Signals and Systems - Electrical Engineering

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

Letting

A(z)= 1 +z−^1 − 4 z−^2 − 4 z−^3 =( 1 +z−^1 )( 1 + 2 z−^1 )( 1 − 2 z−^1 )

we can write

Y(z)= 3

X(z)

A(z)

+

− 1 + 4 z−^1 + 4 z−^2

A(z)

|z|> 2 (9.40)

To determine whether the steady-state response exists or not let us first consider the stability of

the system associated with the given difference equation. The transfer functionH(z)of the system

is computed by letting the initial conditions be zero (this makes the second term on the right

of the above equation zero) so that we can get the ratio of the Z-transform of the output to the

Z-transform of the input. If we do that then

H(z)=

Y(z)

X(z)

=

3

A(z)

Since the poles ofH(z)are the zeros ofA(z), which arez=−1,z=−2, andz=2, then the impulse

responseh[n]=Z−^1 [H(z)] will not be absolutely summable, as required by the BIBO stability,

because the poles ofH(z)are on and outside the unit circle. Indeed, a general form of the impulse

response is

h[n]=[C+D( 2 )n+E(− 2 )n]u[n]

whereC,D, andEare constants that can be found by doing a partial fraction expansion ofH(z).

Thus,h[n] will grow asnincreases and it would not be absolutely summable—that is, the system

is not BIBO stable.

Since the system is unstable, we expect the total response to grow asnincreases. Let us see how we

can justify this. The partial fraction expansion ofY(z), after replacingX(z)in Equation (9.40), is

given by

Y(z)=

2 + 5 z−^1 − 4 z−^3

( 1 −z−^1 )( 1 +z−^1 )( 1 + 2 z−^1 )( 1 − 2 z−^1 )

=

B 1

1 −z−^1

+

B 2

1 +z−^1

+

B 3

1 + 2 z−^1

+

B 4

1 − 2 z−^1

B 1 =Y(z)( 1 −z−^1 )|z− (^1) = 1 =−

1

2

B 2 =Y(z)( 1 +z−^1 )|z− (^1) =− 1 =−

1

6

B 3 =Y(z)( 1 + 2 z−^1 )|z− (^1) =− 1 / 2 = 0
B 4 =Y(z)( 1 − 2 z−^1 )|z− (^1) = 1 / 2 =

8

3