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 steadystate 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 ClosedLoop 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 steadystate 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, LeadLag, 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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