Matrix exponential
Suppose that our state is a vector variable:
x  t   Ax  t 
We can imagine that the solution will have a
“matrix exponential” form:
x  t   x  0 exp  At 
For any square matrix A, the matrix exponential exp(A) is a square matrix function. We
can compute it using Taylor series expansion.
d2 f
t 2
dt
t 0
df
f t   f  0 
dt
t2
exp  At   I  At  A

2

tn n
exp  At    A
n 0 n !
2
In Matlab, exp(A) is computed as expm(A).
In Mathematica, use MatrixExp[A].
t2

2
t 0
tn
A

n!
n
dn f
 n
dt
tn

n!
t 0
Some properties of the matrix exponential
Using Taylor series expansion, one can show the following properties of the matrix
exponential:
exp(0t )  I
exp  A  t1  t2    exp  At1  exp  At2 
d
exp  At   A exp  At   exp  At  A
dt
Other properties of the matrix exponential:
exp( A) exp( A)  I
exp  A exp  B   exp  A  B  only if AB  BA  0
Solution of continuous LTI state equations (vector condition)
x  t   Ax  t   Cu  t 
exp   At  x  t   exp   At  Ax  t   exp   At  Cu  t 
d
exp   At  x  t    exp   At  Cu  t 
dt


exp   At  x  t  0   exp   At  Cu  t  dt
0

exp   A  x    x  0    exp   At  Cu  t  dt
0

x    exp  A  x  0   exp  A   exp   At  Cu  t  dt
0
x    exp  A  x  0    exp  A   t   Cu  t  dt

0
Solution of discrete LTI state equations
x
k 1
 Ax   Cu
k
x   Ax   Cu
1
0
k
0
x 2  Ax1  Cu 1  A2 x 0  ACu  0  Cu 1
x
k 
Ax
k
 0
k 1
  Ak 1 mCu
m 0
m
Relating discrete and continuous representation of a linear system
x  t   exp  Act  x  0   exp  Act   exp   Ac  Cu   d
t
0
t   k  1 
x k 1  x k 1
x
 k 1
 exp  Ac  k  1   x  0   exp  Ac  k  1   
 k 1 
0
exp   Ac  Cc u   d
 exp  Ac  k  1   x  0   exp  Ac  k  1    exp   Ac  Cc u   d
k
0
 exp  Ac  k  1   
 k 1 
k
exp   Ac  Cc u   d
k

 exp  Ac   exp  Ac k   x  0   exp  Ac k    exp   Ac  Cc u   d 


0
 exp  Ac  k  1   
 k 1 
k
 exp  Ac   x   
k
 k 1 
k
 exp  Ac   x k     Ac1

exp   Ac  Cc u   d



exp  A   k  1     
exp Ac   k  1     Cc u   d
c
 exp  Ac   x k     Ac1  I  exp  Ac     Cc u k 
k 1 
k
 C u k 
 c
Assume that u(t) is
constant between the
two sampling
intervals.
Discrete and continuous representation of a linear system
(noise free scenario)
Continuous system
 x  t   Ac x  t   Cc u  t 
Gc  
 y  t   Bc x  t   Dc u  t 
Discrete system
 x k 1  Ad x k   Cd u  k 
Gd   k
 
k 
k 
 y  Bd x  Dd u
  sampling interval
Ad  exp  Ac  
Cd  Ac1  exp  Ac    I  Cc
Bd  Bc
Dd  Dc