Digital Signal Processing Reference
In-Depth Information
The first scalar equation of (13.25) is given by
Á
x
1
[n + 1] = a
11
x
1
[n] + a
12
x
2
[n] +
+ a
1N
x
N
[n]
Á
+ b
11
u
1
[n] +
+ b
1r
u
r
[n].
(13.26)
The
z
-transform of this equation yields
Á
zX
1
(z) - zx
1
[0] = a
11
X
1
(z) + a
12
X
2
(z) +
+ a
1N
X
N
(z)
Á
+ b
11
U
1
(z) +
+ b
1r
U
r
(z).
(13.27)
We will find the complete solution; hence, the initial condition
x
1
[0]
is included. The
second equation in (13.25) is given by
Á
x
2
[n + 1] = a
21
x
1
[n] + a
22
x
2
[n] +
+ a
2N
x
N
[n]
Á
+ b
21
u
1
[n] +
+ b
2r
u
r
[n],
(13.28)
which has the
z
-transform
Á
zX
2
(z) - zx
2
[0] = a
21
X
1
(z) + a
22
X
2
(z) +
+ a
2N
X
N
(z)
Á
+ b
21
U
1
(z) +
+ b
2r
U
r
(z).
(13.29)
The
z
-transform of the remaining equations in (13.25) yield equations
of the same form. We see, then, that these transformed equations may be written in
matrix form as
(n - 2)
z
X
(z) - z
x
[0] =
AX
(z) +
BU
(z).
We wish to solve this equation for
X
(
z
); to do this, we collect all terms containing
X
(
z
) on the left side of the equation:
z
X
(z) -
AX
(z) = z
x
[0] +
BU
(z).
(13.30)
It is necessary to factor
X
(
z
) in the left side to solve this equation. First, the term
z
X
(
z
) is written as
z
IX
(
z
), where
I
is the identity matrix (see Appendix G):
z
IX
(z) -
AX
(z) = (z
I
-
A
)
X
(z) = z
x
[0] +
BU
(z).
(13.31)
This additional step is necessary, since the subtraction of the matrix
A
from the
scalar
z
is not defined; we cannot factor
X
(
z
) directly in (13.30). Equation (13.31)
may now be solved for
X
(
z
):
X
(z) = z(z
I
-
A
)
-1
x
[0] + (z
I
-
A
)
-1
BU
(z).
(13.32)
The solution
x
[
n
] is the inverse
z
-transform of this equation.
Comparing (13.32) and (13.23), we see that the state transition matrix
≥[n]
is
given by
≥[n] =
z
-1
[≥(z)] =
z
-1
[z(z
I
-
A
)
-1
].
(13.33)
