Step-1: Identification of specification of filter

$H_d (e^{jω} )=|H(ω)| e^{-j∝ω} ; -ω_c≤ω≤ω_c$

But N=M=7

We know that,

$∝=\frac{(N-1)}{2}=3$

Therefore,

$H_d (e^{jω} )=e^{-j3ω} ; 0.4π|ω|≤0.6π$

=0 ; otherwise

Window Type: Hamming Window

Step-2: Calculation of IFT of H(ω)

$h_d (n)=\frac{1}{2π} ∫_{-ω_c}^{ω_c)}H_d (ω) e^{jωn} dω$

$=\frac{1}{2π} ∫_{-0.6π}^{-0.4π)}e^{jωn} e^{-j3ω} dω + \frac{1}{2π} ∫_{0.4π}^{0.6π}e^{-j3ω} e^{jωn} dω$

$=\frac{1}{2π} ∫_{-0.6π}^{-0.4π}e^{jω(n-3)} dω + \frac{1}{2π} ∫_{0.4π}^{0.6π}e^{j(n-3)ω} d$

$=\frac{1}{2π} [\frac{e^{j(n-3)ω}}{(j(n-3)}]_{-0.6π}^{-0.4π}+\frac{1}{2π} [\frac{e^{j(n-3)ω}}{(j(n-3)}]_{0.4π}^{0.6π}$

$=\frac{1}{2π} [\frac{(e^{-j0.4π(n-3)}-e^{-j0.6π(n-3))}}{(j(n-3)}+\frac{(e^{j0.6π(n-3)}-e^{j0.4π(n-3))}}{(j(n-3)}]$

$=\frac{1}{π(n-3)} [sin⁡[0.6π(n-3) ]⁡-sin⁡[0.4π(n-3) ] ]$

$h_d (n)=(sin⁡[0.6π(n-3) ]-\frac{[0.4π(n-3) ])}{π(n-3)}$

$h_d (0)=0=h_d (6)$ {by linear phase property}

$h_d (1)=-0.187=h_d (5)$ {by linear phase property}

$h_d (2)=0=h_d (6)$ {by linear phase property}

By L-Hospital’s Rule

$h_d (n)=0.2$

Step-3: Calculation of window Response W(n)

Window Response for hamming is given by

$W(n)=0.54-0.46cos⁡(\frac{2πn}{(N-1))}$

W(0)=0.08=W(6)

W(1)=0.31=W(5)

W(2)=0.77=W(4)

W(3)=1

Step-4: Calculate the impulse Response of filter

Impulse Response is given by

$h(n)=h_d (n)*W(n)$

h(0)=0=h(6)

h(1)=-0.057=h(5)

h(2)=0=h(4)

h(3)=0.2

Step-5: Calculation of filter Transfer function H(z)

$H(z)=∑_{n=0}^{N-1}h(n) z^{-n}$

$∴H(z)=∑_{n=}^6h(n) z^{-n)}$

$∴H(z)=0+(-0.057) z^{-1}+0+0.2z^{-3}+0+(0.057) z^{-5}+0$

$∴H(z)=0-0.057z^{-1}+0+0.2z^{-3}+0-.057z^{-5}+0$

Step-6: Draw the Realization structure