why real part of fft (exp(at)) is negative in some parts in contrary to analytical calculation of fft(exp(at))?
10 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
Hi, I want to calculate the real part of fourier transform of A = exp(at) where "a" is a constant and t = [0 T]. when I calculate it from the analytical definition of the fourier transform, the real( A(f)) >0 ; however, when I use Matlab for some parts real (A(f))<0. In other words, the graphs from two methods are not identical for some ranges. Can you please help me to understand why FFT from Matlab and analytical calculation doesn't match? I did the same procedure for B =exp(-at) and in this case both results had a good agreement. Thank you
0 Commenti
Risposte (2)
Dr. Seis
il 15 Giu 2012
Try plotting the abs result from FFT. Basically, you will not get a purely real, all positive frequency spectrum unless your time domain signal is symmetric and also all positive (or at least nearly all positive).
Kind of a similar situation as here:
0 Commenti
Vedere anche
Categorie
Scopri di più su Fourier Analysis and Filtering in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!