082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

700 Chapter 22 Difference equations and the z transform
Solution F(z) = Z{f[k]} =
∞∑
k=0
kz −k
= 1 z + 2 z 2 + 3 z 3 +···
= 1 z
{1 + 2 z + 3 z 2 +···}
Ifweusethebinomialtheorem(Section6.4)toexpress
(
1− 1 z
) −2
asaninfiniteseries,
we find that
(
1 − 1 ) −2
= 1 + 2 z z + 3 ∣ ∣∣∣
z + ··· provided 1
2 z ∣ < 1,that is |z| > 1
Using thisresultwe see that Z{f[k]}can be written as
F(z) = 1 (
1 − 1 ) −2
z z
and so
F(z) = 1 z
1
(1 −1/z) = z
for |z| > 1
2 (z −1)
2
Example22.11 Findtheztransform ofthe sequencedefinedby
f[k] =Ak
A constant
Solution We find
F(z) = Z{f[k]} =
=A
=
∞∑
Akz −k
k=0
∞∑
k=0
kz −k
Az
(z −1) 2 usingExample 22.10
In the same way as has been done for Laplace transforms, we can build up a library of
sequences and theirztransforms. Some common examples appear in Table 22.2. Note
that inTable22.2aandbareconstants.
Example22.12 UseTable 22.2 tofind theztransformsof
(a) sin 1 2 k
(b) e3k cos2k
Solution Directly fromTable 22.2 wefind
(a) Z{sin 1 2 k} = zsin 1 2
z 2 −2zcos 1 2 +1
(b) Z{e 3k cos2k} =
z 2 −ze 3 cos2
z 2 −2ze 3 cos2 +e 6