16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde Lecture 14 Last time: w(t, r)=w(t-T) ()=」v(t-r)x( Let y(r=w(r)x(t-t)dr For the differential system characterized by its equations of state, specialization to invariance means that the system matrices A, B, C are constants. =Cx For AB. constant. y(1)=Cx(D) x()=Φ(t-1)x()+」c(t-)B(ldr The transition matrix can be expressed analytically in this ca ap(t, r)= A(t, r), where (r,r) This is a matrix form of first order constant coefficient differential equation. The solution is the matrix exponential +A(t-)+A(t-r)2+…+4(1-r) Useful for computing a(t) for small enough t-T Page 1 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 1 of 6 Lecture 14 Last time: wt wt (, ) ( ) τ ⇒ −τ 0 () ( ) ( ) ' Let: () ( ) ( ) t yt wt x d t d d yt w xt d τ τ τ τ τ τ τ τ τ τ −∞ ∞ = − ⎧ = − ⎨ − = ′ ⎩ = − ′ ′′ ∫ ∫ For the differential system characterized by its equations of state, specialization to invariance means that the system matrices ABC , , are constants. x Ax Bu y Cx = + = For ABC , , constant: 0 0 0 () () () ( ) ( ) ( ) ( ) t t y t Cx t x t t t x t t Bu d τ τ τ = =Φ − + Φ − ∫ The transition matrix can be expressed analytically in this case. ( , ) ( , ), where ( , ) d t At I dt Φ =Φ Φ = τ τ ττ This is a matrix form of first order, constant coefficient differential equation. The solution is the matrix exponential. ( ) () 2 2 (, ) 1 1 ( ) ( ) ... ( ) ... 2 ! A t At k k t e e I At A t A t k τ τ τ ττ τ − − Φ = =+ − + − + + − + Useful for computing Φ( )t for small enough t −τ
16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde The solution is y(1)=Cx(1) x(=e(-x()+e(-rBu(o)dr For→∞ x(1)=「e4=)B(r)dr =「e"B(t-r)dr and for a single input, single output (SISO)system, ()=ce If x(t)=e for all past time F(x(t Since w(r)=0 for t<0 for a realizable system, we see that the steady state inusoidal response function, F(o), for a system is the Fourier transform of the weighting function- where the coefficient unity must be used F(o)=w(r)e - dr and w(r) for a stable system is Fourier transformable Then (1)= F(oe/do You can compute the response to any input at all, including transient responses having defined F(o) for all frequencies The static sensitivity of the system is the zero frequency gain, F(O), which is just the integral of the weighting function F(0)=w(r)dr Page 2 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 2 of 6 The solution is 0 0 ( ) ( ) 0 () () () ( ) () t At t A t t y t Cx t x t e x t e Bu t d τ τ − − = = + ∫ For 0t → ∞ : ( ) 0 () ( ) ( ) t A t A x t e Bu d e Bu t d τ τ τ τ τ τ − −∞ ∞ ′ = = − ′ ′ ∫ ∫ and for a single input, single output (SISO) system, ( ) T At wt c e b = If ( ) j t x t e ω = for all past time ( ) 0 0 () ( ) ( ) ( ) () j t j jt yt w e d we de F xt ω τ ωτ ω τ τ τ τ ω ∞ − ∞ − = ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ = ∫ ∫ Since w() 0 τ = for τ < 0 for a realizable system, we see that the steady state sinusoidal response function, F( ) ω , for a system is the Fourier transform of the weighting function – where the coefficient unity must be used. ( ) () j F we d ωτ ω τ τ ∞ − −∞ = ∫ and w( ) τ for a stable system is Fourier transformable. Then 1 () ( ) 2 j t wt F e d ω ω ω π ∞ −∞ = ∫ You can compute the response to any input at all, including transient responses, having defined F( ) ω for all frequencies. The static sensitivity of the system is the zero frequency gain, F(0), which is just the integral of the weighting function. 0 F wd (0) ( ) τ τ ∞ = ∫
16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde Stationary statistics Invariant output statistics implies more than stationary inputs and invariant systems; it also implies that the system has been in operation long enough under the present conditions to have exhausted all transients Input-Output Relations for Correlation and Spectral Density Functions Derive autocorrelation of output in terms of autocorrelation of input y()=」v(r)x(-r)dr y=w(t,)xdt, x w(r)dr, R(r)=yoy(t+r) jdr w(r dt w([)x()x(+T-T2) jdr w(r dt w(t2)R(r+T, -T2) ∫drn(z)dzw2)R2(-z2 R2(r)=x()y(t+r) x(o]w(r)x(+T-T,)dr, ∫v(r)R2(a T-tdT Page 3 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 3 of 6 Stationary statistics Invariant output statistics implies more than stationary inputs and invariant systems; it also implies that the system has been in operation long enough under the present conditions to have exhausted all transients. Input-Output Relations for Correlation and Spectral Density Functions Derive autocorrelation of output in terms of autocorrelation of input 1 11 0 1 1 0 1 1 0 () ( ) ( ) ( ) ( ) yt w xt d y w xd xw d τ τ τ τ τ τ τ ∞ ∞ ∞ = − = = ∫ ∫ ∫ 11 2 2 1 2 0 0 11 2 2 12 0 0 ( ) () ( ) ( ) ( )( )( ) () () ( ) yy xx R yt yt d w d w xt xt dw dw R τ τ τ τ τ τ τ ττ τ τ τ τ ττ τ ∞ ∞ ∞ ∞ = + = − +− = +− ∫ ∫ ∫ ∫ 2 11 2 2 12 0 0 () () ( ) xx y dw dw R τ τ τ τ ττ ∞ ∞ = − ∫ ∫ 1 11 0 1 11 0 ( ) () ( ) () ( ) ( ) () ( ) xy xx R xt yt x t w xt d wR d τ τ τ ττ τ τ ττ τ ∞ ∞ = + = +− = − ∫ ∫
16.322 Stochastic Estimation and Control, Fall 2004 Prof vander Velde Transform to get power density spectrum of output. 下=x(dr F(0)x y(o)=」R(r)edr dr dr, w( dr w(t2)R(r+, -T2)e /er j drR(r+T, -T2 )e et+T-a]dr, w(t) ot dr w(t2)e era In first integral only, let Ir=T+T,-T2 Idr'=dz S,(o)=dr'Rm(r)e"er j dr w(r, Jeer dr, w(r2 )e" jera =Sx()F(-)F() F(o)'S( The power spectral density thus does not depend upon phase properties The cross-spectral density function can be derived similarly, to obtain y(o)=F(o)S(o) Mean squared output in time and frequency domain =R, (0)=dr, w(r dt w(t2)R(c -T2) s(ode F(OF(oS(o)de Generally speaking, with linear invariant systems it is easier to work in the transform domain than the time domain -so we shall commonly use the last expression to calculate the mean squared output of a system. However, control engineers are more accustomed to working with Laplace transforms than with Fourier transforms. By making the change of variables s= jo we can cast this expression in that form Page 4 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 4 of 6 Transform to get power density spectrum of output. 1 12 1 2 0 11 2 2 12 0 0 ( ) 12 11 2 2 0 0 first integral ( ) (0) ( ) () ) () ( ) ( ) () () In first integral o j yy yy j xx j jj xx y xw d F x S R ed d dw dw R e dR e d w e d w e ωτ ωτ ω τ τ τ ωτ ωτ τ τ ω ττ τ τ τ τ τ ττ τ τ ττ τ τ τ τ τ ∞ ∞ − −∞ ∞∞ ∞ − −∞ ∞ ∞∞ − +− − −∞ = = = = ( +− = +− ∫ ∫ ∫∫ ∫ ∫ ∫∫ 1 2 1 2 11 2 2 2 = + nly, let () () () ( ) ()( )() () () j j j yy xx xx xx d d S dR e dw e dw e SFF F S ωτ ωτ ωτ τ ττ τ τ τ ω τ τ ττ ττ ω ωω ω ω ∞ ∞∞ − ′ − −∞ −∞ −∞ ⎧ ′ − ⎨ ⎩ ′ = = ′ ′ = − = ∫ ∫∫ The power spectral density thus does not depend upon phase properties. The cross-spectral density function can be derived similarly, to obtain: () () () xy xx S FS ω = ω ω Mean squared output in time and frequency domain 2 11 2 2 12 0 0 (0) ( ) ( ) ( ) 1 ( ) 2 1 ()( ) () 2 yy xx yy xx y R dw dw R S d FF S d τ τ τ τ ττ ω ω π ω ω ωω π ∞ ∞ ∞ −∞ ∞ −∞ == − = = − ∫ ∫ ∫ ∫ Generally speaking, with linear invariant systems it is easier to work in the transform domain than the time domain – so we shall commonly use the last expression to calculate the mean squared output of a system. However, control engineers are more accustomed to working with Laplace transforms than with Fourier transforms. By making the change of variables s j = ω we can cast this expression in that form
16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde 2jT F(SF(s)Si(s)ds We know that S(o) is even. If it is a rational function of @, and we will work exclusively with rational spectra, it is then a rational function of a. So only even powers of o appear in S(o) and thus Sa= which we may call S(s)is derived from S(o) by replacing o by -s F'(s)is the ordinary transfer function of the system -the Laplace transform of its weighting function. Because w(0)=0, I<0 We shall drop the primes from now on F(S)F(sS.(sds in S(s) Integrating the output spectrum auchy Residue Theorem s-plan fF(s)ds=2r Z(residues of F(s) at the poles enclosed in the contour C) If F(s)has a pole of order m at ==a Res(a) (m-1) (s-a)"F(s) F(s) has a pole of order m at s=a if m is the smallest integer for which Page 5 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 5 of 6 2 1 2 1 () ( ) () 2 j xx j xx s s s ds y FF S j j jj F s F s S s ds j π π ∞ − ∞ ∞ −∞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ = − ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ = − ′′ ′ ∫ ∫ We know that ( ) xx S ω is even. If it is a rational function of ω , and we will work exclusively with rational spectra, it is then a rational function of 2 ω . So only even powers of ω appear in ( ) xx S ω and thus xx s S j ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ which we may call ( ) xx S s is derived from ( ) xx S ω by replacing 2 ω by 2 −s . F s ′( )is the ordinary transfer function of the system – the Laplace transform of its weighting function. Because wt t ( ) 0, 0 = ∫ F s ds j F s = π ∑ If F s( ) has a pole of order m at z a = , ( ) 1 1 1 Res( ) ( ) ( ) 1 ! m m m s a d a s a Fs m ds − − = ⎧ ⎫ = − ⎨ ⎬ ⎡ ⎤ ⎣ ⎦ − ⎩ ⎭ F s( ) has a pole of order m at s a = if m is the smallest integer for which
16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde is finite If F(s) is rational and has a Ist order pole at a F(s) D(s) hen Res(a)m=l=lim[(s-a)F(S)I (a-ba-c Page 6 of 6
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 6 of 6 lim ( ) ( )m s a s a Fs → ⎡ ⎤ − ⎣ ⎦ is finite. If F s( ) is rational and has a 1st order pole at a , ( ) ( ) ( ) ( ) ( )( )... N s F s D s N s s as b = = − − then Res( ) lim ( ) ( ) 1 [ ] ( ) ( )( )... m s a a s aFs N a a ba c = → = − = − −
