function Pe = CPSK(snr_in_db)%UNTITLED1 Summary of this function goes here%  Detailed explanation goes here% snr_in_db=4;fc=50;T=0.15;N=10000;ts=0.0001;VT=0;   %1/2log(lamta)-1/2*int(s0^2-s1^2)A=(5^1/2)/50;  %awgn(x,snr_in_db)的x必須是1mw的功率的（0dbw),故須算出此時幅度大小為多少，A^2/2，故A=                                        % E=1;                     % lamta=1; t=[0:ts:T];  %這就是怎么把 for 轉化成矩陣的方法    S0=A*sin(2*pi*fc.*t); %不能寫s0(t)= ... ,因為括號里的 值必須是整數；     S1=-A*sin(2*pi*fc.*t);                         % SNR=exp(snr_in_db*log(10)/10);                       % sgma=E/sqrt(2*SNR);k=length(t);r=zeros(N,k);numoferr=0;for i=1:N    temp=rand;    if(temp<0.5)        dsource(i)=0;    else        dsource(i)=1;    end    if(dsource(i)==0)        r(i,1:k)= awgn(S0,snr_in_db);    else        r(i,1:k)= awgn(S1,snr_in_db);    end    RT1(i)=int1(r(i,1:k).*S1,ts);% RT1=int(r(i).*S1)是錯的，這樣會只跟第一個量相乘而已    RT2(i)=int1(r(i,1:k).*S0,ts);   test1(i)=int1(S1.*S1,ts);   test2(i)=int1(S0.*S0,ts);    if( RT1(i)-RT2(i)>VT)        R(i)=1;    else        R(i)=0;    end    if(R(i)~=dsource(i))        numoferr=numoferr+1;    endendPe=numoferr/N;% K=S1.*S1;% K1=S0.*S0;% M1=int1(K,ts);% M2=int1(K1,ts);% %deploy int(r(t).*S1(t)&'發送信號' ,and so on% i=[1:N];%    subplot(3,2,1);%    plot(i,RT1);%    xlabel('i');%  ylabel('int(r(t).*S1(t)');%    subplot(3,2,2);%    plot(i,test1);%       xlabel('i');%  ylabel('int(S1(t).*S1(t)');%    subplot(3,2,3);%    plot(i,RT2);%       xlabel('i');%  ylabel('int(r(t).*S0(t)');%    subplot(3,2,4);%    plot(i,test2);%         xlabel('i');%  ylabel('int(S0(t).*S0(t)');%    subplot(3,2,5);%    plot(i,dsource(i));%           xlabel('i');%  ylabel('發送信號');%    subplot(3,2,6);%    plot(i,R(i));%            xlabel('i');%  ylabel('得到信號');% %  plot(i,RT1,i,test1,i,dsource(i))