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 ∞ ∞ ∞∞ −∞ −∞ −∞ −∞ = +−− ′ ′ ′ ′ ′′ ′ ′ ′ ′ ′ ′ ′ ∫∫ ∫∫