Digital Signal Processing Reference
In-Depth Information
We arbitrarily define the elements of
v
[
n
] as follows:
v
1
[n] = x
1
[n];
v
2
[n] = x
1
[n] + x
2
[n].
Thus, from (13.50),
10
11
B
R
v
[n] =
Qx
[n] =
x
[n]
and
10
11
10
-11
P
-1
B
R
B
R
=
Q
=
Q
P
=
.
Hence, the components of
x
[n] =
Pv
[n]
can be expressed as
x
1
[n] = v
1
[n];
x
2
[n] =-v
1
[n] + v
2
[n].
P
=
Q
-1
,
It is seen from this example that, given the vector
v
[
n
] and the transformation we
can solve for the vector
x
[
n
]. Or, given the vector
x
[
n
] and the transformation
Q
, we can solve
for the vector
v
[
n
].
■
Similarity transformation for a second-order system
EXAMPLE 13.11
This example is a continuation of the last example. From (13.52), the system matrices for the
transformed matrices become
10
11
01
-65
10
-11
A
v
=
P
-1
AP
=
B
RB
RB
R
01
-66
10
-11
-11
-12
B
RB
R
B
R
=
=
,
6
10
11
0
1
0
1
B
v
=
P
-1
B
=
B
RB
R
=
B
R
,
and
10
-11
B
R
C
v
=
CP
= [3
1]
= [2
1].
The transformed state equations are then
-11
-12
0
1
B
R
B
R
v
[n + 1] =
A
v
v
[n] +
B
v
u[n] =
v
[n] +
u[n]
6
