Python np.fft.irfft（）中的归一化因子是什么？

我正在使用

np.fft.irfft（）

作为计算Wigner分布程序的一部分。因为我不想要fft的标准化版本，所以我需要标准化因子来“撤销”标准化

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np.fft.irfft（）的归一化因子是什么？

我相信

numpy.fft.irfft

的范数在数组长度上是1。要看下面的程序：

import numpy as np

t = np.array([0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0])
x = np.fft.irfft(t)

print  "Array length: ", len(t), "\t 1/n: ", 1.0/float(len(t)), "\t max x val: ", max(x)


x1 = np.array([0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0])

t1 = np.fft.rfft(x1)

print  "Array length: ", len(t1), "\t 1/n: ", 1.0/float(len(t1)), "\t max t1 val: ", max(t1)

这将产生：

Array length:  128   1/n:  0.0078125     max x val:  0.00787401574803
Array length:  65    1/n:  0.0153846153846   max t1 val:  (1+0j)

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由此我相信irfft的范数是1/n，而rfft的范数是1。这与此一致（请参阅实现细节）。

如果您的答案不在这里：，您可能需要查看源代码，或者设计一些测试用例。逆FFT应该有一个1/N项，它是这样的，并且

x==rfft（irfft（x））

。当您进行正向FFT时，您会得到一个预期的DC项，即输入的总和，如果逆项使您回到原始值，则必须只有1/N“归一化”。您不应该查看

x

的总和，而不是最大值吗？