IFFT for exponential function

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Michael
Michael 2015년 5월 19일
편집: Walter Roberson 2015년 5월 19일
So the inverse Fourier transform of 1/(1+i*s) is exp(-y), I'm trying to verify this using ifft. My code is below and note that I use factor to account for the necessary N/(2pi) prefactor. The shape is normally correct but the amplitude is always off. I use s to denote frequency space and y for the regular function domain. I don't have a massive background on signal processing but if I'm missing something major I can read into it.
NN=2^16;
dy=.01;
ds=2*pi/(dy*NN);
grid=0:ds:(NN-1)*ds;
Lt=(1./(1+1i*20.*grid));
factor=ds*NN/(2*pi);
transform=ifft(Lt)*factor;
freq=0:pi/ds:2*pi/((NN-1)*ds);
density=real(transform);
plot(freq(1:length(freq)),density(1:NN/2));

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