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

22.11 Inversion of z transforms 717
We now split the r.h.s. expression into partial fractions using the standard techniques
discussed inSection 1.7.This gives
F(z)
z
= 1
z −5 + 1
z +4
Multiplying byzgives
F(z) =
z
z −5 + z
z +4
It is now possible to invert this expression for F(z) using the z transform table,
Table 22.2. This gives
f[k] =5 k + (−4) k
22.11.1 Directinversion
Sometimes it is possible to invert a transform F(z) directly by reading off the coefficients.
Example22.28 Find f[k]if
F(z)=1+z −1 +z −2 +z −3 +···
Solution Usingthe definitionoftheztransform wesee that
f[k]=1,1,1,...
Occasionallyitispossible torewriteF(z)toobtain the required form.
Example22.29 Use the binomial theorem to expand
sequence withztransformF(z) =
Solution Usingthe binomial theorem, wehave
(
1 − 1 ) −3 (
=1+(−3) − 1 z z
(
1 − 1 ) −3
up to the term 1 z z4. Hence find the
z 3
(z −1) 3.
)
+ (−3)(−4)
2!
+ (−3)(−4)(−5) (
− 1 3! z
= 1 + 3 z + 6 z + 10
2 z + 15
3 z +··· 4
provided |z| > 1.Since
z 3 ( ) z −1 −3 (
F(z) =
(z −1) = = 1 − 1 ) −3
3 z z
(
− 1 ) 2
z
) 3
+ (−3)(−4)(−5)(−6)
4!
(
− 1 z
) 4
+···