c语言中的dct实现

问题描述

所以我试图在我的 c 代码中实现一个 DCT-ll 函数，但我不确定如何做到这一点？ 我使用这个公式：https://i.stack.imgur.com/UeUBd.png

以及具有这些值的数组：

static float array[] = {0.35,0.35,-0.37,-0.335,-0.285,-0.23,-0.215}

我似乎找不到任何关于如何做到这一点的帮助，我只在 MatLab 中完成过，而且非常简单

close all; clear all; clc ;
N=512; % signal length

M=120; %select the number of DCT coefficients

load mit200
%%
x=ecgsig(1:N,1);

%% DCT transform of signal
y=dct(x);

%% Select first M coefficients
y(M+1:end)=0;

%% reconstructed signal
xrec = idct(y);
plot(x);
hold on;
plot((xrec),'r.');
legend ( 'Original','Recovered');

我能找到的大部分帮助都是这样的：Discrete Cosine Transform DCT implementation C，但它总是使用 8x8 矩阵 --> inMatrix[8][8]?

这是我的代码

#include "contiki.h"

#include "math.h"
#include <string.h>
#include <stdio.h>
#include <stdlib.h>

//Define sizeof - macros
#define SHIFT_AMOUNT 2 // 2^16 = 65536
#define SHIFT_MASK ((1 << SHIFT_AMOUNT) - 1) // 65535 (all LSB set,all MSB clear)
#define pi 3.14
//double static signal[3] = {0.35,0.35};
//static int price = 500 << SHIFT_AMOUNT;
static int dec;
static int frac;
static float s = 0;
static float array[] = {0.35,0.02,-0.21,-0.33,-0.215,-0.2,-0.19,-0.25,-0.255,-0.245,-0.24,-0.235,-0.205,-0.18,-0.165,-0.195,-0.17,-0.155,-0.175,-0.185,-0.14,-0.12,-0.115,-0.13,-0.16,-0.11,-0.125,-0.095,-0.09,-0.08,-0.075,-0.07,-0.065,-0.055,-0.06,-0.04,-0.03,-0.02,-0.015,-0.01,0.015,0.035,0.025,0.04,0.03,0.045,0.05,0.01,-0.005,-0.025,-0.035,-0.045,-0.1,-0.085,-0.15,-0.145,-0.22,-0.225,-0.26,-0.27,-0.28,-0.305,-0.35,-0.34,-0.135,0.075,0.14,0.175,0.135,-0.425,-0.71,-0.915,-1.09,-1255,-1395,-1465,-1505,-1.49,-1.45,-1.39,-1.34,-1.27,-1155,-1025,-0.95,-0.93,-0.91,-0.885,-0.865,-0.84,-0.8,-0.755,-0.715,-0.66,-0.63,-0.58,-0.54,-0.51,-0.465,-0.435,-0.395,-0.365,-0.315,-0.265,0.005,0.055,0.08,0.085,0.09,0.11,0.13,0.15,0.155,0.18,0.185,0.205,0.22,0.235,0.255,0.27,0.29,0.3,0.305,0.325,0.315,0.31,0.345,0.36,0.37,0.375,0.365,0.385,0.41,0.405,0.4,0.395,0.39,0.38,0.355,0.28,0.25,0.21,0.165,0.16,0.145,0.1,-0.105,0.125,0.215,0.32,0.415,0.5,0.605,0.725,0.845,0.94,1.06,1.11,1.1,1005,0.86,0.615,0.24,-0.46,-0.515,-0.44,-0.385,-0.36,-0.32,-0.295,-0.3,-0.325,-0.335};
static int i;

static const int SIZE = 512;


void calcTest(){
  float array2[SIZE];
  for (i = 0; i < SIZE; i++) {
    s = array[i];
    array2[i] = (s * cosf(pi/SIZE * (i + .5) * 1);

    //Print array
    if (s < 0){
      dec = s;
      frac = -(s - dec) * 1000;
      if (dec == 0)
      {
        printf("\n Temp1: -%d.%02u\n",dec,frac);
      }else{
        printf("\n Temp1: %d.%02u\n",frac);
      }           
    }else{
      dec = s;
      frac = (s - dec) * 1000;     
      printf("\n Temp2: %d.%02u\n",(unsigned int)frac);
    }

    //Print array2 which contains the DCT values
    if (s < 0){
      dec = array2[i];
      frac = -(array2[i] - dec) * 1000;
      if (dec == 0)
      {
        printf("\n *Temp1: -%d.%02u\n",frac);
      }else{
        printf("\n *Temp1: %d.%02u\n",frac);
      }           
    }else{
      dec = array2[i];
      frac = (array2[i] - dec) * 1000;     
      printf("\n *Temp2: %d.%02u\n",(unsigned int)frac);
    }
  }
}



//Defining two processors,one for making the 'knock' and one to listen
PROCESS(data_comp,"data_comp");
AUTOSTART_PROCESSES(&data_comp);

/*---------------------------------------------------------------------------*/


PROCESS_THREAD(data_comp,ev,data)
{
  static struct etimer timer;

  PROCESS_BEGIN();  
  
  /* Setup a periodic timer that expires after 10 seconds. */
  etimer_set(&timer,CLOCK_SECOND * 10);
  
  while(1) {
    calcTest();
     /* Wait for the periodic timer to expire and then restart the timer. */
    PROCESS_WAIT_EVENT_UNTIL(etimer_expired(&timer));
    etimer_reset(&timer);
  }

  PROCESS_END();
}

解决方法

要使用公式实现 DCT-II，只需评估公式：

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


#define NumberOf(a) (sizeof (a) / sizeof *(a))


static void DCTII(size_t N,float *y,const float *x)
{
    static const float Pi = 0x3.243f6a8885a308d313198a2e03707344ap0f;

    for (size_t k = 0; k < N; ++k)
    {
        float sum = 0;
        for (size_t n = 0; n < N; ++n)
            sum += x[n] * cosf(Pi/N * (n + .5) * k);
        y[k] = sum;
    }
}


int main(void)
{
    float array[] = {0.35,0.35,-0.37,-0.335,-0.285,-0.23,-0.215};
    float result[NumberOf(array)];

    DCTII(NumberOf(array),result,array);

    for (size_t i = 0; i < NumberOf(result); ++i)
        printf("y[%zu] = %g.\n",i,result[i]);
}

对于更大的长度，这不是最有效的实现方式。