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

696 Chapter 22 Difference equations and the z transform
Solution k =0 k =1
2x[2] −x[1] +x[0] = 0 2x[3] −x[2] +x[1] = −1
x[2] = 1 x[3] = − 3 2
k =2
2x[4] −x[3] +x[2] = −4
x[4] = − 13 4
Asthepreviousexamplesillustrate,todetermineauniquesolutiontoafirst-orderequation
requires one initial condition; for a second-order equation, two initial values are
required.
Engineeringapplication22.5
Numericalsolutionoftheoutputfromadigitallow-passfilter
Recall from Engineering application 22.2 that a low-pass filter is a filter that is designedtoattenuatehighfrequencysignals.Asimpledifferenceequationforadigital
filter isgiven by:
y[n] −ay[n −1] =x[n]
Ifaispositive thenthe filter isalow-passfilter. We will choosea = 0.5 and so
y[n] = 0.5y[n −1] +x[n]
Letusexaminetheresponseofthisfiltertoaunitstepinputappliedatn = 0,thatis
{ 0 n<0
x[n] = n ∈ Z
1 otherwise
Assume the output of the filter is zero prior to the application of the step input, that
isy[n] = 0 forn −1.Fromthe difference equation,wefind
Similarly,
y[0] = 0.5y[−1] +x[0]
= 0.5(0) +1
= 1
y[1] = 0.5y[0] +x[1]
= 0.5(1) +1
= 1.5
andsoon.Whennumericallysolvingadifferenceequation,itisoftenusefultoform
a table with intermediateresults. Table 22.1 shows suchatable.
Figure 22.16 shows the input and output sequences superimposed on the same
graph. The sequence values have been joined to illustrate their trends. Note that the
input signal reaches its final value immediately whereas the output signal takes several
sample intervals to reach its final value. Engineers often refer to this process as
‘smoothing’theinputsignal.WewillseeinChapter23thatrapidlychangingsignals