fir.c

该源码是有限冲击响应滤波器（FIR）算法的实现例子

/////////////////////////////////////////////////
//  Example For ICETEK-VC5416-EDU              //
//          CTR Version : V4                   //
//  Filename: Fir.c                            //
//  Project : Fir.pjt                          //
//  Version : 2.00                             //
//  Write by: Daniel Hawk                      //
//  Company : Realtimedsp Co.Ltd.              //
//                                             //
//  All Rights opened & no Onus  2005.06       //
/////////////////////////////////////////////////
#include <math.h>

#define FIRNUMBER 25
#define SIGNAL1F 1000
#define SIGNAL2F 4500
#define SAMPLEF  10000
#define PI 3.1415926

float InputWave();
float FIR();

float fHn[FIRNUMBER]={ 0.0,0.0,0.001,-0.002,-0.002,0.01,-0.009,
                       -0.018,0.049,-0.02,0.11,0.28,0.64,0.28,
                       -0.11,-0.02,0.049,-0.018,-0.009,0.01,
                       -0.002,-0.002,0.001,0.0,0.0
                     };
float fXn[FIRNUMBER]={ 0.0 };
float fInput,fOutput;
float fSignal1,fSignal2;
float fStepSignal1,fStepSignal2;
float f2PI;
int i;
float fIn[256],fOut[256];
int nIn,nOut;

main()
{
	nIn=0; nOut=0;
	f2PI=2*PI;
	fSignal1=0.0;
	fSignal2=PI*0.1;
	fStepSignal1=2*PI/30;
	fStepSignal2=2*PI*1.4;
	while ( 1 )
	{
		fInput=InputWave();
		fIn[nIn]=fInput;
		nIn++; nIn%=256;
		fOutput=FIR();
		fOut[nOut]=fOutput;
		nOut++;				/* break point */
		if ( nOut>=256 )
		{
			nOut=0;		
		}
	}
}

float InputWave()
{
	for ( i=FIRNUMBER-1;i>0;i-- )
		fXn[i]=fXn[i-1];
	fXn[0]=sin((double)fSignal1)+cos((double)fSignal2)/6.0;
	fSignal1+=fStepSignal1; 
	if ( fSignal1>=f2PI )	fSignal1-=f2PI;
	fSignal2+=fStepSignal2;
	if ( fSignal2>=f2PI )	fSignal2-=f2PI;
	return(fXn[0]);
}

float FIR()
{
	float fSum;
	fSum=0;
	for ( i=0;i<FIRNUMBER;i++ )
	{
		fSum+=(fXn[i]*fHn[i]);
	}
	return(fSum);
}