time signal to fft and back ifft

Hi there

 I am using a time series and calculating fft using example provided by MATLAB

 L = length(b); 

 NFFT = 2^nextpow2(L);

 y = fft(b, NFFT)/L;

 f= Fs/2*linspace(0,1,NFFT/2);

But when i use inverse fourier transform ifft, i do not get real values, i get complex values and the result is not same as the original signal b. I am using y_source = ifft(X,NFFT);

y_source is not same as original signal b.

Could anyone please answer this. Thanks

[Atul]

I forget to tell that in frequency domain i am muting some y values corresponding to some frequencies, so when i calculate real(ifft(X,NFFT)) i do not get back the signal but if i dont mute those very few values then i recover the signal. The thing is i have muted only 12 values out of more than 1/2 million.

[Daniel Shub]

You need to be careful that you keep symmetry. You haven't shown us how you "mute" the values.

Answer 1

[Wayne King]

Note sure why you're dividing by the length of the signal and padding the DFT AND expecting to get the signal back.

    t = 0:0.001:1-0.001;
    x = cos(2*pi*100*t);
    xdft = fft(x);
    xhat = ifft(xdft);
    max(abs(x-xhat))

Answer 2

[ANDRIAMANANTENA Laza]

Hello, I have the same problem with my handling and it blocks the rest of my code.

my y signal is

    y=sin(2*pi*500*t) with t=linspace(0,5/Fs,5) Fs=44100

When I proceed with fft like

    YF= fft(y,NFFT)/length(y)

and try to recover my original signal with

    y2=ifft(YF,NFFT,'symmetric')

I don't have the original signal

Answer 3

[Atul]

Wayne: Because here in example given by MATLAB, they have divided the FFT by L, L=length(x) http://www.mathworks.in/help/matlab/ref/fft.html

could you please elaborate if it indeed needs to be divided by length or not and why. In my opinion there is something wrong with Matlab FFT. FFT from matlab does not matches with FFT from say PITSA. So be careful.