Continuous Time Fourier Transform of a signal matrix

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Anubhav Rohatgi 2012-7-28

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回答： Juhi Maraskole 2020-9-16

I have a matrix of 100 rows and 2 columns. Column 1 consists of time signal at frequency 50Hz (0.02 0.04 0.06...) column 2 is signal whose fft is to be determined. I would like to have a function that determines the Fourier transform of the signal at the frequency determined by the 1st column of the matrix.

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Answer 1

Wayne King 2012-7-28

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编辑：Wayne King 2012-7-28

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If the first column of your matrix is just the time vector with increments of 0.02 seconds, then just take the Fourier transform of the 2nd column

Let X be your matrix

xdft = fft(X(:,2));

If the signal is real-valued, you only need 1/2 the DFT to examine the amplitude spectrum.

The frequency vector can be formed as follows:

freq = 0:50/100:25;

For example:

    t = 0:0.02:(100*0.02)-0.02;
    x = cos(2*pi*10*t)+randn(size(t));
    X(:,1) = t';
    X(:,2) = x';
    % Now X is your matrix
    xdft = fft(X(:,2));
    xdft = xdft(1:length(xdft)/2+1);
    freq = 0:50/100:25;
    plot(freq,abs(xdft))
    xlabel('Hz'); ylabel('Magnitude')

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Anubhav Rohatgi 2012-7-28

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Thanks for your response, but I am not getting the results still here is my code:

    x = data(:,2);
    len = length(x)
    t = 0:0.02:(len*0.02)-0.02;
    xdft = fft(x);
    xdft = xdft(1:length(xdft)/2+1);
    freq = 0:50/len:25;
    plot(freq,abs(xdft))
    xlabel('Hz'); ylabel('Magnitude')

Output is a line of 0 magnitude and on the X axis

and getting a warning:::: Warning: Integer operands are required for colon operator when used as index > In test3 at 9

Datafile can be accessed from :::

https://docs.google.com/open?id=0BzuiGHpNJIxCTXZ1ek9Ob1c5azA

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Answer 2

Wayne King 2012-7-28

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编辑：Wayne King 2012-7-28

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You stated in your original post that your matrix had an even number of elements, 100x2. In fact your matrix is 9079x2. You data also has a mean which is nonzero so that will make the 0 frequency component very large, it's better to remove the mean first.

      x = data(:,2);
      x = detrend(x,0);
      len = length(x)
      t = 0:0.02:(len*0.02)-0.02;
      xdft = fft(x);
      xdft = xdft(1:(length(xdft)+1)/2);
      freq = 0:50/len:25;
      plot(freq,abs(xdft))      
      xlabel('Hz'); ylabel('Magnitude')

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Anubhav Rohatgi 2012-7-28

编辑：Anubhav Rohatgi 2012-7-28

Thanks alot .. you solved my problem. Thank you very very much...

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Answer 3

Anubhav Rohatgi 2012-8-28

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Hi Wayne,

I have tried to solve the problem but the plot I get is not what I want. I have breathing data that is very periodic. I have attached the data link to this

https://docs.google.com/open?id=0BzuiGHpNJIxCTXZ1ek9Ob1c5azA

I want to get frequency peaks at around 3Hz.

Please help me.