16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde Lecture 18 Last time: Semi-free configuration design This is equivalent to H6cr○e D(s) perhaps B). We must stabilize F if it is given as unstabe e desig e (and lote n, s enter the system at the same place. F is fixed C(s) 1+C(S)F(SB(s) so that having the optimum H, we determine C from (s) 1-H(SF(S)B(S) We do not collect H and F together because if F is non-minimum phase, we would not wish to define h by H_(HFopt This leads to an unstable mode which is not observable at the output-thus cannot be controlled by feeding back Associate weighting functions with the given transfer functions F(s)→wF(1) D(s)→>D(1)
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 1 of 5 Lecture 18 Last time: Semi-free configuration design This is equivalent to: Note n s, enter the system at the same place. F is fixed. We design C (and perhaps B). We must stabilize F if it is given as unstable. ( ) ( ) 1 () () () C s H s CsFsBs = + so that having the optimum H , we determine C from ( ) ( ) 1 () () () H s C s H sFsBs = − We do not collect H and F together because if F is non-minimum phase, we would not wish to define H by ( )opt HF H F = This leads to an unstable mode which is not observable at the output – thus cannot be controlled by feeding back. Associate weighting functions with the given transfer functions. () () () () () () H F D Hs w t Fs w t Ds w t → → →
16.322 Stochastic Estimation and Control, Fall 2004 Prof vander Velde If F(s)is unstable, put a stabilizing feed back around it, later associate it with the rest of the system Error analysis We require the mean squared error c(1)=|vn(x1)(-x1)dr o(1)=v(z2)c(1-22)dr2 dr, p(t2) dr,wu()(t-T,-r2) d(O=」w(z3)(t e()=0()-d(1) e(t)2=o(1)2-2o(1)d(1)+d()2 ∫drvn(z)jarw(2)drn()jd(x.)(-x1-)(-2-) =dr, w(r, dr,w, (t2) dr,wn(t,) dr,w (T,R, (T,+T2-T3-r) o(Od(o=j dr, w,(r] dr we, (r i(-=- -, j dr, wo(r. st-r,) dw")Jdw()y(5)(-x=(-5) jdr,wn( dr wE(r,)dr, wD(t,R,(r,+T2-T3) We shall not require d(r) in integral form
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 2 of 5 If F s( ) is unstable, put a stabilizing feedback around it, later associate it with the rest of the system. Error Analysis We require the mean squared error. 1 11 2 22 2 2 1 1 12 3 33 22 2 () ( )( ) () ( )( ) ( ) ( )( ) () ( )( ) () () () () () 2 () () () H F F H D ct w it d ot w ct d d w d w it dt w st d et ot dt et ot otdt dt τ ττ τ ττ τ τ τ τ ττ τ ττ ∞ −∞ ∞ −∞ ∞ ∞ −∞ −∞ ∞ −∞ = − = − = −− = − = − =− + ∫ ∫ ∫ ∫ ∫ 2 2 2 1 1 12 4 4 3 3 34 1 1 2 2 3 3 4 4 12 34 11 22 33 () ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )( ) () () () FH FH HFHF HFH ot d w d w it d w d w it d w d w d w d w it it dw dw dw d τ τ τ τ ττ τ τ τ τ ττ τ τ τ τ τ τ τ τ ττ ττ ττ ττ ττ ∞∞ ∞∞ −∞ −∞ −∞ −∞ ∞∞ ∞ ∞ −∞ −∞ −∞ −∞ ∞ ∞ −∞ −∞ ⎡ ⎤⎡ ⎤ = −− −− ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦ = −− −− = ∫∫ ∫∫ ∫∫∫∫ ∫ ∫ 4 4 1234 ()( ) w R F ii τ τ ττττ ∞ ∞ −∞ −∞ +−− ∫ ∫ 2 2 1 1 12 3 3 3 1 1 2 2 3 3 12 3 1 1 2 2 3 3 123 () () ( ) ( )( ) ( ) ( ) ( ) ( ) ( )( ) ( ) () () () ( ) FH D HFD H F D is otdt d w d w it d w st d w d w d w it st dw dw dw R τ τ τ τ ττ τ τ τ τ τ τ τ τ τ ττ τ τ τ τ τ τ τ τττ ∞∞ ∞ −∞ −∞ −∞ ∞∞ ∞ −∞ −∞ −∞ ∞∞ ∞ −∞ −∞ −∞ ⎡ ⎤⎡ ⎤ = −− − ⎢ ⎥⎢ ⎥ ⎣ ⎦⎣ ⎦ = −− − = +− ∫∫ ∫ ∫∫∫ ∫∫∫ We shall not require 2 d t( ) in integral form
16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde The problem now is to choose wu(t)so as to minimize this e(t), for which we use variational calculus Let vB(1)=0(1)+ov() where wo(t) is the optimum weighting function( to be determined)and dw(t)is an arbitrary variation -arbitrary except that it must be physically realizable Calculate the optimum e and its first and second variations e-=e+se+s =o(1)2+20(1)d(1)+d(1) The optimum e(e for &w(1)=0) e(0o= dr,wo(r,) dr,wr(t2) dr wo(r) dT4WE(T4)R, (T, +T2-T3-t) 2 dr, wo(r,) w (t2) dr wp(t,)R, ([+T2-t3)+d() The first variation in e(t)2is Se('=dt, Sw(r) dr wF(2)dr wo(r3) dt,wE(T),(4,+T2-t3-T) ∫drvs()jdze(a)∫arn(n)Jdt"()尺(+x2-2-) -2 dr, Sw(r dr wr(r2) dr wp(r)R, (r,+T2-t3) In the second term, let: and interchange the order of integration 2nd term=dr sw(r,dri w(2)dr'w'() drw (EaR, (t'+r'-ri'-t2)
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 3 of 5 The problem now is to choose ( ) w t H so as to minimize this 2 e t( ) , for which we use variational calculus. Let: 0 () () () w t w t wt H = + δ where 0 w t( ) is the optimum weighting function (to be determined) and δw t( ) is an arbitrary variation – arbitrary except that it must be physically realizable. Calculate the optimum 2 e and its first and second variations. 2 2 2 22 0 22 2 () 2 () () () ee e e e ot otdt dt =+ + δ δ =+ + The optimum 2 e ( 2 e for δw t() 0 = ): 2 0 10 1 2 2 30 3 4 4 1 2 3 4 2 10 1 2 2 3 3 1 2 3 () ( ) ( ) ( ) ( ) ( ) 2 ( ) ( ) ( ) ( ) () F F ii F D is et d w d w d w d w R d w d w d w R dt τ τ τ τ τ τ τ τ ττττ τ τ τ τ τ τ τττ ∞∞ ∞ ∞ −∞ −∞ −∞ −∞ ∞∞ ∞ −∞ −∞ −∞ = +−− − +− + ∫∫ ∫∫ ∫∫ ∫ The first variation in 2 e t( ) is 2 1 1 2 2 30 3 4 4 1 2 3 4 10 1 2 2 3 3 4 4 1 2 3 4 1 1 2 2 3 3 123 () ( ) ( ) ( ) ( ) ( ) () () () () ( ) 2 () () () ( ) F F ii F F ii F D is et d w d w d w d w R dw dw d w dw R d w dw dw R δ τδ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τδ τ τ τ τ τ τ τ τδ τ τ τ τ τ τ τ τ ∞ ∞ ∞∞ −∞ −∞ −∞ −∞ ∞∞ ∞ ∞ −∞ −∞ −∞ −∞ ∞∞ ∞ −∞ −∞ −∞ = +−− + +−− − +− ∫∫ ∫∫ ∫∫ ∫ ∫ ∫∫∫ In the second term, let: 1 3 2 4 3 1 4 2 τ τ τ τ τ τ τ τ = ′ = ′ = ′ = ′ and interchange the order of integration. 2nd term 1 1 2 2 30 3 4 4 3 4 1 2 () () () () ( ) F F ii d w dw dw dw R τ δτ τ τ τ τ τ τ τ τ τ τ ∞ ∞ ∞∞ −∞ −∞ −∞ −∞ = +−− ′ ′ ′ ′ ′′ ′ ′ ′ ′ ′ ′ ′ ∫∫ ∫∫
16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde but since R,(t'+t4--T2=R,(G+r,) we see that the second term is exactly equal to the first term. Collecting these terms and separating out the common integral with respect to t, gives ∫dn-()drya()R(+2-x) The second variation of e(is 8e(0'=dt, w(r)dr wr(r2) dt, w(t,) dr, wA(T)R,(r, +T2-r, -t By comparison with the expression for o(t), this is seen to be the mean squared output of the system output (output)=8e(0)>0, non-zero input This second variation must be greater than zero, so the stationary point defined by the vanishing of the first variation is shown to be a minimum In the expression for the first variation, Ow(T,)=0 for t, <0 by the requirement that the variation be physically realizable. But Sw(t,) is arbitrary for t, 20,so we can be assured of the vanishing of &e(0 only if the( term vanishes almost everywhere for t, 20. The condition which defines the minimum in e(t) is then jdr,wr(z2)dr, wo(t,) dr,wE(TA)R, (4+t2-t3-z4) dr]wF(r2) dr, wp(r3)R, (T,+T2-T3)=0 for all t. non-real-time Using this condition in the expression for e(0 and remembering that wo(1)=0 for t<o gives the result
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 4 of 5 but since 3412 1234 ( )( ) R R ii ii τ ′′′′ ′′′′ + −− = + −− τττ ττττ we see that the second term is exactly equal to the first term. Collecting these terms and separating out the common integral with respect to 1 τ gives 2 1 1 2 2 30 3 4 4 1 2 3 4 2 2 3 3 123 () 2 ( ) ( ) ( ) ( ) ( ) () () ( ) F F ii F D is et d w d w d w d w R dw dw R δ τδ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τττ ∞ ∞ ∞∞ −∞ −∞ −∞ −∞ ∞ ∞ −∞ −∞ ⎧ = +−− ⎨ ⎩ ⎫ − +− ⎬ ⎭ ∫ ∫ ∫∫ ∫ ∫ The second variation of 2 e t( ) is 2 2 1 1 2 2 3 3 4 4 1234 () ( ) ( ) ( ) ( ) ( ) F F ii δ et d w d w d w d w R τδ τ τ τ τδ τ τ τ τ τ τ τ ∞∞ ∞ ∞ −∞ −∞ −∞ −∞ = +−− ∫∫∫∫ By comparison with the expression for 2 o t( ) , this is seen to be the mean squared output of the system ( )2 2 2 output ( ) 0, non-zero input = > δ e t This second variation must be greater than zero, so the stationary point defined by the vanishing of the first variation is shown to be a minimum. In the expression for the first variation, 1 δw() 0 τ = for 1 τ < 0 by the requirement that the variation be physically realizable. But 1 δw( ) τ is arbitrary for 1 τ ≥ 0 , so we can be assured of the vanishing of 2 δ e t( ) only if the { } term vanishes almost everywhere for 1 τ ≥ 0 . The condition which defines the minimum in 2 e t( ) is then 2 2 30 3 4 4 1 2 3 4 2 2 3 3 123 () () () ( ) () () ( ) 0 F F ii F D is dw dw dw R dw dw R τ τ τ τ τ τ ττττ τ τ τ τ τττ ∞ ∞∞ −∞ −∞ −∞ ∞ ∞ −∞ −∞ +−− − +− = ∫ ∫∫ ∫ ∫ for all 1 τ , non-real-time. Using this condition in the expression for 2 0 e t( ) and remembering that 0 w t() 0 = for t < 0 gives the result
16.322 Stochastic Estimation and Control, Fall 2004 Prof vander Velde (=d(1)2-o( which is convenient for the calculation of e(0) Also since o(0o=d(o-e(0, this says the optimum mean squared output always less than the mean squared desired output. Autocorrelation functions We have arrived at an extended form of the Wiener-Kopf equation which defines the optimum linear system under the ground rules stated before Recall that. R (r)=r()+R(r)+r(r)+r(t) R2(r)=R2(r)+Rn() The free configuration problem is a specialization of the semi-free configuration In this expressic would take F(s)=l, or w(0)=8(0). In that case we have ∫d:d()Jdrw(x)4;(r,R(r r3-4 dz2(2)dr3w(3)R2(x1+22-3)= Wo(T),(T-T)dr, -wp(I,)R (T-T3)dr=0 for T, 20
16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 5 of 5 2 22 0 0 et dt ot () () () = − which is convenient for the calculation of 2 0 e t( ) . Also since 2 22 0 0 ot dt et () () () = − , this says the optimum mean squared output is always less than the mean squared desired output. Autocorrelation Functions We have arrived at an extended form of the Wiener-Kopf equation which defines the optimum linear system under the ground rules stated before. Recall that: () () () () () () () () ii ss sn ns nn is ss ns RRR RR RRR τ ττττ τττ =+++ = + since i sn = + . The free configuration problem is a specialization of the semi-free configuration. In this expression we would take F s() 1 = , or () () wt t F = δ . In that case we have 2 2 30 3 4 4 1 2 3 4 22 3 3 123 03 1 3 3 3 1 3 3 1 () () () ( ) () () ( ) ( ) ( ) ( ) ( ) 0 for 0 ii D is ii D is dd dw d R d dw R wR d w R d τ τ τ τ τδτ τ τ τ τ τδτ τ τ τ τ τ τ τττ τ τττ τ ∞∞ ∞ −∞ −∞ −∞ ∞ ∞ −∞ −∞ ∞ ∞ −∞ −∞ +−− − +− = −− −= ≥ ∫∫ ∫ ∫ ∫ ∫ ∫
