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

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

Marks : 05

Year : MAY 2015

$(i) h[n]=\Bigg{ 1, - \dfrac 12 \Bigg} $ for given function, $H(z) = 1 – 1/2.Z^{-1} \ = \dfrac {Z-1}Z$ This is ALL ZERO system. Hence, this is $(ii) H[z]=\dfrac {z^{-1}-a}{1-az^{-1}}$ 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.