
Equating like powers of z and solving for e
aT
and vT,
e
2aT
¼ 9 ) e
aT
¼ 3(4:435)
2(e
aT
cos vT) ¼3 ) cos (vT) ¼
1
2
) vT ¼
p
3
(4:436)
The quadratic numerator in F(z) in Equation 4.433 must be expressed as a linear combination of
the standard numerator forms in the last two rows of Table 4.4, that is,
2z
2
11z ¼ c
1
(e
aT
sin v T)z þ c
2
[z
2
(e
aT
cos vT)z](4:437)
Solving for c
1
and c
2
in Equation 4.437 leads to c
1
¼16
ffiffiffiffiffiffiffiffi
3=9
p
, c
2
¼ 2: F(z) is now written in a
form where Table 4.4 can be used to find f
k.
F(z) ¼
1
7
c
1
(e
aT
sin vT)z
z
2
2(e
aT
cos vT)z þ e
2aT
þ
c
2
[z
2
(e
aT
cos vT)z]
z
2
2(e
aT
cos vT)z þ e
2aT
þ
2
7
z
z 1
(4:438)
f (k)
1
7
[c
1
e
akT
sin kvT þ c
2
e
aT
cos k