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

22.6 Numerical solution of difference equations 697
Table22.1
Numerical solution of a difference
equation.
n x[n] y[n −1] y[n]
−1 0 0 0
0 1 0 1
1 1 1 1.5
2 1 1.5 1.75
3 1 1.75 1.88
4 1 1.88 1.94
5 1 1.94 1.97
6 1 1.97 1.98
7 1 1.98 1.99
8 1 1.99 2.00
9 1 2.00 2.00
10 1 2.00 2.00
2
1
0 12345678910
Figure22.16
Input and output sequences forthe
low-pass filter.
y[n]
x[n]
n
tend to be richer in high frequencies than those that change more slowly. The effect
of the low-pass filter is to filter out these high frequencies and so the output from
the filter changes more slowly than the input. In Chapter 23 we will also examine a
continuous low-pass filter which does the same jobforcontinuous signals.
Numerical solution of difference equations is often the only feasible method of obtainingasolutionformanypracticalengineeringsystems.Thisisbecausetheinputterms
are usually obtained as a result of sampling an input signal and so cannot be expressed
analytically. However, in some cases it is possible to express the input analytically and
then an analytical solution to the difference equation may be feasible. While analytical
methods analogous to those applied to differential equations are available,ztransform
techniquesaremorepopularwithengineersandsotheseareintroducedinthefollowing
sections.
EXERCISES22.6
1 Given
x[n+2]+x[n+1]−x[n]=2
x[0] = 3 x[1] = 5
findx[2],x[3],x[4]andx[5].
2 If
z[n]z[n −1] =n 2 z[1] = 7
findz[2],z[3]andz[4].
3 Determinex[2]andx[3]given
(a) 2x[n +2] −5x[n +1] = 4n,x[1] = 2
(b) 6x[n] −x[n −1] +2x[n −2]
=n 2 −n,x[0]=1,x[1]=2
(c) 3x[n −1] +x[n −2] −9x[n −3]
= (n−1) 2 ,x[0] =3,x[1] =2
4 Calculate the firstfive termsofthe following
difference equations with the given initialconditions:
(a) x[n +1] −x[n] = 2,x[0] = 3
(b) x[n +2] +x[n +1] −x[n] = 4,
x[0] = 5,x[1] = 7
(c) x[p+2]−x[p+1]+2x[p] =2,
x[0] = 1,x[1] = 1