要求：輸入波形為一個低頻率的正弦波與一個高頻的正弦波疊加，通過觀察頻域和時域圖輸入波形中的低頻波形通過了FIR濾波器，而高頻部分則大部
分被濾除。
請問：怎么修改頻率，程序中fStepSignal1=2*PI/30;
fStepSignal2=2*PI*1.4;是什么意思
單片機源程序如下:
#include "myapp.h"
#include "ICETEK-VC5509-EDU.h"
#include "scancode.h"
#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);
}
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