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Design an ideal low-pass filter with N = 11 with frequency response,

Design an ideal low-pass filter with N = 11 with frequency response,

$$H_d\left(e^{j v}\right)= \begin{cases}1, & \text { for }-\frac{\pi}{2} \leq \omega \leq \frac{\pi}{2} \\ 0, & \text { for } \frac{\pi}{2} \leq|\omega| \leq \pi\end{cases}$$

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Solution:

$H_d(\omega)=\left\{\begin{array}{lc}\\ 1, & \text { for }-\frac{\pi}{2} \leq \omega \leq \frac{\pi}{2} \\\\ 0, & \text { for } \frac{\pi}{2} \leq|\omega| \leq \pi\\ \end{array}\right.$

The filter coefficients are given by,

\begin{aligned} h_d(n) & =\frac{1}{2 \pi} \int_{-\pi}^\pi H_d(\omega) e^{j \omega n} d \omega \\\\ & =\frac{1}{2 \pi} \int_{-\pi / 2}^{\pi / 2}(1) e^{j \omega n} d \omega=\frac{1}{2 \pi}\left[\frac{e^{j \omega n}}{j n}\right]_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \\\\ & =\frac{1}{n \pi}\left[\frac{e^{j \frac{n \pi}{2}}-e^{-j \frac{n \pi}{2}}}{2 j}\right] \\\\ & =\frac{1}{n \pi} \sin \frac{n \pi}{2} \text { for } n \neq 0\\ \end{aligned}\\

$\begin{array}{ll} h_d(n)=\frac{1}{2} \text { for } n=0 \text { [using L'Hospital rule] } \\\\ h_d(0)=\frac{1}{2}, & h_d(1)=\frac{1}{\pi} \sin \frac{\pi}{2}=\frac{1}{\pi}=h_d(-1) \\\\ h_d(2)=\frac{1}{2 \pi} \sin \pi=0=h_d(-2), & h_d(3)=\frac{1}{3 \pi} \sin \frac{3 \pi}{2}=-\frac{1}{3 \pi}=h_d(-3) \\\\ h_d(4)=\frac{1}{4 \pi} \sin 2 \pi=0=h_d(-4), & h_d(5)=\frac{1}{5 \pi} \sin \frac{5 \pi}{2}=\frac{1}{5 \pi}=h_d(-5) \end{array}$

Assuming the window function,

$w(n)= \begin{cases}1, & \text { for }-5 \leq n \leq 5 \\ 0, & \text { otherwise }\end{cases}\\$

We have,

$h(n)=h_d(n) \cdot w(n)=h_d(n)\\$

\begin{aligned} & h(0)=\frac{1}{2}, h(1)=\frac{1}{\pi}=h(-1), \quad h(2)=0=h(-2), h(3)=-\frac{1}{3 \pi}=h(-3), \\\\ & h(4)=0=h(-4), h(5)=\frac{1}{5 \pi}=h(-5)\\ \end{aligned}\\

\begin{aligned} H(z) & =z^{-(N-1) / 2}\left[h(0)+\sum_{n=1}^{(N-1) / 2} h(n)\left[z^{-n}+z^n\right]\right]=z^{-5}\left[h(0)+\sum_{n=1}^5 h(n)\left[z^{-n}+z^n\right]\right] \\\\ & =z^{-5}\left[h(0)+h(1)\left[z+z^{-1}\right]+h(3)\left[z^3+z^{-3}\right]+h(5)\left[z^5+z^{-5}\right]\right] \\\\ & =h(5)+h(3) z^{-2}+h(1) z^{-4}+h(0) z^{-5}+h(1) z^{-6}+h(3) z^{-8}+h(5) z^{-10} \\\\ & =\frac{1}{5 \pi}-\frac{1}{3 \pi} z^{-2}+\frac{1}{\pi} z^{-4}+\frac{1}{2} z^{-5}+\frac{1}{\pi} z^{-6}-\frac{1}{3 \pi} z^{-8}+\frac{1}{5 \pi} z^{-10}\\ \end{aligned}\\

Therefore, the coefficients of the realizable digital filter are:

\begin{aligned}\\ & h(0)=\frac{1}{5 \pi}=h(10), h(1)=0=h(9), h(2)=-\frac{1}{3 \pi}=h(8), \\\\ & h(3)=0=h(7), h(4)=\frac{1}{\pi}=h(6), h(5)=\frac{1}{2}\\ \end{aligned}\\