基于ICETEK–VC5509-AE实验箱DSP算法有限冲激响应滤波器(FIR)算法实验 -

标题: 基于ICETEK–VC5509-AE实验箱DSP算法有限冲激响应滤波器(FIR)算法实验 [打印本页]

作者: cnbr 时间: 2023-11-2 21:53
标题: 基于ICETEK–VC5509-AE实验箱DSP算法有限冲激响应滤波器(FIR)算法实验
要求：输入波形为一个低频率的正弦波与一个高频的正弦波叠加，通过观察频域和时域图输入波形中的低频波形通过了FIR滤波器，而高频部分则大部
分被滤除。
请问：怎么修改频率，程序中fStepSignal1=2*PI/30;

fStepSignal2=2*PI*1.4;是什么意思

单片机源程序如下:
#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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