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Series Filter Loop Error

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S
S am 31 Jan. 2024
Kommentiert: Mathieu NOE am 27 Feb. 2024
I am trying to make a loop that goes through different filters in cascade, so the output of one is the input to the next. I want to use the built in matlab gammatone function to do this. I have been stuck here for a bit. Thank you for your time!
fs = 20e3;
numFilts = 32;
BW = 100; %Filter Bandwidth
filter_number = 5;
CenterFreqs = logspace(log10(50), log10(8000), numFilts);
CF1 = CenterFreqs - BW/2; %Lower cutoff frequency
CF2 = CenterFreqs + BW/2; %Upper cutoff frequency
t = linspace(0,2*pi,200);
input = sin(t) + 0.25*rand(size(t));
for ii = 1:filter_number
output = gammatone(input, CenterFreqs, fs)
end
Unrecognized function or variable 'gammatone'.
[h{ii},f] = freqz(gammatone{ii}.Coefficients,1,4*8192,fs);
Error: (I do not have a line 32?, also I did not make gammatone so I don't think there is anything wrong with it)
Not enough input arguments.
Error in gammatone (line 32)
A = 10^((q_factor) / 20);
Error in t131 (line 13)
output = gammatone(input, CenterFreqs, fs)
  15 Kommentare
Mathieu NOE
Mathieu NOE am 13 Feb. 2024
I will transform my comment above into a answer so you can accept it (and tx for that ! )
for the g factor , we don't need it inside the gammatone function itself as in the main code we are doing it in these lines :
(up to you to choose a 0 or -3 dB peak amplitude of the filters)
% scale the IR amplitude so that the max modulus is 0 dB
a = 1/max(abs(tmp));
% % or if you prefer - 3dB
% g = 10^(-3 / 20); % 3 dB down from peak
% a = g/max(abs(tmp));
Mathieu NOE
Mathieu NOE am 13 Feb. 2024
FYI, the lines of code above could also be written this way if you prefer
% scale the IR amplitude to match a required Bode plot magnitude
g = 1; % the max modulus is 0 dB
% g = 10^(-3 / 20); % or if you prefer - 3dB
a = g/max(abs(tmp));

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Mathieu NOE
Mathieu NOE am 7 Feb. 2024
Verschoben: Mathieu NOE am 13 Feb. 2024
hello again
so I wanted to learn a bit about gammatone filters , so I read this , which I suppose i also from where you started your code
there are a few modifications I made in the code , see the original lines commented are replaced by new code
so the function output is the IR of the filter , we can use it latter on (not yet coded) to filter any time signal
IMHO, now the IR and Bode plots are correct
here we plot only the first 5 filters , even though the CF are computed for 32 filters
fs = 20e3;
numFilts = 32; %
% BW = 100; %Filter Bandwidth
filter_number = 5;
order = 4;
CenterFreqs = logspace(log10(50), log10(8000), numFilts);
% CF1 = CenterFreqs - BW/2; %Lower cutoff frequency
% CF2 = CenterFreqs + BW/2; %Upper cutoff frequency
% t = linspace(0,2*pi,200);
% input = sin(t) + 0.25*rand(size(t));
figure(1)
hold on
for ii = 1:filter_number
IR = gammatone(order, CenterFreqs(ii), fs);
[tmp,f] = freqz(IR,1,1024*2,fs);
% scale the IR amplitude so that the max modulus is 0 dB
a = 1/max(abs(tmp));
% % or if you prefer - 3dB
% g = 10^(-3 / 20); % 3 dB down from peak
% a = g/max(abs(tmp));
IR = IR*a;
[h{ii},f] = freqz(IR,1,1024*2,fs); % now store h{ii} after amplitude correction
plot(IR)
end
title('Impulse Response');
hold off
figure(2)
hold on
for ii = 1:filter_number
plot(f,20*log10(abs(h{ii})))
end
title('Bode plot');
set(gca,'XScale','log');
xlabel('Freq(Hz)');
ylabel('Gain (dB)');
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function IR = gammatone(order, centerFrequency, fs)
% Design a gammatone filter
earQ = 9.26449;
minBW = 24.7;
% erb = (centerFrequency / earQ) + minBW;
erb = ((centerFrequency/earQ)^order + (minBW)^order)^(1/order);% we use the generalized
% function (depending on order)
% B = 1.019 .* 2 .* pi .* erb; % no, the 3pi factor is implemented twice in your code
B = 1.019 * erb;
% g = 10^(-3 / 20); % 3 dB down from peak % what is it for ? see main code above
f = centerFrequency;
tau = 1. / (2. .* pi .* B);
% gammatone filter IR
t = (0:1/fs:10/f);
temp = (t - tau) > 0;
% IR = (t.^(order - 1)) .* exp(-2 .* pi .* B .* (t - tau)) .* g .* cos(2*pi*f*(t - tau)) .* temp;
IR = ((t - tau).^(order - 1)) .* exp(-2*pi*B*(t - tau)).* cos(2*pi*f*(t - tau)) .* temp;
end
  4 Kommentare
S
S am 26 Feb. 2024
@Mathieu NOE I just posted it as a new question in case you wanted me to repost!
Mathieu NOE
Mathieu NOE am 27 Feb. 2024
hello @S
ok , I'll have a look in your other post !

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