4-1
Chapter 4
State-Space Solutions and
Realizations

Discuss solutions of state-space and study how to
transform transfer functions into state equations
長庚大學電機系
4-2
Outline

Solution of linear time-invariant (LTI) state equations

Equivalent state equations

Realizations

Solution of linear time-varying (LTV) equations

Equivalent time-varying equations

Time-varying realizations
長庚大學電機系
4-3
Solution of LTI State Equations
Consider the LTI state-space equation
 x(t )  Ax(t )  Bu(t )

 y(t )  Cx(t )  D(t )

Solution in s-domain (by Laplace transform)
x(t )  Ax(t )  Bu(t )
 sxˆ ( s)  x(0)  Axˆ ( s)  Buˆ ( s(Take
)
Laplace transform)
 xˆ ( s )  ( sI  A) 1[ x(0)  Buˆ ( s)]
 yˆ ( s )  C ( sI  A) 1[ x(0)  Buˆ ( s)]  Duˆ ( s)
 C ( sI  A) 1 x(0)  [C ( sI  A) 1 B  D]uˆ ( s)
長庚大學電機系
4-4

Solution in t-domain:
x  Ax  Bu
 e  At x  e  At ( Ax  Bu )
(Multiply integrating factor)
d  At
 (e x)  e  At Bu
dt
e
 At
t
x(t )  x(0)   e  A Bu ( )d
0
t
 x(t )  e x(0)   e A(t  ) Bu ( )d
At
0
t
 y (t )  C[e x(0)   e A(t  ) Bu ( )d ]  Du (t )
At
0
長庚大學電機系
Computation of e

At
4-5
At
How to compute e ?
At
 By Laplace transform: e
 L1[(sI  A)1 ]
 Function of square matrix:
k

(
At
)
At
 Power series method: e  
k!
k 0
 Jordan form:
A  QJQ1  e At  QeJt Q1
長庚大學電機系
4-6
 0 1 
Example: Compute e , A  

 1 2 
At

By Laplace transform method:
 ( s  2)
1
2

s

1
(
s

1)


1
( sI  A)  
 
 1
 1 s  2 
 ( s  1) 2

t

(1

t
)
e
 e At  L1[( sI  A) 1 ]  
t
te

1 
( s  1) 2 

s 
( s  1) 2 
te  t 
t 
(1  t )e 
長庚大學電機系
4-7

By Jordan form method:
1 1 
 1 1 
1
A  QJQ , Q  
, J  

1
0
0

1




t
t

 1
e
te
At
Jt 1
 e  Qe Q  Q 
Q
t 
 0 e 
 (1  t )e  t
te  t 

t
t 
(1  t )e 
 te
長庚大學電機系
4-8

By function of square matrix method:
Let f ( )  et and h( )   0  1
 ( A)  {1, 1}
 f (1)  e t  h(1)   0  1
f '(1)  te t  h '(1)  1
  0  (1  t )e t , 1  te t
 e At  f ( A)  h( A)   0 I  1 A
 (1  t )e t

t
te

te t 
t 
(1  t )e 
長庚大學電機系
4-9
Example: From the previous example, the solution of
0 1
0
x
x   u

1 2 
1
is
t
x(t )  e x(0)   e A(t  ) Bu ( )d
At
0
(1  t )e t

t
te

t
 ( t  )


t

(
t


)
e
u
(

)
d


te 
0


x
(0)


 t

 ( t  )
(1  t )e t 
  [1  (t   )]e
u ( )d 
 0

長庚大學電機系
Stability of Zero-Input Response
4-10
Let A  QJQ1. Then
 Zero-input response
xzi (t )  e At x(0)  QeJt Q1x(0)
k 1
 Since
 t
t
t
t 
e Jk (  )t
e






te
e 



tet


et

(k  1)!
 (i)  ( A)  C   x(t )  0 as t  
(ii)    ( A) with   C   x(0)  x(t )  
(iii)  ( A)  C   j  axisand each eigenvalue at
jw-axis has index 1  All solution are bounded.
(iv)  (A)  C-  j  axis and   jw-axis having
index > 1   solution which is not bounded.
長庚大學電機系
4-11
Discretization
Consider  x  Ax  Bu



 y  Cx  Du
Suppose that u(t )  u(kT ) for kT  t  (k  1)T
Define u[k ]: u(kT )
kT
AkT
x[k ] : x( kT )  e x(0)   e A( kT  ) Bu ( ) d
0
 x[k  1]  e
A ( k 1)T
x(0)  
( k 1)T
0
 e x[k ]  
AT
( k 1)T
kT
e A( kT T  ) Bu ( ) d
e A( kT T  ) Bu ( )d
T
 e x[k ]  (  e A d ) Bu[k ]
AT
0
(by letting   kT  T   and the assumption on u(t ))
長庚大學電機系
4-12

Thus, if input changes value only at discrete-time instance
kT and we compute only the responses at t=kT then system
 x  Ax  Bu

 y  Cx  Du
becomes
 x[k  1]  Ad x[k ]  Bd u[k ]

 y[k ]  Cd x[k ]  Dd u[k ]


with
T
Ad  e , Bd  ( e A d ) B, Cd  C, Dd  D
AT
T
Bd  (  e A d ) B
0
0
T
 (  ( I  A  ...)d ) B
0
 (TI  AT 2 / 2! ...) B
 A1 (e AT  I ) B
if A is nonsingular
長庚大學電機系
4-13
Solution of Discrete-time Equation
 x[k  1]  Ax[k ]  Bu[k ]

 y[k ]  Cx[k ]  Du[k ]
Consider

Note that,
x[1]  Ax[0]  Bu[0]
x[2]  Ax[1]  Bu[1]
 A2 x[0]  ABu[0]  Bu[1]
In general,
k 1
x[k ]  A x[0]   Ak 1m Bu[m]
k
m0
k 1
 y[k ]  CAk x[0]   CAk 1m Bu[m]  Du[k ]
m 0
長庚大學電機系
4-14
Stability of Zero-Input Response
Let A  QJQ1  Am  QJ mQ1
 m


J km ( )  

 0




m
(m  k  2) m  k 1 

(k  1)!

 , m  k 1

m




If i ( A)  1, i  Am x[0]  0, as m , x[0].
If i  i ( A)  1 or i ( A)  1 with index  1
 Am x[0] is unbounded for some x[0] as m goes to
infinity.
長庚大學電機系
4-15
Equivalent State Equations

To transform the system into a canonical form or as
simple as possible.

Consider
 x  Ax  Bu

 y  Cx  Du
Let x  Px, P is a nonsingular matrix
 x  Ax  Bu

 y  Cx  Du
A  PAP1, B  PB, C  CP1, D  D
the two systems are called equivalent and x  Px
is called an equivalence transformation.
長庚大學電機系
4-16

( )  det( I  A)  det( PP 1  PAP 1 )
 det( P) det( I  A) det( P 1 )
 det( I  A)  ( )
i.e., equivalent state equation have the same
characteristic polynomial and thus the same set of
eigenvalues.

Gˆ  C ( sI  A) 1 B  D
 CP 1[ P ( sI  A) P 1 ]1 PB  D
 CP 1 P ( sI  A) 1 P 1 PB  D
 C ( sI  A) 1 B  D  Gˆ ( s )
i.e., transfer matrices are preserved under equivalence
transformation.
長庚大學電機系
4-17

Two state equation are said to be zero-state equivalent
if they have the same transfer matrix. In such a case,
D  C (sI  A)1 B  D  C (sI  A)1 B


 D   CA Bs  D   CAi 1Bs i
i 1
i
i 1
i 1

Theorem: Two LTI system {A,B,C,D} and {A, B, C, D}
are zero-state equivalent (or have the same transfer
matrix if and only if D  D and CAi B  CAi B, i  0,1,...

Clearly, two equivalent LTI systems implies that they
are zero-state equivalent.
長庚大學電機系
4-18
Realization
1
Question: Given (A,B,C,D)  Gˆ (s)  C(sI  A) B  D .
Conversely, given Gˆ (s) and suppose that the LTI system
is lumped, how to find state-space representation
(A,B,C,D)?
ˆ
 A transfer matrix G( s) is said to be realizable if there
exist {A,B,C,D} such that Gˆ (s)  C(sI  A)1 B  D
{A,B,C,D} is called a realization of Gˆ (s).

If the realization problem is solvable, then the state
space description is not unique (discussed later)
長庚大學電機系
4-19
Theorem: A q  p transfer matrix Gˆ (s) is realizable
, each entry of Gˆ (s) is a proper rational function.
長庚大學電機系
4-20

The realization discussed above (in the proof of the
last theorem) is called controllable canonical form.
A dual realization called observable canonical form
is described below:
 1 I q

  2 I q
A

  r 1 I q
  I
r q

C   Iq
0q
Iq
0q
0q
Iq
0q
0q
0q
0q
0q 
 N1 

0q 
 
N2 

, G 
 

Iq 
 
Nr 


0q 
0q  , D  Gˆ ()
長庚大學電機系
4-21
Solutions of LTV Equations
Consider x(t )  A(t ) x(t ), A(t )  Rnn
0
 A(t )  C Solution ! for any given initial state.
0
We assume hereafter that A(t )  C


The solution set is a vector space of dimension n
X (t ) : ( x1 (t ),..., xn (t ))  Rnn is called a fundamental
matrix of x(t )  A(t ) x(t ) if each column xi (t ) is a
solution, and {x1 (t )...xn (t )} is linearly independent.
Note that, there are infinite many fundamental matrices
長庚大學電機系
4-22

The fundamental matrix (t, t0 )  X (t ) X 1 (t0 ) is called
the state transition matrix.
 (t , t0 ) is also a fundamental matrix.
 (t , t0 ) is unique no matter what fundamental
matrix X (t ) is chosen.
(Reason: Let Y (t )  X (t )C  Y (t )Y 1 (t0 )  X (t ) X 1 (t0 ))

(t , t )  I

(t, t0 )(t0 , )  (t, )

The unique solution of

1 (t, t0 )  (t0 , t )
 x  A(t ) x
is  (t , t0 ) x0

 x(t0 )  x0
長庚大學電機系
0 0
x(t ) or
Example: consider x(t )  

 t 0
x1 (t )  0  x1  x1 (0)
 x1 (t )  0

 x2 (t )  tx1 (t )
1
x2 (t )  tx1  x2 (t )  x1 (0)t 2  x2 (0)
2
 1 
1
 x(t )   1 2  is a solution with x(0)   
 t 
 0
2 
0
 0
x(t )    is a solution with x(0)   
1
1
0
 1
is a fundamental
 X (t )   2

t / 2 1
4-23
matrix since
it has linearly independent columns.

 1
(t , t0 )  X (t ) X 1 (t0 )   1 2
 t
2
1
0
 1
1
0


1 2


 t
10 1

2
0

1

長庚大學電機系
4-24

Let (t, t0 ) be the state transition matrix of x(t )  A(t ) x(t ).
Then the unique solution of
 x(t )  A(t ) x(t )  B(t )u(t )
is

 y(t )  C (t ) x(t )  D(t )u (t )
t

x(t )  (t , t0 ) x(t0 )   (t , ) B( )u ( )d
t0


 y (t )  C (t )(t , t ) x(t )  C (t ) t (t , ) B( )u ( )d  D(t )u (t )
0
0
t0

zero-input response

zero state response
t
 yzs (t )   [C (t )(t , ) B( )  D(t ) (t   )]u ( )d
t0

 G(t , )  C (t )(t , ) B( )  D(t ) (t   ) At  A
If A ñ constant matrix  (t, )  e e  e A(t  )
the unique solution of x(t )  Ax(t )  Bu(t ) is
t
A(t t0 )
x(t )  e
x(t0 )   e A(t  ) Bu( )d
t0
長庚大學電機系
4-25
Discrete-Time Case

 x[k  1]  A[k ]x[k ]  B[k ]u[k ]
Consider 
 y[k ]  C[k ]x[k ]  D[k ]u[k ]
 x[k0  2]  A[k0  1]x[k0  1]  B[k0  1]u[k0  1]
 A[k0  1] A[k0 ]x[k0 ]  A[k0  1]B[k0 ]u[k0 ]
 B[k0  1]u[k0  1]
x[k0  3]  A[k0  2] A[k0  1] A[k0 ]x[k0 ]
(k0  3, k0 )
(k0  3, k0  1)
 A[k0  2]A[k0  1]B[k0 ]u[k0 ]
 A[k0  2]B[k0  1]u[k0  1]
 B[k0  2]u[k0  2]
(k0  3, k0  2)
長庚大學電機系
4-26

For k  k0 define
[k  1, k0 ]  A[k ][k , k0 ], [k0 , k0 ]  I .



[k , k0 ]  A[k 1] A[k0 ], for k  k0
[k , k0 ]  [k , k1 ][k1, k0 ], for k  k1  k0
The solution of
x[k  1]  A[k ]x[k ]  B[k ]u[k ]
is x[k ]  [k , k0 ]x[k0 ] 

k 1
 [k , m  1]B[m]u[m]
m k0
for k  k0
If we define [k , k0 ]  0 for k  k0
y[k ]  C[k ]x[k ]  D[k ]u[k ]  C[k ][k , k0 ]x[k0 ]

k
 (C[k ][k , m  1]B[m]  D[m] [k  m])u[m]
m  k0
G[ k ,m ]
長庚大學電機系
4-27
Equivalent Time-Varying Equation
x  A(t ) x  B(t )u
Consider 
.........................................(V1)
 y  C (t ) x  D(t )u


Let x  P(t ) x. Assume that P(t )is nonsingular, P(t ) and
are continuous.
 x  A(t ) x  B (t )u
.........................................(V2)

 y  C (t ) x  D(t )u
1
where A(t )  [ P(t ) A(t )  P(t )]P (t )
B (t )  P(t ) B(t )
C (t )  C (t ) P 1 (t )
D(t )  D(t )

The two systems are called equivalent. And P(t ) is
called an equivalence transformation.
長庚大學電機系
4-28

Recall that System (V1) has impulse response
G(t, )  C(t ) X (t ) X 1 ( ) B( )  D(t ) (t  )
The impulse response for System (V2) is
G (t , )  C (t ) X (t ) X 1 ( ) B ( )  D(t ) (t   )
 C (t ) P 1 (t ) P(t ) X (t ) X 1 ( ) P 1 ( ) P( ) B( )  D(t ) (t   )
 G (t ,  )
Thus, the impulse response matrix is invariant under
any equivalence transformation.
長庚大學電機系