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Design 6th linear phase LPF with cutoff frequency $\pi/2$ using blackman window.
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Let 'M' be the FIR filter length.

Order of FIR filter $=\mathrm{M}-1=6$

$\therefore \mathrm{M}=7$

Given, Cut-off frequency $\omega_{c}=\frac{\pi}{2} ;$ Gain $C=1$

Now, $\alpha=\frac{M-1}{2}=\frac{7-1}{2}=3$

Let Desired Frequency Response of LPF be

$H_{d}\left(e^{j w}\right)=\left\{\begin{array}{cc}{C e^{-j \alpha \omega}} & {|\omega| \leq \omega_{c}} \\ {0} & {\omega_{c} \leq|\omega| \leq \pi}\end{array}\right.$

$\therefore H_{d}\left(e^{j w}\right)=\left\{\begin{array}{cc}{1 e^{-j 3 \omega}} & {|\omega| \leq \pi / 2} \\ {0} & {\pi / 2 \leq|\omega| \leq \pi}\end{array}\right.$

Desired Impulse Response

$h_{d}(n)=\frac{1}{2 \pi} \int_{-\pi}^{\pi} H_{d}(\omega) e^{j \omega n} d \omega$

$=\frac{1}{2 \pi}\left\{\int_{-\pi}^{-\pi / 2} 0+\int_{-\pi / 2}^{\pi / 2} e^{-3 j \omega} e^{j \omega n} d \omega+\int_{\pi / 2}^{\pi} 0\right\}$

$=\frac{1}{2 \pi} \int_{-\pi / 2}^{\pi / 2} e^{(n-3) j \omega} d \omega \cdots(1)$

$=\frac{1}{2 \pi}\left[\frac{e^{(n-3) j \omega}}{(n-3) j}\right]_{-\pi / 2}^{\pi / 2}$

$=\frac{1}{2 \pi} \times \frac{1}{(n-3) j}\left[e^{(n-3) j \pi / 2}-e^{-(n-3) j \pi / 2}\right]$

$=\frac{1}{(n-3) \pi} \sin \frac{(n-3) \pi}{2}(\mathrm{n} \neq 3)$

Put $n=3$ in $(1), h_{d}(3)=\frac{1}{2 \pi} \int_{-\pi / 2}^{\pi / 2} e^{0} d \omega$

$=\frac{1}{2 \pi}[\omega]_{-\pi / 2}^{\pi / 2}$

$=\frac{1}{2 \pi}\left[\frac{\pi}{2}-\left(\frac{-\pi}{2}\right)\right]$

$=0.5$

$\therefore$ Desired Impulse Response

$h_{d}(n)=\left\{\begin{array}{cl}{\frac{\sin (n-3) \pi / 2}{(n-3) \pi}} & {n \neq 3} \\ {0.5} & {n=3}\end{array}\right.$

For linear phase FIR filters,

$h_{d}(n)=h_{d}(M-1-n)=h_{d}(7-1-n)=h_{d}(6-n)$

$\therefore h_{d}(6)=h_{d}(0) ; h_{d}(5)=h_{d}(1) ; h_{d}(4)=h_{d}(2)$

$h_{d}(n)$ is of infinite duration. To make it of finite duration we multiply it with Blackman window function

$W(n)=\left\{\begin{array}{cc}{0.42-0.5 \cos \frac{\pi n}{\alpha}+0.08 \cos \frac{2 \pi n}{\alpha}} & {0 \leq n \leq 6} \\ {0} & {\text { else }}\end{array}\right.$

$\therefore$ Finite Impulse Response $h(n)=h_{d}(n) \times W(n)$

$\therefore$ The filter coefficients are

$\begin{array}{|c|c|c|c|}\hline \mathbf{n} & {h_{d}(n)} & {W(n)} & {h(n)} \\ \hline 0 & {-0.1061} & {0} & {0.0000} \\ \hline 1 & {0.0000} & {0.13} & {0.0000} \\ \hline 2 & {0.3183} & {0.63} & {0.2005} \\ \hline 3 & {0.5000} & {1} & {0.5000} \\ \hline 4 & {0.3183} & {0.63} & {0.2005} \\ \hline 5 & {0.0000} & {0.13} & {0.0000} \\ \hline 6 & {-0.1061} & {0} & {0.0000} \\ \hline\end{array}$

$\therefore$ Transfer Function of Digital FIR Filter is,

$H(z)=\sum_{n=0}^{7-1} h(n) \cdot z^{-n}$

$=0+0+0.2005 z^{-2}+0.5 z^{-3}+0.2005 z^{-4}+0+0$

$\therefore H(z)=0.2005\left(z^{-2}+z^{-4}\right)+0.5 z^{-3}$