FFT / IFFT question .

I have a 4000 points time history rectangular pulse described as

t=linspace(0,0.4,4000);

for i=1:4000 if t(i)>=0&t(i)<=0.01 r(i)=1e5; else r(i)=0; end end

in the frequency domain i have :

y=r; L=length(y); Fs=L/0.4; NFFT = L; % Next power of 2 from length of y Y = fft(y,NFFT); f = Fs/2*linspace(0,1,NFFT/2);

The issue is :

If i use ifft(Y) i got the original time history (obviously) - no problem

but if i take half of the frequency domain vector and use ifft with the symmetric argument i got messed up time hystory . WHY ????

J=Y(1:NFFT/2+1); TD=ifft(J, 'symmetric');

TD is not the original time domain pulse . What am i missing here ?

Answer 1

[Wayne King]

Hi Marcelo, Even if you use 'symmetric' option, you still have to provide the entire frequency vector input. 'symmetric' just takes care of the rounding errors that can yield nonzero imaginary parts in the inverse Fourier transform of a conjugate symmetric Fourier representation.

You should input Y and not J. By inputtting

J = Y(1:/NFFT/2+1), you are giving ifft() an input vector that is not conjugate symmetric.

[Marcelo]

Thanks for your answer Wayne .

I didn`t realize that . What am i supposed to do if i have only half of the spectrum ? add fliplr(conj(Y)) to the original frequency vector before use ifft with symmetric argument ?

[Wayne King]

Hi Marcelo, yes, but be careful with 0 and the Nyquist, they only occur once, so you don't want to repeat DC twice for example.

Answer 2

[Knut]

Actually:

>> ifft([1 2 3 2 1], 'symmetric')
ans =
2.2000   -0.5236   -0.0764   -0.0764   -0.5236

>> ifft([1 2 3 NaN inf], 'symmetric')
ans =
2.2000   -0.5236   -0.0764   -0.0764   -0.5236

[Wayne King]

Hi Knut, that is correct, but that is not what the OP was doing. He was trying to do this.

rng default;

x = randn(8,1);

xdft = fft(x);

% Now truncate the DFT to only include DC up to and including the Nyquist

ifft(xdft(1:5),'symmetric');

That does not work. The OP was attempting to just provide ifft() from DC to the Nyquist.

[Knut]

Ah, I made the same error myself once. The "logical" way for the symmetric option to work would be that you only supplied the non-redundant vector, and I think that is how the underlying FFTW library works anyways. But in MATLAB it seems that we have to make a dummy vector 2x its needed length, before it gets internally stripped (?) down to 1x prior to calling FFTW.

I was hoping to speedup an application by doing minimal preprocessing before the call to IFFT, when only the non-redundantcoefficients were available to me.