Fast Fourier Transform of a real function

조회 수: 2 (최근 30일)
amin rasoulof
amin rasoulof 2015년 7월 7일
답변: Walter Roberson 2015년 7월 8일
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

이 질문에 답변하려면 로그인하십시오.

답변 (1개)

Walter Roberson
Walter Roberson 2015년 7월 8일
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)

카테고리

Help Center 및 File Exchange에서 Fourier Analysis and Filtering에 대해 자세히 알아보기

태그

제품

Community Treasure Hunt

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

Start Hunting!

Translated by