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 $Q(\omega) \rightarrow$ Phase response.

We define phase delay \& group delay of a filter as,

$ T_p=\frac{-\theta(\omega)}{\omega} \quad \& \quad T_g=\frac{-d \theta(\omega)}{\partial \omega}...(4)\\ $

For FIR filters with linear phase we can define $Q(\omega)=-\alpha \omega$

$ -\pi \leqslant \omega \leqslant \pi \quad........(5) $

where $\alpha \rightarrow$ constant phase delay.

Substituting $\mathrm{eq}^n$ (5) in (4) we get.

$ T_p=\frac{\alpha \omega}{\omega} \quad \&, T_g=\alpha \frac{d \omega}{\partial \omega} $

$ T_p=T_g=\alpha....(6) $

This equation shows $\alpha$ is independent of frequency.