Design a filter with $\mathrm{H}_{\mathrm{d}}\left(\mathrm{e}^{\mathrm{jw}}\right)=\mathrm{e}^{-\mathrm{j} 3 \mathrm{w}},-\pi / 4 \leq \mathrm{w} \leq \pi / 4$

$ 0 \quad, \pi / 4\lt|\mathrm{w}|\lt\pi\\ $

Using Hanning window with $\mathrm{N}=7$

Solution:

$ H_d\left(e^{j \omega}\right)=e^{-j 3 \omega}\\ $

The frequency response is having a term $e^{-j \omega(N-1) / 2}$ which gives $h(n)$ symmetrical about $n=\frac{N-1}{2}=3$, we get a causal sequence.

$ \text { We have, } \begin{aligned}\\ h_d(n) &=\frac{1}{2 \pi} \int_{-\pi / 4}^{\pi / 4} e^{-j 3 \omega} \cdot e^{j \omega s} d \omega \\\\ &=\frac{1}{2 \pi} \int_{-\pi / 4}^{\pi / 4} e^{j(n-3) \omega} d \omega \\\\ &=\frac{\sin \pi / 4(n-3)}{\pi(n-3)}\\ \end{aligned}\\ $

For $n=7$, we have,

$ \begin{aligned}\\ &h_d(0)=h_d(6)=0.075 . \\\\ &h_d(1)=h_d(5)=0.159 .\\ \\ &h_d(2)=h_d(4)=0.22 \\\\ &h_d(3)=0.25 .\\ \end{aligned}\\ $

The non-causal windows sequence is,

$ \begin{aligned}\\ w_{H n}(n) &=0.5+0.5 \cos \frac{2 \pi n}{N-1} \text { for }-(N-1) / 2 \leq n \leq \frac{N-1}{2} \\\\ &=0, \quad \text { otherwise. }\\ \end{aligned}\\ $

For $N=7$.

$W_{H n}(n)=0.5+0.5 \cos \frac{2 \pi n}{N-1}$ for $-3 \leq n \leq 3$

0, otherwise.

$ \begin{aligned}\\ &\omega_{A n}(0)=0.5+0.5=1 \\\\ &W_{A n}(-1)=W_{H+n}(1)=0.5+0.5 \cos \pi / 3=0.75 \\\\ &W_{A n}(-2)=W_{A n}(2)=0.5+0.5 \cos 2 \pi / 3=0.25 \\\\ &W_{A n}(-3)=0.5+0.5 \cos \pi=0 .\\ \end{aligned}\\ $

The causal window sequence can be obtained by shifting the sequence $w_{A_m}(n)$ to right by 3 Samples.

$ \begin{aligned}\\ &\text { in.c. } W_{A n}(0)=W_{H_n}(b)=0, W_{H_n}(1)=W_{\text {Hn }}(5)=0.25\\\\ &W_{\text {An }}(2)=W_{\text {An }}(3)=0.75, \quad W_{\text {An }}(3)=1\\ \end{aligned}\\ $

The filter Coefficients using tanning windows are,

$ \begin{aligned}\\ &h(n)=h_d(n) w_{\text {An }}(0)=\text { for } 0 \leq n \leq 6 . \\\\ &h(0)=h(6)=h_d(0) w_{\text {tn }}(0)=(0.075)(0)=0 .\\ \\ &h(1)=h(5)=h_d(1) \cdot w_{\text {tn }}(1)=(0.159)(0.25)=0.03975 \\\\ &h(2)=h(4)=h_d(2) \cdot w_{\text {An }}(2)=(0.22)(0.75)=0.165 \\\\ &h(3)=h_d(3) \cdot w_{\text {An }}(3)=(0.25)(1)=0.25 .\\ \end{aligned}\\ $