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Presentation On FREQUENCY RESPONSE PLOT (NYQUIST PLOT-CONTROL ENGINEERING)

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Published in: Engineering
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Nyquist plot is a plot used to verify stability of the system. From a Nyquist plot, we can tell a number of closed-loop poles on the right half plane. If there is any closed-loop pole on the right half plane, the system goes unstable. If there is no closed-loop pole on the right half plane, the system is stable.

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  1. Frequency Response Method NYQUIST PLOT INC 341
  2. Knowledge Before Studying Nyquist Criterion R(s) + Il(s) unstable if there is any pole on RHP (right half plane) G(s) INC 341
  3. Open-loop system: NG(s) NH (S) (s) DH(s Characteristic equation: NGNH DGDH + NGNH DGDH DGDH poles of G(s)H(s) and 1 are the same Closed-loop system: G (s)DH (S) + N G (s)NH (S zero of 1 is pole of T(s) INC 341
  4. (s — l)(s Zero - 3)(s —4) — 6)(s Zero - Poles - a,b,c,d Poles - G(s) Zero — ? ? ? ? Poles - To know stability, we have to know a,b,c,d INC 341
  5. Stability from Nyquist plot From a Nyquist plot, we can tell a number of closed-loop poles on the right half plane. — If there is any closed-loop pole on the right half plane, the system goes unstable. — If there is no closed-loop pole on the right half plane, the system is stable. INC 341
  6. Nyquist Criterion Nyquist plot is a plot used to verify stability of the system. mapping contour function F (s) mapping all points (contour) from one plane to another by function F(s). INC 341
  7. (s (s Contour A PI s-plane INC 341 1m F-plane Q' Contour B
  8. Contour A as-plan c Contour A Iin Contour B s-plane s-plane Contour A FCS) IV(s) FCS) FCS) F-pl an c (a) 1m F-plane 1 (b) 1m F-plane (s (s (s (s (s Re Re Contour B Contour B Re Contour B Re Contour B Re R 1 1 Contour A s-plane Contour A s-plane Zl PI (c) 1 (d) (e) 11m F-plane Irn F-plane Pole/zero inside the contour has 360 deg. angular change. Pole/zero outside contour has 0 deg. angular change. Move clockwise around contour, zero inside yields rotation in clockwise, pole inside yields rotation in counterclockwise INC 341
  9. Characteristic equation s-plane Contour A x 1m 1 + Gil-plane R Contour B N = p-Z N = # of counterclockwise direction about the origin P = # of poles of characteristic equation inside contour = # of poles of open-loop system z = # of zeros of characteristic equation inside contour = # of poles of closed-loop system INC 341
  10. Characteristic equation Increase size of the contour to cover the right half plane s-plane More convenient to consider the open-loop system (with known pole/zero) INC 341
  11. Nyquist diagram of G(s)H(s) 'Open-loop system' Mapping from characteristic equa to open-loop system by shifting to the left one step Z = # of closed-loop poles inside the right half plane P = # of open-loop poles inside the right half plane N = # of counterclockwise revolutions around -1 INC 341
  12. s-plane O O s-plane O O O zeros of 1 + G(s)H(s) (a) (b) 1m Gil-plane 1 1m Gil-plane -1 Re Test radius Re poles of closed-loop system Location not known INC 341 poles of 1 + G(s)fl(s) x poles of G(s)H(s) Location is known
  13. Properties Of NyqUiSt plot If there is a gain, K, in front of open-loop transfer function, the Nyquist plot will expand bv a factor of K. Nyquist Diagram < 0.5 -0.5 2 1.5 1 -1.5 -2 2/(s+0.5) Il(s+0.5) 0.5/(s+0.5) 1.5 2 2.5 Real Axis 3 3.5 4 INC 341
  14. Nyquist plot Nyquist Diagram 2.51(s.2) Real Axis INC 341 Open loop system has pole at Closed-loop system has pole at 1 1+G(S) (s -1) If we multiply the open-loop with a gain, K, then we can move the closed-loop pole's position to the left-half plane
  15. Nyquist plot example (cont.) New look of open-loop system: Corresponding closed-loop system: Evaluate value of K for stability 1
  16. Adjusting an open-loop gain to guarantee stability R(s) + (a) Step I: sketch a Nyquist Diagram Step Il: find a range of K that makes the system stable! INC 341
  17. How to make a Nyquist plot? Easy way by Matlab nyquist — Nyquist: bode - Bode: INC 341
  18. Step l: make a Nyquist plot Starts from an open-loop transfer function (set Set s = jo and find frequency response — At dc, - Find at which the imaginary part equals zero INC 341
  19. s2 +8 s +15 02 -k 8 j (D -k 15 (15 — 02) -k 02 —6jo+8 (8—02) —6 jo (15 — 02) + 8 j CD (8 — -k (8 — 02) — (8 — 02) -k 6 jCD (15 - - - 4802 + j(1540 -1403) (8 02) 2 +6202 Need the imaginary term = O, Substitute 0=Jii back in to the transfer function And get GO) = -1.33 - 48(11) (8 —11)2 -k 62 (11) INC 341 — 540 412 -1.31
  20. s -plane (b) INC 341 At dc, s=o, o = Il -1.33 At imaginary part-O 1m B' (c) Gif-plane 1 A' 15
  21. Step Il: satisfying stability condition P = 2, N has to be 2 to guarantee stability Marginally stable if the plot intersects -1 For stability, 1.33K has to be greater than 1 K > 1/1.33 K > 0.75 INC 341
  22. Example Evaluate a range of K that makes the system stable 1m Contour s-plane -2 INC 341 x -1 x Gil-plane 1 20 1 (a) (b)
  23. Step I: find frequency at which imaginary part Set s = jot) G(jo) ((jO)2 -k -k -k 2) — 02) — jo(6 — 02) — 02) 2 +02 (6 — 02) 2 At O, A/G the imaginary part = O Plug o back in the transfer function and get G = -0.05 INC 341
  24. Step Il: consider stability condition P = 0, N has to be 0 to guarantee stability Marginally stable if the plot intersects -1 For stability, 0.05K has to be less than 1 K < 1/0.05 K < 20 INC 341
  25. Gain Margin and Phase Margin Gain margin is the change in open-loop gain (in dB), required at 180 of phase shift to make the closed-loop system unstable. Phase margin is the change in open-loop phase shift, required at unity gain to make the closed-loop system unstable. GM/PM tells how much system can tolerate before going unstable!!! INC 341
  26. GM and PM via 1m Gil-plane -1 1 a Nyquist diagram 1 Unit circle Phase difference before instability Phase margin Nyquist plot 1 Re Q' Gain difference before instability Gain margin — GM = 20 log a INC 341
  27. GM and PM via Bode Plot M (dB) O dB Phase (degrees) 1800 Gain plot Phase plot log (D log o •The frequency at which the phase equals 180 degrees is called the phase crossover frequency OG •The frequency at which the magnitude equals 1 is called the gain crossover frequency CDØ gain crossover frequency INC 341 phase crossover frequency
  28. Example Find Bode Plot and evaluate a value of K that makes the system stable The system has a unity feedback with an open-loop transfer function G(s) — First, let's find Bode Plot ofG(s) by assuming that K=40 (the value at which magnitude plot starts from O dB) INC 341
  29. -20 -40 dB/dec —60 80 -60 dB/dec IOO 120 0.01 0.1 Frequency (rad/s) 10 —450/dec -900/dec -135 -180 -900/dec -225 -270 0.01 0.1 IOO —450/dec 100 10 Frequency (rad/s) At phase = -180, w = 7 rad/sec, magnitude = -20 dB INC 341
  30. GM>O, system is stable!!! Can increase gain up 20 dB without causing instability (20dB = 10) Start from K = 40 with K < 400, system is stable INC 341
  31. Closed-loop transient and closed-loop frequency responses R(s) + INC 341 C(s) '2nd system' 2 on s(s + 2 Con) S2 -k 240 n S -k On
  32. Damping ratio and closed-loop frequency response 10 20 log Mp -10 -15 -20 24 1-42 -on 1-242 log op log OBW Log frequency (rad/s) Magnitude Plot of closed-loop system INC 341
  33. Response speed and closed-loop frequency response — On (l — ) + 444 — -k 2 BW BW BW (I — ) + 444 — 442 + 2 (1 — ) -k 444 — 442 + 2 = frequency at which magnitude is 3dB down from value at dc (O rad/sec), or M INC 341
  34. Find from Open-looo Freauencv Resoonse 0 -1 —4 O —5 -8 Closed-loop magnitude Nichols Charts 280 260 240 220 200 180 -160 140 120 100 80 Open-loop phase (degrees) From open-loop frequency response, we can find at the open-loop frequency that the magnitude lies between -6dB to -7.5dB (phase between -135 to -225) INC 341
  35. Relationship between damping ratio and phase margin Of open-loop frequency response Phase margin of open-loop frequency response Can be written in terms of damping ratio as following 1 = tan INC 341 24 — -k 1 -k 444
  36. Example Open-loop system with a unity feedback has a bode Ot below, approximate settling time and peak time 0 " ( 6 ? 一 つ ム 一 6 0 4 8 っ ( 6 0 一 の 召 Ⅳ = 37 洋 80 一 06 1 2 3 4 Frequency (rad/s) ー 100 ー 120 ー 140 ー 160 ー 180 ー 200 720 -240 1 2 3 4 5 6 7 8 9 10 INC 341 PT & BP
  37. Solve for PM - INC 341 35 BW = tan — 2ζ2 H- ζ = Ο.32 (1 — ) H- 4ζ4 444 — 442 +2 ΡΤ & ΒΡ

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