Systems Analysis and Control

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1 Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain

2 Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering Phase Margin Lag Compensators Change in steady-state error M. Peet Lecture 24: Control Systems 2 / 27

3 Phase and Gain Margin Recall: 3 indicators Gain Margin Phase Margin Bandwidth From Phase Margin and Closed-Loop Bandwidth: Percent Overshoot Peak Time Rise Time Settling Time Percent Overshoot: ζ is from Φ M only %OS = e ( πζ 1 ζ ) 2 M. Peet Lecture 24: Control Systems 3 / 27

4 Phase and Gain Margin Given ζ, we need ω BW to find T r, T s and T p T s = f 1(ζ) ω BW T p = f 2(ζ) ω BW T r = f 3(ζ) ω BW M. Peet Lecture 24: Control Systems 4 / 27

5 Phase and Gain Margin This is all analysis. Design Problem: Achieve 1% Overshoot T r = 2s T s = 1s Step 1: Translate into Φ M and ω BW constraints. Get desired ζ from 1% Overshoot ζ = =.57 ( ln ) %OS 1 π 2 + ln 2 ( %OS 1 ) M. Peet Lecture 24: Control Systems 5 / 27

6 Phase and Gain Margin The desired ζ yields a desired phase margin. Φ M,desired = ζdes 1 = 57 The toughest constraint will be Rise Time: T r < 2s ω BW = f 3(ζ) T r > 2.12 = M. Peet Lecture 24: Control Systems 6 / 27

7 Phase and Gain Margin We can also look at settling time: T s < 1s ω BW = f 1(ζ) T s > 8 1 =.8 Therefore: We want Phase margin of Φ M = 57 Bandwidth of ω BW > 1 M. Peet Lecture 24: Control Systems 7 / 27

8 Lead Compensators So far we only know how to find Φ M and ω BW, but not influence them. Φ M = 35 at ω c = 2.13 ω BW = 3.7 Question: How can we increase Φ M and decrease ω BW? Magnitude (db) Bode Diagram Gm = 1.2 db (at 4.24 rad/s), Pm = 35 deg (at 2.13 rad/s) 15 9 Phase (deg) Answer: Increase gain at ω > ω BW,desired So that G(ıωBW,desired ) = 7dB. Increase phase by 3 at crossover frequency Frequency (rad/s) Lead Compensation M. Peet Lecture 24: Control Systems 8 / 27

9 Lead Compensators Question: How do we design our lead compensator? In root locus, we had pole placement. Look at the Two Lead Compensators: D(s) = k s + 1 s and D 2 (s) = k s + 1 s Bode Diagram 4 3 Magnitude (db) Phase (deg) Frequency (rad/sec) What are the differences? M. Peet Lecture 24: Control Systems 9 / 27

10 Lead Compensators Consider the generalized form of lead compensation α determines how much phase is added α < 1 for lead compensation α > 1 for lag compensation T determines where the phase is added. D(s) = k T s + 1 αt s + 1 We want extra phase at the crossover frequency Note that magnitude is also added at high frequency. Could increase the crossover frequency. Changes the phase margin. M. Peet Lecture 24: Control Systems 1 / 27

11 Lead Compensators Question: Where is the phase added? Find the point of Maximum Phase. The Phase contribution of the lead compensator is Φ(ω) = D(ıω) = T (ıω + 1) (αt ıω + 1) = tan 1 (T ω) tan 1 (αt ω) To find peak phase contribution, set D(ıω) ω =. Which means Dividing by 1 α, we get Φ ω = (T ω) 2 T (αt ω) 2 αt = 1 + (αt ω) 2 (1 + (T ω) 2 )α = 1 α + (α 1)(αT 2 ω 2 ) = 1 αt 2 ω 2 = or ω max = 1 T α M. Peet Lecture 24: Control Systems 11 / 27

12 Lead Compensators So we get that the maximum phase contribution is at ω max = 1 T 1 T α. Convert to a log ω Bode plot: log ω max = 1 2 The maximum occurs at the average of log 1 T [ log 1 T + log 1 ] αt and log 1 αt. 9 Phase (deg) Frequency (rad/sec) log(1/t) log(1/ αt) ( log(1/ T) + log(1/ αt) ) 2 M. Peet Lecture 24: Control Systems 12 / 27

13 Lead Compensators Reconsider our example: T = 1 α = 1 1 D(s) = k s + 1 s Phase (deg) Frequency (rad/sec) ω = 1 ω = 1 ω max = Maximum Phase occurs at ω = 1 T α = M. Peet Lecture 24: Control Systems 13 / 27

14 Lead Compensators Reconsider our example: T = 1 α = 1 1 D(s) = k s + 1 s Phase (deg) Frequency (rad/sec) ω = 1 ω = 1 ω max =1 Maximum Phase occurs at ω = 1 T α = 1 M. Peet Lecture 24: Control Systems 14 / 27

15 Lead Compensators So if we want to add phase at ω c, then we need This is not definitive T α = 1 ω c Depends on how much phase we want to add. Now consider the case where we want to add 3 of phase margin. Question: How much phase does a lead compensator add? We already know the frequency of peak phase. Make this the crossover frequency Independent of T! Φ max = tan 1 1 α tan 1 ( αω) ( ) 1 α = sin α M. Peet Lecture 24: Control Systems 15 / 27

16 Lead Compensators We can plot Φ max vs. 1 α. For 3 phase, we want 1 α = 3. Lets add 3 of phase at ω c = 2.13 T = 1 3 = ω c α 2.13 =.814 M. Peet Lecture 24: Control Systems 16 / 27

17 Lead Compensators This gives us a lead compensator: D(s) = T s + 1 αt s + 1 =.814s s Bode Diagram Magnitude (db) Phase (deg) Frequency (rad/s) M. Peet Lecture 24: Control Systems 17 / 27

18 Lead Compensators Add this to the original plot to get 4 Bode Diagram Gm = 8.79 db (at 6.15 rad/s), Pm = 38.6 deg (at 3.44 rad/s) 2 Magnitude (db) Phase (deg) Frequency (rad/s) New Phase margin is New ω c = 4.79, new ω BW = 9. Crossover frequency is increased, which reduces the phase margin! M. Peet Lecture 24: Control Systems 18 / 27

19 Lead Compensators To avoid changing the crossover frequency, we control the gain. D(s) = k T s + 1 αt s + 1 We want D(ω c ) = 1. Since we have T = 1 ω c α, this yields 1 α ı + 1 D(ıω c ) = k αı + 1 Thus to preserve the crossover frequency, we set α + 1 k = 1 α = + 1 = M. Peet Lecture 24: Control Systems 19 / 27

20 Lead Compensators Apply the final controller to get 4 Bode Diagram Gm = 13.6 db (at 6.15 rad/s), Pm = 65 deg (at 2.13 rad/s) Magnitude (db) Phase (deg) Frequency (rad/s) New Phase margin is 65. New ω c = 2.13, new ω BW = 4. New phase margin is on target - bandwidth increased slightly. M. Peet Lecture 24: Control Systems 2 / 27

21 Lag Compensators What about steady-state error? Want to increase G() No effect on Φ M or ω BW Bode Diagram 3 2 Magnitude (db) Phase (deg) Frequency (rad/sec) Ignore the lead compensator. Increase G() by 15dB. No change at ω c or ω BW e ss = G() M. Peet Lecture 24: Control Systems 21 / 27

22 Lag Compensators Question: How do we increase G() without changing ω BW? We want to reduce gain at ω =. Want no effect on Φ M or G() Look at the Two Lag Compensators: D(s) = 2 log D() = 15 2 log D(ıω BW ) = s s 1 s and D 2 (s) = s Bode Diagram Magnitude (db) Phase (deg) Frequency (rad/sec) M. Peet Lecture 24: Control Systems 22 / 27

23 Lag Compensators Lets use the form: α determines how much phase is added D(s) = T s + 1 αt s + 1 α > 1 for lag compensation As before, min phase is given by ω = 1 T α Center this point at low frequency Change in magnitude at high frequency is 2 log α(t s + 1) 2 log(αt s + 1) If T is large, this is just. At low frequency, gain is 2 log α M. Peet Lecture 24: Control Systems 23 / 27

24 Lag Compensators For our problem, set T =.1 and then we want so α = 1.75 = log α = 15 D(s) = s s Bode Diagram 15 Magnitude (db) Phase (deg) Frequency (rad/sec) M. Peet Lecture 24: Control Systems 24 / 27

25 Lag Compensators The system with lag compensation: 15 Bode Diagram Gm = 1.2 db (at 4.23 rad/sec), Pm = 34.8 deg (at 2.13 rad/sec) 1 Magnitude (db) Phase (deg) Frequency (rad/sec) Bandwidth: ω BW = 3.7 Phase Margin: Φ M = 35 M. Peet Lecture 24: Control Systems 25 / 27

26 Lag Compensators Combine the lead and lag compensators: 15 Bode Diagram Gm = 12.4 db (at 6.92 rad/sec), Pm = 51.1 deg (at 2.78 rad/sec) 1 Magnitude (db) Phase (deg) Frequency (rad/sec) Φ M = 51 ω BW = 4.78 M. Peet Lecture 24: Control Systems 26 / 27

27 Summary You Have Learned: Classical Control Systems Exam Material: Root Locus Drawing Asymptotes, Break Points, etc. Compensation Gain, Lead-Lag, Notch Filters Bode and Frequency Response What is frequency response? Drawing Bode Plot Closed Loop Dynamics Compensation Nyquist Plot Drawing and Concepts Stability Margins M. Peet Lecture 24: Control Systems 27 / 27

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