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Design a FIR filter using window method for following specification. Use hamming window of length.

$H(e^{jω} )=e^{-j3ω} ; 0≤|ω|≤\frac{3π}{4}$

=0 ; otherwise

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Step-1: Identify the specification of filter

$N=7 ∝=3 ω_c=\frac{-3π}{4}≤ω≤\frac{3π}{4}$

Window Type: Hamming

Step-2: Calculate the Inverse Fourier Transform of H(ω)

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

$=\frac{1}{2π} ∫_{\frac{(-3π)}{4}}^{\frac{3π}{4}}e^{-j3ω} e^{jωn} dω$

$=\frac{1}{2π} ∫_{\frac{(-3π)}{4}}^{\frac{3π}{4}}e^{j(n-3)ω} dω$

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

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

$h_d (n)=\frac{sin⁡\frac{3π}{4}(n-3)}{(π(n-3)}$

$h_d (0)=0.075=h_d (6 )$ {by linear phase property}

$h_d (1)=-0.159=h_d (5 )$ {by linear phase property}

$h_d (2)=0.225=h_d (4 )$ {by linear phase property}

By L-Hospital’s Rule

$h_d (3)=0.75$

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

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

W(0)=0.08=W(6)

W(1)=0.31=W(5)

W(2)=0.77=W(4)

W(3)=1

Step-4: Calculation of window Response h(n)

h(0)=0.006=W(6)

h(1)=-0.049=W(5)

h(2)=0.173=W(4)

h(3)=0.75

Step-5: Calculate of filter Transfer Function

$H(z)=∑_{n=0}^{N-1}h(n) z^{-n}$

$H(z)=0.006-0.049z^{-1}+0.173z^{-2}+0.75z^{-3}+0.173z^{-4}-0.049z^{-5}+0.006z^{-6}$

Step-6: Realization Structure