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BIOMEDICAL INSTRUMENTATION

BIOMEDICAL INSTRUMENTATION. 2012103843 REPORT #9 원지혜. Non-faded Display & QRS Detector. Non-faded Display. static int p=0; static int displayBuffer [1000]; displayBuffer [p]= newdata ; if((++p)==1000) p=0;. 0. 1. 2. 999. 998. erase. ring buffer 구현. paint. QRS Detector.

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BIOMEDICAL INSTRUMENTATION

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  1. BIOMEDICAL INSTRUMENTATION 2012103843 REPORT #9 원지혜

  2. Non-faded Display & QRS Detector Non-faded Display static int p=0; static intdisplayBuffer[1000]; displayBuffer[p]=newdata; if((++p)==1000) p=0; 0 1 2 999 998 erase ring buffer 구현 paint QRS Detector integration ECG Signal LPF HPF ( )2 (미분) BPF Digital Filter Thresholding Decision

  3. LPF & HPF H(Z) = = = y[n]-2y[n-1]+y[n-2]=x[n]-2x[n-6]+x[n-12] 1. LPF(Low Pass Filter) int QRS_LPF(int x) //y[n]=x[n]-2x[n-6]+x[n-12]+2y[n-1]-y[n-2] { static int y1=0, y2=0, x[26], p=12; int y; x[p]=x[p+13]=x; y=x[p]-(x[p+6]<<1)+x[p+12]+(y1<<1)-y2; y2=y1; y1=y; if(--p<0) p=12; return (y>>5); LPF의 출력을 HPF에 입력하여야 함. p 2.HPF(High Pass Filter) int QRS_HPF(int x) //y(n)=x[n-16]-1/32(l[n-1]+x[n]-x[n-32]) { static int l1=0, x[66], p=32; int l; x[p]=x[p+33]=x; l=l1+x[p]-x[n+32]; l1=l; if(--p<0) p=32; return(x[p+16]-(l>>5)); } HPF의 출력을 뒤의 필터의 입력으로 씀. 현재 것까지 13개 필요. 효율적이기 위해 두 배 사용 low pass filter (초기값. 뒤에서부터 저장)

  4. Differentiation & Square x2=x1; x1=x; return(y); } main() { int y1,y2…… d=getdata; y1=QRS_LPF(d); y2=QRS_HPF(y1); y3=QRS_Derivative(y2); } 4. Square(제곱) ① intQRS_Square { return(); } ② intQRS_Abs(int) { if(x<0) return(-x); else return(x); } 3. Differentiation(미분) intQRS_Derivative(int x) //y[n]=x[n]-x[n-1] //y[n]=1/8(2x[n]+x[n-1]-x[n-3]-2x[n-4]) ①간단한 Derivatitive { static int x1=0; int y; y=x-x1;(가장 간단) x1=x; return(y); } ②5 Point Difference { static int x1,x2,x3,x4; int y; y=((x<<1)+x1-x3-(x4<<1))>>3; x4=x3; x3=x2; ①보다 ②가 신호의 변하는 경향을 시간적으로 좀 더 넓게 반영한다. ->신호의 변화 추이에 더 좋다. ① 은 시간이 너무 오래 걸림. 따라서 ② 의 방법으로 구현해본다.

  5. Integration & two channel에서의 low pass filter 5. Integration(적분) int QRS_MWI(int x) (QRS 폭은 보통 100ms) //y(n)=1/32(x[n]+x[n-1]+……+x[n-31]) { static int x[32],p=0; static long sum=0; long ly; int y; if(++p==32) p=0; sum -=x[p]; sum+=x; x[p]=x; ly=(sum>>5); if(ly>32400) y=32400; else y=(int)ly; return y; } d1=GetDatach1(); d2=GetDatach2(); l1=MyLPF1(d1); l2=MyLPF2(d2); int MyLPF1(int x) { static int x1,x2,x3,x4; x4=x3; x3=x2; x2=x1; x1=x; return (y); } MyLPF2에 대해서도 같은 방법으로 함수를 만듦. ※LPF 함수를 두 개 쓰는 이유 하나를 쓰면 두 채널에 하나의 필터를 연결한 것이 됨. 따라서 필터링이 중첩되므로 문제가 됨. 참고) 여기서의 적분= 최근 몇 개의 데이터를 더하는 것 (일정한 폭을 가지는 window를 signal 들어올 때 마다 가지고 다니면서 더함)

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