- 1 • THE Z-TRANSFORM 341
Solution This follows trivially from Equation (9-1) X(z) = 3[xn) =
L:=O XnZ-n = 1 + L::'=l Qz- n = 1.
- EXAMPLE 9.2 The z-transform of the unit-step sequence
{lforn>O
Xn = u[n) = 0 for n < 0
is X(z) = ,.:_ 1.
Solutio n From Equation (9-1) X (z)
L::=-0 (z-
1
t = i-~- 1 = ,.:.1."'oo Lm=oXnZ -n "'oo Lm=Oz -n
- EXAMPLE 9.3 The z-transform of the sequence Xr. = bn is X (z) = .:.b.
- EXAMPLE 9.4 The z- transform of the exponential sequence Xn = e'"' is
X (z) = ,.: ••.
Solution From the definition X(z) = L::=oXnz- n
"°' 00 (••n) "°'00 (••)n l z
L....n=O zn = L....n=O 7 = 1-s! , = ~·
9.1.2 Properties of the z-transform
Given that 3[xn] = .3[x[n)] = X(z) and 3[Yn] = 3[y(nJI = Y(z):
(i) Linearity. 3[c1Xn + C2Yn) = 3[c1x(n] + c2y[nJI = c1X(z) + c2Y(z).
(ii) Delay Shift. 3 [x [n - N )u[n - NJJ = X(z)z- N.
(iii) Advance Shift. 3[x[n +NJ] = zN (X(z) - x[OJ - x[l]z-^1 - x[2Jz-^2 -
... - x (N -l )z-N+^1 ), or
.3[xn+N) = zN(X (z) - Xo -X1z-^1 - x2z -^2 - ... - XN- l Z-N+l).
(iv) Multiplication by n. 3[nxn) = 3 [nx[n)J = - z:fzX(z).• EXAMPLE 9.5 The z-transform of the sequence x,.. = cos(an) is given by
X( ) _ z(z- cos(a))
Z - z•-2i cos(a) + 1 ·