fft(x), why divide with L?
조회 수: 22 (최근 30일)
이전 댓글 표시
Hello, everybody.
Let me just introduce myself, I am kind of newbie of fft world!
Currently reading books and searching thru the internet, but I really can't find out how below things work.
the first example, it says;
Fs = 1000; % Sampling frequency
T = 1/Fs; % Sampling period
L = 1500; % Length of signal
t = (0:L-1)*T; % Time vector
S = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
X = S + 2*randn(size(t));
and for this signal, do fft.
Y = fft(X);
and
P2 = abs(Y/L);
I really don't know why divide Y with L?
the FFT formula I know, there is no dividing with L.... Why should I divide with L??
Please anyone can kindly tell me...?
Thanks!
댓글 수: 0
채택된 답변
Star Strider
2022년 10월 12일
The Fourier transform uses repeated summations of the original vector to create the transform vector. Dividing my the length of the original vector normalises the summations, resulting in an accurate estimate of the various frequency components. This is discussed more completely in the Wikipedia article on the Discrete Fourier transform in the section on Motivation.
추가 답변 (1개)
dpb
2022년 10월 12일
편집: dpb
2022년 10월 12일
The short answer is "Because!" <VBG>
The story behind the answer is in the reference document for the FFTW routine that MATLAB uses to compute the FFT.
The link is in the documentation "References" section and going there will take you to everything (and more) that you could ever possibly want to know about FFTW.
But, the answer to your Q? (which just expounds on "Because!" a little) is in the FAQ list at <FFTW FAQ 3.10>.
Another <conversation> within the last few days also points out that for correct energy conservation do NOT divide by the total N of the FFT but by the L of the input signal when N>L for zero-padding of the FFT for higher resolution/interpolation in the frequency domain.
참고 항목
카테고리
Help Center 및 File Exchange에서 Spectral Measurements에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!