$i) h(n)={1,\frac{-1}{2}}$ $ii) H(z)=\frac{(z^{-1}-a)}{(1-az^{-1} )}$

Solution:

$h(n)=1, \frac{-1}{2}$

Taking Z transform

$H(z)=\sum_{n=0}^{1} h(n) z^{-n}$

$=1-\frac{1}{2} z^{-1}$

$Put Z=e^{j w}$

$H(w)=1-\frac{1}{2} e^{-j w}$

$=1-\frac{1}{2}[\cos w-j \sin w]$

$H(w)=1-\frac{1}{2} \cos w-j \sin \frac{w}{2}$

From pass band it is observed that given filter is High pass filter.

$ii) H(z)=\frac{(z^{-1}-a)}{(1-az^{-1})}$

$Put z=e^{jw}$

$H(w)=\frac{(e^{-jw}-a)}{(1-ae^{-jw} )}$

$=\frac{((cos⁡w-jsin w)-a)}{(1-a(cos⁡w-jsin w))}$

$=\frac{((cos⁡w-a)-j sin⁡w)}{((-a cos⁡w)+ja sin⁡w )}$

Assuming a=1

$H(w)=\frac{((cos⁡ω-1)-j sin⁡w)}{((1-cos⁡w)+jsin w)}$

All the value of w gives same response, Hence it is all pass filter.