16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
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.
H ( s) =
C ( s)
1 + C ( s) F ( s ) B( s )
so that having the optimum H , we determine C from
H ( s)
C ( s) =
1 − H ( s) F ( 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
( HF )opt
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 ( s ) → wH (t )
F ( s ) → wF (t )
D ( s ) → wD (t )
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16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
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.
∞
c(t ) =
∫w
H
(τ 1 )i (t − τ 1 )dτ 1
−∞
∞
o( t ) =
∫w
F
(τ 2 )c(t − τ 2 )dτ 2
−∞
∞
=
∞
∫ dτ w
2
F
−∞
∞
d (t ) =
∫w
D
(τ 2 ) ∫ dτ 1wH (τ 1 )i (t − τ 1 − τ 2 )
−∞
(τ 3 ) s(t − τ 3 )dτ 3
−∞
e( t ) = o( t ) − d ( t )
e(t )2 = o(t )2 − 2o(t )d (t ) + d (t )2
∞
∞
⎡∞
⎤⎡ ∞
⎤
o(t ) = ⎢ ∫ dτ 2 wF (τ 2 ) ∫ dτ 1wH (τ 1 )i (t − τ 1 − τ 2 ) ⎥⎢ ∫ dτ 4 wF (τ 4 ) ∫ dτ 3wH (τ 3 )i (t − τ 3 − τ 4 ) ⎥
−∞
−∞
⎣ −∞
⎦⎣ −∞
⎦
2
∞
=
∫ dτ w
1
H
−∞
∞
=
∫
−∞
∞
∞
∞
−∞
−∞
−∞
∞
∞
∞
−∞
−∞
(τ 1 ) ∫ dτ 2 wF (τ 2 ) ∫ dτ 3wH (τ 3 ) ∫ dτ 4 wF (τ 4 )i (t − τ 1 − τ 2 )i (t − τ 3 − τ 4 )
dτ 1wH (τ 1 ) ∫ dτ 2 wF (τ 2 ) ∫ dτ 3wH (τ 3 ) ∫ dτ 4 wF (τ 4 )Rii (τ 1 + τ 2 − τ 3 − τ 4 )
−∞
⎡
⎤⎡ ∞
⎤
o(t )d (t ) = ⎢ ∫ dτ 2 wF (τ 2 ) ∫ dτ 1wH (τ 1 )i (t − τ 1 − τ 2 ) ⎥⎢ ∫ dτ 3wD (τ 3 ) s(t − τ 3 ) ⎥
−∞
⎣ −∞
⎦⎣ −∞
⎦
∞
∞
=
∫
−∞
∞
∞
=
∫ dτ w
1
−∞
∞
∞
−∞
−∞
∞
∞
−∞
−∞
dτ 1wH (τ 1 ) ∫ dτ 2 wF (τ 2 ) ∫ dτ 3wD (τ 3 )i (t − τ 1 − τ 2 ) s(t − τ 3 )
H
(τ 1 ) ∫ dτ 2 wF (τ 2 ) ∫ dτ 3wD (τ 3 )Ris (τ 1 + τ 2 − τ 3 )
We shall not require d (t ) 2 in integral form.
Page 2 of 5
16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
The problem now is to choose wH (t ) so as to minimize this e(t ) 2 , for which we
use variational calculus.
Let:
wH (t ) = w0 (t ) + δ w(t )
where w0 (t ) is the optimum weighting function (to be determined) and δ w(t ) is
an arbitrary variation – arbitrary except that it must be physically realizable.
Calculate the optimum e 2 and its first and second variations.
e 2 = e02 + δ e 2 + δ 2 e 2
e 2 = o(t ) 2 + 2o(t )d (t ) + d (t ) 2
The optimum e 2 ( e 2 for δ w(t ) = 0 ):
∞
e(t )02 =
∫
∞
−∞
∞
∞
dτ 1w0 (τ 1 ) ∫ dτ 2 wF (τ 2 ) ∫ dτ 3w0 (τ 3 ) ∫ dτ 4 wF (τ 4 )Rii (τ 1 + τ 2 − τ 3 − τ 4 )
−∞
−∞
∞
∞
−∞
−∞
−∞
∞
−2 ∫ dτ 1w0 (τ 1 ) ∫ dτ 2 wF (τ 2 ) ∫ dτ 3wD (τ 3 )Ris (τ 1 + τ 2 − τ 3 ) + d (t ) 2
−∞
2
The first variation in e(t ) is
δ e( t ) 2 =
∞
∞
∫
−∞
∞
∞
−∞
−∞
−∞
∞
∞
dτ 1δ w(τ 1 ) ∫ dτ 2 wF (τ 2 ) ∫ dτ 3w0 (τ 3 ) ∫ dτ 4 wF (τ 4 )Rii (τ 1 + τ 2 − τ 3 − τ 4 )
∞
∞
+ ∫ dτ 1w0 (τ 1 ) ∫ dτ 2 wF (τ 2 ) ∫ dτ 3δ w(τ 3 ) ∫ dτ 4 wF (τ 4 )Rii (τ 1 + τ 2 − τ 3 − τ 4 )
−∞
−∞
−∞
−∞
∞
∞
∞
−∞
−∞
−∞
−2 ∫ dτ 1δ w(τ 1 ) ∫ dτ 2 wF (τ 2 ) ∫ dτ 3wD (τ 3 )Ris (τ 1 + τ 2 − τ 3 )
In the second term, let:
τ 1 = τ 3′
τ 2 = τ 4′
τ 3 = τ 1′
τ 4 = τ 2′
and interchange the order of integration.
∞
2nd
term =
∞
∫ dτ ′δ w(τ ′) ∫ dτ ′ w
1
−∞
1
2
−∞
F
∞
∞
−∞
−∞
(τ 2′ ) ∫ dτ 3′w0′ (τ 3′ ) ∫ dτ 4′ wF (τ 4′ )Rii (τ 3′ + τ 4′ − τ 1′ − τ 2′ )
Page 3 of 5
16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
but since Rii (τ 3′ + τ 4′ − τ 1′ − τ 2′ ) = Rii (τ 1′ + τ 2′ − τ 3′ − τ 4′ ) 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 τ 1 gives
∞
∞
⎧∞
δ e(t ) = 2 ∫ dτ 1δ w(τ 1 ) ⎨ ∫ dτ 2 wF (τ 2 ) ∫ dτ 3w0 (τ 3 ) ∫ dτ 4 wF (τ 4 )Rii (τ 1 + τ 2 − τ 3 − τ 4 )
−∞
−∞
−∞
⎩ −∞
∞
2
∞
⎫
τ
τ
d
w
(
)
∫−∞ 2 F 2 −∞∫ dτ 3wD (τ 3 )Ris (τ 1 + τ 2 − τ 3 ) ⎬⎭
∞
−
The second variation of e(t ) 2 is
δ 2 e( t ) 2 =
∞
∫
−∞
∞
∞
∞
−∞
−∞
−∞
dτ 1δ w(τ 1 ) ∫ dτ 2 wF (τ 2 ) ∫ dτ 3δ w(τ 3 ) ∫ dτ 4 wF (τ 4 )Rii (τ 1 + τ 2 − τ 3 − τ 4 )
By comparison with the expression for o(t ) 2 , this is seen to be the mean squared
output of the system
( output )
2
= δ 2 e(t ) 2 > 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, δ w(τ 1 ) = 0 for τ 1 < 0 by the requirement
that the variation be physically realizable. But δ w(τ 1 ) is arbitrary for τ 1 ≥ 0 , so
we can be assured of the vanishing of δ e(t ) 2 only if the
{ } term vanishes almost
everywhere for τ 1 ≥ 0 . The condition which defines the minimum in e(t )2 is then
∞
∫ dτ w
2
−∞
F
∞
∞
(τ 2 ) ∫ dτ 3w0 (τ 3 ) ∫ dτ 4 wF (τ 4 )Rii (τ 1 + τ 2 − τ 3 − τ 4 )
−∞
−∞
∞
∞
−∞
−∞
− ∫ dτ 2 wF (τ 2 ) ∫ dτ 3wD (τ 3 )Ris (τ 1 + τ 2 − τ 3 ) = 0
for all τ 1 , non-real-time.
Using this condition in the expression for e(t )02 and remembering that w0 (t ) = 0
for t < 0 gives the result
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16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
e(t )02 = d (t )2 − o(t )02
which is convenient for the calculation of e(t )02 .
Also since o(t )02 = d (t )2 − e(t )02 , 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:
Rii (τ ) = Rss (τ ) + Rsn (τ ) + Rns (τ ) + Rnn (τ )
Ris (τ ) = Rss (τ ) + Rns (τ )
since i = s + n .
The free configuration problem is a specialization of the semi-free configuration.
In this expression we would take F ( s ) = 1 , or wF (t ) = δ (t ) . In that case we have
∞
∞
∞
∫ dτ d (τ ) ∫ dτ w (τ ) ∫ dτ δ (τ
2
−∞
2
3
−∞
0
3
4
4
)Rii (τ 1 + τ 2 − τ 3 − τ 4 )
−∞
∞
∞
−∞
−∞
− ∫ dτ 2δ (τ 2 ) ∫ dτ 3wD (τ 3 )Ris (τ 1 + τ 2 − τ 3 ) =
∞
∫
−∞
w0 (τ 3 )Rii (τ 1 − τ 3 )dτ 3 −
∞
∫w
D
(τ 3 )Ris (τ 1 − τ 3 )dτ 3 = 0 for τ 1 ≥ 0
−∞
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