Thread: Help with Arrays and Moving Averages

Hello all,

I'm writing a code that generates sine waves, adds random noise, and then cleans it up by averaging those values. I've got it so it averages every 3 values using a sliding point window, however I need to write it as an array and to tell the truth, I'm stumped. I understand what to do...I just can't think of how to implement that in C.

Code:

/* Alexander DeTrano
 This program generates normal sine waves then creates noise using random
 numbers and the "cleans up the wave" by a series of averages and then plots
 them on screen.  When N = 40000 a high pitched noise is created intitally and
 then when the wave with noise is added, a static is picked  up, then the
 averaged values clean it up and reduce some static.
*/
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <time.h>
#define N 40000
double r, pi, f = 1500, w, t ; 
int i; int n=N;
void create_sine(double y[], int n)
{
   for ( i = 0; i < n; ++i )
   {
      t = i/8000.0;
      y[i] =  sin(w*t);
   }
}
void print_array(double y[], int n)
{
   for ( i=0; i<n; ++i )
   {
      printf("%f\n",y[i]);
   }
}
void add_noise(double y[], int n)
{
 for( i=0; i<n; ++i)
  {
  r = 2*(rand()/(RAND_MAX+1.0)) - 1;
  t = i/8000.0;
  y[i]=r + sin(w*t);
  }
} 
void avgm( double y[], int n, int m)
/* this is my old function using constant values for the moving point window.
 * Basically a0 a1 and a2 need to go inside an array that's size m and have
   them averaged similar to how they are averaged here  */

{ double a0=0, a1=0, a2=0;
  for (i=0;i<n;++i)
   {
    a0=a1;
    a1=a2;
    a2=y[i];
    y[i]=(a0+a1+a2)/3.0;
   }
}
 
int main( void)
{
   int seed; double y[N]; int m;
   void create_sine(double y[], int n);
   void print_array(double y[], int n);
   void add_noise(double y[], int n);
   void avgm(double y[], int n, int m);
   pi = 4*atan(1);
   w = 2*pi*f;
   seed = time(0);
   srand(seed);
   create_sine( y, N);  
   print_array( y, N);
   add_noise(   y, N);
   print_array( y, N);
   avgm( y, N, 3);
    print_array( y, N);   
return 0;
}

Any help would be greatly appreciated!

Cheers,
Alexander

What is it that you don't know how to do exactly?


Quzah.

Use an array of pointers to type. Or a pointer to a pointer to type. Say you know your run length will be 100, but each of the elements in the run will range from 0 to 10. Now you could do an array of 100x10, or, you could use an array of pointers to type, and allocate each number / list as you run across them:

Code:

int *pa[ 100 ];
size_t x;
int n;

for( x = 0; x < 100; x++ )
{
    puts( "Enter the number of items here." );
    if( scanf( "&#37;d", &n ) == 1 )
    {
        pa[ x ] = malloc( sizeof( int ) * n );
        readnumbers( pa[ x ], n ); /* simplified for post length */
    }
}

If you're not sure how long your run is (where I used 100), or how long each is, you can prompt for that too, by using a pointer to pointer instead:

Code:

int **pa;
...
puts( "How many?" );
n = getnumber();
pa = malloc( n * sizeof *pa ); /* allocate the run length */
... now allocate each one as I did earlier ...

Then all you have to do is remember how long each sub-list is. You could use another array to store the length of each, or you could have allocated an extra element in each one (say the first one), and use it to store its length:

Code:

for( x = 0; x < array[ 0 ][ y ]; x++ )
{
    printf( "array[ %d ][ y ] is %d\n", x + 1, array[ x + 1 ][ y ] );
}

Where this would be nested in a loop that ran through the length of the run. Now were I doing it this way, if my "usable" length was 3, I would have actually 4 element, but the value of the first spot, which was storing the length, wouldn't count itself as part of the length. So in this case, it would store 3, instead of 4.

Search the forum for multi-dimensional arrays or dynamic arrays or something like that, and you should find plenty of hits on how to make and free them, because it gets asked all the time.


Quzah.