Lecture 11
∗
October 11, 2012
1
Last Time
ẋ = Ax + Bu
Time Domain
y = Cx
ŷ = C(Is − A)−1 Bˆ(u)
2
Frequency Domain
Why is it Called Frequency Domain?
mẍ + Kx = u

q 
k
0
0
ẋ1
m  x1

q
+
u(t)
=
x
1
ẋ2
k
2
0
−
m
y(t) = x1 (t)
x1
= 1, 0
x2
When the input is u = 0, the output is easily found from the matrix
exponential (Peano-Baker Series)
y(t) = Asin(ωt) + Bcose(ωt)
q
k
where ω = m

s
sI − A = q
k
m
∗ This
q 
k
− m
= s
ω
s
−ω
s
work is being done by various members of the class of 2012
1
Dynamic Systems: State-space control
c(sI − A)−1 =
2
1
2
s + ω2
s
ω
−ω
s
Using from above (A)(B)(C)

q 
k
0
0
m
A= q
B=
C = 1, 0
1
k
− m
0


s
ω
2
2
2
2
0
−1
s
+ω
s
+ω


c(sI − A) = 1, 0
ω
s
− s2 +ω2 s2 +ω2 1
ω
s2 + ω 2
Let u(t) = sin(wo t). Consider the system forced by u.
ŷ = g(s)û(s)
=
ω
ωo
s2 + ω 2 s2 + ωo2
=λ
ωo
ω
β
s2 + ω 2 s2 + ωo2
where
=λ=
3
ωo
ω
β= 2
ωo2 − ω 2
ω − ωo2
Changing Basis
Changing Basis: The effect on time and frequency doman representations,
Given
ẋ = Ax
(1)
We have seen that solutions are conveniently represented by
eAt
 λ t
e 1

 0

 0
= P
 .


 .
0
0
0
0
e λ2 t
0 eλ3 t
.
.
.
.
0
0
.
.
.
.
.
.
.
.
.
.
.
.

. 0
. 0 


. 0 
P −1
. . 

. . 

. eλn t
Dynamic Systems: State-space control
3
In the case that A is normal. How do the representations of
ẋ = Ax + Bu
y = Cx
(t)
and
ŷ = c(sI − A)−1 B û
look under a change of basis
Let z = T x, where T is any invertable matrix, Then
ż = T ẋ
= T (Ax + Bu)
= T AT −1 z + T Bu
y = Cx = CT −1 z
In terms of the state variable z, the system is written as
ż = Āz + B̄u
y = C̄z
Ā = T AT −1
B̄ = T B
C̄ = CT −1
The frequency domain representation of the transform system is
ŷ = C̄(Is − Ā)−1 B̄ û
= CT −1 (Is − T AT −1 )T B û
= CT −1 (T T −1 s + T AT −1 )T B û
= CT −1 (T [Is − A]T −1 )T B û
= CT −1 (T [Is − A]−1 T −1 )T B û
C(Is − A)−1 B û
Dynamic Systems: State-space control
4
4
Realization Theory
Realization Theory: How do we get a state space representation from a transfer
function?
Case 1: SISO (single input single output) system of the form
g(s) =
1
sn
+ an−1
sn−1
+ ..... + ao
this relates inputs to outputs by ŷ = g(s)û
(sn + an−1 sn−1 + ..... + ao )ŷ(s) = û(s)
Taking the inverse laplace transforms
dn y
dn−1 y
+
a
+ ..... + ao y(t) = u(t)
n−1
dtn
dtn−1
First orderizing this in the obvious way.
x1 = y
x2 = ẏ
.
.
xn = y n−1
 
0
ẋ1
 ẋ2   0
  
 . = .
  
 .   0
−ao
ẋn

   
0
x1
.
0
 x2  0
.
0 
   
   
.
0 
  .  +  .  u(t)
 .  0

.
1
1
. −an−1
xn
 
x1
 x2 
 

y = x1 = 1, ......, 0 
 . 
 . 
xn
1
0
.
0
−a1
0
1
.
0
−a2
Case 2: SISO system of the form
g(s) =
bn−1 S n−1 + ...... + Bo
+ an−1 sn−1 + ..... + ao
sn
Dynamic Systems: State-space control
5
Again we wish to find a system such that
ŷ = g(s)û
or in other words
g(s) =
ŷ(s)
u(s)
To find the state space representation, introduce an intermediate variable x̂,
and write
g(s) =
ŷ x̂
x̂ û
we write these factors as
1
x̂
= n
n−1
û
s + an−1 s
+ ..... + ao
ŷ
= bn−1 S n−1 + ...... + Bo
x̂
The state space representation of x
  
ẋ1
0
1
0
 ẋ2   0
0
1
  
 . = .
.
.
  
 .   0
0
0
−ao −a1 −a2
ẋn
in terms of u has been seen to be
   
x1
.
0
0
 x2  0
.
0 
   
   
.
0 
  .  +  .  u(t)

 .  0
.
1
. −an−1
1
xn
x = x1
The time domain rendering of y is then
y(t) = bn−1
dn−1
x + bo x(t)
dtn−1
= bn−1 xn + ........ + bo x1
The overall system is
  
ẋ1
0
 ẋ2   0
  
 . = .
  
 .   0
ẋn
−ao
1
0
.
0
−a1
0
1
.
0
−a2
   
.
0
x1
0
 x2  0
.
0 
   
   
.
0 
  .  +  .  u(t)
.
1   .  0
. −an
xn
1
Dynamic Systems: State-space control
6


x1
 x2 

y(t) = bo , b1 , ........bn−1 
 . 
xn
Starting from the same transfer function we perform long division
an−1
+ bn−2 −bsn−1
+ .....
2
(sn + an−1 sn−1 + ..... + ao ) bn−1 S n−1 + ...... + Bo
bn − 1sn−1 + bn−1 an−1 sn−2 + ....
bn−2 − bn−1 an−1 )sn−2 + ....
bn−1
s
Proper rational functions, in this way, admit Taylor series expansions about
infinity
λ2
λ3
λ1
+
+
+ ....
s
s
s
what about MIMO (multiple input multiple output) systems?
g(s) =
G(s) = C(sI − A)−1 B
−1
1
A
= C I−
B
s
s
For s sufficiently large, every entry in the matrix A
is small hence we may
s
appeal to writing
A
I−
s
−1
A
= f( )
s
where
f (r) =
1
1−r
1
= 1 + r + r2 + ....
1−A
−1
A
A A2
I−
= I + + 2 + ....
s
s
s
The
G(s) =
Note that
−1
1
A
CB
CAB
CA2 B
C I−
B=
+
+
+ .....
2
s
s
s
s
s3
Dynamic Systems: State-space control
L−1
1
sk+1
7
=
tk
k!
Hence the inverse Laplace transform of G(s) is
2 2
3 3
A t
A t
CB + C(At)B + C
B+C
B
2
3!
= CeAt B
Theorem: Let G(s) be a q × m matrixof rational functions such that the
degrees of the numerators exceed the degrees of the denominators for each entry.
Then there exist constant matrices A, B, C such that
G(s) = C(Is − A)−1 B.
Proof let P (s) be the monic polynomial that is the least common multiple
(l.c.m) of the denominators (monic → leading coefficient = 1, l.c.m. : if f (s) =
(s − λ)(s − β)(s − γ), g(s) = (s − β)(s − δ), then the l.c.m is (s − λ)(s − β)(s −
γ)(s − δ); Then
P (s)G(s) = E − 0 + E1 s + E2 s2 + Er−1 sn−1
Where r = degree of P
Let Om be the m x m zero matrix, let I x m
Define A to be the rm x rm matrix

Om
Im
Om
 Om
O
Im
m

.
.
.
A=

 Om
Om
Om
−po Im −p1 Im −p2 Im
be the m x m identity matrix.

.
Om
.
Om 


.
.

.
Im 
. −pn−1 Im
where
P (s) = sn + Pr−1 sn−1 + ....... + P1 s + po
let


Om
Om 



B=
 .  , C = Eo , E1 , ...., Em
Om 
Im
Now we will show that
G(s) = C(Is − A)−1 B
Dynamic Systems: State-space control
8
Note: that (Is − A)−1 B is the solution x̂ to the matrix equation
(Is − A)X = B
(∗)
Partition


x̂1
 x̂2 
 

x̂ = 
 . 
 . 
x̂n
Compatible with A (i.e. x̂k is m-dimensional)
then (*) may be written


x̂1
Om
 x̂2   Om
 

s
 .  .
 .   Om
x̂n
−po Im
Im
Om
.
Om
−p1 Im
Om
Im
.
Om
−p2 Im
   
.
Om
x̂1
0
 x̂2   0 
.
Om 
   
 .  +  . 
.
.
   
.
Im   .   0 
. −pn−1 Im
Im
x̂n
component wise this is
sx̂i = x̂i+1 i = 1, ...., r − 1
sx̂r = −Po x̂1 − P1 x̂2 ..... − Pr−1 x̂r + Im
= (−Po − P 1s − ..... − Pr−1 S n−1 )x̂1 + Im
the last equation may be rendered as
S n x̂1 = (−Po − P 1s − ..... − Pr−1 S n−1 )x̂1 + Im
x̂ =
1
I −m
P (s)
Now note that


x̂1
 x̂2 
 
−1

C(Is − A) B = C 
 . 
 . 
x̂n
= Eo x̂1 + E1 x̂2 + .... + En−1 x̂r
= Eo x̂1 + E1 sx̂2 + .... + En−1 sn−1 x̂r
Dynamic Systems: State-space control
9
= (Eo + E1 s + .... + En−1 sn−1 )x̂1
=

Om
 Om

A=
 .
 Om
−po Im
1
(Eo + E1 s + .... + En−1 sn−1 ) = G(s)
P (s)
Im
Om
.
Om
−p1 Im
Om
Im
.
Om
−p2 Im



Om
.
Om
Om 
.
Om 



 B =  .  , C = Eo , E1 , ...., Em
.
.



Om 
.
Im 
Im
. −pn−1 Im
Standard controllable realization of G(s)