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

720 Chapter 22 Difference equations and the z transform
Dividing both sides byzgives
Y(z)
z
=
2
(z −2)(z −3)
Expressing the r.h.s.inpartialfractions yields
Y(z)
z
Y(z)=
= 2
z −3 − 2
z −2
2z
z −3 − 2z
z −2
Inverting gives the solution tothe difference equation:
y[k] = 2(3 k ) −2(2 k )
EXERCISES22.12
1 Useztransformsto solvethe following difference
equations:
(a) x[k +1] −3x[k] = −6,x[0] = 1
(b) 2x[k +1] −x[k] = 2 k ,x[0] = 2
(c) x[k +1] +x[k] = 2k +1,x[0] = 0
(d) x[k +2] −8x[k +1] +16x[k] = 0,
x[0] = 10,x[1] = 20
(e) x[k +2] −x[k] = 0,x[0] = 0,x[1] = 1
2 Solvethe difference equation
x[k +2] −3x[k +1] +2x[k] = δ[k]
subjectto the conditionsx[0] =x[1] = 0.
3 Solvethe difference equation
y[k +2] +3y[k +1] +2y[k] = 0
subjectto the conditionsy[0] = 0,y[1] = 1.
4 Solvethe difference equation
x[k +2] −7x[k +1] +12x[k] =k
subjectto the conditionsx[0] = 1,x[1] = 1.
Solutions
1 (a) x[k] = 3 −2(3 k )
2 k
(b)
3 + 5(1/2)k
3
(c) k (d) 10(4 k ) −5(k4 k )
(e)
u[k] − (−1) k
2
May be expressed asx[k] =
{ 0 k even
1 kodd
2 (2 k−1 −1)u[k −1]
3 (−1) k − (−2) k
4
k
6 + 5 11
u[k] +
36 4 (3)k − 17
9 (4)k
REVIEWEXERCISES22
1 State
(i) the order
(ii) the independentvariable
(iii) the dependentvariable
(iv) whether linearornon-linear
foreachofthe following equations:
(a) x[n] +x[n −2] = 6
(b) y[k+1]+ky[k−1]−k=0
(c) (y[z] +1)y[z +1] =z 2
(d) z[n] −z[n −1] =n 2 z[n −2]