One-Sided Z-Transform Inverse 557

9.5 One-Sided Z-Transform Inverse 557

rational. That term equals

B(1 − 0.5z −1 ) + Cz −1

B Cz −1

( 1 − 0.5z −1 ) 2 = 1 − 0.5z −1 + ( 1 − 0.5z −1 ) 2

which is very similar to the expansion for multiple poles in the inverse Laplace transform. Once we find the values of B and C, the inverse Z-transforms are obtained from the Z-transforms table. A simple method to obtain the coefficients B and C is to first obtain C by multiplying the two sides

of Equation (9.39) by (1 − 0.5z −1 ) 2 to get Y(z)(1 − 0.5z −1 ) 2 = B(1 − 0.5z −1 ) + Cz −1

and then letting z −1 = 2 on both sides to find that

−1 ) Y(z)(1 − 0.5z 2

The B value is then obtained by choosing a value for z −1 that is different from 1 or 0.5 to compute Y(z). For instance, assume you choose z −1 = 0 and that you have found A and C, then

Y(z)| z −1 =0 =A+B=1

from which B = −3. The complete response is then

y[n] = 4u[n] − 3(0.5) n − 0.5n(0.5) u[n]

■ Example 9.19

Find the complete response of the difference equation y[n] + y[n − 1] − 4y[n − 2] − 4y[n − 3] = 3x[n]

n≥0

y[−1] = 1 y[−2] = y[−3] = 0 x[n] = u[n]

Determine if the discrete-time system corresponding to this difference equation is BIBO stable or not, and the effect this has in the steady-state response.

Solution

Using the time-shifting and linearity properties of the Z-transform, and replacing the initial conditions, we get

Y(z)[1 + z −1

−2 − 4z −3 − 4z ] = 3X(z) + [−1 + 4z −1

+ 4z −2 ]

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

Letting A(z) = 1 + z −1 − 4z −2 − 4z −3 = (1 + z −1 )( 1 + 2z −1 )( 1 − 2z −1 ) we can write

X(z)

−1 + 4z −1 + 4z −2

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

Y(z) = 3

A(z) +

|z| > 2

A(z)

the system associated with the given difference equation. The transfer function H(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

Y(z)

H(z) =

X(z)

A(z)

Since the poles of H(z) are the zeros of A(z), which are z = −1, z = −2, and z = 2, then the impulse response h[n] = Z −1 [H(z)] will not be absolutely summable, as required by the BIBO stability, because the poles of H(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) ]u[n]

where C, D, and E are constants that can be found by doing a partial fraction expansion of H(z). Thus, h[n] will grow as n increases 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 as n increases. Let us see how we can justify this. The partial fraction expansion of Y(z), after replacing X(z) in Equation (9.40), is given by

2 + 5z −1 − 4z −3 Y(z) = ( 1−z −1 )( 1+z −1 )( 1 + 2z −1 )( 1 − 2z −1 )

B 1 = Y(z)(1 − z ) | z −1 =1 =−

B 2 = Y(z)(1 + z −1 ) | z −1 =−1 =−

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

B 4 = Y(z)(1 − 2z −1 ) | z −1 =1/2 =