$(i) h[n]={1,-\dfrac 12} \ (ii) H[z]=\dfrac {z^{-1}-a}{1-az^{-1}}$

Mumbai University > EXTC > Sem 6 > Discrete Time Signal Processing

Marks : 05

Year : MAY 2015

$$(i)h[n]={1,-\dfrac 12}$$

for given function,

$H(z) = 1 – 1/2.Z^{-1} \\ = \dfrac {Z-1}Z $

This is ALL ZERO system.

Hence, this is High pass filter.

$(ii) H[z]=\dfrac {z^{-1}-a}{1-az^{-1}} $

Given $H (z) = \dfrac {Z^{-1}-a}{1-a.Z^{-1}} \\ H (z) = \dfrac {1-a.z}{ z-a} =\dfrac {a.z+1}{z-a}\\ H (z) =-a \dfrac {z - \dfrac 1a}{z-a} $

Pole; z=a , Zero; z=1/a

Here, poles are inverse of zeros, hence filter is ALL PASS IIR FILTER.