Automatic Control

# Automatic Control

## Automatic Control

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##### Presentation Transcript

1. Automatic Control Review

2. Please Return Loan Clickers to the MEG office after Class!Today! FINAL EXAM: Wednesday December 8        8:00 AM to 10:00 a.m.

3. Feedback Terminology In Block diagrams, we use not the time domain variables, but their Laplace Transforms. Always denote Transforms by (s)!

4. State-Variable Form Deriving differential equations in state-variable form consists of writing them as a vector equation as follows: whereis the output and u is the input

5. Transfer Function

6. Fourier Transform: Let period T  infinity The interval between Discrete frequencies  0 The Fourier series becomes the Fourier Transform

7. where Compare with the definition of the Laplace Transform

8. Bode Magnitude Plot 1. Note K and wb 2. Draw |F| from low freq to wb 3. Draw |F| from wb , slope -1/decade

9. Bode Phase Plot 1. Phase = -450 at wb 2. Draw f from 0 to wb/10, slope =0 3. Draw f from wb/10 freq to 10*wb 4. Min Phase is -900 from 10*wb

10. Decibels • An alternate unit of Magnitude or Gain • Definition: xdB = 20* lg(x) • dB Notation is widely used in Filter theory and Acoustics

11. Bode Plot Construction G(s) = 2/(s)(s+1) 1. Construct each Element plot 2. Graphical Summation Gain = 2. Integrator Slope = -1 Slope = -2 Integrator Phase = -90 deg.

12. Bode Plot of 1/(s(s+1)): Matlab Plot

13. Bode Plot Construction G(s) = 5*(s+1)/(10s+1)(100s+1) 1. Construct each Element plot Slope = -1 K = 5 Slope = -2 2. Graphical Summation: Complete plot. Note beginning and final values Slope = -1

14. Phase Plot Construction 2. Graphical Summation of phase angles. Note beginning and final phase values. Here: f = 0 at w = 0, and f = -90 final angle G(s) = 5*(s+1)/(10s+1)(100s+1) K = 5 Initial Phase is zero to 0.001, follows the first Phase up to 0.01 Final phase: Constant - 90 deg - 90 deg./decade +45 deg./decade 0 deg./decade

15. Bode Plot Construction: Matlab Plot

16. Given: An open-loop system At w = 0.1, the Magnitude is approximately • (A) 1 • (B) 0.1 • (C) 0.01 • (D) 0 • (E) 1/(1000)

17. Given: An open-loop system At w = 0.1, the Magnitude is approximately • (A) 1 • (B) 0.1 • (C) 0.01 • (D) 0 • (E) 1/(1000)

18. Given: An open-loop system At w =1, the phase angle is approximately • (A) 0 degrees • (B) -45 degrees • (C) -135 degrees • (D) -180 degrees • (E) -90 degrees

19. Given: An open-loop system At w =1, the phase angle is approximately • (A) 0 degrees • (B) -45 degrees • (C) -135 degrees • (D) -180 degrees • (E) -90 degrees

20. Bode Lead Design 1. Select Lead zero such that the phase margin increases while keeping the gain crossover frequency as low as reasonable. 2. Adjust Gain to the desired phase margin.

23. Bode Lead Design f-margin = 51 deg. K = 100

24. Bode Lag Design 1. All other design should be complete. Gain K and phase margin are fixed 2. Select Lag zero such that the phase margin does not drop further. (Slow) 3. Steady State Gain should now be about 10 times larger than without Lag.

25. Bode Lag Design

26. Lag compensator |p| = 0.1*z G(s) = Construct each Element plot Slope = 0 Slope = -1 Note Break Frequencies Slope = 0 Gain = 0.1 Slope = 0 Phase = 0 Slope = 0 Slope = 0

27. Bode Lag Design

28. Bode Lag Design f-margin = 39 deg. K = 10

29. Lead Design Example (a) P-control for phase margin of 45 degrees. Controller gain K = 0.95

30. (b) Lead-control for phase margin of 45 degrees. Lead zero and pole in RED. Initial design: Lead is too slow Lead is too slow. Lead Zero should be near the phase margin. Here: Place Lead zero around 3 rad/s.

31. (b) Lead-control for phase margin of 45 degrees. Lead zero and pole in RED. Improved design: Lead zero at 3, pole at 30 rad/s Final step: adjust gain K such that |F| = 0 dB at wcr. Result: The controller gain is now K = 3.4 (4 times better than P-control) Lead zero at 3. Lead pole at 30. New gain crossover at 5 rad/s

32. Bode Lead and Lag Design: General placement rules