《随机预算与调节》（英文版）Lecture 17 Last time: Ground rules for filtering and control system design

16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde Lecture 17 Last time: Ground rules for filtering and control system design General system desired output error wa(t) w.(t) signal system output n(t) Real Syste System parameters are contained in w, (()and w,(O) Desired output is generated by taking the signal through the desired operator The difference between the actual output and the desired output is the error, whose mean squared value we want to minimize. We require a stable and realizable system. The error e(t) is given by e()=∫w、x)(-z)7-w()(-)dr+∫v,()m(-r)r ve(z1)s(t-1)dr1+」wn(x1)m(-x1)dr1 where w,(r)=w(r)-w(T Using the replacement we(T,) reduces the number of terms in Ree (r) from nine to Page 1 of 7

16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 1 of 7 Lecture 17 Last time: Ground rules for filtering and control system design General system System parameters are contained in ( ) w t s and ( ) w t n . Desired output is generated by taking the signal through the desired operator. The difference between the actual output and the desired output is the error, whose mean squared value we want to minimize. We require a stable and realizable system. The error e t( ) is given by: 1 11 1 11 1 11 1 11 1 11 () ( )( ) ( )( ) ( ) ( ) ( )( ) ( )( ) sd n e n et w st d w st d w nt d w st d w nt d τ ττ τ ττ τ ττ τ ττ τ ττ ∞∞ ∞ −∞ −∞ −∞ ∞ ∞ −∞ −∞ = −− −+ − = −+ − ∫∫∫ ∫ ∫ where () () () www e sd τ = − τ τ . Using the replacement 1 ( ) we τ reduces the number of terms in ( ) Ree τ from nine to four

16.322 Stochastic Estimation and Control, Fall 2004 Prof. VanderⅤelde R(r)=e(De(t+r) drld2w2(1川2(z2)R2(+1-z2) dr, dr,we(t,)wn (t2)R(t+T1-T2) ∫ dr,dr;w,、)"(x2)Rn(z+-2) +dr dr w,( ) w, (t2)R(r+T, -t2) Se(s)=r(r)e dr =F2(-s)F2(s)S(s) +FCSF(s)S,(s) +F(SF(sS(s) F,GSF(s),( where F(S)=F(s)-F4(s) We know the configuration so we can work out the form of each of the transforms f Computation of the transfer functions F(s) and Fn(s) many system configur especially with minor feedback loops, manipulating the diagram into a standard input-output form is very time- conserving and subject to errors Usually just tracing signals around the loops is easiest Example: Find F(s) s=0 s(TS+1) Page 2 of 7

16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 2 of 7 1 2 1 2 12 1 2 1 2 12 1 2 1 2 12 1 2 1 2 12 ( ) ()( ) () () ( ) () () ( ) () () ( ) () () ( ) ee e e ss e n sn n e ns n n nn R etet d dw w R d dw w R d dw w R d dw w R τ τ τ τ τ τ ττ τ τ τ τ τ ττ τ τ τ τ τ ττ τ τ τ τ τ ττ τ ∞ ∞ −∞ −∞ ∞ ∞ −∞ −∞ ∞ ∞ −∞ −∞ ∞ ∞ −∞ −∞ = + = +− + +− + +− + +− ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ () () ( ) () () ( ) () () ( ) () () ( ) () () j ee ee e e ss e n sn n e ns n n nn Ss R e d F sF sS s F sF sS s F sF sS s F sF sS s ωτ τ τ ∞ − −∞ = = − + − + − + − ∫ where () () () Fs Fs F s e sd = − . We know the configuration so we can work out the form of each of the transforms F . Computation of the transfer functions ( ) F s e and ( ) F s n For many system configurations especially with minor feedback loops, manipulating the diagram into a standard input-output form is very time￾conserving and subject to errors. Usually just tracing signals around the loops is easiest. Example: Find ( ) F s n

16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde Don't try to manipulate this by block diagram algebra into the form G1(s) G25 Rather: y KviI, y+n s(tS+1) Now the rest is just algebra KFl-Kpy--y Multiply by s and collect terms (Gs+s2+KFKps+K, K,)y=Ksn 2=n n ts'+s+Kks+kk Now integrate to get e, using s= jo Two methods Cauchys Theorem 2 :2rj>Res(poles of See(s)in LHP 2Res(poles of Se(s)in LHP) Tabulated Integral (applicable only to rational functions See=Sss+(Ss+(Sns+(s Page 3 of 7

16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 3 of 7 Don’t try to manipulate this by block diagram algebra into the form: Rather: ( 1) F I P K K y Ky y n ss s τ ⎡ ⎤ ⎛ ⎞ = − +− + ⎢ ⎥ ⎜ ⎟ + ⎣ ⎦ ⎝ ⎠ Now the rest is just algebra, 2 I F P K s y sy K K y y n s τ ⎡ ⎤ += − − + ⎢ ⎥ ⎣ ⎦ Multiply by s and collect terms. ( ) 3 2 3 2 FP FI F F FP FI s s K K s K K y K sn y Ks n s s KKs KK τ τ ++ + = = ++ + Now integrate to get 2 e , using s j = ω : 2 1 ( ) 2 j ee j e S s ds π j ∞ − ∞ = ∫ Two methods: Cauchy’s Theorem: ( ) ( ) 2 1 2 Res poles of ( ) in LHP 2 Res poles of ( ) in LHP ee ee e j Ss j S s π π = = ∑ ∑ Tabulated Integral (applicable only to rational functions): ( ) ( ) () 1 ()( ) 2 () ( ) ee ss sn ns nn j n j SS S S S csc s I ds π j d sd s ∞ − ∞ =+ + + − = − ∫

16.322 Stochastic Estimation and Control, Fall 2004 Prof vander Velde Just work out the coefficients c(s)and d (s)and use the table to find the integral n OS+()Sn must be integrated together. For the rest of the semester, we will only deal with signals and noise which are uncorrelated, so these terms will be Having S (o)integrated to get an expression for e(p, P2,,Pn), where p,are the design parameters, we must determine the optimum set of values of the p Differentiate with respect to other variables than the parameters ae2 If you have one parameter, just plot and find the minimum. For multiple parameters, the optimization becomes quite complex If the signal, noise and disturbance can be considered uncorrelated, then and we can express these components of e in any direct or convenient way. Example: a servo d(t) st K 1 o(t) Signal s(t)is a member of a stationary ensemble: S(O)=4 0-+a Disturbance d(n) is a member of the ensemble of constant functions d()=D Noise n(t) is a member of a stationary white noise ensemble: Sm(O)=M Page 4 of 7

16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 4 of 7 Just work out the coefficients c s( ) and d s( ) and use the table to find the integral n I . () () sn ns S S + must be integrated together. For the rest of the semester, we will only deal with signals and noise which are uncorrelated, so these terms will be zero. Having ( ) ee S ω integrated to get an expression for ( ) 2 1 2 , ,..., n e pp p , where i p are the design parameters, we must determine the optimum set of values of the i p . Differentiate with respect to other variables than the parameters. 2 1 2 2 0 0 e p e p ∂ = ∂ ∂ = ∂ # If you have one parameter, just plot and find the minimum. For multiple parameters, the optimization becomes quite complex. If the signal, noise and disturbance can be considered uncorrelated, then 2 222 s n d eeee =++ and we can express these components of 2 e in any direct or convenient way. … Example: A servo Signal s t( ) is a member of a stationary ensemble: 2 2 ( ) ss A S a ω ω = + Disturbance d t( ) is a member of the ensemble of constant functions: 2 dt D ( ) = Noise n t( ) is a member of a stationary white noise ensemble: ( ) nn S N ω =

16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde Desired output is the signal Find k which minimizes the steady-state mean squared error The inputs s(o), d(t), n(r)are all independent. s and n have e zero mean Therefore the three inputs are uncorrelated We'll need the error(signal minus desired) transfer function and the noise transfer function F(s)= K (s)=1 F(s)=F(s)-Fesired(s) K S K stK For the F F(SFGsS(s)ds S s)(a-s) The integrand F()S(s) (K+s)(K-s)(a+s(a-s VAC (K+s)(a+s)(K-S(a-s) (a+K)s+ s+(a+K-s)+ak d(sd(s) Page 5 of 7

16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 5 of 7 Desired output is the signal. Find K which minimizes the steady-state mean squared error. The inputs st dt nt ( ), ( ), ( ) are all independent. s and n have zero mean. Therefore, the three inputs are uncorrelated. 2 222 s d n eeee =++ We’ll need the error (signal minus desired) transfer function and the noise transfer function. For the signal: ( ) 1 s K s K F s K s K s = = + + () 1 F s desired = () () () 1 e s desired K s Fs Fs F s sK sK − = − = −= + + For the noise: ( ) n K F s s K − = + 2 1 () ( ) () 2 j s e e ss j e F s F s S s ds π j ∞ − ∞ = − ∫ ( ) ( ) 2 2 ( ) ss A AA S s a s as as = = − +− The integrand: ( ) ( ) ( ) ( ) ( )( ) ( )( ) 2 2 () ( ) () ( ) ( ) ( ) ( )( ) () ( ) () ( ) e e ss s s AA F sF sS s Ks Ks as as As A s K sas K sas As A s s a K s aK s a K s aK cs c s ds d s − − = + −+− ⎡ ⎤ − = ⎢ ⎥ ++ −− ⎣ ⎦ ⎡ − ⎤ = ⎢ ⎥ + + + + + −+ ⎣ ⎦ − = −

16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde dodd. 2d.d A 2(a+K) For the disturbance: Steady state response 0 to constant input d O 0(s)= K O=lim(sO(s) lim l 3-0(s+Ks) K O F(s)=F(s)-Fdesired(s) K s+K +K K F(S= K+sK-s Using Cauchy,s Theorem: e?=Res( pole at s=-k) KN 1 Page 6 of7

16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 6 of 7 22 2 2 10 02 1 2 012 12 2 2 2( ) s cd cd c e I ddd dd A a K + == → = + For the disturbance: Steady state response O to constant input d ( ) 0 0 2 2 2 2 1 1 1 1 ( ) lim ( ) 1 lim ss s s ss O s d sK K s d O s s Ks O sO s d d s s Ks K d D O K K → → = = + + = + = ⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ + = = () () () 1 ( ) ( ) e s desired n ee n Fs Fs F s K s sK sK K F s s K K K Ss N K sK s = − − = −= + + − = + − − = + − Using Cauchy’s Theorem: ( ) 2 2 Res pole at 1 ( )2 n e sK K N KN K K = =− = = +

16.322 Stochastic Estimation and Control, Fall 2004 Prof vander velde ++-NK K+a K 2 2D dK =1Ax3-2D(K+a)2+1M2(k+ay=0 order polynomial in K=0 You should check the stability of your solution at your solution point. For stability in this example, k>0 Semi-Free Configuration design (Free configuration design is a special case of this. Fixed transfer you have to deal with. You must drive the plant, F(S). We will design a compensator C(s), which can be in a closed loop configuration, perhaps with something in the feed back path, B(s) Page 7 of7

16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 7 of 7 ( ) 2 2 2 2 3 3 2 32 1 1 2 2 1 2 1 2 0 2 1 1 2( ) ( ) 0 2 2 5 order polynomial in 0 th A D e NK Ka K A de D N dK K K a AK D K a NK K a K = ++ + − = −+ = + = − ++ += = = You should check the stability of your solution at your solution point. For stability in this example, K > 0 . Semi-Free Configuration Design (Free configuration design is a special case of this.) Fixed transfer you have to deal with. You must drive the plant, F s( ) . We will design a compensator C s( ), which can be in a closed loop configuration, perhaps with something in the feedback path, B( )s