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 () s wt and () n wt. 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 ()et is given by: 111 111 1 11 111 1 11 () ()( ) ()( ) ()( ) ()( ) ()( ) sdn en et wstd wstd wntd wst d wnt d τ ττ τ ττ τ ττ τττ τττ ∞∞∞ ?∞ ?∞ ?∞ ∞∞ ?∞ ?∞ =???+? =?+? ∫∫∫ ∫∫ where () () () esd wwwτ ττ=?. Using the replacement 1 () e w τ reduces the number of terms in () ee R τ from nine to four. 16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 2 of 7 121 2 12 121 2 12 121 2 12 121 2 12 () ()( ) () () ( ) () () ( ) () () ( ) () () ( ) ee ees ensn nens nnn Retet ddwwR ddww R ddwwR ddww R ττ τ ττ τ τττ τ ττ τ τττ τ ττ τ τττ τ ττ τ τττ ∞∞ ?∞ ?∞ ∞∞ ?∞ ?∞ ∞∞ ?∞ ?∞ ∞∞ ?∞ ?∞ =+ ? ++ ? ++ ∫∫ ∫∫ ∫∫ ∫∫ () () ()()() ()() () ( ) () () ( ) () () j ee ee ees ensn nens nnn Ss R e d FsFsSs FsFsSs FsFsSs FsFsSs ωτ τ τ ∞ ? ?∞ = =? +? +? +? ∫ where () () () esd Fs Fs Fs=?. We know the configuration so we can work out the form of each of the transforms F . Computation of the transfer functions () e Fs and () n Fs 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 () n Fs 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) FI P KK yKyyn ss sτ ?? ?? =?+?+ ?? ?? + ?? ?? Now the rest is just algebra, 2 I FP K sy sy K Ky y n s τ ? ? += ? ? + ? ? ? ? Multiply by s and collect terms. () 32 32 FP FI F F FP FI ssKKsKKyKsn yK nssKKsKK τ τ ++ + = = ++ + Now integrate to get 2 e , using sjω= : 2 1 () 2 j ee j eSsd jπ ∞ ?∞ = ∫ Two methods: Cauchy’s Theorem: () () 2 1 2 Res poles of ( ) in LHP 2 Res poles of ( ) in LHP ee ee ej Ss j Ss 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 Id jdsdsπ ∞ ?∞ =+ + + ? = ? ∫ 16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 4 of 7 Just work out the coefficients ()cs and ()ds and use the table to find the integral n I . () () snns SS+ 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 12 , ,..., n epp 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 2222 s nd eeee=++ and we can express these components of 2 e in any direct or convenient way. … Example: A servo Signal ()st is a member of a stationary ensemble: 22 () ss A S a ω ω = + Disturbance ()dt is a member of the ensemble of constant functions: 2 ()dt D= Noise ()nt is a member of a stationary white noise ensemble: () nn SNω = 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. 2222 s dn eeee=++ We’ll need the error (signal minus desired) transfer function and the noise transfer function. For the signal: () 1 s K K s Fs K sK s == + + () 1 desired Fs= () () () 1 e s desired Ks Fs Fs F s sK sK ? =? =?= + + For the noise: () n K Fs sK ? = + 2 1 () ( ) () 2 j sees j eFsFSds jπ ∞ ?∞ =? ∫ ()() 22 () ss AAA Ss as asas == ?+? The integrand: ()()()() ()()()() 22 () ( ) () () () () ()() () ( ) () ( ) ee s ss AA FsF sS s KsKsasas As A s Ksas Ksas As A s saKsaKsaKsaK cs c s dsd s ? ?= +?+? ?? ? = ?? ++ ?? ?? ? ? ? = ? ? ++ + ++ ?+ ? ? ? = ? 16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 6 of 7 22 2 2 10 02 1 2 012 12 22 2( ) s cd cd c eI ddd dd A aK + == → = + For the disturbance: Steady state response O to constant input d () 0 0 2 2 22 1 1 1 1 () lim ( ) 1 lim ss s s ss O s K dsK s d Os sKs OsOs dd s sKs K dD O KK → → == + + = + = ?? == ?? + ?? == () () () 1 () () e s desired n ee n Fs Fs F s Ks sK sK K Fs sK KK Ss N KsKs =? ? =?= ++ ? = + ?? = +? Using Cauchy’s Theorem: () 2 2 Res pole at 1 ()2 n esK KN KN KK ==? == + 16.322 Stochastic Estimation and Control, Fall 2004 Prof. Vander Velde Page 7 of 7 () 2 2 2 2 3 3232 1 1 2 2 1 21 2 0 2 11 2( ) ( ) 0 22 5 order polynomial in 0 th A D eNK KaK A de D N dK K Ka 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, 0K > . 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, ()Fs. We will design a compensator ()Cs, which can be in a closed loop configuration, perhaps with something in the feedback path, ()B s .