Question: Design the symmetric FIR LPF where desired frequency response is given
0

as,

$\operatorname{Hd}(w)=\left\{\begin{array}{ll}{e^{(-jw)}} & \\ 0 \end{array}\right.$ for $|\mathrm{w}| \leq \mathrm{w}_{0}$ otherwise

symmetric fir lpf • 22 views
 modified 4 weeks ago  • written 4 weeks ago by Ankit Pandey • 70
0

Solution:

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

we have $\alpha=1,$ Gain $\mathrm{C}=1$ and Cut-off frequency $\omega_{c}=w_{0}$

But $\alpha=\frac{M-1}{2}=1$

$\therefore$ Filter length $(\mathrm{M})=3$

$\therefore$ Desired Impulse Response

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

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

$=\frac{1}{2 \pi} \int_{-w_{0}}^{w_{0}} e^{(n-1) j \omega} d \omega \cdots(1)$

$=\frac{1}{2 \pi}\left[\frac{e^{(n-1) j \omega}}{(n-1) j}\right]_{-w_{0}}^{w_{0}}$

$=\frac{1}{2 \pi} \times \frac{1}{(n-1) j}\left[e^{(n-1) j w_{0}}-e^{-(n-1) j w_{0}}\right]$

$=\frac{1}{2 \pi} \times \frac{1}{(n-1) j} \times 2 j \sin (n-1) w_{0}$

$=\frac{1}{(n-1) \pi} \sin (n-1) w_{0}(n \neq 1)$

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

$=\frac{1}{2 \pi}[\omega]_{-w_{0}}^{w_{0}}$

$=\frac{1}{2 \pi}\left[w_{0}-\left(-w_{0}\right)\right]=\frac{w_{0}}{\pi}$

$\therefore$ Desired Impulse Response

$h_{d}(n)=\left\{\begin{array}{cl}{\frac{\sin (n-1) w_{0}}{(n-1) \pi}} & {n \neq 1} \\ {w_{0} / \pi} & {n=1}\end{array}\right.\cdots(2)$

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

For linear phase FIR filters,

$h_{d}(n)=h_{d}(M-1-n)=h_{d}(3-1-n)=h_{d}(2-n)$

$\therefore h_{d}(2)=h_{d}(0) \cdots(3)$

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

For $n=0, h(0)=h_{d}(0) \times W(0)$

$=\frac{\sin (0-1) w_{0}}{(0-1) \pi} \times W(0)=W(0) \frac{\sin w_{0}}{\pi} \cdots(4)$

For $n=1, h(1)=h_{d}(1) \times W(1)=\frac{w_{0}}{\pi} \times W(1)$

$=W(1) \frac{w_{0}}{\pi}$ (From 2)

For $n=2, h(2)=h(0)=W(0) \frac{\sin w_{0}}{\pi}(\text { From } 3 \& 4)$

$\therefore$ Transfer Function of Digital FIR Filter is

$H(z)=\sum_{n=0}^{3-1} h(n) \cdot z^{-n}$

$=W(0) \frac{\sin w_{0}}{\pi} \cdot z^{0}+W(1) \frac{w_{0}}{\pi} \cdot z^{-1}+W(0) \frac{\sin w_{0}}{\pi} \cdot z^{-2}$

$=W(0) \frac{\sin w_{0}}{\pi}\left(1+z^{-2}\right)+W(1) \frac{w_{0}}{\pi} \cdot z^{-1}$

$\therefore H(z)=\frac{1}{\pi}\left[W(0) \sin w_{0}\left(1+z^{-2}\right)+W(1) w_{0} \cdot z^{-1}\right]$