Help with fourier transform

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aurc89
aurc89 el 18 de Sept. de 2014
Comentada: aurc89 el 19 de Sept. de 2014
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

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Geoff Hayes
Geoff Hayes el 18 de Sept. de 2014
Editada: Geoff Hayes el 18 de Sept. de 2014
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');

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