Question

MATLAB中fft的频率轴怎么计算

Answer 1

假设你的信号是 8个点，采样频率是 100Hz。

那么，该信号的频率是50Hz，那么频率轴每个间隔是 50/(8-1), 设为df
那么，频率轴是 0 df 2*df 3*df 4*df
也就是说，对于8个点的信号，你会得到频率间隔是 50/(8-1), 可以得到 8/2+1个频率点。
也就是说，对于N个点的信号，你会得到频率间隔是 50/(N-1), 可以得到 N/2+1个频率点。
注意，N是2的某次幂

Answer 2

下面是MATLAB演示程序中的一段关于FFT的代码。
First create some data. Consider data sampled at 1000 Hz. Start by forming a time axis for our data, running from t=0 until t=.25 in steps of 1 millisecond. Then form a signal, x, containing sine waves at 50 Hz and 120 Hz.

t = 0:.001:.25;
x = sin(2*pi*50*t) + sin(2*pi*120*t);

%%
% Add some random noise with a standard deviation of 2 to produce a noisy
% signal y. Take a look at this noisy signal y by plotting it.

y = x + 2*randn(size(t));
plot(y(1:50))
title('Noisy time domain signal')

%%
% Clearly, it is difficult to identify the frequency components from looking at
% this signal; that's why spectral analysis is so popular.
%
% Finding the discrete Fourier transform of the noisy signal y is easy; just
% take the fast-Fourier transform (FFT).

Y = fft(y,256);

%%
% Compute the power spectral density, a measurement of the energy at various
% frequencies, using the complex conjugate (CONJ). Form a frequency axis for
% the first 127 points and use it to plot the result. (The remainder of the 256
% points are symmetric.)

Pyy = Y.*conj(Y)/256;
f = 1000/256*(0:127); =====================【问】这个频率轴是怎么运算的？？为什么要在前面*1000/256?还有为什么只取前面一半的点？？
plot(f,Pyy(1:128))
title('Power spectral density')
xlabel('Frequency (Hz)')

%%
% Zoom in and plot only up to 200 Hz. Notice the peaks at 50 Hz and 120 Hz.
% These are the frequencies of the original signal.

plot(f(1:50),Pyy(1:50))
title('Power spectral density')
xlabel('Frequency (Hz)')