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C++ 使用FFTW并计算得到的傅里叶级数_C++_Fft_Fftw_Dft - Fatal编程技术网

C++ 使用FFTW并计算得到的傅里叶级数

C++ 使用FFTW并计算得到的傅里叶级数,c++,fft,fftw,dft,C++,Fft,Fftw,Dft,我是FFTW的新手。我想把一个函数分解成傅里叶级数。到目前为止,我还没有做到。我的代码如下: // 1) Create discretizations for my function 'my_function' int N = 100; // number of discretizations float x_step = (x_end - x_start) / ((float)(N - 1)); fftw_complex *Input, *Output; fftw_plan

我是FFTW的新手。我想把一个函数分解成傅里叶级数。到目前为止,我还没有做到。我的代码如下:

  // 1) Create discretizations for my function 'my_function'
  int N = 100; // number of discretizations
  float x_step = (x_end - x_start) / ((float)(N - 1));
  fftw_complex *Input, *Output;
  fftw_plan my_plan;
  Input = (fftw_complex*)fftw_malloc(sizeof(fftw_complex) * N);
  Output = (fftw_complex*)fftw_malloc(sizeof(fftw_complex) * N);
  float x = x_start;
  ForIndex(i, N) {
    Input[i][0] = my_function(x);
    cout << "Input[" << i << "]=" << Input[i][0] << endl;
    Input[i][1] = 0;
    x += x_step;
  }

  my_plan = fftw_plan_dft_1d(N, Input, Output, FFTW_FORWARD, FFTW_ESTIMATE);

  fftw_execute(my_plan);

  // 3) Evaluation - this is the part I am confused with
  // I should get something very close to 'my_function' when I plot, shouldn't I?
  ForIndex(i, N) {
    float sum = 0;
    float x = (float)i;
    sum = Output[0][0] / 2.0f;
    for (int k = 1; k < N; k++) {
      // Fourier series
      // http://en.wikipedia.org/wiki/Fourier_series
      float s = 2.0f*(float)M_PI * (float)k * x / (float)N;
      sum += Output[k][0] * std::cos(s) + Output[k][1] * std::sin(s);
      // I also tried
      // sum += Output[k][0] * std::sin(s + Output[k][1]);
      // to no avail
    }
    cout << "For x=" << x << ", y=" << (sum / (float)N) << endl;
  }
//1)为我的函数“我的函数”创建离散化
int N=100;//离散化次数
浮动x_步进=(x_结束-x_开始)/(浮动)(N-1));
fftw_复数*输入,*输出;
fftw_计划我的计划;
输入=(fftw_复合体*)fftw_malloc(sizeof(fftw_复合体)*N);
输出=(fftw_复合体*)fftw_malloc(sizeof(fftw_复合体)*N);
浮动x=x_开始;
ForIndex(i,N){
输入[i][0]=my_函数(x);

cout您非常接近预期结果!获得正确输出需要两点:

  • 第一项
    Output
    是输入信号的平均值。因此它是
    sum=Output[0][0];
    ,而不是
    sum=Output[0][0]/2.0f;
  • i
    是复数,
    i*i==-1
    因此,当虚部相乘时,必须在结果中添加减号。因此:

    sum += Output[k][0] * std::cos(s) - Output[k][1] * std::sin(s);
    
下面是一段经过更正的代码,将由
g++main.cpp-o main-lfftw3-lm
编译:

#include <fftw3.h>
#include <iostream>
#include <cmath>
#include <stdlib.h>

using namespace std;

float my_function(float x){
    return x;
}
int main(int argc, char* argv[]){

    // 1) Create discretizations for my function 'my_function'
    int N = 100; // number of discretizations
    float x_end=1;
    float x_start=0;
    int i;
    float x_step = (x_end - x_start) / ((float)(N - 1));
    fftw_complex *Input, *Output;
    fftw_plan my_plan;
    Input = (fftw_complex*)fftw_malloc(sizeof(fftw_complex) * N);
    Output = (fftw_complex*)fftw_malloc(sizeof(fftw_complex) * N);
    float x = x_start;
    for(i=0;i<N;i++){
        Input[i][0] = my_function(x);
        cout << "Input[" << i << "]=" << Input[i][0] << endl;
        Input[i][1] = 0;
        x += x_step;
    }

    my_plan = fftw_plan_dft_1d(N, Input, Output, FFTW_FORWARD, FFTW_ESTIMATE);

    fftw_execute(my_plan);

    for(i=0;i<N;i++){ 
      printf("%d %g+%gi\n",i,Output[i][0],Output[i][1]);
    }

    // 3) Evaluation - this is the part I am confused with
    // I should get something very close to 'my_function' when I plot, shouldn't I?
    for(i=0;i<N;i++) {
        float sum = 0;
        float x = (float)i;
        sum = Output[0][0];
        for (int k = 1; k < N; k++) {
            // Fourier series
            // http://en.wikipedia.org/wiki/Fourier_series
            float s = 2.0f*(float)M_PI * (float)k * x / (float)N;
            sum += Output[k][0] * std::cos(s) - Output[k][1] * std::sin(s);
            // I also tried
            // sum += Output[k][0] * std::sin(s + Output[k][1]);
            // to no avail
        }
        cout << "For i=" << i <<" x="<<x_start+x_step*i<< " f(x)="<<my_function(x_start+x_step*i)<<" y=" << (sum / (float)N) << endl;
    }

}
my_plan2 = fftw_plan_dft_1d(N, Output, input, FFTW_BACKWARD, FFTW_ESTIMATE);