Fast Fourier Transform of a real function

3 vues (au cours des 30 derniers jours)
amin rasoulof
amin rasoulof le 7 Juil 2015
Réponse apportée : Walter Roberson le 8 Juil 2015
Hi! I got a question about the FFT properties; why FFT of a real function is not a real function as it is a property of Fourier transform? For example FFT of a gaussian function should be a gaussian, however, abs(fft(gaussian)) is a gaussian. thanks

Connectez-vous pour répondre à cette question.

Réponses (1)

Walter Roberson
Walter Roberson le 8 Juil 2015
That is not a property of the Fourier transform. The fft of a real function is generally complex and is always hermitian. The complex part is absent only if everything is in phase (e.g., the sum of sines)

Catégories

En savoir plus sur Fourier Analysis and Filtering dans Help Center et File Exchange

Tags

Produits

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by