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

718 Chapter 22 Difference equations and the z transform
we have
F(z)=1+ 3 z + 6 z 2 + 10
z 3 + 15
z 4 +···
ThusF(z) can beinverted directly togive
f[k] = 1,3,6,10,15,... thatis,f[k] =
(k +2)(k +1)
2
k 0
EXERCISES22.11
1 Find the inverseztransformsofthe following:
4z z 2 +2z
(a) (b)
z −4 3z 2 −4z−7
z +1 2z 3 +z
(c)
(z −3)z 2 (d)
(z−3) 2 (z−1)
2 Find the inverseztransformsof
(a)
2z
(z −2)(z −3)
(c) 1− 2 z +
z
(z −3)(z −4)
(b)
(d)
3 Express
(e)
ez
(ez −1) 2 4 If
z 2
(z 2 − 1 9 )
2z 2
(z −1)(z −0.905)
F(z) =
(z +1)(2z −3)(z −2)
z 3
in partial fractions andhence obtain its inversez
transform.
F(z) =
find f[k].
10z
(z −1)(z −2)
Solutions
1 (a) 4(4 k )
(b)
13(7/3) k
30
− (−1)k
10
(c) u[k −2] 4(3)k−2 − δ[k −2]
3
May be written as:
(d)
3 k−2 u[k −2] +3 k−3 u[k −3]
19k(3 k )
6
+ 3u[k]
4
2 (a) −2(2 k ) +2(3 k )
+ 5(3k )
4
(b) e −k k
(c) δ[k] −2δ[k −1] +4 k −3 k
(1/3) k + (−1/3) k
(d)
2
(e) 21.05u[k] −19.05(0.905) k
3 2 − 5 z − 1 z 2 + 6 z 3
f[0]=2,f[1]=−5,f[2]=−1,
f[3]=6,f[k]=0 k4
4 10(2 k −u[k])
22.12 THEzTRANSFORMANDDIFFERENCEEQUATIONS
InChapter21wesawhowusefultheLaplacetransformcanbeinthesolutionoflinear,
constantcoefficient,ordinarydifferentialequations.Similarlytheztransformhasarole
toplayinthe solutionofdifference equations.