16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
Last time: Ground rules for filtering and control system design
System parameters are contained in s
( ) and w t .
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:
= ∫
∞
−∞ w s
τ s t − ) d
1
− ∫
∞
−∞ w d
τ s t − ) d
1
+ ∫
∞
−∞ w n
τ n t −
1
)
1 w n
τ n t −
1
) = ∫
∞
−∞ w e
τ s t where w e
τ = w s
− ) d
1
+
− w d
Using the replacement w e four.
∫
∞
−∞
.
1
τ reduces the number of terms in R ee
( ) from nine to
Page 1 of 7

16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
R ee
τ = ( ) ( + τ )
=
∞
−∞
∫ ∫
∞
−∞
τ e
τ w τ R ss
( τ τ τ
2
)
+
∞
−∞
∫ ∫
∞
−∞
τ e
τ w τ R sn
( τ τ τ
2
)
+
∞
−∞
∫ ∫
∞
−∞
τ n
τ w τ R ns
( τ τ τ
2
) ee
=
+
∞
−∞
∫ ∫
∞
−∞
τ n
τ w τ R nn
( τ τ τ
2
∞
−∞
R ee
τ e − j ωτ d τ
= F e
( ) ( ) ss
( )
+ F e
( ) ( ) sn
( )
+ F n
( ) ( ) ns
( )
+ F n
( ) ( ) nn
( ) where F s = F s − F s .
)
We know the configuration so we can work out the form of each of the transforms F .
( )
F s
For many system configurations especially with minor feedback loops, manipulating the diagram into a standard input-output form is very timeconserving and subject to errors.
Usually just tracing signals around the loops is easiest.
F s
Page 2 of 7

16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
Don’t try to manipulate this by block diagram algebra into the form:
Rather: y =
( τ
K
F
+ 1)
⎡
⎣
−
⎝
K
I s
Now the rest is just algebra,
τ + =
F
⎡
⎣
− K y −
K s
I
⎤
⎦
(
Multiply by s and collect terms.
τ s 3 + s 2 + +
)
=
⎞
⎠
⎤
⎦ y n
=
τ s 3 + s 2 + +
I
Now integrate to get e 2 , using s = j ω : e 2 =
1
2 π j j ∞
∫ ee
Two methods:
Cauchy’s Theorem: e 2 =
2
1
π j
2 π j
∑ (
S s
)
(
S s = ∑ )
Tabulated Integral (applicable only to rational functions):
S ee
= S ss
+ ( )
S sn
+ ( )
S ns
+ ( )
S nn
I n
=
1
2 π j j ∞
∫
−
− ds
Page 3 of 7

16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde
Just work out the coefficients ( ) and d s and use the table to find the integral I n
.
( )
S sn
+ ( )
S ns
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 S ee
ω integrated to get an expression for
( p n
)
, where p i
are the design parameters, we must determine the optimum set of values of the p i
.
Differentiate with respect to other variables than the parameters.
∂ e 2
∂ p
1
= 0
∂ e 2
∂ p
2
= 0
#
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 e 2 = e s
2 + e 2 n
+ e 2 d and we can express these components of e 2 in any direct or convenient way. …
Signal ( ) is a member of a stationary ensemble:
Disturbance ( )
S ss
ω =
ω 2
A
+ a 2 d t is a member of the ensemble of constant functions:
2 = D
Noise ( ) is a member of a stationary white noise ensemble: S nn
ω = N
Page 4 of 7

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 ( ), ( ), ( ) are all independent. s and n have zero mean.
Therefore, the three inputs are uncorrelated. e 2 = e s
2 + e 2 d
+ e 2 n
We’ll need the error (signal minus desired) transfer function and the noise transfer function.
For the signal:
K s
F desired
=
1 + s
K
( ) 1 s s =
=
K
+
( ) = s
( ) − F desired
=
K − s
For the noise: n
=
− K e s
2 =
1
2 π j j ∞
∫ ( ) ( ) ss
( ) ss
=
A a 2 − s 2
=
A A
( ) ( )
The integrand:
( ) ( ) ss
( ) =
− s s
K s − +
A
−
A
( ) ( ) ( ) ( )
=
=
=
( +
As
)( +
⎡
) ( − )(
−
− )
⎤
⎥
⎦ s 2 + ( +
As
) +
⎡
⎢
⎣
2 + ( +
−
)( ) aK
−
−
⎤
⎥
⎦
Page 5 of 7

16.322 Stochastic Estimation and Control, Fall 2004
Prof. Vander Velde e s
2 = I
2
= c d
0
+ c d
2
2
→ c 2
1
2
=
A
2( )
For the disturbance: Steady state response O to constant input d
1
O d
=
1 + s
K s
=
1
=
1 d
O ss
= lim s → 0
( sO s
)
= lim s → 0 s
1
+ d ⎞
= d s K s K
O 2 ss
= d
K
2
2
=
D
K 2
( ) = s
( ) − F desired
=
K − s
=
− K n n
=
− K − K
N ee
Using Cauchy’s Theorem: e 2 n
= ( s = − K
)
=
(
=
1
) 2
KN
Page 6 of 7

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