
(c) a
1
¼(1 þa) ¼(1 þ0.5) ¼1.5, a
2
¼a b ¼0.5 0.25 ¼0.25, a
3
¼b ¼0.25
zI A ¼ z
100
010
001
2
6
6
4
3
7
7
5
010
001
a
3
a
2
a
1
2
6
6
4
3
7
7
5
(4:546)
¼
z 10
0 z 1
a
3
a
2
z þ a
1
2
6
6
4
3
7
7
5
¼
z 10
0 z 1
0:25 0:25 z 1:5
2
6
6
4
3
7
7
5
(4:547)
F(z) ¼ z(zI A)
1
(4:548)
Inverting (zI A) followed by multiplication by z results in
F(z) ¼
z
z
3
1:5z
2
þ 0:25z þ 0:25
z
2
1:5z þ 0:25 z 1:51
0:25 z(z 1:5) z
0:25z 0:25(z þ 1) z
2
2
6
6
4
3
7
7
5
(4:549)
From Equation 4.532, with
u
k
¼0, k ¼0, 1, 2, . . . the solution for y
k
is
y
k
¼ CF
k
x
0
(4:550)
where the initial state
x
0
¼
y
3
y
2
y
1
2
6
6
4
3
7
7
5
¼
0
0
P
0
2
6
6
4
3
7
7
5
The transition matrix F
k
is obtained by inverse z-transforming F(z) in Equation 4.549. The last
column of F
k
is all ...