Review

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

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

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

Transfer Function

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

where Compare with the definition of the Laplace Transform

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

Bode Phase Plot 1. Phase = at  b 2. Draw  from 0 to  b /10, slope =0 3. Draw  from  b /10 freq to 10*  b 4. Min Phase is from 10*  b

Decibels An alternate unit of Magnitude or Gain Definition: x dB = 20* lg(x) dB Notation is widely used in Filter theory and Acoustics xlg(x)X(db)

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

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

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

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

Bode Plot Construction: Matlab Plot

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

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

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

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

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.

Bode Lead Design

 -margin = 51 deg. K = 100

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.

Bode Lag Design

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

Bode Lag Design

 -margin = 39 deg. K = 10

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

(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.

(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 Lead zero at 3. Lead pole at 30. New gain crossover at 5 rad/s Final step: adjust gain K such that |F| = 0 dB at  cr. Result: The controller gain is now K = 3.4 (4 times better than P- control)

Bode Lead and Lag Design: General placement rules