Solution:

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

where $h[n] \rightarrow$ Impulse response of filter. The Fourier transform of h[n] is,

$ H(\omega)=\sum_{n=0}^{N-1} h[n] e^{-j \omega n} \quad-\text...... { (2) } $

which is periodic in frequency with period $2 \pi$

$ H(\omega)=\pm|H(\omega)| e^{j Q(\omega)}...(3) $

Where $|H(w)| \rightarrow$ Magnitude response $Q(\omega) …