(Window Design) Without using MATLAB, use the window design method with a Hammin

Question

(Window Design) Without using MATLAB, use the window design method with a Hamming window to find an expression for the coefficients {h[n]}^36 _n = 0 of a length-37 FIR filter that approximates an ideal bandpass filter with cutoff frequencies pi/4 and 37pi/4. Plot the magnitude and phase responses of this filter on a linear scale (not the default logarithmic scale!). From these plots, estimate the passband ripple, the stopband ripple, and the number of zeros that lie on the unit circle (when assessing the stopband and passband ripple, please assume omega_s1 = 0.17pi, omega_p1 = 0.33pi, omega_p2 = 0.67pi, and omega_s2 = 0.83pi).

Explanation / Answer

For Hamming Window Approximate width of main lobe =8pi/M

here n=0 to 36 & M=37

So far we have determined the window type and its length. Using the equation describing a Hamming window, we find the window as

w[n]={0.540.46cos(2n/M)           0nM

           0                                      otherwise}

M=37

w[n]=0.540.46cos(2n/M)

Consider a band-pass filter with the low cut-off and high cut-off of

c,l=pi/4 and c,u=3pi/4, respectively.

hd,bandpass[n]=(c,u/)sinc(c,un/)(c,l/)sinc(c,ln/)

In this example, the ideal impulse response will be

hd,bandpass[n]=0.75sinc(0.75n)0.25sinc(0.25n)

Apply a time shift of M/2 and multiply hd,lowpass[n] by w[n]

To have a causal linear-phase response, we need to apply a time shift equal to M/2 in the ideal impulse response and multiply the result by w[n]. Therefore, we find

h[n]=[0.75sinc(0.75n)0.25sinc(0.25n)][0.540.46cos(2n/M)]

Using this equation magnitude & phasor response of this filter can be plotted. From these plots , passband ripple , stopband ripple, the number of zeroes that lie on unit circle can be calculated.

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