C++ 中的基数 3,5 FFT

Question

我正在尝试在 C++ 中实现 A Radix-5,Radix-3 FFT，我已经设法编写了一个 Radix-2，但是当涉及到 Radix 3 或 5 时我有某种类型的错误，假设我做了一个 FFT 3 个样本，这将显示正确的结果，但是如果我执行 3 * 3 的 9 FFT，它不会显示正确的结果。

我最初从 Python 获取代码，它在那里工作，我试图简单地将它“复制”到 C++，这是原始的 Python 代码：

def fft(x):
    """
    radix-2,3,5 FFT algorithm
    """
    N = len(x)
    if N <= 1:
        return x
    elif N % 2 == 0:
        # For multiples of 2 this formula works
        even = fft(x[0::2])
        odd =  fft(x[1::2])
        T = [np.exp(-2j*np.pi*k/N)*odd[k] for k in range(N//2)]
        return [even[k] + T[k] for k in range(N//2)] + \
               [even[k] - T[k] for k in range(N//2)]
    elif N % 3 == 0:
        # Optional, implementing factor 3 decimation
        p0 = fft(x[0::3])
        p1 = fft(x[1::3])
        p2 = fft(x[2::3])
        # This will construct the output output without the simplifications
        # you can do explorint symmetry
        
        for k in range(N):
            return [p0[k % (N//3)] +
                    p1[k % (N//3)] * np.exp(-2j*np.pi*k/N) + 
                    p2[k % (N//3)] * np.exp(-4j*np.pi*k/N)]
    elif N % 5 == 0:
        #factor 5 decimation
        p0 = fft(x[0::5])
        p1 = fft(x[1::5])
        p2 = fft(x[2::5])
        p3 = fft(x[3::5])
        p4 = fft(x[4::5])

        return [p0[k % (N//5)] +
                p1[k % (N//5)] * np.exp(-2j*np.pi*k/N) + 
                p2[k % (N//5)] * np.exp(-4j*np.pi*k/N) + 
                p3[k % (N//5)] * np.exp(-6j*np.pi*k/N) +
                p4[k % (N//5)] * np.exp(-8j*np.pi*k/N)
               for k in range(N)]

x = [1,1.00071,1.00135,1.00193,1.00245,1.0029,1.00329,1.00361,1.00387]
assert(np.allclose(fft(x), np.fft.fft(x)))

这是我的 C++ 代码（fft.hpp）：

#define _USE_MATH_DEFINES
#pragma once
#include <cmath>
#include <vector>
#include <complex>


using std::vector;
using std::complex;

vector<complex<float>> slicing(vector<complex<float>> vec, unsigned int X, unsigned int Y, unsigned int stride)
{
    // To store the sliced vector
    vector<complex<float>> result;

    // Copy vector using copy function()
    int i = X;
    while (result.size() < Y)
    {
        result.push_back(vec[i]);
        i = i + stride;
    }
    // Return the final sliced vector
    return result;
}

void fft(vector<complex<float>>& x)
{
    // Check if it is splitted enough

    const size_t N = x.size();
    if (N <= 1)
        return;

    else if (N % 2 == 0)
    {
        //Radix-2
        vector<complex<float>> even = slicing(x, 0, N / 2, 2); //split the inputs in even / odd indices subarrays
        vector<complex<float>>  odd = slicing(x, 1, N / 2, 2);

        // conquer
        fft(even);
        fft(odd);

        // combine
        for (size_t k = 0; k < N / 2; ++k)
        {
            complex<float> t = std::polar<float>(1.0, -2 * M_PI * k / N) * odd[k];
            x[k] = even[k] + t;
            x[k + N / 2] = even[k] - t;
        }
    }
    else if (N % 3 == 0)
    {
        //Radix-3
        //factor 3 decimation
        vector<complex<float>> p0 = slicing(x, 0, N / 3, 3);
        vector<complex<float>> p1 = slicing(x, 1, N / 3, 3);
        vector<complex<float>> p2 = slicing(x, 2, N / 3, 3);

        fft(p0);
        fft(p1);
        fft(p2);

        for (int i = 0; i < N; i++)
        {
            complex<float> temp = p0[i % (int)N / 3];
            temp += (p1[i % (int)N / 3] * std::polar<float>(1.0, -2 * M_PI * i / N));
            temp += (p2[i % (int)N / 3] * std::polar<float>(1.0, -4 * M_PI * i / N));
            x[i] = temp;
        }
    }
    else if (N % 5 == 0)
    {
        //Radix-5
        //factor 5 decimation
        vector<complex<float>> p0 = slicing(x, 0, N / 5, 5);
        vector<complex<float>> p1 = slicing(x, 1, N / 5, 5);
        vector<complex<float>> p2 = slicing(x, 2, N / 5, 5);
        vector<complex<float>> p3 = slicing(x, 3, N / 5, 5);
        vector<complex<float>> p4 = slicing(x, 4, N / 5, 5);

        fft(p0);
        fft(p1);
        fft(p2);
        fft(p3);
        fft(p4);
        for (int i = 0; i < N; i++)
        {
            complex<float> temp = p0[i % (int)N / 5];
            temp += (p1[i % (int)N / 5] * std::polar<float>(1.0, -2 * M_PI * i / N));
            temp += (p2[i % (int)N / 5] * std::polar<float>(1.0, -4 * M_PI * i / N));
            temp += (p3[i % (int)N / 5] * std::polar<float>(1.0, -6 * M_PI * i / N));
            temp += (p4[i % (int)N / 5] * std::polar<float>(1.0, -8 * M_PI * i / N));
            x[i] = temp;
        }
    }
}

和 main.cpp：

#define _USE_MATH_DEFINES
#include <stdio.h>
#include <iostream>
#include "fft.hpp"


typedef vector<complex<float>> complexSignal;

int main()
{
    complexSignal abit;
    int N = 9;
    abit.push_back({1,0});
    abit.push_back({1.00071 ,0 });
    abit.push_back({1.00135 ,0 });
    abit.push_back({1.00193 ,0 });
    abit.push_back({1.00245 ,0 });
    abit.push_back({1.0029 ,0 });
    abit.push_back({1.00329 ,0 });
    abit.push_back({1.00361 ,0 });
    abit.push_back({1.00387 ,0 });
    //abit.push_back({ 1.0029 ,0 });
    std::cout << "Before:" << std::endl;
    for (int i = 0; i < N; i++)
    {
        //abit.push_back(data[0][i]);
        std::cout << abit[i] << std::endl;
    }
    std::cout << "After:" << std::endl;
    fft(abit);
    for (int i = 0; i < N; i++)
    {
        std::cout << abit[i] << std::endl;
    }
    return 0;
}

我从 CPP 得到以下输出：

(9.02011,0)
(5.83089,-4.89513)
(0.700632,-3.98993)
(-0.000289979,0.000502368)
(-0.00218513,0.000362784)
(-0.00179241,0.00139188)
(-0.000289979,-0.000502368)
(0.000175771,-0.00354373)
(-0.003268,-0.00558837)

虽然我应该得到：

(9.020109999999999+0j)
(-0.0032675770104925446+0.005588577982060319j)
(-0.0023772289746976797+0.0024179090499282354j)
(-0.0022250000000012538+0.0011691342951078987j)
(-0.002185194014811494+0.00036271471530890747j)
(-0.0021851940148113033-0.00036271471530980844j)
(-0.0022249999999994774-0.0011691342951105632j)
(-0.002377228974696629-0.0024179090499291786j)
(-0.00326757701049002-0.005588577982061138j)

如您所见，只有第一个结果是正确的。

Answer 1

在 Python 中它说k % (N//3) ，在 C++ 中它说i % (int)N / 3 。 这些不一样！ 请参阅C++ 中的运算符优先级。 您需要使用括号以正确的顺序执行操作： i % ((int)N / 3) 。

Pepijn Kramer关于将-2 * M_PI * i / N更改为-2.0 * M_PI * static_cast<double>(i)/static_cast<double>(N)是一种很好的做法，在我看来，但不应该在这种情况是因为从左到右的评估以及在浮点上下文中将整数自动提升为浮点数。 尽管如此，确保所有常量在需要的地方都是浮点数不会花费太多精力，并且可以防止难以发现的错误。