Page 215 - Schaum's Outline of Theory and Problems of Signals and Systems

204 THE Z-TRANSFORM AND DISCRETE-TIME LTI SYSTEMS [CHAP. 4

Thus,

z z
y(z) = ay-\ - +K
z-a (z-a)(z-b)
Using partial-fraction expansion, we obtain
z K z z
Y,(z) =ay-,- +- -
z-a b-~(~z-b z-a
Taking the inverse z-transform of Y,(z), we get

b a
Y[n] =ay- ,anu[n] + K-bnu[n] - K-anu[n]
b-a b-a

which is the same as Eq. (2.158).

4.38. For each of the following difference equations and associated input and initial
conditions, determine the output y[n I:

(a) y[nl - fy[n - 11 =x[nl, with x[n] = (:In, y[- 11 = 1
(b) 3y[n] - 4y[n - 11 + y[n - 21 =An], with x[n] = (i)", y[ - 11 = 1, y[- 21 = 2

Taking the unilateral z-transform of the given difference equation, we get

Y,(z) - +{z-'Y,(z) +y[- I]} =X,(z)
Substituting y[- 11 = 1 and X,(z) into the above expression, we get

Thus,

Hence,

y[n) = 7(;)"+' - 2(f)n n2 -1
z
(b) x[n] ++ X,(z) = ---7
2-2
Taking the unilateral z-transform of the given difference equation, we obtain

3Y,(z) - 4{z- '5 ( z)+Y[-I]}+{~- 2Y, ( z) +z-'y[- 11 +~[-2]} =XI(Z)