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

714 Chapter 22 Difference equations and the z transform
Example22.23 Find the sequence whoseztransformis
Solution
1
z −1 = 1 z
z z
z −1 =z−1
z −1
From Table 22.2 wehave
Z{u[k]} =
z
z −1
So from the second shift property, wehave
Z{u[k −1]} =z −1 z
z −1
1
z −1 .
The required sequence isthereforeu[k −1].
1
Example22.24 Findthe sequencewhoseztransformis
z 2 (z −1) 2.
1
Solution The expression does not appear in the table of transforms, but we observe
z 2 (z −1)
2
that
1
z 2 (z −1) = 1 z
2 z 3 (z −1) 2
z
and does appear. Itfollows from Table 22.2 that
(z −1)
2
z
Z{k} =
(z −1) 2
From the second shift property,z −3 z
istheztransform of (k −3)u[k −3].
(z −1)
2
22.10.4 Thecomplextranslationtheorem
Z{e −bk f[k]} =F(e b z) whereF(z)istheztransform of f[k].
Example22.25 Given thattheztransform ofcos(ak) is
z(z −cosa)
z 2 −2zcosa+1
find theztransform of e −2k cos(ak).
Solution Since b = 2, the complex translation theorem states that we replace z by e 2 z in the z
transformF(z).
F(e 2 z) =
e2 z(e 2 z −cosa)
e 4 z 2 −2e 2 zcosa+1
istherefore the required transform.