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

716 Chapter 22 Difference equations and the z transform
and write
{ } z
Z −1 z −2
= 2 k
where Z −1 denotes the inverseztransform.Also,
3
z −2 = 3 z
zz −2
= 3z −1 z
z −2
Using the second shift property wesee that
Z{3(2) k−1 u[k −1]} = 3 z
so that
{ } z +3
Z −1 z −2
z
z −2
{ } { z
= Z −1 + Z −1
z −2
3z −1
}
z
z −2
= 2 k +3(2) k−1 u[k −1]
{ 1 k=0
=
2 k +3(2) k−1 k=1,2,...
{ 1 k=0
=
2(2) k−1 +3(2) k−1 k = 1,2,...
{ 1 k=0
=
5(2) k−1 k =1,2,...
Often it is necessary to split a complicated expression into several simpler ones, using
partialfractions,beforeinversioncanbecarriedout.Also,ifweexaminetheztransform
table, Table 22.2, wenoticethat nearlyall ofthe entrieshave azterminthe numerator.
For this reason it is convenient to divide a complicated expression byzbefore splitting
it into partialfractions. We illustrate this techniqueby meansofanexample.
Example22.27 Findthe sequencewhoseztransformis
F(z) =
2z 2 −z
(z −5)(z +4)
Solution The first stage istodivide the expression forF(z)byz. We have
F(z) =
F(z) =
F(z)
z
=
2z 2 −z
(z −5)(z +4)
z(2z −1)
(z −5)(z +4)
2z−1
(z −5)(z +4)