written 22 months ago by |
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.