Lecture 6 Describing function analysis

Transcription

1 Lecture 6 Describing function analysis Today s Goal: To be able to Derive describing functions for static nonlinearities Predict stability and existence of periodic solutions through describing function analysis Material: Slotine and Li: Chapter 5 Chapter 14 in Glad & Ljung Chapter 7.2 (pp.28 29) in Khalil (Chapter 8 in Adaptive Control by Åström & Wittenmark) Lecture notes

2 Course Outline Lecture 1-3 Lecture 2-6 Lecture 7-8 Lecture 9-13 Lecture 14 Modelling and basic phenomena (linearization, phase plane, limit cycles) Analysis methods (Lyapunov, circle criterion, describing functions) Common nonlinearities (Saturation, friction, backlash, quantization) Design methods (Lyapunov methods, Backstepping, Optimal control) Summary

3 Example: saturated sinusoidals input signal output signal time [s] time [s] Sine Wave Saturation Scope The effective gain (the ratio sat(asinωt) ) varies with the Asinωt input signal amplitude A.

4 Motivating Example r e u y G(s) 2 1 u y G(s)= 4 andu=sat(e) gives stable oscillation forr=. s(s+1) 2 How can the oscillation be predicted? Q: What is the amplitude/topvalue ofuandy? What is the frequency?

5 Motivating Example r e u y G(s) 2 1 u y G(s)= 4 andu=sat(e) gives stable oscillation forr=. s(s+1) 2 How can the oscillation be predicted? Q: What is the amplitude/topvalue ofuandy? What is the frequency?

6 Recall the Nyquist Theorem Assume G(s) stable, and k is positive gain. The closed-loop system is unstable if the point 1/k is encircled by G(iω) The closed-loop system is stable if the point 1/k is not encircled by G(iω) e u y k G(s) 1/k G(iω)

7 Heuristic reasoning: Motivating Example (cont d) For what frequency and what amplitude is "the loop gain"f G= 1? Introduce N(A) as an amplitude dependent approximation of the nonlinearity f( ). G(iω) e u y f( ) G(s) 1/N(A) A y=g(iω)u G(iω)N(A)y G(iω)= 1 N(A)

8 Motivating Example (cont d) 2 1/N(A) 4 6 G(iω) Introduce N(A) as an amplitude dependent gain-approximation of the nonlinearity f( ). Heuristic reasoning: For what frequency and what amplitude is "the loop gain"n(a) G(iw)= 1? The intersection of the 1/N(A) and the Nyquist curveg(iω) predicts amplitude and frequency.

9 Motivating Example (cont d) 2 1/N(A) 4 6 G(iω) Introduce N(A) as an amplitude dependent gain-approximation of the nonlinearity f( ). Heuristic reasoning: For what frequency and what amplitude is "the loop gain"n(a) G(iw)= 1? The intersection of the 1/N(A) and the Nyquist curveg(iω) predicts amplitude and frequency.

10 Motivating Example (cont d) 2 1/N(A) 4 6 G(iω) Introduce N(A) as an amplitude dependent gain-approximation of the nonlinearity f( ). Heuristic reasoning: For what frequency and what amplitude is "the loop gain"n(a) G(iw)= 1? The intersection of the 1/N(A) and the Nyquist curveg(iω) predicts amplitude and frequency.

11 How do we derive the describing function N(A)? Does the intersection predict a stable oscillation? Are the estimated amplitude and frequency accurate?

12 Fourier Series Every periodic functionu(t)=u(t+t) has a Fourier series expansion u(t)= a 2 + (a n cosnωt+b n sinnωt) n=1 = a 2 + a 2 n+b 2 n sin[nωt+arctan(a n/b n )] n=1 where ω=2π/t and. a n = 2 T T u(t)cosnωtdt b n = 2 T T u(t)sinnωtdt Note: Sometimes we make the change of variablet φ/ω

13 The Fourier Coefficients are Optimal The finite expansion solves û k (t)= a k 2 + (a n cosnωt+b n sinnωt) min û n=1 2 T [ u(t) ûk (t) ] 2 dt T if{a n,b n } are the Fourier coefficients.

14 The Key Idea e u y N.L. G(s) Assumee(t)=Asinωt andu(t) periodic. Then u(t)= a 2 + a 2 n+b 2 n sin[nωt+arctan(a n/b n )] n=1 If G(inω) G(iω) forn=2,3,... anda =, then y(t) G(iω) a 2 1 +b2 1 sin[ωt+arctan(a 1/b 1 )+argg(iω)] Find periodic solution by matching coefficients in y = e.

15 Definition of Describing Function The describing function is N(A,ω)= b 1(ω)+ia 1 (ω) A e(t) N.L. u(t) e(t) N(A,ω) û 1 (t) IfGis low pass anda =, then û 1 (t)= N(A,ω) Asin[ωt+argN(A,ω)] can be used instead ofu(t) to analyze the system. Amplitude dependent gain and phase shift!

16 Idea: Use the describing function to approximate the part of the signal coming out from the nonlinearity which will survive through the low-pass filtering linear system. e(t)=asinωt=im(ae iωt ) e(t) u(t) N.L. u(t)= a 2 + (a n cosnωt+b n sinnωt) n=1 e(t) N(A,ω) u 1 (t) u 1 (t)=a 1 cos(ωt)+b 1 sin(ωt) =Im(N(A,ω)Ae iωt ) where the describing function is defined as N(A,ω)= b 1(ω)+ia 1 (ω) A = U(iω) N(A,ω)E(iω)

17 Existence of Limit Cycles G(iω) e u y f( ) G(s) 1/N(A) A y=g(iω)u G(iω)N(A)y G(iω)= 1 N(A) The intersections ofg(iω) and 1/N(A) give ω anda for possible limit cycles.

18 Describing Function for a Relay H u 1.5 e u e H φ a 1 = 1 π 2π 2π u(φ)cosφdφ= π b 1 = 1 π u(φ)sinφdφ= 2 π Hsinφdφ= 4H π The describing function for a relay is thusn(a)= 4H πa.

19 Describing Function for Odd Static Nonlinearities Assumef( ) and ( ) are odd static nonlinearities (i.e., f( e)= f(e)) with describing functionsn f andn. Then, ImN f (A,ω)=, coeff.(a 1 ) N f (A,ω)=N f (A) N αf (A)= αn f (A) N f+ (A)=N f (A)+N (A)

20 Limit Cycle in Relay Feedback System r e u y G(s) /N(A) G(iω) G(s)= 3 (s+1) 3 with feedback u= sgny 3/8= 1/N(A)= πa/4 A=12/8π.48 G(iω)= 3/8 ω 1.7, T=2π/ω=3.7

21 Limit Cycle in Relay Feedback System (cont d) The prediction via the describing function agrees very well with the true oscillations: 1.5 y u G filters out almost all higher-order harmonics.

22 Describing Function for a Saturation u H e D D e.4.2 u.2.4 H.6.8 φ Lete(t)=Asinωt=Asinφ. First seth=d. IfA Dthen N(A)=1, ifa>dthen for φ (,π) { Asinφ, φ (,φ ) (π φ u(φ)=,π) D, φ (φ,π φ ) where φ =arcsind/a.

23 Describing Function for a Saturation (cont d) a 1 = 1 π b 1 = 1 π 2π 2π φ u(φ)cosφdφ= u(φ)sinφdφ= 4 π = 4A sin 2 φdφ+ 4D π π = A ) (2φ +sin2φ π π/2 π/2 u(φ)sinφdφ φ sinφdφ

24 Describing Function for a Saturation (cont d) IfH=D N(A)= 1 ) (2φ +sin2φ, A D π ForH =Dthe rulen αf (A)= αn f (A) gives N(A)= H ) (2φ +sin2φ, A D Dπ N(A) forh=d=1 NOTE: dependance of A shows up in φ =arcsind/a

25 3 minute exercise: What oscillation amplitude and frequency do the describing function analysis predict for the Motivating Example?

26 Solution: Find ω anda such thatg(iω) N(A)= 1; AsN(A) is positive and real valued, find ω s.t. argg(iω)= π= π 2 2arctan ω 1 = π= ω=1. which gives a period time of 6.28 seconds. G(1.i) N(A)=2 N(A)=1= N(A)=.5. To find the amnplitude A, either ) Alt. 1 Solve (numerically)n(a)= 1 1 π (2φ +sin2φ =.5, where φ =arcsin(1/a) Alt. 2 From the diagram ofn(a) one can finda N(A) forh=d=

27 The Nyquist Theorem Assume G(s) stable, and k is positive gain. The closed-loop system is unstable if the point 1/k is encircled by G(iω) The closed-loop system is stable if the point 1/k is not encircled by G(iω) e u y k G(s) 1/k G(iω)

28 How to Predict Stability of Limit Cycles AssumeG(s) stable. For a givena=a : A increases if the point 1/N f (A ) is encircled byg(iω) A decreases otherwise e u y f G(s) G(iω) 1/N(A) A stable limit cycle is predicted

29 How to Predict Stability of Limit Cycles G(Ω) 1/N(A) An unstable limit cycle is predicted An intersection with amplitudea is unstable ifa<a gives decreasing amplitude anda>a gives increasing.

30 Stable Periodic Solution in Relay System r e u y G(s) G(iω) 1/N(A) G(s)= (s+1)2 (s+1) 3 with feedback u= sgny gives one stable and one unstable limit cycle. The left most intersection corresponds to the stable one.

31 Periodic Solution in Relay System The relay gainn(a) is higher for smalla: Big amplitudes Small relay gain No encirclement Shrinking amplitudes Growing relay gain One encirclement Growing amplitudes Shrinking relay gain Small amplitudes High relay gain No encirclement Shrinking amplitudes Growing relay gain Stable periodic orbit Unstable periodic orbit

32 Σ Automatic Tuning of PID Controller Period and amplitude of relay feedback limit cycle can be used for autotuning. PID A T u Process y Relay 1

33 Describing Function for a dead-zone relay D u e e.4.2 u D Lete(t)=Asinωt=Asinφ. Then for φ (,π) {, φ (,φ ) u(φ)= D, φ (φ,π φ ) where φ =arcsind/a (ifa D)

34 Describing Function for a dead-zone relay cont d. a 1 = 1 π b 1 = 1 π = 4D π 2π 2π u(φ)cosφdφ= u(φ)sinφdφ= 4 π cosφ = 4D π π/2 1 D 2 /A 2 φ Dsinφdφ {, A<D N(A)= 4 1 D πa 2 /A 2, A D

35 Plot of Describing Function for dead-zone relay N(A) ford= Notice that N(A) 1.3/A for large amplitudes

36 Pitfalls Describing function analysis can give erroneous results. DF analysis may predict a limit cycle, even if it does not exist. A limit cycle may exist, even if DF analysis does not predict it. The predicted amplitude and frequency are only approximations and can be far from the true values.

37 Example The control of output powerx(t) from a mobile telephone is critical for good performance. One does not want to use too large power since other channels are affected and the battery length is decreased. Information about received power is sent back to the transmitter and is used for power control. A very simple scheme is given by ẋ(t)= αu(t) u(t)= signy(t L), α, β> y(t)=βx(t). Use describing function analysis to predict possible limit cycle amplitude and period of y. (The signals have been transformed so x=corresponds to nominal output power) y(t)=βx(t) x(t) y(t) y(t L) α x(t) s β e sl y(t L) u(t)

38 The system can be written as a negative feedback loop with G(s)= e sl α β s and a relay with amplitude 1. The describing function of a relay satisfies 1/N(A)= πa/4 hence we are interesting ing(iω) on the negative real axis. A stable intersection is given by π= argg(iω)= π/2 ωl which gives ω= π/(2l).this gives πa 4 = G(iω) = α β ω =2Lα β π and hencea=8lα β/π 2. The period is given by T=2π/ω=4L. (More exact analysis gives the true values A= α βl andt=4l, so the prediction is quite good.)

39 Accuracy of Describing Function Analysis Control loop with frictionf= sgny: F Friction y ref _ C u _ G y Corresponds to G 1+GC = s(s b) s 3 +2s 2 +2s+1 with feedback u= sgny The oscillation depends on the zero ats=b.

40 Accuracy of Describing Function Analysis 1.2 b=1/3 1.8 b=4/ DF predicts period times and ampl.(t,a) b=4/3 =(11.4,1.) and(t,a) b=1/3 =(17.3,.23) y b=4/3 Simulation: y b=1/ (T,A) b=4/3 =(12,1.1) (T,A) b=1/3 =(22,.28) Accurate results only if y is sinusoidal!

41 Analysis of Oscillations A summary There exist both time-domain and frequency-domain methods to analyze oscillations. Time-domain: Poincaré maps and Lyapunov functions Rigorous results but hard to use for large problems Frequency-domain: Describing function analysis Approximate results Powerful graphical methods

42 Today s Goal To be able to Derive describing functions for static nonlinearities Predict stability and existence of periodic solutions through describing function analysis

43 Next Lecture Saturation and antiwindup compensation Lyapunov analysis of phase locked loops Friction compensation