Signals and Systems - Electrical Engineering

9.5 One-Sided Z-Transform Inverse 553

To obtain a closed-form solution, we use the Z-transform. Taking the one-sided Z-transform of the
two sides of the equation, we get

Z(y[n])=Z(ay[n−1])+Z[x[n])

Y(z)=a(z−^1 Y(z)+y[−1])+X(z)

Solving forY(z)in the above equation, we obtain

Y(z)=

X(z)

1 −az−^1

+

ay[−1]

1 −az−^1

(9.37)

where the first term depends exclusively on the input and the second depends exclusively on the
initial condition. If the inputx[n] and the initial conditiony[−1] are given, we can then find the
inverse Z-transform to obtain the complete solutiony[n] of the form

y[n]=yzs[n]+yzi[n]

where the zero-state responseyzs[n] is due exclusively to the inputx[n] and zero initial conditions,
and the zero-input responseyzi[n] is the response due to the initial conditiony[−1] with zero
input.

In this simple case we can obtain the complete solution for any inputx[n] and any initial condition
y[−1]. Indeed, by expressing 1/( 1 −az−^1 )as its Z-transform sum—that is,

1

1 −az−^1

=

∑∞

k= 0

akz−k

Equation (9.37) becomes

Y(z)=

∑∞

k= 0

X(z)akz−k+ay[−1]

∑∞

k= 0

akz−k

=X(z)+aX(z)z−^1 +a^2 X(z)z−^2 +···+ay[−1]( 1 +az−^1 +a^2 z−^2 +···)

Using the time-shift property we then get the complete solution,

y[n]=x[n]+ax[n−1]+a^2 x[n−2]+···+y[−1]a( 1 +aδ[n−1]+a^2 δ[n−2]+···)

=

∑∞

k= 0

akx[n−k]+ay[−1]

∑∞

k= 0

akδ[n−k] (9.38)

for any inputx[n], initial conditiony[−1], anda.