Solution:

The transfer function of the filter is given by,

$\begin{aligned}\\ & H(z)= \sum_{n-0}^{N-1} h(n) z^{-n} \\\\ &= h(0)+h(1) z^{-1}+h(2) z^{-2}+h(3) z^{-3}+h(4) z^{-4}+h(5) z^{-5}+h(6) z^{-6} \\\\ &+h(7) z^{-7}+h(8) z^{-8} \\\\ & \alpha=\frac{N-1}{2}=\frac{9-1}{2}=4 .\\ \end{aligned}\\ $

$ \begin{aligned}\\ H(z)= & z^{-4}\left[h(0) z^4+h(1) z^3+h(2) z^2+h(3) z^1+h(4) z^0+h(5) z^{-1}+h(6) z^{-2}\right. \\\\ & …