclement Nor: Z = P + NTo have closed-loop stable: need Z = 0, i.e. N = –P

  24. 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: • No encirclement, N = 0. • Open-loop stable, P = 0 • Z = P + N = 0, no unstable poles in closed-loop, stable

  25. Example:

  26. 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

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

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

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

  30. 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. Find Nyquist criterion to determine closed-loop stability

  31. 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 withK = 1? ________

  32. 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? ______

  33. 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 ∞?

  34. 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: _______

  35. Correct Incorrect

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

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

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

  39. 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.

  40. Incorrect

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