why real part of fft (exp(at)) is negative in some parts in contrary to analytical calculation of fft(exp(at))?
10 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
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 Kommentare
Antworten (2)
Dr. Seis
am 15 Jun. 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 Kommentare
Siehe auch
Kategorien
Mehr zu Fourier Analysis and Filtering finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!