16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Lecture 14 Last time: w(t ,τ ) ⇒ w(t − τ ) t y (t ) = ∫ w(t − τ ) x(τ )dτ −∞ ⎧ τ ' = t −τ Let: ⎨ ⎩ − dτ = dτ ′ ∞ y (t ) = ∫ w(τ ′) x (t − τ ′)dτ ′ 0 For the differential system characterized by its equations of state, specialization to invariance means that the system matrices A, B, C are constants. x = Ax + Bu y = Cx For A, B, C constant: y (t ) = Cx (t ) t x (t ) = Φ (t − t0 ) x (t0 ) + ∫ Φ (t − τ ) Bu (τ )dτ t0 The transition matrix can be expressed analytically in this case. d Φ (t ,τ ) = AΦ (t ,τ ), where Φ (τ ,τ ) = I dt This is a matrix form of first order, constant coefficient differential equation. The solution is the matrix exponential. Φ (t ,τ ) = e A( t −τ ) e A( t −τ ) = I + A(t − τ ) + 1 2 1 A (t − τ ) 2 + ... + Ak (t − τ ) k + ... 2 k! Useful for computing Φ (t ) for small enough t − τ . Page 1 of 6 16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde The solution is y (t ) = Cx (t ) t x (t ) = e A( t −t0 ) x (t0 ) + ∫ e A( t −τ ) Bu (t )dτ t0 For t0 → ∞ : t x (t ) = ∫e A( t −τ ) Bu (τ )dτ −∞ ∞ = ∫ e Aτ ′ Bu (t − τ ′)dτ ′ 0 and for a single input, single output (SISO) system, w(t ) = c T e At b If x (t ) = e jωt for all past time ∞ y (t ) = ∫ w(τ )e jω ( t −τ ) dτ 0 ⎡∞ ⎤ = ⎢ ∫ w(τ )e − jωτ dτ ⎥ e jωt ⎣0 ⎦ = F (ω ) x (t ) Since w(τ ) = 0 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. F (ω ) = ∞ ∫ w(τ )e − jωτ dτ −∞ and w(τ ) for a stable system is Fourier transformable. Then ∞ 1 w( t ) = F (ω )e jωt d ω ∫ 2π −∞ 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. ∞ F (0) = ∫ w(τ )dτ 0 Page 2 of 6 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 (t ) = ∫ w(τ 1 ) x (t − τ 1 )dτ 1 0 ∞ y = ∫ w(τ 1 ) xdτ 1 0 ∞ = x ∫ w(τ 1 )dτ 1 0 R yy (τ ) = y (t ) y (t + τ ) ∞ ∞ = ∫ dτ 1w(τ 1 ) ∫ dτ 2 w(τ 2 ) x (t − τ 1 ) x (t + τ − τ 2 ) 0 0 ∞ ∞ = ∫ dτ 1w(τ 1 ) ∫ dτ 2 w(τ 2 ) Rxx (τ + τ 1 − τ 2 ) 0 0 ∞ ∞ 0 0 y 2 = ∫ dτ 1w(τ 1 ) ∫ dτ 2 w(τ 2 ) Rxx (τ 1 − τ 2 ) Rxy (τ ) = x (t ) y (t + τ ) ∞ = x (t ) ∫ w(τ 1 ) x (t + τ − τ 1 )dτ 1 0 ∞ = ∫ w(τ 1 ) Rxx (τ − τ 1 )dτ 1 0 Page 3 of 6 16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Transform to get power density spectrum of output. ∞ y = x ∫ w(τ )dτ 0 = F (0) x S yy (ω ) = ∞ ∫R yy (τ )e − jωτ dτ −∞ ∞ = ∫ −∞ ∞ ∞ 0 0 dτ ∫ dτ 1w(τ 1 ) ∫ dτ 2 w(τ 2 ) Rxx (τ + τ 1 − τ 2 )e − jωτ1 ∞ = ∞ ∞ − jω (τ +τ −τ ) − jωτ jωτ ∫−∞ dτ Rxx (τ + τ 1 − τ 2 )e 1 2 ∫0 dτ 1w(τ 1 )e 1 ∫0 dτ 2 w(τ 2 )e 2 first integral ⎧τ ′=τ +τ 1 − τ 2 In first integral only, let ⎨ ⎩ dτ ′ = dτ S yy (ω ) = ∞ ∫ −∞ ∞ dτ ′Rxx (τ ′)e − jωτ ′ ∫ dτ 1w(τ 1 )e jωτ1 −∞ ∞ ∫ dτ 2 w(τ 2 )e − jωτ 2 −∞ = S xx (ω ) F ( −ω ) F (ω ) 2 = F (ω ) S xx (ω ) The power spectral density thus does not depend upon phase properties. The cross-spectral density function can be derived similarly, to obtain: S xy (ω ) = F (ω ) S xx (ω ) Mean squared output in time and frequency domain ∞ ∞ 0 0 y = R yy (0) = ∫ dτ 1w(τ 1 ) ∫ dτ 2 w(τ 2 ) Rxx (τ 1 − τ 2 ) 2 = 1 2π ∞ ∫S yy (ω )d ω −∞ ∞ 1 = ∫ F (ω ) F (−ω )S xx (ω )dω 2π −∞ 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 = jω we can cast this expression in that form. Page 4 of 6 16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde y2 = = 1 2π j∞ ⎛ s ⎞ ⎛ s ⎞ ⎛ s ⎞ ds F ⎜ ⎟F ⎜ − ⎟ S xx ⎜ ⎟ ⎝ j⎠ ⎝ j⎠ ⎝ j⎠ j − j∞ 1 2 jπ ∫ ∞ ∫ F ′( s) F ′(− s)S ′ ( s)ds xx −∞ We know that S xx (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 ω 2 . So only ⎛s⎞ even powers of ω appear in S xx (ω ) and thus S xx ⎜ ⎟ which we may call S xx ( s ) is ⎝ j⎠ derived from S xx (ω ) by replacing ω 2 by − s 2 . F ′( s ) is the ordinary transfer function of the system – the Laplace transform of its weighting function. Because w(t ) = 0, t < 0 . We shall drop the primes from now on. j∞ 1 y2 = F ( s ) F ( − s ) S xx ( s )ds 2π j −∫j∞ ω 2 = − s 2 ⎫⎪ ⎬ in S xx ( s ) ω 4 = s 4 ⎪⎭ Integrating the output spectrum General method Cauchy Residue Theorem >∫ F ( s)ds = 2π j ∑ ( residues of F ( s) at the poles enclosed in the contour C ) C If F ( s ) has a pole of order m at z = a , Res( a ) = ⎧ d m −1 ⎫ 1 m ⎨ m −1 ⎡⎣( s − a ) F ( s ) ⎤⎦ s =a ⎬ ( m − 1)! ⎩ ds ⎭ 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 lim ⎡( s − a ) F ( s ) ⎤ ⎦ s →a ⎣ is finite. m If F ( s ) is rational and has a 1st order pole at a , N ( s) F ( s) = D( s) N ( s) = ( s − a )( s − b)... then Res( a ) m =1 = lim [( s − a ) F ( s )] s→a = N (a ) ( a − b)( a − c )... Page 6 of 6