Signals and Systems - Electrical Engineering

9.5 One-Sided Z-Transform Inverse 563

so thatB=−A=−2. The inverse is then found to be

x[n]= 2 u[n]

︸︷︷︸

causal

+

[

− 2 (n+^1 )u[−n−1]+ 2 (n+^2 )nu[−n−1]

]

︸ ︷︷ ︸

anti-causal n

nExample 9.22

Find all the possible impulse responses connected with the following transfer function of a discrete

filter having poles atz=1 andz=0.5:

H(z)=

1 + 2 z−^1 +z−^2

( 1 −0.5z−^1 )( 1 −z−^1 )

determine the cases when the filter is BIBO stable.

Solution

As a function ofz−^1 this function is not proper since both its numerator and denominator are of

degree 2. After division we have the following partial fraction expansion:

H(z)=B 0 +

B 1

1 −0.5z−^1

+

B 2

1 −z−^1

There are three possible regions of convergence that can be attached toH(z):

n R 1 :|z|>1 so that the corresponding impulse responseh 1 [n]=Z−^1 [H(z)] is causal with a

general form

h 1 [n]=B 0 δ[n]+[B 1 (0.5)n+B 2 ]u[n]

The pole atz=1 makes this filter unstable, as its impulse response is not absolutely summable.

n R 2 :|z|<0.5 for which the corresponding impulse responseh 2 [n]=Z−^1 [H(z)] is anti-causal

with a general form

h 2 [n]=B 0 δ[n]−(B 1 (0.5)n+B 2 )u[−n−1]

The region of convergenceR 2 does not include the unit circle and so the impulse response is

not absolutely summable(H(z)cannot be defined atz=1 because it is not in the region of

convergenceR 2 ). The impulse responseh 2 [n] grows asnbecomes smaller and negative.

n R 3 : 0.5<|z|<1, which gives a two-sided impulse responseh 2 [n]=Z−^1 [H(z)] of general

form

h 3 [n]=B 0 δ[n]+B 1 ((0.5)nu[n]

︸ ︷︷ ︸

causal

−B 2 u[−n−1]

︸ ︷︷ ︸

anti-causal

Again, this filter is not stable. n