Question

我正在借助FFT对两个整数信号进行卷积，但是不知何故我做不到。我不确定FFT的实现是否正确。尤其是数学部分。

大修改：我现在发布了所有代码。抱歉，我不从此开始。我确定错误只是在FFT部分，但可能还有更多我忽略的问题。我知道代码很乱而且不干净。一切都是零散的，可以用更简单，更简洁的方式进行编程，但是我正在一点一点地进行测试。至于输入，它从命令行读取两个信号。建立为一个数字，表示信号有多大，信号以整数数组eq 2表示：[1，-1]和10：[0,0,0,1,1,1,1,0,0， 0]。然后，应通过对两个信号执行FFT，对信号进行卷积，然后进行逐位乘法。对结果信号进行逆FFT。再次打印其长度，然后打印包含整数的数组。打印本身正确，但是结果数组中的值不正确。我希望现在再次变得更加清楚，对不起，谢谢您的帮助。

#include <stdio.h>
#include <stdlib.h>
#include <math.h> 
#include <complex.h>

double PI;

int *readSignal(int *len) {    //reads the signal
    int *x;
    char c;
    scanf("%d:", len);
    x = calloc(*len, sizeof(int));
    do c = getchar(); while (c != '[');
    if (len > 0) {
       scanf("%d", &x[0]);
       for (int i=1; i < *len; i++) scanf(",%d", &x[i]);
    }
    do c = getchar(); while (c != ']');
    return x;
}

void printSignal(int len, int *x) { //prints the signal
    printf("%d: [", len);
    if (len > 0) {
        printf("%d", x[0]);
        for (int i=1; i < len; i++) 
            printf(",%d", x[i]);
        }
        printf("]\n");
    }

void *padSignal(int len, int lenSig, int *x) {  //ensures that the signal is of size 2^n by padding it with 0's
    int *padded;
    padded = calloc(len, sizeof(int));
    for (int i=0; i < lenSig; i++) {
        padded[i] = x[i];
    }
    return padded;
}

void fft(double complex signal[], int length, int power) {
    if (length == 1) {
        return;
    }
    double complex *signalODD = calloc((length/2+1), sizeof(double complex));
    double complex *signalEVEN = calloc((length/2+1), sizeof(double complex));
    int index1 = 0;
    int index2 = 0;
    for(int i = 0; i < length; i++) {
        if(i % 2 ==0) {
            signalEVEN[index1] = signal[i];
            index1++;
        }
        else {
            signalODD[index2] = signal[i];
            index2++;
        }
    }
    fft(signalEVEN,length/2, power+1);
    fft(signalODD,length/2, power+1);
    for(int i = 0; i<length/2-1; i++) {
        signal[i] = signalEVEN[i] + cexp((I*2*PI*i)/length)*signalODD[i];
        signal[i+length/2] = signalEVEN[i]-cexp((I*2*PI*i)/length)*signalODD[i];
    }
    free(signalODD);
    free(signalEVEN);
}

void ifft(double complex signal[], int length, int power) {
    if (length == 1) {
        return;
    }
    double complex *signalODD = calloc((length/2+1), sizeof(double complex));
    double complex *signalEVEN = calloc((length/2+1), sizeof(double complex));
    int index1 = 0;
    int index2 = 0;
    for(int i = 0; i < length; i++) {
        if(i % 2 ==0) {
            signalEVEN[index1] = signal[i];
            index1++;
        }
        else {
            signalODD[index2] = signal[i];
            index2++;
        }
    }
    fft(signalEVEN,length/2, power+1);
    ifft(signalODD,length/2, power+1);
    for(int i = 0; i<length/2-1; i++) {
        signal[i] = signalEVEN[i] + cexp((I*-2*PI*i)/length)*signalODD[i];
        signal[i+length/2] = signalEVEN[i]-cexp((I*-2*PI*i)/length)*signalODD[i];
    }
    free(signalODD);
    free(signalEVEN);
}


int checkPowerofTwo(double len) { //checks for the closed power of 2
    double x = 1;
    while(len > pow(2,x)) {
        x++;
    }
    return pow(2,x);
}

int main(int argc, char *argv[]) {
  int lenH, *H;
  int lenX, *X;
  int *paddedX;
  int *paddedH;
  double length;
  H=readSignal(&lenH); //reads in the signal H
  X=readSignal(&lenX); //reads in signal X

  length = lenH+lenX-1;
  paddedH=padSignal((length),lenH,H); //pads the signal to the length
  paddedX=padSignal((length),lenX,X); // pads the signal to the length

  double complex *signalX = calloc(length, sizeof(double complex)); //creats a complex signal X and fills it with paddedX
  for (int i = 0; i<length; i++) {
      signalX[i] = paddedX[i];
  }
  double complex *signalH = calloc(length, sizeof(double complex)); // same for H
  for (int i = 0; i<length; i++) {
      signalH[i] = paddedH[i];
  }
  fft(signalX, length, 1); //performs the fast fourier transform on X
  fft(signalH,length, 1); // performs the fast fourier transfom on H
  double complex *signalY = calloc(length, sizeof(double complex)); //makes complex signal Y
  for (int i = 0; i<length; i++) { //performs the convolution
      signalY[i] = signalX[i]*signalH[i];
  }
  ifft(signalY, length,1);

  int *output = calloc(length, sizeof(int)); //creates the final output signal
  for (int i = 0; i<length; i++) {
      output[i] = creal(signalY[i]);
  }
  printSignal(length,output);

  free(signalX);
  free(signalH);
  free(signalY);
  free(H);
  free(X);
  free(paddedH);
  free(paddedX);
  free(output);

  return 0;
}

Answer 1

在：

if(i % 2 ==0 && i != 0)

您为什么要排除i == 0？在if(i % 2 ==0)和fft中将其都更改为ifft。

在fft和ifft中，以下行：

for(int i = 0; i<lenght/2-1; i++) {

应为：

for(int i = 0; i<lenght/2; i++) {

在ifft中，递归意外地使用了fft：

fft(signalEVEN,lenght/2, power+1);
fft(signalODD,lenght/2, power+1);

将其更改为ifft。

calloc调用不需要那么多空间：

calloc((lenght/2 + 1), sizeof(double complex));

可以是：

calloc((lenght/2), sizeof(double complex));

此外，的正确拼写是“ length”。

在修复了这些问题后，fft和ifft例程似乎可用于某些肤浅的情况。

实施Cooley FFT的麻烦

1 个答案: