1d linear convolution in ANSI C code?

问题

Rather than reinvent the wheel, I wonder if anyone could refer me to a 1D linear convolution code snippet in ANSI C? I did a search on google and in stack overflow, but couldn't find anything in C I could use.

For example, for Arrays A, B, and C, all double-precision, where A and B are inputs and C is output, having lengths len_A, len_B, and len_C = len_A + len_B - 1, respectively.

My array sizes are small and so any speed increase in implementing fast convolution by FFT is not needed. Looking for straightforward computation.

回答1:

Here's how:

#include <stddef.h>
#include <stdio.h>

void convolve(const double Signal[/* SignalLen */], size_t SignalLen,
              const double Kernel[/* KernelLen */], size_t KernelLen,
              double Result[/* SignalLen + KernelLen - 1 */])
{
  size_t n;

  for (n = 0; n < SignalLen + KernelLen - 1; n++)
  {
    size_t kmin, kmax, k;

    Result[n] = 0;

    kmin = (n >= KernelLen - 1) ? n - (KernelLen - 1) : 0;
    kmax = (n < SignalLen - 1) ? n : SignalLen - 1;

    for (k = kmin; k <= kmax; k++)
    {
      Result[n] += Signal[k] * Kernel[n - k];
    }
  }
}

void printSignal(const char* Name,
                 double Signal[/* SignalLen */], size_t SignalLen)
{
  size_t i;

  for (i = 0; i < SignalLen; i++)
  {
    printf("%s[%zu] = %f\n", Name, i, Signal[i]);
  }
  printf("\n");
}

#define ELEMENT_COUNT(X) (sizeof(X) / sizeof((X)[0]))

int main(void)
{
  double signal[] = { 1, 1, 1, 1, 1 };
  double kernel[] = { 1, 1, 1, 1, 1 };
  double result[ELEMENT_COUNT(signal) + ELEMENT_COUNT(kernel) - 1];

  convolve(signal, ELEMENT_COUNT(signal),
           kernel, ELEMENT_COUNT(kernel),
           result);

  printSignal("signal", signal, ELEMENT_COUNT(signal));
  printSignal("kernel", kernel, ELEMENT_COUNT(kernel));
  printSignal("result", result, ELEMENT_COUNT(result));

  return 0;
}

Output:

signal[0] = 1.000000
signal[1] = 1.000000
signal[2] = 1.000000
signal[3] = 1.000000
signal[4] = 1.000000

kernel[0] = 1.000000
kernel[1] = 1.000000
kernel[2] = 1.000000
kernel[3] = 1.000000
kernel[4] = 1.000000

result[0] = 1.000000
result[1] = 2.000000
result[2] = 3.000000
result[3] = 4.000000
result[4] = 5.000000
result[5] = 4.000000
result[6] = 3.000000
result[7] = 2.000000
result[8] = 1.000000

回答2:

Not tested, but it seems like it would work...

void conv(const double v1[], size_t n1, const double v2[], size_t n2, double r[])
{
    for (size_t n = 0; n < n1 + n2 - 1; n++)
        for (size_t k = 0; k < max(n1, n2); k++)
            r[n] += (k < n1 ? v1[k] : 0) * (n - k < n2 ? v2[n - k] : 0);
}

Tip: If it takes less time to reinvent a wheel than to find one, do consider the former.

回答3:

Since, we are taking convolution of 2 finite length sequences, hence the desired frequency response is achieved if circular convolution is performed rather than linear convolution. A very simple implementation of circular convolution will achieve the same result as the algorithm given by Alex.

#define MOD(n, N) ((n<0)? N+n : n)
......
......

for(n=0; n < signal_Length + Kernel_Length - 1; n++)
{
    out[n] = 0;
    for(m=0; m < Kernel_Length; m++)
    {
        out[n] = h[m] * x[MOD(n-m, N)];
    }
}

回答4:

I used @Mehrdad's approach, and created the following anwer:

void conv(const double v1[], size_t n1, const double v2[], size_t n2, double r[])
{
    for (size_t n = 0; n < n1 + n2 - 1; n++)
        for (size_t k = 0; k < max(n1, n2) && n >= k; k++)
            r[n] += (k < n1 ? v1[k] : 0) * (n - k < n2 ? v2[n - k] : 0);
}

There's problem with index exceeding lower bound when in second loops k gets bigger than n, so, guess there should be extra condition to prevent that.

来源：https://stackoverflow.com/questions/8424170/1d-linear-convolution-in-ansi-c-code

标签

convolution