Finding Fundamental Frequency of an Audio Signal

Hi all,

I am writing a code that takes input from a .wav file of a guitar playing a single note and am working on displaying the note being played. To do this, I must find the fundamental frequency of the audio file. I have been trying to find the fundamental frequency of an F chord, but the frequency I am finding is 220 Hz, which is the fundamental frequency of an A chord. My main struggles have been with the "pwelch" function. Can anyone recommend any changes to the parameters of the function? Or would my best bet to be to use a different method? Any help is greatly appreciated.

clc; clear;

%Get the audio file 
[y,Fs] = audioread('chord-F.wav');  %y = samples, Fs = sample rate
sound(y,Fs); %Listen to audio file

%Get time of the signal and plot the signal
samples = length(y);
x = samples/Fs;    
t = linspace(0, x, samples);
subplot(2,1,1),plot(t,y), ylabel('Amplitude'), xlabel('Time (secs)');


%Converting signal to frequency estimate
    %Plot the frequency spectrum
    %pwelch = power spectral density estimate = (x, window, ...
    % noverlap, f1, fs)
f1 = 0:(Fs/samples):(Fs/2-(Fs/samples));    %find the frequencies of the signal
[Pxx, f] = pwelch(y, gausswin(Fs), Fs/2, f1/4, Fs);
subplot(2,1,2), plot(f,Pxx), ylabel('PSD'), xlabel('Frequency (Hz)');

    %Get the frequency estimate (spectral peak)
[~, loc] = max(Pxx);
freq_est = f(loc);
title(['Frequency Estimate = ', num2str(freq_est), 'Hz']);

[Frencita Tellis]

i tried to use the audiread function but iam not able to read my audiofile. Someone please help.

[Eugenia Angela Salsabillah]

Have you solved this? So, how would exactly know the frequency of the sound?

Answer 1

[Mathieu NOE]

hello

would you try your wav file on my little code ? see below

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 8192;    % 
NOVERLAP = round(0.75*NFFT);
w = hanning(NFFT);     % Hanning window / Use the HANN function to get a Hanning window which has the first and last zero-weighted samples.

% spectrogram dB scale
spectrogram_dB_scale = 100;  % dB range scale (means , the lowest displayed level is XX dB below the max level)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% options 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% if you are dealing with acoustics, you may wish to have A weighted
% spectrums 

% option_w = 0 : linear spectrum (no weighting dB (L) )
% option_w = 1 : A weighted spectrum (dB (A) )
option_w = 0;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%[data,Fs]=wavread('Approach_Gear_Drop_Aft Ctr.wav ');  %(older matlab)
% or 
[data,Fs]=audioread('myWAVaudiofile.wav'); %(newer matlab)

channel = 1;

signal = data(:,channel);
samples = length(signal);
dt = 1/Fs;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 1 : averaged FFT spectrum
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

[sensor_spectrum, freq] = pwelch(signal,w,NOVERLAP,NFFT,Fs);

% convert to dB scale (ref = 1)
sensor_spectrum_dB = 20*log10(sensor_spectrum);

% apply A weigthing if needed
if option_w == 1
    pondA_dB = pondA_function(freq);
    sensor_spectrum_dB = sensor_spectrum_dB+pondA_dB;
    my_ylabel = ('Amplitude (dB (A))');
else
    my_ylabel = ('Amplitude (dB (L))');
end


figure(1),semilogx(freq,sensor_spectrum_dB);grid
title(['Averaged FFT Spectrum  / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(freq(2)-freq(1)) ' Hz ']);
xlabel('Frequency (Hz)');ylabel(my_ylabel);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 2 : time / frequency analysis : spectrogram demo
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

[sg,fsg,tsg] = specgram(signal,NFFT,Fs,w,NOVERLAP);  

% FFT normalisation and conversion amplitude from linear to dB (peak)
sg_dBpeak = 20*log10(abs(sg))+20*log10(2/length(fsg));     % NB : X=fft(x.*hanning(N))*4/N; % hanning only

% apply A weigthing if needed
if option_w == 1
    pondA_dB = pondA_function(fsg);
    sg_dBpeak = sg_dBpeak+(pondA_dB*ones(1,size(sg_dBpeak,2)));
    my_title = ('Spectrogram (dB (A))');
else
    my_title = ('Spectrogram (dB (L))');
end

% saturation of the dB range : 
% saturation_dB = 60;  % dB range scale (means , the lowest displayed level is XX dB below the max level)
min_disp_dB = round(max(max(sg_dBpeak))) - spectrogram_dB_scale;
sg_dBpeak(sg_dBpeak<min_disp_dB) = min_disp_dB;

% plots spectrogram
figure(2);
imagesc(tsg,fsg,sg_dBpeak);colormap('jet');
axis('xy');colorbar('vert');grid
title([my_title ' / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(fsg(2)-fsg(1)) ' Hz ']);
xlabel('Time (s)');ylabel('Frequency (Hz)');


function pondA_dB = pondA_function(f)
	% dB (A) weighting curve
	n = ((12200^2*f.^4)./((f.^2+20.6^2).*(f.^2+12200^2).*sqrt(f.^2+107.7^2).*sqrt(f.^2+737.9^2)));
	r = ((12200^2*1000.^4)./((1000.^2+20.6^2).*(1000.^2+12200^2).*sqrt(1000.^2+107.7^2).*sqrt(1000.^2+737.9^2))) * ones(size(f));
	pondA = n./r;
	pondA_dB = 20*log10(pondA(:));
end

[Kaleb Guillot]

Thank you for your response!

However, I do not believe that this is entirely the solution that I am looking for. I am merely looking for an estimation of the fundamental frequency of a recording of a guitar note to display the particular note being played, based off of this diagram:

while when I try out your code I produce these graphs:

[Mathieu NOE]

hello

as far as I see , the first peak of the fft is around 175 Hz which is more or less what the theory says (table)

so what is the problem ?

[Kaleb Guillot]

How should I go about indexing out that first peak in order to find the corresponding fundamental frequency?

[Mathieu NOE]

hello

yes, yiu can use findpeaks to get the peaks of the fft and take the first peak frequency

[Abhijit Sardar]

sir how to find second and third harmonics after this? also i am trying to find peak frequency for ultrasonic wave signal will this code work for that.I have amplitude data of 2000 data points only so least hanning window would be 1024 i would be loosing some data( I am not familiar with signal processing).

[Mathieu NOE]

hello

with the function findpeaks you can find multiple peaks from a fft spectrum - see the doc to see the different options.

for the ultrasonic file, yes , the code works too. If you have 2000 data points, you can go up to nfft = 2000, to get the max frequency resolution, but there will be no averaging of course - this may not be critical if your time data are not corrupted with noise.

[Mangesh KAle]

Hello, How did you calculate the nfft=8192

[Syed Abdullah]

I think that might be the sampling frequency of audio signal