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Liang on 18 Sep 2020
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Answered: Preston Pan on 1 Jul 2022
Accepted Answer: Star Strider
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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
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Bas Zweers on 22 Sep 2020
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⋮
You need to use fftshift for the correct frequency plot
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Accepted Answer
Star Strider on 18 Sep 2020
Open in MATLAB Online
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.
Also, since this is a discrete filter, the freqz function will do what you want:
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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Liang on 18 Sep 2020
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https://support.mathworks.com/matlabcentral/answers/595918-building-low-pass-filter-with-sinc-function#comment_1014835
Edited: Liang on 18 Sep 2020
Thanks, it worked for me!
However, I am still figuring out why if K is too small (i.e. 2 or 3) that we apparently get a high-pass filter. My assignment specifically states that we need to design an ideal low-pass filter with a cut-off frequency of pi/2.
So if anyone has some insight please let me know.
Star Strider on 18 Sep 2020
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https://support.mathworks.com/matlabcentral/answers/595918-building-low-pass-filter-with-sinc-function#comment_1014841
As always, my pleasure!
I have not investigated the sinc function in that respect. Taking its Fourier transform could provide significant insight (analytically if possible, numerically if that is all that is available). The Fourier transform will reveal it to be a square wave (actually rectangle function) in the frequency domain. Mapping the time-domain parameters to the frequency domain parameters to produce the desired design remains the issue.
Liang on 22 Sep 2020
Direct link to this comment
https://support.mathworks.com/matlabcentral/answers/595918-building-low-pass-filter-with-sinc-function#comment_1020214
UPDATE:
Apparently, I made a mistake in the mathematical procedure to come up with my sinc low-pass filter. The pi in the sinc function shouldn't be there. Now everything is working correctly. Thanks again for the help.
Star Strider on 23 Sep 2020
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As always, my pleasure!
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More Answers (1)
Preston Pan on 1 Jul 2022
Open in MATLAB Online
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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