C-programming. The program is to view a sequence of 30 points of speech data. The display will be shown in 5 columns with 6 data each. Then the program will show the statistic of its average power, average magnitude and how many zero crossings. Zero crossing is the event when the multiplication of two consecutive data result in negative number. /*––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––*/ /* Homework #9 program */ /* */ /* This program computes average power, average magnitude */ /* and zero crossings from a speech signal. */ #include #include #define MAXIMUM 50 int main(void) { /* Declare variables */ int k=0, npts=30; double speech[MAXIMUM]={0.000000,-0.023438,-0.031250,-0.031250,-0.039063,-0.039063,-0.023438,0.000000,0.023438,0.070313,-0.039063,-0.039063,0.046875,0.101563,0.117188,0.101563,0.070313,0.054688,0.023438,0.000000,-0.031250,-0.039063,-0.070313,-0.070313,-0.070313,-0.070313,-0.062500,-0.046875,-0.039063,-0.031250}; /* Declare the function prototypes */ /* Compute and print statistics. */ printf("\n SPEECH DATA "); print(,,);//complete the statement printf("\n"); printf(" SPEECH STATISTICS : \n"); printf(" average power: %f \n",);//complete the statement printf(" average magnitude: %f \n",);//complete the statement printf(" zero crossings: %d \n",);//complete the statement /* Exit program. */ return 0; } /* This function prints the data in the number of desired */ /* column numcols */ void print(double x[], int npts, int numcols) { int i,j; printf("\n "); for(j=0;j
C-
The program is to view a sequence of 30 points of speech data. The display will be shown in 5 columns with 6 data each. Then the program will show the statistic of its average power, average magnitude and how many zero crossings.
Zero crossing is the event when the multiplication of two consecutive data result in negative number.
/*––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––*/
/* Homework #9 program */
/* */
/* This program computes average power, average magnitude */
/* and zero crossings from a speech signal. */
#include <stdio.h>
#include <math.h>
#define MAXIMUM 50
int main(void)
{
/* Declare variables */
int k=0, npts=30;
double speech[MAXIMUM]={0.000000,-0.023438,-0.031250,-0.031250,-0.039063,-0.039063,-0.023438,0.000000,0.023438,0.070313,-0.039063,-0.039063,0.046875,0.101563,0.117188,0.101563,0.070313,0.054688,0.023438,0.000000,-0.031250,-0.039063,-0.070313,-0.070313,-0.070313,-0.070313,-0.062500,-0.046875,-0.039063,-0.031250};
/* Declare the function prototypes */
/* Compute and print statistics. */
printf("\n SPEECH DATA ");
print(,,);//complete the statement
printf("\n");
printf(" SPEECH STATISTICS : \n");
printf(" average power: %f \n",);//complete the statement
printf(" average magnitude: %f \n",);//complete the statement
printf(" zero crossings: %d \n",);//complete the statement
/* Exit program. */
return 0;
}
/* This function prints the data in the number of desired */
/* column numcols */
void print(double x[], int npts, int numcols)
{
int i,j;
printf("\n ");
for(j=0;j<numcols;j++)
printf("-----------");
printf("\n");
for(j=0;j<numcols;j++)
{
printf(" %2d to %2d",((npts/numcols)*j)+1,((npts/numcols)*j)+(npts/numcols));
}
printf("\n ");
for(j=0;j<numcols;j++)
printf("-----------");
printf("\n");
for(i=0;i<npts/numcols;i++)
{
for(j=0;j<numcols;j++)
printf("%11.6lf",x[((npts/numcols)*j)+i]);
printf("\n");
}
}
/*––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––*/
/* This function returns the average power of an array x */
/* with npts elements. */
double ave_power(double x[],int npts)
{
/* Declare and initialize variables. */
int k;
double sum=0;
/* Determine average power. */
for (k=0; k<=npts-1; k++)
sum += x[k]*x[k];
/* Return average power. */
return sum/npts;
}
/*––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––*/
/* This function returns the average magnitude of an array x */
/* with npts elements. */
double ave_magn(double x[],int npts)
{
/* Declare and initialize variables. */
int k;
double sum=0;
/* Determine average power. */
for (k=0; k<=npts-1; k++)
sum += fabs(x[k]);
/* Return average magnitude. */
return sum/npts;
}
/*––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––*/
/* This function returns a count of the number of zero */
/* crossings in an array x with npts values. */
int crossings(double x[],int npts)
{
/* Declare and initialize variables. */
int count=0, k;
/* Determine number of zero crossings. */
for (k=0; k<=npts-2; k++)
if (x[k]*x[k+1] < 0)
count++;
/* Return number of zero crossings. */
return count;
}
/*––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––*/
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