Problem Set 2 Solutions 1

Transcription

1 Massachusetts Institute of echnology Department of Electrical Engineering and Computer Science 6.45: MULIVARIABLE CONROL SYSEMS by A. Megretski Problem Set Solutions Problem. For each of the statements below decide whether it is true or false. For a true statement, sketch a proof. For a false statement, give a counterexample. (a) H norm of a stable finite order C LI state space model is never larger than times its H-Infinity norm. his statement is false. o see this, consider the stable first order LI C system with transfer function G(s) = G a (s) =, s + a where a > is a real parameter. hen G a = sup G(jω) = sup =, a + ω a ω R ω R ( ) / G = e at a H dt = a (we used the fact that g a (t) = e at, where t, is the impulse response of G a ). Hence, as a, H-Infinity norm of G a becomes arbitrarily large relative the H norm of G a. Version of February, 4

2 (b) H-Infinity norm of a stable finite order C LI state space model is never larger than times its H norm. his statement is false. o see this, use G a from (a) with a. (c) H norm of a stable finite order D LI state space model is never larger than times its H-Infinity norm. he statement is true for SISO models, because G = H π G(e jω ) dω π π π max { G(e jω ) }dω = G. π π ω However, for MIMO systems, the statement is false. o see this, consider G = G(z) which is an n-by-n identity matrix: its H-Infinity norm equals while the H norm equals n. (d) H-Infinity norm of a stable finite order D LI state space model is never larger than times its H norm. his statement is false. o see this, consider the stable first order LI D system with transfer function H(z) = H a (z) =, a/z where a (, ) is a real parameter. hen H a = sup H(e jω ) =, a ω R ( ) / k H a H = a = a k= (we used the fact that h a [k] = a k, where k, is the impulse response of H a ). Hence, as a, H-Infinity norm of H a becomes arbitrarily large relative the H norm of H a. Problem. For continuous time (non-li) systems S a with scalar input f = f (t) and scalar output g = g(t), described below, find their L gains, as functions of parameter a >. Support your answer with arguments (typically,

3 3 to show that L gain of a system equals, say,.5, one has to find input/output pairs of infinite energy (i.e. not converging to zero) for which the asymptotic (as time converges to infinity) output-to-input energy ratio is arbitrarily close to.5 =.5, and, in addition, to show that the asymptotic energy ratio cannot be larger than.5). (a) g(t) = a sin(f (t)); L gain equals a. o see that the L gain cannot be larger, note that for all real y. Hence sin(y) y g(t) dt = a sin(f (t)) dt a f (t) dt. o see that the L gain cannot be smaller, consider input f (t) e, where π > is a small parameter. hen, for every γ we have {γ f (t) g(t) }dt = (γ π a sin (π)), which will converge to minus infinity as unless γ π a sin (π). Since this inequality must be satisfied for all π whenever γ is larger than the L gain, we have γ a. (b) g(t) = f (at) sin(t); For a L gain equals / a. For a > the gain is infinite (and the system is not causal). o see that L gain does not exceed / a for a, note that g(t) dt = f (at) sin (t)dt a f (at) dt = f (t) dt f (t) dt. a a

4 4 o see that L gain is not smaller than / a for a, consider { k )π π, (k + f (t) = /, t [(k + )π + π], k {,,... },, otherwise. (he main idea is that f (t) should be zero when sin(t) is not close to, and the energy of f (t) on the interval [a, ] should converge to zero as, while the total energy of f should be infinite.) hen {γ f (t) g(t) }dt γ f (t) dt cos (π) f(at) dt o show that L gain is infinite for a >, consider { r k, t [(k + )π π, (k + )π + π], k {,,... }, f (t) =, otherwise, o see that L gain does not exceed + a, note that ( ) ax + y ( + a) a x + y x, y, ( cos (π) ) a = γ f (t) dt + γ f (t) dt, a which converges to as unless γ cos (π)/a. Since π > can be arbitrarily small, g /a for a. where r is a parameter. It is easy to see that, when r is sufficiently large compared to γ, the integrals converge to minus infinity as. (c) g(t) = af (t) f (t ). L gain equals + a. {γ f (t) g(t) }dt a and hence g(t) dt ( + a)a f (t) dt + ( + a) f (t ) dt ( + a) f (t) dt + f (t) dt. o see that L gain is not smaller than + a, consider the case when f (t), and hence g(t) ( + a).

5 5 Problem.3 Continuous time signal q = q(t) is the output of a pure double integrator system with input f = f (t), and g(t) = q(t) + bf (t), where b > is a known constant (do the calculations for b =. and b = ). Find an LI filter F = F (s) which takes g = g(t) as an input and outputs an estimate qˆ = qˆ(t) of q = q(t), which is good in one of the following interpretations: (a) Assuming that f = [f ; f ] is white noise, minimize the asymptotic value of the variance of the estimation error e = q qˆ. (b) Minimize the L gain from f = [f ; f ] to the estimation error e = q qˆ with accuracy percent. For a system with state space equations and sensor output a standard observer has the format ẋ(t) = A x(t) + B w(t) y (t) = C x(t) + D w(t), x(t) ˆ = A x(t) ˆ + L(C x(t) ˆ y (t)), where matrix L is chosen in such way that A + LC is a Hurwitz matrix. Note that here ˆx is a result of applying an LI transformation to y. Hence d(t) = y (t) C x(t) ˆ is a result of applying an LI transformation to y (t), and, reciprocally, y (t) is a result of applying an LI transformation to d(t). herefore, designing an LI filter F with input y output v = F y, to be an optimal estimate of a state component q(t) = C x(t) can be reduced to designing an LI filter F d with input d(t) and output v d = F d d, to be a good estimate of q d (t) = q(t) C x(t): the relation between v d and v will be ˆ v(t) = C x(t) ˆ + v d (t). he relation between w, d, and q d is given by ė = (A + LC )e + (B + LD )w, d = C e + Dw, q d = C e,

6 6 where e(t) = x(t) ˆx(t) plays the role of system state. he task of finding an optimal or suboptimal estimator for q d based on measuring d can be formulated as a standard LI feedback optimization setup with A = A +LC, B = B +LD, B =, C = C, D =, D = I, C = C, D = D, D =. o implement the filter optimization approach using MALAB, we use a design SIMULINK model ps 3a.mdl, w /s^ plant (s+)/(s^+s+) observer y 3 u e w b Gain handled by M-function ps 3.mdl: function [Fh,Fhi,Eh,Ehi]=ps_3(b) % function ps_3(b) % % solution for Problem.3 in 6.45/Spring 4 if nargin<, b=; end assignin( base, b,b); s=tf( s ); assignin( base, s,s); load_system( ps_3a ); [a,b,c,d]=linmod( ps_3a ); close_system( ps_3a ); [ar,br,cr,dr]=ssdata(minreal(ss(a,b,c,d))); p=pck(ar,br,cr,dr); nmeas=; ncon=;

7 7 ricmethd=; quiet=; [kh,gh]=hsyn(p,nmeas,ncon,ricmethd,quiet); [ah,bh,ch,dh]=unpck(kh); [ag,bg,cg,dg]=unpck(gh); Kh=ss(ah,bh,ch,dh); G=(s+)/(s^+s+); disp( H controller: ) Fh=tf(minreal(Kh*(G-)+G)) Eh=tf(minreal(ss(ag,bg,cg,dg))); gmin=; gmax=norm(eh,inf); tol=.; epr=e-; epp=e-6; [khinf,ghinf]=hinfsyn(p,nmeas,ncon,gmin,gmax,tol,ricmethd,epr,epp,quiet); [ahi,bhi,chi,dhi]=unpck(khinf); [ag,bg,cg,dg]=unpck(ghinf); Ehi=tf(minreal(ss(ag,bg,cg,dg))); Khi=ss(ahi,bhi,chi,dhi); disp( H-Infinity controller: ) Fhi=tf(minreal(Khi*(G-)+G)) Note the need for using minreal.m: MALAB does not eliminate uncontrollable/unobservable states automatically.