Deterministic STR
1,The idea
2,Pole placement design
3,Indirect STR
4,Continuous time STR
5,Direct STR
6,Disturbances with known properties
7,Conclusions
The idea
Automate estimation and control
Process parameters
Controller
design
Estimation
Controller
Process
Controller
parameters
Reference
Input Output
Specification
Self-tuning regulator
The separation principle
#0F Estimate parameters
#0F Design controller as if the estimates were
correct,#28The Certainty Equivalence
Principle#29
#0F Direct and indirect controllers
Pole Placement Design
#0F Process Model
#0F Model Following
#0F Causality Conditions
#0F Examples
#0F Interpretation of polynomial A
o
#0F Relations to Model Following
#0F Summary
Process Model
A#28q#29y#28t#29=B#28q#29#28u#28t#29+v#28t#29#29
Alternatively
A
#03
#28q
,1
#29y#28t#29=B
#03
#28q
,1
#29#28u#28t,d
0
#29+v#28t,d
0
#29#29
Relative degree d
0
= degA#28q#29,degB
Same notation for continuous time systems
Ay#28t#29=B#28u#28t#29+v#28t#29#29
c#0D K,J,#C5str#F6m and B,Wittenmark 1
Closed Loop System
Process
Ay#28t#29=B#28u#28t#29+v#28t#29#29
Controller
Ru#28t#29=Tu
c
#28t#29,Sy#28t#29
Controller
u
Process
y
B
A
u
c
Ru = Tu
c
- Sy
v
S
Closed loop system
y#28t#29=
BT
AR+BS
u
c
#28t#29+
BR
AR+ BS
v#28t#29
u#28t#29=
AT
AR+BS
u
c
#28t#29,
BS
AR+ BS
v#28t#29
Closed loop characteristic polynomial
AR+BS = A
c
Model Following
Closed loop response
y#28t#29=
BT
AR+ BS
u
c
#28t#29
Desired response
A
m
y
m
#28t#29=B
m
u
c
#28t#29
Perfect model following
BT
AR+BS
=
BT
A
c
=
B
m
A
m
Cancellations in LHS,B = B
+
B
,
B
m
= B
,
B
0
m
A
c
= A
o
A
m
B
+
R = R
0
B
+
#281#29
Hence
AR
0
+B
,
S = A
o
A
m
= A
0
c
T = A
o
B
0
m
Causality Conditions
Controller
Ru = Tu
c
,Sy
Hence
degS #14 degR
degT #14 degR
Diophantine Equation
AR+BS = A
c
has many solutions
R = R
0
+QB
S = S
0
,QA
We have degA#3EdegB
degA
c
#14 2degA,1
Example 3.1 Cancellation of Zero
G#28s#29=
1
s#28s+1#29
Sampling with h =0:5
H#28q#29=
b
0
q+b
1
q
2
+a
1
q+a
2
=
0:1065q+0:0902
q
2
,1:6065q+0:6065
B
m
#28q#29
A
m
#28q#29
=
b
m0
q
q
2
+a
m1
q +a
m2
=
0:1761q
q
2
,1:3205q+0:4966
B
+
#28q#29=q+b
1
=b
0
B
,
#28q#29=b
0
B
0
m
#28q#29=b
m0
q=b
0
#282#29
Choose A
o
=1
#28q
2
+a
1
q+a
2
#29#011+b
0
#28s
0
q+s
1
#29=q
2
+a
m1
q+a
m2
a
1
+b
0
s
0
=a
m1
a
2
+b
0
s
1
=a
m2
#283#29
s
0
=
a
m1
,a
1
b
0
s
1
= a
m2
,a
2
b
0
#284#29
c#0D K,J,#C5str#F6m and B,Wittenmark 2
Example 3.2 - No Cancellation
H#28q#29=
b
0
q+b
1
q
2
+a
1
q+a
2
=
0:1065q+0:0902
q
2
,1:6065q+0:6065
We have B
+
=1and B
,
= B = b
0
q + b
1
H
m
#28q#29=#0C
b
0
q+b
1
q
2
+a
m1
q+a
m2
=
b
m0
q+b
m1
q
2
+a
m1
q+a
m2
Diophantine equation
#28q
2
+a
1
q + a
2
#29#28q +r
1
#29+#28b
0
q+b
1
#29#28s
0
q +s
1
#29
=#28q
2
+a
m1
q+a
m2
#29#28q +a
o
#29
Put q =,b
1
=b
0
solve for r
1
r
1
=
X
0
b
2
0
+X
1
b
0
b
1
+ X
2
b
2
1
b
2
1
,a
1
b
0
b
1
+a
2
b
2
0
where
X
0
= a
o
a
m2
X
1
= a
2
,a
m2
,a
o
a
m1
X
2
= a
o
+a
m1
,a
1
#285#29
Then solve for s
0
and s
1
.
Example 3.3 - Continuous time
G#28s#29=
b
s#28s+a#29
Choose
A
o
#28s#29=s+a
o
and
B
m
#28s#29
A
m
#28s#29
=
!
2
s
2
+2#10!s+!
2
Diophantine equation
s#28s + a#29#28s +r
1
#29+b#28s
0
s+s
1
#29
=#28s
2
+2#10!s+!
2
#29#28s +a
o
#29
Example 3.3 - Cont.
Identi#0Ccation of coe#0Ecients of powers of s
a+ r
1
=2#10!+a
o
ar
1
+bs
0
= !
2
+2#10!a
o
bs
1
= !
2
a
o
#286#29
If b 6=0
r
1
=2#10!+a
o
,a
s
0
=
a
o
2#10!+!
2
,ar
1
b
s
1
=
!
2
a
o
b
#287#29
Interpretation of PolynomialA
0
What happens if we solve the problem with
state space methods?
What is the interpretation of pole-zero cancel-
lations?
c#0D K,J,#C5str#F6m and B,Wittenmark 3
Relations to Model Following
The control lawis
Ru = Tu
c
,Sy
Notice that
T
R
=
A
o
B
0
m
R
=
#28AR
0
+B
,
S#29B
0
m
A
m
R
=
AB
m
BA
m
+
SB
m
RA
m
Hence
u =
T
R
u
c
,
S
R
y =
AB
m
BA
m
u
c
+
SB
m
RA
m
u
c
,
S
R
y
=
AB
m
BA
m
u
c
,
S
R
#28y,y
m
#29
#288#29
u y
S S
u
c
A
B
S
R
B
A
- 1
y
m
B
m
A
m
Summary
#0F Pole placement gives linear equations
#0F When do solutions exist?
#0F When do they not exist?
#0F What is the key di#0Eculty?
Indirect STR
1,Estimation
2,An Algorithm
3,Examples
Estimation
Process model #28No disturbances!#29
y#28t#29=,a
1
y#28t,1#29,a
2
y#28t,2#29,:::,a
n
y#28t,n#29
+b
0
u#28t,d
0
#29+:::+b
m
u#28t,d
0
,m#29
#289#29
Introducing
#12
T
=#28a
1
a
2
::,a
n
b
0
::,b
m
#29
'
T
#28t,1#29 = #28,y#28t,1#29#01#01#01,y#28t,n#29
u#28t,d
0
#29:::u#28t,d
0
,m#29#29
#2810#29
The model becomes a regression
y#28t#29='
T
#28t,1#29#12
Estimates given by
^
#12#28t#29=
^
#12#28t,1#29 +K#28t#29"#28t#29
"#28t#29=y#28t#29,'
T
#28t,1#29
^
#12#28t,1#29
K#28t#29=
P#28t,1#29'#28t,1#29
#15 +'
T
#28t,1#29P#28t,1#29'#28t,1#29
P#28t#29=
1
#15
,
I,K#28t#29'
T
#28t,1#29
#01
P#28t,1#29
#2811#29
c#0D K,J,#C5str#F6m and B,Wittenmark 4
An Indirect STR
1,Data A
o
and A
m
and n
2,Estimate parameters of model by RLS
3,Obtain polynomials R and S by solving
Diophantine Equation
4,Compute Polynomial T = A
o
B
0
m
5,Compute control signal from
Ru = Tu
c
,Sy
6,Go to 2
Example 3.4 - Cancellation of Zero
Process model
y#28t#29+a
1
y#28t,1#29 + a
2
y#28t,2#29
= b
0
u#28t,1#29 + b
1
u#28t,2#29
Control law
u#28t#29+r
1
u#28t,1#29 = t
0
u
c
#28t#29,s
0
y#28t#29,s
1
y#28t,1#29
Simulation
0 2040608010
1
0
1
0 2040608010
4
2
0
2
Time
Time
u
c
y
u
0 5 10 15 20
2
1
0
0 5 10 15 20
0.0
0.1
0.2
Time
Time
^a
2
^a
1
^
b
1
^
b
0
Example 3.5 - No Cancellation
Process model
y#28t#29+a
1
y#28t,1#29 + a
2
y#28t,2#29
= b
0
u#28t,1#29 + b
1
u#28t,2#29
Control law
u#28t#29+r
1
u#28t,1#29 = t
0
u
c
#28t#29,s
0
y#28t#29,s
1
y#28t,1#29
c#0D K,J,#C5str#F6m and B,Wittenmark 5
Simulation
0 2040608010
1
0
1
0 2040608010
4
2
0
2
Time
Time
u
c
y
u
0 100 200 300 400 500
2
1
0
0 100 200 300 400 500
0.0
0.1
0.2
Time
Time
^a
2
^a
1
^
b
1
^
b
0
Simulation
Process
G#28s#29=
b
s#28s+a#29
Desired response
G
m
#28s#29=
!
2
s
2
+2#10!s+!
2
Observer polynomial A
o
#28s#29=s+a
o
with
a
o
=2.
Controller structure
u#28t#29=,
s
0
p+s
1
p+r
1
y#28t#29+
t
0
#28p+a
o
#29
p+r
1
u
c
#28t#29
where
r
1
=2#10!+a
o
,a #2812#29
s
0
=
a
o
2#10!+!
2
,ar
1
b
#2813#29
s
1
=
!
2
a
o
b
#2814#29
t
0
=
!
2
b
#2815#29
c#0D K,J,#C5str#F6m and B,Wittenmark 6
1,The idea
2,Pole placement design
3,Indirect STR
4,Continuous time STR
5,Direct STR
6,Disturbances with known properties
7,Conclusions
The idea
Automate estimation and control
Process parameters
Controller
design
Estimation
Controller
Process
Controller
parameters
Reference
Input Output
Specification
Self-tuning regulator
The separation principle
#0F Estimate parameters
#0F Design controller as if the estimates were
correct,#28The Certainty Equivalence
Principle#29
#0F Direct and indirect controllers
Pole Placement Design
#0F Process Model
#0F Model Following
#0F Causality Conditions
#0F Examples
#0F Interpretation of polynomial A
o
#0F Relations to Model Following
#0F Summary
Process Model
A#28q#29y#28t#29=B#28q#29#28u#28t#29+v#28t#29#29
Alternatively
A
#03
#28q
,1
#29y#28t#29=B
#03
#28q
,1
#29#28u#28t,d
0
#29+v#28t,d
0
#29#29
Relative degree d
0
= degA#28q#29,degB
Same notation for continuous time systems
Ay#28t#29=B#28u#28t#29+v#28t#29#29
c#0D K,J,#C5str#F6m and B,Wittenmark 1
Closed Loop System
Process
Ay#28t#29=B#28u#28t#29+v#28t#29#29
Controller
Ru#28t#29=Tu
c
#28t#29,Sy#28t#29
Controller
u
Process
y
B
A
u
c
Ru = Tu
c
- Sy
v
S
Closed loop system
y#28t#29=
BT
AR+BS
u
c
#28t#29+
BR
AR+ BS
v#28t#29
u#28t#29=
AT
AR+BS
u
c
#28t#29,
BS
AR+ BS
v#28t#29
Closed loop characteristic polynomial
AR+BS = A
c
Model Following
Closed loop response
y#28t#29=
BT
AR+ BS
u
c
#28t#29
Desired response
A
m
y
m
#28t#29=B
m
u
c
#28t#29
Perfect model following
BT
AR+BS
=
BT
A
c
=
B
m
A
m
Cancellations in LHS,B = B
+
B
,
B
m
= B
,
B
0
m
A
c
= A
o
A
m
B
+
R = R
0
B
+
#281#29
Hence
AR
0
+B
,
S = A
o
A
m
= A
0
c
T = A
o
B
0
m
Causality Conditions
Controller
Ru = Tu
c
,Sy
Hence
degS #14 degR
degT #14 degR
Diophantine Equation
AR+BS = A
c
has many solutions
R = R
0
+QB
S = S
0
,QA
We have degA#3EdegB
degA
c
#14 2degA,1
Example 3.1 Cancellation of Zero
G#28s#29=
1
s#28s+1#29
Sampling with h =0:5
H#28q#29=
b
0
q+b
1
q
2
+a
1
q+a
2
=
0:1065q+0:0902
q
2
,1:6065q+0:6065
B
m
#28q#29
A
m
#28q#29
=
b
m0
q
q
2
+a
m1
q +a
m2
=
0:1761q
q
2
,1:3205q+0:4966
B
+
#28q#29=q+b
1
=b
0
B
,
#28q#29=b
0
B
0
m
#28q#29=b
m0
q=b
0
#282#29
Choose A
o
=1
#28q
2
+a
1
q+a
2
#29#011+b
0
#28s
0
q+s
1
#29=q
2
+a
m1
q+a
m2
a
1
+b
0
s
0
=a
m1
a
2
+b
0
s
1
=a
m2
#283#29
s
0
=
a
m1
,a
1
b
0
s
1
= a
m2
,a
2
b
0
#284#29
c#0D K,J,#C5str#F6m and B,Wittenmark 2
Example 3.2 - No Cancellation
H#28q#29=
b
0
q+b
1
q
2
+a
1
q+a
2
=
0:1065q+0:0902
q
2
,1:6065q+0:6065
We have B
+
=1and B
,
= B = b
0
q + b
1
H
m
#28q#29=#0C
b
0
q+b
1
q
2
+a
m1
q+a
m2
=
b
m0
q+b
m1
q
2
+a
m1
q+a
m2
Diophantine equation
#28q
2
+a
1
q + a
2
#29#28q +r
1
#29+#28b
0
q+b
1
#29#28s
0
q +s
1
#29
=#28q
2
+a
m1
q+a
m2
#29#28q +a
o
#29
Put q =,b
1
=b
0
solve for r
1
r
1
=
X
0
b
2
0
+X
1
b
0
b
1
+ X
2
b
2
1
b
2
1
,a
1
b
0
b
1
+a
2
b
2
0
where
X
0
= a
o
a
m2
X
1
= a
2
,a
m2
,a
o
a
m1
X
2
= a
o
+a
m1
,a
1
#285#29
Then solve for s
0
and s
1
.
Example 3.3 - Continuous time
G#28s#29=
b
s#28s+a#29
Choose
A
o
#28s#29=s+a
o
and
B
m
#28s#29
A
m
#28s#29
=
!
2
s
2
+2#10!s+!
2
Diophantine equation
s#28s + a#29#28s +r
1
#29+b#28s
0
s+s
1
#29
=#28s
2
+2#10!s+!
2
#29#28s +a
o
#29
Example 3.3 - Cont.
Identi#0Ccation of coe#0Ecients of powers of s
a+ r
1
=2#10!+a
o
ar
1
+bs
0
= !
2
+2#10!a
o
bs
1
= !
2
a
o
#286#29
If b 6=0
r
1
=2#10!+a
o
,a
s
0
=
a
o
2#10!+!
2
,ar
1
b
s
1
=
!
2
a
o
b
#287#29
Interpretation of PolynomialA
0
What happens if we solve the problem with
state space methods?
What is the interpretation of pole-zero cancel-
lations?
c#0D K,J,#C5str#F6m and B,Wittenmark 3
Relations to Model Following
The control lawis
Ru = Tu
c
,Sy
Notice that
T
R
=
A
o
B
0
m
R
=
#28AR
0
+B
,
S#29B
0
m
A
m
R
=
AB
m
BA
m
+
SB
m
RA
m
Hence
u =
T
R
u
c
,
S
R
y =
AB
m
BA
m
u
c
+
SB
m
RA
m
u
c
,
S
R
y
=
AB
m
BA
m
u
c
,
S
R
#28y,y
m
#29
#288#29
u y
S S
u
c
A
B
S
R
B
A
- 1
y
m
B
m
A
m
Summary
#0F Pole placement gives linear equations
#0F When do solutions exist?
#0F When do they not exist?
#0F What is the key di#0Eculty?
Indirect STR
1,Estimation
2,An Algorithm
3,Examples
Estimation
Process model #28No disturbances!#29
y#28t#29=,a
1
y#28t,1#29,a
2
y#28t,2#29,:::,a
n
y#28t,n#29
+b
0
u#28t,d
0
#29+:::+b
m
u#28t,d
0
,m#29
#289#29
Introducing
#12
T
=#28a
1
a
2
::,a
n
b
0
::,b
m
#29
'
T
#28t,1#29 = #28,y#28t,1#29#01#01#01,y#28t,n#29
u#28t,d
0
#29:::u#28t,d
0
,m#29#29
#2810#29
The model becomes a regression
y#28t#29='
T
#28t,1#29#12
Estimates given by
^
#12#28t#29=
^
#12#28t,1#29 +K#28t#29"#28t#29
"#28t#29=y#28t#29,'
T
#28t,1#29
^
#12#28t,1#29
K#28t#29=
P#28t,1#29'#28t,1#29
#15 +'
T
#28t,1#29P#28t,1#29'#28t,1#29
P#28t#29=
1
#15
,
I,K#28t#29'
T
#28t,1#29
#01
P#28t,1#29
#2811#29
c#0D K,J,#C5str#F6m and B,Wittenmark 4
An Indirect STR
1,Data A
o
and A
m
and n
2,Estimate parameters of model by RLS
3,Obtain polynomials R and S by solving
Diophantine Equation
4,Compute Polynomial T = A
o
B
0
m
5,Compute control signal from
Ru = Tu
c
,Sy
6,Go to 2
Example 3.4 - Cancellation of Zero
Process model
y#28t#29+a
1
y#28t,1#29 + a
2
y#28t,2#29
= b
0
u#28t,1#29 + b
1
u#28t,2#29
Control law
u#28t#29+r
1
u#28t,1#29 = t
0
u
c
#28t#29,s
0
y#28t#29,s
1
y#28t,1#29
Simulation
0 2040608010
1
0
1
0 2040608010
4
2
0
2
Time
Time
u
c
y
u
0 5 10 15 20
2
1
0
0 5 10 15 20
0.0
0.1
0.2
Time
Time
^a
2
^a
1
^
b
1
^
b
0
Example 3.5 - No Cancellation
Process model
y#28t#29+a
1
y#28t,1#29 + a
2
y#28t,2#29
= b
0
u#28t,1#29 + b
1
u#28t,2#29
Control law
u#28t#29+r
1
u#28t,1#29 = t
0
u
c
#28t#29,s
0
y#28t#29,s
1
y#28t,1#29
c#0D K,J,#C5str#F6m and B,Wittenmark 5
Simulation
0 2040608010
1
0
1
0 2040608010
4
2
0
2
Time
Time
u
c
y
u
0 100 200 300 400 500
2
1
0
0 100 200 300 400 500
0.0
0.1
0.2
Time
Time
^a
2
^a
1
^
b
1
^
b
0
Simulation
Process
G#28s#29=
b
s#28s+a#29
Desired response
G
m
#28s#29=
!
2
s
2
+2#10!s+!
2
Observer polynomial A
o
#28s#29=s+a
o
with
a
o
=2.
Controller structure
u#28t#29=,
s
0
p+s
1
p+r
1
y#28t#29+
t
0
#28p+a
o
#29
p+r
1
u
c
#28t#29
where
r
1
=2#10!+a
o
,a #2812#29
s
0
=
a
o
2#10!+!
2
,ar
1
b
#2813#29
s
1
=
!
2
a
o
b
#2814#29
t
0
=
!
2
b
#2815#29
c#0D K,J,#C5str#F6m and B,Wittenmark 6