手动FFT不能给我与FFT相同的结果 [英] Manual fft not giving me same results as fft

问题描述

import numpy as np
import matplotlib.pyplot as pp

curve = np.genfromtxt('C:\Users\latel\Desktop\kool\Neuro\prax2\data\curve.csv',dtype =     'float', delimiter = ',')
curve_abs2 = np.empty_like(curve)
z = 1j
N = len(curve)
for i in range(0,N-1):
    curve_abs2[i] =0
    for k in range(0,N-1):
        curve_abs2[i] += (curve[i]*np.exp((-1)*z*(np.pi)*i*((k-1)/N)))

for i in range(0,N):
    curve_abs2[i] = abs(curve_abs2[i])/(2*len(curve_abs2))

#curve_abs = (np.abs(np.fft.fft(curve)))
#pp.plot(curve_abs)
pp.plot(curve_abs2)
pp.show()

#后面的代码为我提供了3个值.但这只是...不同

The code behind # gives me 3 values. But this is just ... different

错误的^^这段代码: http://www.upload.ee/image/3922681/Ex5problem.png

Wrong ^^ this code: http://www.upload.ee/image/3922681/Ex5problem.png

使用numpy.fft.fft()进行更正: http://www.upload. ee/image/3922682/Ex5numpyformulas.png

Correct using numpy.fft.fft(): http://www.upload.ee/image/3922682/Ex5numpyformulas.png

推荐答案

有几个问题:

1. 您正在为curve_abs2的元素分配复杂的值，因此应声明为复杂的，例如curve_abs2 = np.empty_like(curve, dtype=np.complex128). (我建议使用名称curve_fft代替curve_abs2.)

1. You are assigning complex values to the elements of curve_abs2, so it should be declared to be complex, e.g. curve_abs2 = np.empty_like(curve, dtype=np.complex128). (And I would recommend using the name, say, curve_fft instead of curve_abs2.)

在python中，range(low, high)给出了序列[low, low + 1, ..., high - 2, high - 1]，因此必须使用range(0, N)(如果需要，可以将其简化为range(N))，而不是range(0, N - 1).

In python, range(low, high) gives the sequence [low, low + 1, ..., high - 2, high - 1], so instead of range(0, N - 1), you must use range(0, N) (which can be simplified to range(N), if you want).

您的公式中缺少2的因数.您可以使用z = 2j来解决此问题.

You are missing a factor of 2 in your formula. You could fix this by using z = 2j.

在内部循环中求和的表达式中，您将curve索引为curve[i]，但这应该是curve[k].

In the expression that is being summed in the inner loop, you are indexing curve as curve[i], but this should be curve[k].

在该表达式中，您无需从k中减去1，因为k循环的范围是0到N-1.

Also in that expression, you don't need to subtract 1 from k, because the k loop ranges from 0 to N - 1.

因为k和N是整数，并且您正在使用Python 2.7，所以表达式(k-1)/N中的除法将是整数除法，并且所有k都将得到0.要解决此问题和上一个问题，可以将该术语更改为k / float(N).

Because k and N are integers and you are using Python 2.7, the division in the expression (k-1)/N will be integer division, and you'll get 0 for all k. To fix this and the previous problem, you can change that term to k / float(N).

如果您解决了这些问题，则在第一个双循环结束时，数组curve_abs2(现在为复杂数组)应与np.fft.fft(curve)的结果匹配.不会完全一样，但是差异应该很小.

If you fix those issues, when the first double loop finishes, the array curve_abs2 (now a complex array) should match the result of np.fft.fft(curve). It won't be exactly the same, but the differences should be very small.

您可以使用numpy向量化计算完全消除该双重循环，但这是另一个问题.

You could eliminate that double loop altogether using numpy vectorized calculations, but that is a topic for another question.

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