Building Low-pass filter with Sinc function
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Liang
le 18 Sep 2020
Réponse apportée : Preston Pan
le 1 Juil 2022
Dear Community,
I am trying to build a low-pass filter by using a sinc function for my homework assignment. I then use convolution to later filter an audio sample with this filter. However, when I plot the filter in a bode plot it looks like a high-pass filter. Can anyone tell me what I'm doing wrong?
Thanks in advance!
%% Downsampled by K with low-pass filter
% Build filter
clear all; close all
K = 2;
fs = 1600;
N = 51;
n = (-(N-1)/2:1:(N-1)/2);
h = (1/K) * sinc((pi/K)*n);
% Plot frequency response filter
[H, H_vec] = fftFreq(h, fs, 1 );
figure
plot(H_vec*2*pi/fs, abs(H))
filt_tf =tf(h,1,1/fs,'Variable','z^-1');
figure
bode(filt_tf)
function [ X , f ] = fftFreq( data , fs, w )
% Number of FFT points
NFFT = length( data );
% calculate FFT
X = fft(data .* w);
% calculate frequency spacing
df = fs/NFFT;
% calculate unshifted frequency vector
f = (0:(NFFT-1)) * df;
end
1 commentaire
Réponse acceptée
Star Strider
le 18 Sep 2020
I am not exactly certain what the problem is from a theoretical prespective (I will leave it to you to explore that), however the sinc pulse is too narrow. Increase ‘K’ to 4 or more, and you get a lowpass result.
figure
freqz(h,1,2^16,fs)
If you are going to use it as a FIR discrete filter, do the actual filtering with the filtfilt function for the best results.
.
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Preston Pan
le 1 Juil 2022
Consider removing the pi in the argument of sinc. I get that scaling is necessary to respect the fourier scaling relationship and preserve unit gain in the passband but I think that would just be rect(K*t) <--> 1/|K| * sinc(f/K).
When I removed it and did
h=(1/K)*sinc(n/K)
the filter produced the desired behavior.
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