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

22.8 Sampling a continuous signal 703
Wecanapplytheztransformdirectlytoacontinuousfunction, f (t),ifweregardthe
function as having been sampled at discrete intervals of time. Consider the following
example.
Example22.13 (a) Find theztransformof f (t) = e −t sampled att = 0,0.1,0.2,....
(b) Find theztransform of f (t) = e −t sampled att = 0,0.01,0.02,....
(c) Express the sequences obtained in(a)and (b) as series of weighted impulses.
Solution (a) The sequence of sampled values is
that is,
1,e −0.1 ,e −0.2 ,...
e −0.1k k ∈ NandT = 0.1
From Table 22.2 wefind theztransform ofthis sequence is
(b) The sequence of sampled values is
that is,
1,e −0.01 ,e −0.02 ,...
z
z −e −0.1.
e −0.01k k ∈ N and T = 0.01
z
From Table 22.2 we find the z transform of this sequence is
z −e−0.01. We note
that modifying the sampling interval,T, alters theztransform even though we are
dealing with the same function f (t).
(c) WhenT = 0.1 we have, from Equation (22.8),
∞∑
f ∗ (t) = e −0.1k δ(t −0.1k)
k=0
=1δ(t)+e −0.1 δ(t −0.1)+e −0.2 δ(t −0.2)+···
Notethatanadvantageofexpressingthesequenceasaseriesofweightedimpulsesis
thatinformationconcerningthetimeofoccurrenceofaparticularvalueiscontained
inthe corresponding δ term.
WhenT = 0.01, we have
∞∑
f ∗ (t) = e −0.01k δ(t −0.01k)
k=0
= 1δ(t) +e −0.01 δ(t −0.01) +e −0.02 δ(t −0.02) + ···
Example22.14 (a) Findtheztransform ofthe continuousfunction f (t) = cos3t sampledatt =kT,
k∈N.
(b) WritedownthefirstfourtermsofthesampledsequencewhenT = 0.2,andexpress
the sequenceasaseries ofweighted impulses.