I want a FFT plot of a .mat file at 8000Hz (in 3 seconds) to find the dominant frequencies but I'm not getting an peaks. Whats wrong with the codes?Thank you very much for any help!

if true
  % code
end
Here is the code I have but getting no peaks:
Fs = 8000; % Sampling frequency. 
t = 0:1/Fs:3-1/Fs;
x = cos(2*pi*1000*t)+rand(size(t));
N = length(x);
xdft = fft(x);
xdft = xdft(1:N/2+1);
psdx = (1/(Fs*N)) * abs(xdft).^2;
psdx(2:end-1) = 2*psdx(2:end-1);
freq = 0:Fs/length(x):Fs/2;

plot(freq,10*log10(psdx)) grid on title('Periodogram Using FFT') xlabel('Frequency (Hz)') ylabel('Power/Frequency (dB/Hz)')

[Star Strider]

When I run your code, I see the expected peak at 1 kHz.

[Kristy Kwok]

Hi Star, Thank you for your reply. I wasnt sure if my coding was correct. The peak at 1000Hz is because I put 1000 in this line: x = cos(2*pi*1000*t)+rand(size(t)); I have tried to put 800 and the peak will be at 800Hz. I am not sure how to confirm if the code is correct.

Answer 1

[Star Strider]

My pleasure.

Your code is correct if it produces the result you expect. You have calculated the ‘freq’ frequency vector correctly. I slightly revised your code to plot three different frequencies, at 800, 1000, and 1200 Hz. You can see all of them clearly in the plot (and you can add as many frequencies to the ‘Fvct’ vector as you want):

Fs = 8000; % Sampling frequency. 
t = 0:1/Fs:3-1/Fs;
Fvct = [800 1000 1200];
x = cos(2*pi*Fvct(:)*t);
x = sum(x) + rand(size(t));
N = length(x);
xdft = fft(x);
xdft = xdft(1:N/2+1);
psdx = (1/(Fs*N)) * abs(xdft).^2;
psdx(2:end-1) = 2*psdx(2:end-1);
freq = 0:Fs/length(x):Fs/2;

plot(freq,10*log10(psdx))
grid on
title('Periodogram Using FFT')
xlabel('Frequency (Hz)')
ylabel('Power/Frequency (dB/Hz)')

A slightly different implementation is given in the R2015a documentation on the fft (link) function.

[Kristy Kwok]

Thank you so much for your help Star! I had a look at the following link https://au.mathworks.com/help/signal/ug/power-spectral-density-estimates-using-fft.html.

All the examples used on there have some similar like"Create a signal consisting of a 100 Hz sine wave in N(0,1) additive noise. " and therefore the codes that used there were x = cos(2*pi*100*t) + randn(size(t)); whereas for my case, I have a .mat file with a guitar tone and only given the sampling frequency at 8000Hz for 3 seconds. Therefore I was not sure the coding. Can I please ask Star, for the 800, 1000,1200 frequencies, are they just random frequencies to check the the dominant frequencies on the FFT?

[Star Strider]

My pleasure.

‘... are they just random frequencies to check the the dominant frequencies on the FFT?’

Yes. They are designed to demonstrate that your code works as you intend it to, and will plot all the frequencies correctly.

[Kristy Kwok]

Thanks Star!(You really a STAR!) But I might come across a silly but I am really confused here! If I use random frequencies on the FFT codes, how do I find out the real dominant frequencies of the .mat file I have?

[Star Strider]

Thank you!

Substitute the data in your .mat file for your ‘x’ vector. Be certain to assign the sampling frequency for your signal to your ‘Fs’ variable.

Note that sound files (for example using the audioread function) are by definition column vectors, and usually have 2 columns (left channel and right channel). You may have to change your code to be compatible with that convention. Use the size function to determine the characteristics of your data.

You can create a time vector for your imported sound file as:

t = linspace(0, length(x), length(x))/Fs;

[Kristy Kwok]

Thank you for the instruction but it doesnt work for my .mat file :(

[Star Strider]

It works for it for me:

D = load('spectral_4.mat');
x = D.y;
Fs = 8000; % Sampling frequency. 
N = length(x);
t = linspace(0, N, N)/Fs;
xdft = fft(x);
xdft = xdft(1:N/2+1);
psdx = (1/(Fs*N)) * abs(xdft).^2;
psdx(2:end-1) = 2*psdx(2:end-1);
freq = 0:Fs/length(x):Fs/2;

plot(freq,10*log10(psdx))
grid on
title('Periodogram Using FFT')
xlabel('Frequency (Hz)')
ylabel('Power/Frequency (dB/Hz)')

Be certain that ‘Fs’ is correct for your signal. It is not in of your ‘spectral_4.mat’ file. It should be included.

Your signal is actually about 4 seconds long in your file. There is 1 second of silence followed by about 3 seconds of sound.

[Kristy Kwok]

Thank you very much Star! I can see now a better looking periodogram then my previous one. May I please ask what am I looking at here? I really appreciate your help! Many many thanks!

[Star Strider]

My pleasure!

You are looking at a simplified power spectral density of your plucked guitar string. The principal frequency appears to be at about 293.8 Hz, known as ‘D_4’.

[Kristy Kwok]

I cannot thank you enough Star! Thank you very much for all your help. Have a great week ahead! :D

[Star Strider]

As always, my pleasure!

I hope you do as well!