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

706 Chapter 22 Difference equations and the z transform
22.9.1 Mappingthesplanetothezplane
When designing an engineering system it is often useful to consider ansplane representationofthesystem.Thecharacteristicsofasystemcanbequicklyidentifiedbythe
positionsofthepolesandzerosaswesawinChapter21.Engineerswilloftenmodifythe
system characteristics by introducing new poles and zeros or by changing the positions
ofexistingones.Unfortunately,itisnotconvenienttousethesplanetoanalysediscrete
systems. For a sampled signalEquation (22.9) yields
∞∑
F ∗ (s)=L{f ∗ (t)} = f[k]e −skT
k=0
ThecontinuoussignalsandsystemsthatwereanalysedinChapter21hadLaplacetransformsthatweresimpleratiosofpolynomials
ins.Thiswasoneofthemainreasonsfor
usingLaplacetransformstosolvedifferentialequations;theproblemwasreducedtoone
ofreasonablystraightforwardalgebraicmanipulation.HerewehaveaLaplacetransform
thatisverycomplicated.Infactitcanhaveaninfinitenumberofpolesandzeros.Tosee
this consider the following example.
( ) πt
Example22.16 The continuous signal f (t) = cos is sampled at 1 second intervals starting from
2
t=0.
(a) Find the Laplace transformof the sampled signal f ∗ (t).
(b) Show thatF ∗ (s)has an infinity of poles.
(c) Find theztransformof the sampled signaland show thatthis has justtwo poles.
( ) πt
Solution (a) The continuous signal f (t) = cos sampled at 1 second intervals gives rise to
2
the sequence 1,0, −1,0,1,0, −1,.... Consequently, fromEquation (22.9)
∞∑
∞∑
F ∗ (s)=L{f ∗ (t)} = f[k]e −skT = f[k]e −sk sinceT = 1
thatis,
k=0
k=0
F ∗ (s)=1+0−e −2s +0+e −4s +0−e −6s +···
This is a geometric progression with common ratio −e −2s and hence its sum to
infinity is
thatis,
1
1 − (−e −2s )
F ∗ 1
(s) =
1 +e −2s
(b) PolesofF ∗ (s)willoccurwhen1 +e −2s = 0.Writings = σ +jω weseethatpoles
will occur when e −2(σ+jω) = −1. Since −1 can be written as e j(2n−1)π ,n ∈ Z (see
Chapter 9), wesee thatpoles will occur when
e −2σ−2jω = e j(2n−1)π