51工具盒子英文:
Fast Fourier Transform in Python?
问题 {#heading}
我有这个具有振荡的数据集：
y = array([ 9.88706879e-05, -1.80853647e-05, 2.42572582e-05,
1.12215205e-04, 1.32105126e-04, 1.13424614e-05, -1.58262175e-04, -2.62013276e-04,
-2.58070932e-04, -1.53975865e-04, -8.19357356e-05, -1.55734157e-04,
-2.90791620e-04, -3.70294471e-04, -3.46855608e-04, -2.23495910e-04,
-1.35441615e-04, -2.11411786e-04, -4.21891416e-04, -6.77753516e-04,
-8.09657243e-04, -6.97948704e-04, -5.01935670e-04, -4.20075723e-04,
-5.28464040e-04, -8.14942203e-04, -1.03669983e-03, -9.76604755e-04,
-7.50889655e-04, -5.34882634e-04, -4.06928662e-04, -3.96093220e-04,
-4.31306957e-04, -4.25399844e-04, -3.26933980e-04, -1.32440493e-04,
5.40550849e-06, -4.87299567e-05, -2.04672372e-04, -3.15870097e-04])
x = array([-25, -20, -15, -10, -5, 0, 5, 10, 15, 20, 25, 30, 35,
40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100,
105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165,
170])
如何在Python中测量快速傅里叶变换（Fast Fourier Transform）？
这是我的数据外观：
这是FFT的输出：
N = np.shape(x)
T = (200/N)*10e-12 # 时间步长（秒）
xf = fftfreq(N,1/T)[:N//2]
yf = fft(y)
plt.figure()
plt.plot(xf, 2.0/N*np.abs(yf[0:N//2]))
plt.xlabel('频率（Hz）')
plt.ylabel('f(t)')
我想知道这个脚本是否正确。 英文:
I have this dataset with oscillations:
y = array([ 9.88706879e-05, -1.80853647e-05, 2.42572582e-05,
1.12215205e-04,
1.32105126e-04, 1.13424614e-05, -1.58262175e-04, -2.62013276e-04,
-2.58070932e-04, -1.53975865e-04, -8.19357356e-05, -1.55734157e-04,
-2.90791620e-04, -3.70294471e-04, -3.46855608e-04, -2.23495910e-04,
-1.35441615e-04, -2.11411786e-04, -4.21891416e-04, -6.77753516e-04,
-8.09657243e-04, -6.97948704e-04, -5.01935670e-04, -4.20075723e-04,
-5.28464040e-04, -8.14942203e-04, -1.03669983e-03, -9.76604755e-04,
-7.50889655e-04, -5.34882634e-04, -4.06928662e-04, -3.96093220e-04,
-4.31306957e-04, -4.25399844e-04, -3.26933980e-04, -1.32440493e-04,
5.40550849e-06, -4.87299567e-05, -2.04672372e-04, -3.15870097e-04])
`x = array([-25, -20, -15, -10, -5, 0, 5, 10, 15, 20, 25, 30, 35,
40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100,
105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165,
170])
`
How can I measure the Fast Fourier Transform in Python?
That is how my data look:
and this is the output of the FFT:
N = np.shape(x)
T = (200/N)*10e-12 #time step in sec
xf = fftfreq(N,1/T)[:N//2]
yf = fft(y)
plt.figure()
plt.plot(xf, 2.0/N*np.abs(yf[0:N//2]))
plt.xlabel('freq.(Hz)')
plt.ylabel('f(t)')
I am wondering if the script is correct.
答案1 {#1}
得分: 1
以下是翻译好的部分：
-
10e-12不是10^-12；它是10^-11。请使用1e-12表示10^-12。 -
N = np.shape(x)会导致N成为一个元组，在代码的其他部分不能正常工作。所以给定的代码会出现TypeError错误。
但这并不能解决你的问题。我对FFT不太熟悉，但简化你的代码并结合一些直觉（可能是不正确的），我可以得到以下代码，看起来结果是正确的：
N = len(x)
dt = 5e-12
xf = fftfreq(N, dt)[:N//2]
yf = fft(y)
plt.figure()
plt.plot(xf, 2.0/N * np.abs(yf)[0:N//2])
plt.xlabel('freq.(Hz)')
plt.ylabel('f(t)')
（这里的"魔法"值 5e-12 可以通过多种方式获得。一种方式是 dt = (x[1] - x[0]) * 1e-12，假设 x 间隔相等（这应该是正确的），并将其转换为秒。）
这将生成下面的图像。在大约3e10 Hz附近有一个小峰值，这相当于大约1 / 3e10 = 3.3e-11秒或33皮秒的周期。根据你的问题中的x-y数据图，这看起来是正确的。
英文:
There are two practical mistakes in your original code:
-
10e-12is not 10^-12; it's 10^-11. Use1e-12for 10^-12 -
N = np.shape(x)results inNbeing a tuple, which shouldn't work in the other parts of the code. So the given code fails with aTypeError.
But that won't fix your problem. I'm not too familiar with FFTs, but simplifying your code a bit, combined with some intuition (which might be incorrect), I can get the following with seemingly correct results:
N = len(x)
dt = 5e-12
xf = fftfreq(N, dt)[:N//2]
yf = fft(y)
plt.figure()
plt.plot(xf, 2.0/N * np.abs(yf)[0:N//2])
plt.xlabel('freq.(Hz)')
plt.ylabel('f(t)')
(The "magic" 5e-12 can be obtained in numerous ways. One way is dt = (x[1] - x[0]) * 1e-12, assuming equal spacing for x (which it should have anyway) and converting to seconds.)
That yields the image below. There is a small peak around 3e10 Hz, which equals a period of about 1 / 3e10 = 3.3e-11 seconds or 33 picoseconds. Which seems about right, looking at the x-y data graph in your question.
