Signals and Systems - Electrical Engineering

556 C H A P T E R 9: The Z-Transform

Just as with the Laplace transform, the steady-state response of a difference equation

y[n]+

∑N

k= 1

aky[n−k]=

∑M

m= 0

bmx[n−m]

is due to simple poles ofY(z)on the unit circle. Simple or multiple poles inside the unit circle give a transient,

while multiple poles on the unit circle or poles outside the unit circle create an increasing response.

nExample 9.18

Solve the difference equation

y[n]=y[n−1]−0.25y[n−2]+x[n] n≥ 0

with zero initial conditions andx[n]=u[n].

Solution

The Z-transform of the terms of the difference equation gives

Y(z)=

X(z)

1 −z−^1 +0.25z−^2

=

1

( 1 −z−^1 )( 1 −z−^1 +0.25z−^2 )

=

z^3

(z− 1 )(z^2 −z+0.25)

|z|> 1

Y(z)has three zeros atz=0, a pole atz=1, and a double pole atz=0.5. The partial fraction

expansion ofY(z)is of the form

Y(z)=

A

1 −z−^1

+

B( 1 −0.5z−^1 )+Cz−^1

( 1 −0.5z−^1 )^2

(9.39)

where the terms of the expansion can be found in the Z-transforms table. Within some constants,

the complete response is

y[n]=Au[n]+[B(0.5)n+Cn(0.5)n]u[n]

The steady state is thenyss[n]=A(corresponding to the pole on the unit circlez=1) since the

other two terms, corresponding to the double polez=0.5 inside the unit circle, make up the

transient response. The value ofAis obtained as

A=Y(z)( 1 −z−^1 )

∣

∣z− (^1) = 1 = 4
To find the complete responsey[n] we find the constants in Equation (9.39). Notice in Equa-
tion (9.39) that the expansion term corresponding to the double polez=0.5 has as numerator a
first-order polynomial, with constantsBandCto be determined, to ensure that the term is proper