How can I plot the fft of a pressure data?

Hello. I have a question to all. I really want to know to plot fft my pressure data. So, I tried to do FFT, but it didn`t go well. If there is someone can catch mistake, then please may I get some advise? I attached my file!

 [fname,pathname] = uigetfile('.csv','Select CSV File to Load, Plot, Compute RMS & FFT');
    disp([pathname fname])

    data = csvread([pathname fname]);

    [N,m] = size(data);
    t = data(:,1);         %time in seconds
    x = data(:,2);         %array of data
    Fs = 1/(t(2)-t(1));

    figure(1)
    plot(t,x)
    xlabel('Time (s)');
    ylabel('Pressure');
    title(chamber pressure);
    grid on;

    freq = 0:Fs/length(x):Fs/2;           %frequency array for FFT
    xdft = fft(x);                        %Compute FFT
    xdft = 1/length(x).*xdft;             %Normalize
    xdft(2:end-1) = 2*xdft(2:end-1); 

    figure(3)
    plot(freq,abs(xdft(1:floor(N/2)+1)))
    xlabel('Frequency (Hz)');
    ylabel('Fourier spectrum');
    title(['FFT' chamber pressure]);
    grid on;

Answer 1

[Star Strider]

I continue to recommend the R2015a documentation for fft (link). Specifically not the code between the first (top) two plot figures.

[abstine33]

Dear Star Strider. Thanks to your answer. Figure(1) is just plot of the chamber pressure. Thanks, I will try documentation for fft. After that, If I still get question, I reply on you.

[Star Strider]

My pleasure.

I posted my Answer before you posted your code and data. Your Fourier transform code produces the correct result, so there is no need to change it.

The problem is that the d-c offset is 1.2157 and the rest of your amplitude data are on the order of 1E-004. Use semilogy instead of plot (or subtract the mean from ‘x’ before doing the fft) to see the rest of it.

Example:

figure(3)
semilogy(freq,abs(xdft(1:floor(N/2)+1)))
xlabel('Frequency (Hz)');
ylabel('Fourier spectrum');
title(['FFT' ' chamber pressure']);
grid on;

Your Fourier transform code and plot otherwise produces the same results mine does.

[abstine33]

I really appreciate to you to answer nicely. And your code is more simple than mine. :)

I can also knew code about semilogy thanks to you. However, as you mentioned I need to subtract the mean from 'x'.

So, may I get advise about how can I subtract the mean from 'x'? Thank you. -abstine33-

[Star Strider]

My pleasure.

I would do it as:

Fn = Fs/2;

FTx = fft(x-mean(x))/N;
Fv = linspace(0, 1, fix(N/2)+1)*Fn;
Iv = 1:length(Fv);

figure(4)
plot(Fv, abs(FTx(Iv))*2)
grid
set(gca, 'Ylim',[0 2.5E-003])

So the ‘FTx’ assignment changes so the mean of ‘x’ is subtracted from every element of ‘x’ before taking the Fourier transform. (There is nothing special about my Fourier transform calculation, it is the way I am used to coding it. It gives the same result as yours.)

It reveals that you have a peak at about 2700 Hz and a second peak at about 2600 Hz, and several other peaks that seem to be sub-harmonics.

Use the Signal Processing Toolbox findpeaks function to determine the exact frequencies and amplitudes. The findpeaks function has several name-value pair options that make it easier to use. See the documentation for it for details.

[abstine33]

OH MY GOD! You are genius!!

Thank you so much to make FFT plot!!!! :)

I think the peaks on 2700~2600 Hz is caused by the Helmholz. But as your advise, I will try to use the findpeaks.

Again, thank you all your kind answers!

[Star Strider]

Thank you!

My pleasure!

Answer 2

[Youssef Essakhi]

xdft = 1/length(x).*xdft;             %Normalize
  xdft(2:end-1) = 2*xdft(2:end-1);

plz i don't understand this link . can you explain to me . thank you