英文:
Fast Fourier Transform in Python?
问题
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?
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
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-12
is not 10^-12; it's 10^-11. Use1e-12
for 10^-12 -
N = np.shape(x)
results inN
being 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.
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