matlab code for musical note recognition based on frequency

Question

varun taliparambe vitel 2020-6-14

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/547824-matlab-code-for-musical-note-recognition-based-on-frequency

评论： Geetesh Mokhare 2021-3-29

i am working on a project to identify the musical note based on frequencies using window methods i need help to write the code and understanding it . ihave a code but its to advanced to understand......

clear;
clc;
close all;
% Note Recognition
%% Note Initialization
mainNames = char('C', 'C#', 'D', 'D#', 'E', 'F', 'F#', 'G', 'G#', 'A', 'A#', 'B');
names = char('A0', 'A#0' ,'B0');
A0 = 27.5;
ind = 4;
for i = 0:87
    data(i+1) = A0 * (2^(1/12))^i;
    if i>2
         a = [mainNames( rem((ind - 4), 12)+1,:)  num2str(fix((ind - 4)/12)+1)];
         names = char(names, a);
         ind = ind + 1;
    end
end
% Band Initialization
bands(1) = 20;
for i = 1:87
    bands(i+1) = +7+(data(i) + data(i+1))/2;
end
bands(89) = 4500;
%%
fileName = 'c.wav';  % File name
[y, Fs] = audioread(fileName);            % Read audio file
y = (y(:,1) + y(:,2))*4;                % Decrease 2 channels to 1 
%y = (y(:,1));
y(1:2:end) = 0;   %from big vector take only odd index values         % Do decimation(decreace the no of samples)
frameLength = 4410*2; % 2 GÃ¼zel oldu
endPart = frameLength*ceil(length(y)/frameLength);  % complete the last frame %ceil is to round off to nearest integer
y(length(y)+1 : endPart) = 0;
f = linspace(1,Fs,frameLength); %creates linearly spaced vector 
%%
harmonics = 0;
for i = 1:round(length(y)/frameLength)      % For each frame
    % Divide audio into frames
    frames(i, 1:frameLength) = y( (frameLength*(i-1)+1):(frameLength*i) )';
    
    frame = y( (frameLength*(i-1)+1):(frameLength*i) )';
    frame = frame .* hamming(length(frame))';   % Hamming Window
    fframe = abs(fft(frame));                   % FFT
    
    fp = sum(fframe);
    p = sum(abs(frame));
    b = true;
    if(p < 200 || fp < 1000) % Put a threshold for processing
       b = false; 
    end
    
    % Bands
    for i=1:88
        freqBand(i) = mean(fframe( round(bands(i)/(Fs/frameLength) ):round(bands(i+1)/(Fs/frameLength))))^2;
    end
    
    % Plotting
    subplot(3,1,1)
    stem(freqBand)
    subplot(3,1,2)
    plot(fframe)
    xlim([0,500])
    subplot(3,1,3)
    plot(fframe)
    ylim([-1 1])
    
    hold off
    pause(0.1)
    sound(fframe,Fs)
    
    
    % Desicion
    m = find(freqBand == max(freqBand(:)));
    if(b) disp(names(m,:));     % Print the result
    else disp('.'); end
    
    if(b)
        index = 1;
        for i = 1:88
            if(freqBand(i) > 2000)
                harmonics(index) = i;
                index = index+1;
            end 
        end
    
    end
end

this is the code.....

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Paavan Kumar Dornadula 2020-7-20

how to verify is above code is correct or not?

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Answer 1

Thiago Henrique Gomes Lobato 2020-6-14

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/547824-matlab-code-for-musical-note-recognition-based-on-frequency#answer_450999

To really undestand the method, first try to define how exactly you gonna get the frequencies. Your main steps are :

Define your notes and which frequency do they have (Line 5-23)
Read your sound data and define the window size to evaluate the sound (Line 24-33). Here your code has some flaws. Ex: Decimation change the sampling rate, so you can't simply do it to "decrease number of samples" and don't compensate it afterwards. If you want a mono channel from a stereo recording you normally take the mean, not the sum times 4.
For each evaluation window, take the FFT and check which note has the frequency closest to the maximum amplitude (Line 34-86). In your code the FFT is summed in bands so each note has its "own" band and then you can just check the strongest band.

The algorithm is quite simple, but the way it was implemented in your code is indeed not so optimal. I would suggest you to try to implement it yourself in a more clean way based in those main steps. This should also help you to better internalize it.