MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn moreOpportunities for recent engineering grads.

Apply Today**New to MATLAB?**

Asked by Anubhav Rohatgi
on 28 Jul 2012

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.

*No products are associated with this question.*

Answer by Wayne King
on 28 Jul 2012

Edited by Wayne King
on 28 Jul 2012

Accepted answer

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')

Anubhav Rohatgi
on 28 Jul 2012

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

Answer by Wayne King
on 28 Jul 2012

Edited by Wayne King
on 28 Jul 2012

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')

Anubhav Rohatgi
on 28 Jul 2012

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

Answer by Anubhav Rohatgi
on 28 Aug 2012

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.

## 0 Comments