离散傅立叶变换的傅立叶级数

问题描述

我正在尝试通过离散傅立叶变换重新创建一个函数.在Matlab中，可以这样进行:

I'm trying to recreate a function from a discrete fourier transform. In Matlab it would be done like this:

function [y] = Fourier(dft,x)
n = length(dft);
y = cos(pi*(x+1)'*(0:n-1))*real(dft)+sin(pi*(x+1)'*(0:n-1))*imag(dft)
end

我在Python中的尝试失败了，因为我不知道如何正确地将所有系数相加

My attempt in Python is falling flat because I don't know how to add up all the coefficients correctly

def reconstruct(dft, x):
n = len(dft)
y = ([(coeff.real)*np.cos(np.pi*x*nn) + (coeff.imag)*np.cos(np.pi*x*nn) for coeff in dft for nn in range(0,n)])

但这是不正确的，因为我需要对n求和并将这些和加在一起.我要去哪里?

But this isn't correct because I need to sum over n and add those sums together. Where am I off?

我要重新创建的方程式如下:

The equation I am trying to recreate is below:

推荐答案

您正在运行两个嵌套循环，而不是一个.试试这个:

You were running two nested loops instead of one. Try this:

y = ([(dft[nn].real)*np.cos(np.pi*x*nn) + (dft[nn].imag)*np.cos(np.pi*x*nn) for nn in range(0,n)])

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