EE611
Deterministic Systems
Realizations, State-Transition Matrices
Kevin D. Donohue
Electrical and Computer Engineering
University of Kentucky
Realizations
Every LTI system has an input-output description of the
form:
  s u
 s
y  s= G
If system is also lumped, state-space descriptions also
exist:
ẋ t=A x tB ut
y t =C x tD ut 
G  s 
which are referred to as realization of the transfer
matrix
.
G  s  is realizable if ∃ a finite-dimensional state equation
{A,B,C,D} ∋ G  s = DC  s I−A −1 B 
Transfer Matrix Realizations
A transfer matrix G s is realizable iff G s is proper
rational matrix.
Consider an element of G s that is not proper (M>0):
M
M −1
g p ,q  s  =d 0 s d 1 s ...d M −1 sd M 
b1 sn−1b2 sn−2...bn−1 sbn
sn a1 s n−1 a2 sn−2...a n−1 sa n
How would the realization equations below have to
change to accommodate the not proper G s ?
ẋ t=A x tB ut 
y t =C x tD ut
Example
Find the state equations for the circuit using 2 different
state definitions. Show the transfer matrix is the same.
Assume u(t) input and y(t) output.
u(t)
R
L
C
+
y(t)
-
1
LC
g  s=
R
1
2
s  s
L
LC
Realize a Strictly Proper TF
For a proper TF, long division can be applied to
decompose it into a constant term (for the d scalar) and
a strictly proper TF expressed below:
g sp  s=
1 sn−1 2 sn−2... n−1 sn
sn 1 sn−1 2 s n−2...n−1 s n
The op-amp circuit for this TF can be realized as:
β1
u(t)
β2
β3
1/S
−α1
1/S
−α2
−α3
1/S
−αn
.....
1/S
βn
y(t)
Realize a Strictly Proper TF
The state equations for the strictly proper TF:
g sp  s=
1 sn−1 2 sn−2... n−1 sn
sn 1 sn−1 2 s n−2...n−1 s n
are written from:
x 1= ẋ 2 x 2 = ẋ 3 ... x n−1= ẋ n
ẋ 1=u−1 x1 −2 x 2 ...−n−1 x n−1 − n x n
y=1 x 12 x 2 ...n−1 x n−1 n x n
[
−1 − 2
0
ẋ= 1
⋮
⋮
0
0
] []
... −n−1 −n
1
...
0
0 x 0 u
0
⋮
⋮
⋮
0
...
1
0
y= [  1  2 ...  n−1 n ] x
Transfer Matrix Realizations
In general for a proper q x p transfer matrix G s its
realization can be expressed as follows.
Find a common denominator d(s) for all element in G s
and divide through by the denominator (if necessary) to
separate each rational polynomial into a constant and
strictly proper rational polynomial
 s= G∞

 s
G
G
sp
Then expand the strictly proper part into:
1
r−1
r−2

G sp  s=
N
s
N
s
... Nr−1 sNr ]
[
1
2
d  s
where r is the order of d(s).
Transfer Matrix Realizations
A realization of G scan therefore be written as:
[
] []
−1 I p − 2 I p ... − r−1 I p − r I p
Ip
Ip
0
...
0
0
ẋ=
x 0 u
⋮
⋮
⋮
⋮
⋮
⋮
0
0
0
...
Ip
0

y=[ N1 N 2 ... Nr−1 Nr ] xG∞u
Solution of Linear Time-Varying State Equation
Given a linear-time varying state-space equation:
ẋ t=A t x tB t ut 
y t =Ct x tD t ut 
the solution can be written as:
t
x t= t ,t 0 x t 0 ∫ t , B ud 
t0
t
y t =Ct  t , t 0  xt 0 Ct ∫  t , B ud Dtut 
t0
where  t , is called the state-transition matrix
State Transition Matrix
A fundamental matrix for the homogeneous equation
ẋ t=A t x t
is a matrix X(t) ∋ its columns are unique solutions of:
Ẋ t =A t  X t 
and X(t) is nonsingular for all t.
Then for any fundamental matrix of ẋ t=A t x t
the state transition matrix is given by:
 t , t 0 =X t  X−1 t 0 
and is the unique solution of:
̇ t , t 0 =A tt , t 0 
for initial condition
 t 0, t 0 =I
Examples
Find the state transition matrices for to the following:
[ ]
ẋ t=
[
0 0
x t
t 0
]
1 exp−t
ẋ t=
xt 
0
−3
Special Case Solution
For the case when
A t 

t
∫ A d 
t0

=
t
∫ A d 
t0

A t 
Then the solution for the state transition matrix becomes:
̇ t 0, t =exp

t

∞
∫ A  d  =∑
t0
k =0
1
k!

t
∫ A  d 
t0
k

Discrete Time Case
The state transition matrix can be computed through recursion
for the discrete time case. Given
 k 1, k 0 =A [k ] k , k 0 
Substitute repeatedly into
 k , k 0 =A [k −1]k −1, k 0 
 k , k 0 =A [k −1] A [k −2] k −2, k 0 
 k , k 0 =A [k −1] A [k −2]..... A [k 0 ] k 0, k 0 
Then the solution for the state transition matrix becomes:
 k , k 0 =A [k −1] A [k −2]..... A [k 0 ]
Useful Matrix Relationships
−1
d A t
d A t  −1
−1
=−A t 
A t
dt
dt
∣A B∣=∣A∣∣B∣
−1
 A B  =B−1 A−1
T
 A B  =BT AT
tr  A B  =tr  B A 
Homework U6.1
Find the op-amp circuit and state-space realization for
the proper rational TF given below:
2 s33 s
g s= 3
s 2 s2 5 s25