352 CHAPTER 9 • z -TRANSFORMS AND APPLICAT IONS( c) Use residues to find the solution.( d) Use convolution to find the solution.Solution
(a) The first step is to solve the equation Yh[n + 1) - ayh[n) = 0. The
trial solution is Yh[n) = r", and substitution produces r"+l -ar" = 0,
from which we obtain the characteristic equation r - a = O. The rootis r = a and the homogeneous solution is
The second step is to find a particular solution to the nonhomogeneous
equationYp{n + 1] -ayp[n ] = b.
Since the right-hand side is a constant we try yp[n ] = c, and substi-
tution produces c - ac = b. Hence c = 1 ~... The general solution
isy[n] = Yh[n ] + y,,[n]
=C1a " +--b.
1-aThen y[O) = c1 a^0 + 1 ~a. = Yo can be solved for c 1 = "~ 1 +Yo- Substi-
tuting this in the previous expression for y[n] yieldsy[n] = (-b- + Yo)a" - b
a-1 a - 1
=yoa n +--^1 a "b - --^1 b
a - 1 a - 1
a" -1
= yoa"+--b.a-1
(b) Start with the difference equation y [n + 1) = ay[n ) + b. Take the
z-transform of both sides
z
z(Y(z)-Yo)= aY(z ) + b-.z-1
Solve for Y(z) and get Y(z) = b(;~fJ°(!:.:~·. Then expand and obtain
z ab 1 b 1
Y(z ) =Yo(- )+-(- ) - - ( - ).
z - a a - 1 z-a a-1 z - 1