
Substituting (z 1)=T for s in Equation 4.783 leads to
Dz(z)
Dz
com
(z)
¼
^
K
C
K
a
T
4
l
1
z (l
1
l
0
T)
z
5
þ g
4
z
4
þ g
3
z
3
þ g
2
z
2
þ g
1
z þ g
0
(4:784)
where
g
4
¼5 þ m
4
T
g
3
¼10 4m
4
T þ m
3
T
2
g
2
¼10 þ 6m
4
T 3m
3
T
2
þ m
2
T
3
g
1
¼ 5 4m
4
T þ 3m
3
T
2
2m
2
T
3
þ m
1
T
4
g
0
¼1 þ m
4
T m
3
T
2
þ m
2
T
3
m
1
T
4
þ m
0
T
5
9
>
>
>
>
>
>
>
=
>
>
>
>
>
>
>
;
(4:785)
To simulate the altitude response to a step input command of magnitude Dz
com
¼A, we need the
difference equation relating Dz
k
and (Dz
com
)
k
. Cross multiplying Equation 4.784 after multiplying
numerator and denominator by z
5
gives
(1 þ g
4
z
1
þ g
3
z
2
þ g
2
z
3
þ g
1
z
4
þ g
0
z
5
)Dz(z)
¼
^
K
C
K
a
T
4
[l
1
z
4
( l
1
l
0
T)z
5
]Dz
com
(z)(4:786)
Invert z -transforming both sides of Equation 4.786 ...