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System type, steady state tracking

System type, steady state tracking. C(s). G p (s). Type 0: magnitude plot becomes flat as w  0 phase plot becomes 0 deg as w  0 Kv = 0, Ka = 0 Kp = flat magnitude height near w  0. Asymptotic straight line, -20dB/ dec. Kv in dB. Kv in Value.

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System type, steady state tracking

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  1. System type, steady state tracking C(s) Gp(s)

  2. Type 0: magnitude plot becomes flat as w 0 phase plot becomes 0 deg as w 0 Kv = 0, Ka = 0 Kp = flat magnitude height near w 0

  3. Asymptotic straight line, -20dB/dec Kv in dB Kv in Value Type 1: magnitude plot becomes -20 dB/dec as w 0 phase plot becomes -90 deg as w 0 Kp = ∞, Ka = 0 Kv = height of asymptotic line at w = 1 = w at which asymptotic line crosses 0 dB horizontal line

  4. Asymptotic straight line, -40dB/dec Ka in dB Sqrt(Ka) in Value Type 2: magnitude plot becomes -40 dB/dec as w 0 phase plot becomes -180 deg as w 0 Kp = ∞, Kv = ∞ Ka = height of asymptotic line at w = 1 = w2at which asymptotic line crosses 0 dB horizontal line

  5. Margins on Bode plots In most cases, stability of this closed-loop can be determined from the Bode plot of G: • Phase margin > 0 • Gain margin > 0 G(s)

  6. 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

  7. Stability from Nyquist plot G(s) • Get completeNyquist plot • Obtain the # of encirclement of “-1” • # (unstable poles of closed-loop) Z= # (unstable poles of open-loop) P + # encirclement N To have closed-loop stable: need Z = 0, i.e. N = –P

  8. Here we are counting only poles with positive real part as “unstable poles” • jw-axis poles are excluded • Completing the NP when there are jw-axis poles in the open-loop TF G(s): • If jwo is a non-repeated pole, NP sweeps 180 degrees in clock-wise direction as w goes from wo- to wo+. • If jwo is a double pole, NP sweeps 360 degrees in clock-wise direction as w goes from wo- to wo+.

  9. 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

  10. 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? ________

  11. 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? ______

  12. 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 ∞?

  13. 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: _______

  14. Correct Incorrect

  15. 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?

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

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

  18. 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.

  19. Incorrect

  20. Example: G(s) stable, P = 0 • Get conjugatefor ω < 0 • Connect ω = 0–to ω = 0+.Needs to goone full circlewith radius ∞.Two choices.

  21. Choice a) : N = 0 Z = P + N = 0 closed-loopstable Incorrect!

  22. Choice b) : N = 2 Z = P + N= 2 Closedloop has two unstable poles Correct!

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

  24. Example: G(s) has one unstable pole • P = 1, no unstable zeros • Get conjugate • Connectω = 0–to ω = 0+.How?One unstablepole/zeroIf connect in c.c.w.

  25. # encirclement N = ? If “–1” is to the left of A i.e. A > –1 then N = 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

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

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

  28. If A < –1 < 0 :N = ______Z = P + N = ______ stability of c.l. : ______ If B < –1 < A : A=-0.2, B=-4, C=-20N = ______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)

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

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