# Raktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Basic Feedback Analysis & Design

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1 AERO 422: Active Controls for Aerospace Vehicles Basic Feedback Analysis & Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University

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3 Routh s Stability Criterion Given characteristic equation Factor out any roots at the origin and multiply by - if needed D G (s) = s n + a s n + + a n s + a n, How to determine stability? (without using MATLAB :) Necessary condition: All coefficients of characteristic polynomial be positive, ie a i > 0 Any coefficient missing ( = 0) or negative, then poles are outside LHP Necessary and Sufficient condition: System is stable iff all the elements in the first column of Routh array are positive AERO 422, Instructor: Raktim Bhattacharya 3 / 28

4 Routh s Stability Criterion Routh Array where Row n s n a 2 a 4 Row n s n a a 3 a 5 Row n 2 s n 2 b b 2 b 3 Row n 3 s n 3 c c 2 c 3 Row 2 s 2 Row s Row 0 s 0 [ ] a2 det a a 3 b = a [ ] a a det 3 b b 2 c = b [ ] a4 det a a 5 b 2 = a [ ] a a det 5 b b 3 c 2 = b There are two special cases! See web-appendix of textbook AERO 422, Instructor: Raktim Bhattacharya 4 / 28

5 Stabilizing Gain Example One parameter r + e u y C P y design parameter Given P (s) = s+ s(s )(s+6) and C(s) = K Characteristic Equation Numerator of + P C No pole zero cancellation s 3 + 5s 2 + (K 6)s + K = 0 Routh s Table Stabilizing Gains s 3 K 6 s 2 5 K s 4K 30 5 s 0 K 4K 30 > 0, K > 0 5 K > 75, K > 0 AERO 422, Instructor: Raktim Bhattacharya 5 / 28

6 Stabilizing Gain Example 2 Two parameters r + e u y C P y Given P (s) = (s+)(s+2), C(s) = K + K I s Characteristic Equation Numerator of + P C No pole zero cancellation Routh s Table s 3 + 3s 2 + (2 + K)s + K I = 0 s K s 2 3 K I s 6+3K K I 3 s 0 K I Stabilizing Gains K > K I /3 2, K I > 0 AERO 422, Instructor: Raktim Bhattacharya 6 / 28

7 Stabilizing Gain Example Two parameters (contd) r + e u y C P y Stabilizing Gains K > K I /3 2, K I > 0 Given P (s) = (s+)(s+2), C(s) = K + K I s Step Response 4 K=,KI= K=0,KI=5 2 Amplitude Time (seconds) What about performance? AERO 422, Instructor: Raktim Bhattacharya 7 / 28

8 Step Response Time Domain Performance Specification 5 Step Response Amplitude Time Second Order System: poles = σ ± jω d, ω n = σ 2 + ω 2 d, ζ = σ/ω n M p = e πζ/ ζ 2 t r = 8 t s = 46 ω n σ AERO 422, Instructor: Raktim Bhattacharya 8 / 28

9 Step Response Time Domain Performance Specification Second Order Systems Desired Location of Poles M p = e πζ/ ζ 2 t r = 8 t s = 46 ω n σ ω n 8/t r ζ ζ(m p ) σ 46/t s Adjust K, K I to satisfy additional performance related constraints Controller gain tuning AERO 422, Instructor: Raktim Bhattacharya 9 / 28

10 Routh s Stability Criterion Conditions that ensure Re p i < α, for α > 0 Given polynomial s n + a s n + + a n s + a n = 0 Modify Routh s stability criterion to ensure Re p i < α, for α > 0 Replace s := q α and substitute in polynomial to get q n + b q n + + b n q + b n = 0 Apply Routh s criterion to the polynomial in q AERO 422, Instructor: Raktim Bhattacharya 0 / 28

11 Routh s Stability Criterion Example r + e u y C P y Given P (s) = s 2 +4s+ and controller C(s) = K + K 2 /s Find range of values for K, K 2 such that all poles are left of α Characteristic equation: s 3 + 4s 2 + (K + )s + K 2 Substitute s := q α, with α = Apply Routh s criterion q 3 + q 2 + (K 4)q K + K AERO 422, Instructor: Raktim Bhattacharya / 28

12 Routh s Stability Criterion Example (contd) Polynomial in q q 3 + q 2 + (K 4)q K + K Routh s Table s 3 K 4 s 2 K 2 K + 2 s 2K K s 0 K 2 K Inequalities 2K K 2 6 > 0, K 2 K + 2 > 0 AERO 422, Instructor: Raktim Bhattacharya 2 / 28

13 Routh s Stability Criterion Example (contd) K K K K (a) 2K K 2 6 > 0, K 2 K + 2 > 0, Poles left of (b) K K 2 /4 + > 0, K 2 > 0, Poles left of 0 AERO 422, Instructor: Raktim Bhattacharya 3 / 28

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15 Disturbance Rejection d r + u + + e y C P y A Let P (s) := (s/p +)(s/p 2 +), and C(s) := K Total response Y (s) = = P C + P C R(s) + P + P C D(s) AK (s/p + )(s/p 2 + ) + AK R(s) + A (s/p + )(s/p 2 + ) + AK D(s) Steady state value with feedback lim sy (s) = AK s 0 + AK ( ) lim sr(s) + s 0 A + AK ( ) lim sd(s) s 0 AERO 422, Instructor: Raktim Bhattacharya 5 / 28

16 Robustness to Plant Uncertainty r + e u y C P y Transfer function from reference to output G yr (s) = P C + P C = AK (s/p + )(s/p 2 + ) + AK Suppose A A + δa What is the effect on T (s) = G yr (s)? steady state gain AERO 422, Instructor: Raktim Bhattacharya 6 / 28

17 Robustness to Plant Uncertainty (contd) T ss = δt ss = dt ss da δa AK + AK K = ( + AK) 2 δa ( ) ( AK = + AK + AK ) δa A = δt ss T ss = δa + AK A AERO 422, Instructor: Raktim Bhattacharya 7 / 28

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19 Analysis of Steady State Error r + e u y C P y E(s) = + P C R(s) = e ss = lim s s 0 + L(s) = lim s 0 + L(s) s k Investigate e ss 0 for various values of k s k+ Value of k r(t) 0 (t) e ss = constant = Type 0 t e ss = constant = Type I 2 t 2 /2! e ss = constant = Type II AERO 422, Instructor: Raktim Bhattacharya 9 / 28

20 Steady State Error Type Zero r + e u y C P y Type 0 System Constant steady state error to step reference k = 0 e ss = lim s 0 + L(s) s k = lim s 0 + L(s) = + K p Position Error Constant K p K p = lim L(s) s 0 AERO 422, Instructor: Raktim Bhattacharya 20 / 28

21 Steady State Error Type One r + e u y C P y Type System Constant steady state error to ramp reference k = e ss = lim s 0 + L(s) = lim s 0 + L(s) s k s = lim s 0 sl(s) = K v Velocity Error Constant K v K v = lim sl(s) s 0 AERO 422, Instructor: Raktim Bhattacharya 2 / 28

22 Steady State Error Type Two r + e u y C P y Type 2 System Constant steady state error to parabolic reference k = 2 e ss = lim s 0 + L(s) = lim s 0 + L(s) s k s 2 = lim s 0 s 2 L(s) = K a Acceleration Error Constant K a K a = lim s 2 L(s) s 0 AERO 422, Instructor: Raktim Bhattacharya 22 / 28

23 Steady State Error Summary r + e u y C P y Various Constants K p = lim s 0 L(s) Steady State Errors K v = lim s 0 sl(s) K a = lim s 0 s 2 L(s) Step Ramp Parabola Type 0 +K p Type I 0 K v Type II 0 0 K a AERO 422, Instructor: Raktim Bhattacharya 23 / 28

24 Steady State Error Summary (contd) r + e u y C P y Quickly identify ability to track polynomials Robustness property higher type tracks lower order polynomials Can be extended to study G yd and other transfer functions AERO 422, Instructor: Raktim Bhattacharya 24 / 28

25 Truxal s Formula Let T (s) := G yr (s) be given by T (s) = K Πm i= (s z i) Π n i= (s p i) Most common case: e ss to step is zero = Type I system DC gain lim s 0 T (s) = System error E(s) = R(s)( T (s)) System error due to ramp e ss = lim se(s) = lim s( T (s)) s 0 s 0 s 2 = lim T (s) s 0 s Using L Hopital s rule dt e ss = lim s 0 ds = K v for type I systems AERO 422, Instructor: Raktim Bhattacharya 25 / 28

26 Truxal s Formula Contd Let T (s) := G yr (s) be given by Using L Hopital s rule T (s) = K Πm i= (s z i) Π n i= (s p i) dt e ss = lim s 0 ds = K v for type I systems K v is related to the slope of T (s) at origin AERO 422, Instructor: Raktim Bhattacharya 26 / 28

27 Truxal s Formula Contd Let T (s) := G yr (s) be given by Rewrite Truxal s Formula T (s) = K Πm i= (s z i) Π n i= (s p i) dt e ss = lim T(0) = s 0 ds T d = lim log T (s) s 0 ds d = lim s 0 ds ( ) m n log(k) + log(s z i ) log(s p i ) K v = m i= i= z i n p i= i i= AERO 422, Instructor: Raktim Bhattacharya 27 / 28

28 Truxal s Formula Design Implication K v = m i= z i n p i= i Observe effect of pole/zero location on /K v Useful for design of dynamic compensators Example Third order type I system has closed-loop poles 2 ± 2j, 0 The system has one zero Where should it be for K v = 0? AERO 422, Instructor: Raktim Bhattacharya 28 / 28

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