给出采样率FSample并转换块大小N，您可以计算频率分辨率deltaF，采样间隔deltaT和总捕获时间capT使用以下关系:

Given sampling rate FSample and transform blocksize N, you can calculate the frequency resolution deltaF, sampling interval deltaT, and total capture time capT using the relationships:

deltaT = 1/FSample = capT/N
deltaF = 1/capT = FSample/N

还请记住，FFT将值从0返回到FSample，或者等效地从-FSample/2返回到FSample/2.在您的绘图中，您已经将-FSample/2放到0部分. NumPy包括一个帮助程序函数，可以为您计算所有这些信息: fftfreq .

Keep in mind also that the FFT returns value from 0 to FSample, or equivalently -FSample/2 to FSample/2. In your plot, you're already dropping the -FSample/2 to 0 part. NumPy includes a helper function to calculate all this for you: fftfreq.

对于deltaT = 1 hour和N = 576的值，您得到deltaF = 0.001736 cycles/hour = 0.04167 cycles/day，从-0.5 cycles/hour到0.5 cycles/hour.因此，如果您在bin 48(和bin 528)处有一个幅度峰值，则它对应于48*deltaF = 0.0833 cycles/hour = 2 cycles/day.

For your values of deltaT = 1 hour and N = 576, you get deltaF = 0.001736 cycles/hour = 0.04167 cycles/day, from -0.5 cycles/hour to 0.5 cycles/hour. So if you have a magnitude peak at, say, bin 48 (and bin 528), that corresponds to a frequency component at 48*deltaF = 0.0833 cycles/hour = 2 cycles/day.

通常，您应在计算FFT之前对您的时域数据应用窗口函数，以减少光谱泄漏. Hann窗口几乎从来都不是一个坏选择.您也可以使用rfft功能跳过输出的-FSample/2, 0部分.因此，您的代码将是:

In general, you should apply a window function to your time domain data before calculating the FFT, to reduce spectral leakage. The Hann window is almost never a bad choice. You can also use the rfft function to skip the -FSample/2, 0 part of the output. So then, your code would be:

ft = np.fft.rfft(signal*np.hanning(len(signal)))
mgft = abs(ft)
xVals = np.fft.fftfreq(len(signal), d=1.0) # in hours, or d=1.0/24 in days
plot(xVals[:len(mgft)], mgft)