Help with fourier transform

Hello! I'm using this function to perform a simple Fourier Transform:

function four=FourierDir(t,s,nu)

% number of points in the time domain
N=length(t);
Nf=length(nu);

% sampling step in the time domain
Dt=diff(t);
Dt(N)=Dt(N-1);

four=zeros(1,Nf);

for ii=1:Nf

      four(ii)=sum(Dt.*s.*exp(-1i*2*pi.*t.*nu(ii)));

end;

How can I perform the same operation but using the matlab function 'fft' , given the same input parameters t,s,nu? s is the set of data I want to transform along t direction, nu is a vector I define for the number of points.

Thanks for the help

Answer 1

[Geoff Hayes]

Suppose the following (I'm assuming that you have done something similar for t, s, and nu)

 Fs = 8192;             % sampling rate (Hz)
 t  = 0:1/Fs:1-1/Fs;    % time period (seconds)
 s  = sin(2000*2*pi*t); % 2000 Hz signal
 N  = 8192;             % N-point FFT (block size)
 nu = 0:1:N-1;          % number of points

We use your Discrete Fourier Transform function (is that what it is?) to get

 dftY = FourierDir(t,s,nu);

and plot the results

 fHz  = 0:Fs/N:(N-1)*Fs/N;
 figure;
 plot(fHz,abs(dftY),'b');

Note that in the plot, we see the signal frequency at 2000 Hz with a height of 0.5 (not quite the amplitude of the signal).

Now if we do something similar with the MATLAB fft function as

 yFft = fft(s,N);

and plot on the same figure

 hold on;
 plot(fHz,abs(yFft)*(2/N),'g'); % multiply by 2/N since input signal is real

we see that the FFT'd signal has a frequency of 2000 Hz and an amplitude of 1.0 (the 2/N factor gives us the amplitude of the input signal).

So all that is different between the MATLAB fft output and yours is a factor of two. I guess it all depends on what your function, FourierDir, is returning. If we want the fft data to be the same, then we could just do

 hold on;
 yFft = fft(s,N)*(2/N)/2;  % now, yFft should be same as yDft
 plot(fHz,abs(yFft),'g');

[aurc89]

Thanks!