Continuous Time Fourier Transform of a signal matrix

7 Ansichten (letzte 30 Tage)
Anubhav Rohatgi
Anubhav Rohatgi am 28 Jul. 2012
Beantwortet: Juhi Maraskole am 16 Sep. 2020
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.

Akzeptierte Antwort

Wayne King
Wayne King am 28 Jul. 2012
Bearbeitet: Wayne King am 28 Jul. 2012
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')
  1 Kommentar
Anubhav Rohatgi
Anubhav Rohatgi am 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 :::

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (3)

Wayne King
Wayne King am 28 Jul. 2012
Bearbeitet: Wayne King am 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')
  1 Kommentar
Anubhav Rohatgi
Anubhav Rohatgi am 28 Jul. 2012
Bearbeitet: Anubhav Rohatgi am 28 Jul. 2012
Thanks alot .. you solved my problem. Thank you very very much...

Melden Sie sich an, um zu kommentieren.


Anubhav Rohatgi
Anubhav Rohatgi am 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.

Juhi Maraskole
Juhi Maraskole am 16 Sep. 2020
x(t) = e-AT

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by