Question

DON'T USE MATLAB to design a digital filter using the window method to meet the following requirements:
Stop band attenuation: ≥43 dB and <49 dB
Pass band edge frequency: 3.2 kHz
Stop band: 0 to 2.4 kHz
Maximum pass band ripple: 0.0546 dB
Phase response: Linear.
Sampling frequency: 16 kHz
Determine the first 6 filter coefficients.

NB: Please don't use MATLAB to get the answers. do it manually!

Answer 1

FIR High pass filter design
(a)
Fs=16000 Hz
fp=3200Hz
fs=2400 Hz;
fc=(fp+fs)/2=(3200+2400)/2= 2800Hz

Cut off frequency,
wc=2*pi*fc/Fs = 2*pi*2800/16000 = 0.35 pi

Transition width =2*pi*( fp-fs)/Fs=2*pi*( 3200-2400)/16000= 0.1pi
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(b) For the stop band attenuation between 43 db and 49db ,Hamming window is needed.

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(c) The main lobe width the window is 8pi/N

N=8 pi/ transition width = 8*pi/0.1*pi = 80

Choose N=81

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(d)
The desired impulse response of FIR low pass filter is

hd(n) =sin(wc(n-a))/(pi(n-a))
where wc = 0.35 pi
n=0 to 80
a=(N-1)/2 =(81-1)/2= 40
hd(40) =wc/pi=0.35pi/pi =0.35.

hd(n) =sin(0.35 *pi(n-40))/(pi(n-40))

The first 6 ideal filter coefficients are

hd(n) = [0 -0.0073 -0.0068 0.0013 0.0084 0.0064 ...]
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e)
Tha Hamming window function is
wH(n) = 0.54-(0.46*cos(2*pi*n/(N-1)) where N=81
      = 0.54-(0.46*cos(2*pi*n/80))

The first 6 hamming window samples are

wH(n)=[0.08 0.0814 0.0857 0.0927 0.1025 0.1150 ...]
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f)
The impulse response h(n) is h(n)=hd(n) wH(n)
The product of hd(n) and wH(n) is

h(n)= [0 -0.5921 -0.5805 0.1248 0.8621 0.7396....] *(10^(-3))

Calculations: