自编64qam

自编64qam -...

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#include <stdlib.h> #include <stdio.h> #include <math.h> #define N 2400 #define PI 3.1415926 void rf(int A[(N/2)+1],int B[(N/6)+1]) */ {int i,n; n=1; for(i=1;i<=N/2 {if(A[i++]==0) {if(A[i++]==0) {if(A[i++]==0) {B[n++]=-7;} else {B[n++]=-5;} } else {if(A[i++]==0) {B[n++]=-1;} else {B[n++]=-3;} } } else {if(A[i++]==0) {if(A[i++]==0) {B[n++]=7;} else {B[n++]=5;} } else {if(A[i++]==0) {B[n++]=1;} else {B[n++]=3;} } } } } int f3(float A[(N/6)-62-63],int */ {int n,i; n=0; for(i=1;i<=((N/6)-63-63);i++) {if(A[i]>6) {B[++n]=7;} if((A[i]>4)±±(A[i]<6)) {B[++n]=5;} if((A[i]>2)±±(A[i]<4)) {B[++n]=3;} if((A[i]>0.5)±±(A[i]<2)) {B[++n]=1;} if((A[i]>-2)±±(A[i]<-0.5)) {B[++n]=- 1;} if((A[i]>-4)±±(A[i]<-2)) {B[++n]=-3;} if((A[i]>-6)±±(A[i]<-4)) {B[++n]=-5;} if(A[i]<-6) {B[++n]=-7;} } return n; } main() {FILE *fpS,*fp,*fpI1,*fpQ1; FILE *fpRI3,*fpRQ3,*fpRI2,*fpRQ2; /*FILE *fpI,*fpQ;*/ int S[N+1],Q[(N/2)+1],I[(N/2)+1]; int Q1[(N/6)+1],I1[(N/6)+1]; float R[65],RQ[(N/6)-62],RI[(N/6)-
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This note was uploaded on 04/24/2010 for the course COMUNICATI 1 taught by Professor 1 during the Spring '10 term at Xiamen University.

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&egrave;‡&ordf;&ccedil;&frac14;–64qam -...

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