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 .