Question

我理解接收信号方程表示为y=hx+n。我在MCCDMA Rayleigh和AWGN频道中有一个从发送器到接收器的传输脚本。从一些论坛上看，我发现距离和路径损耗指数可以包括在频道增益中h = h* d^(-v/2);

其中d =距离和v =路径损失指数

以下是我的MATLAB代码的一半，其中包括我的瑞利信道中的距离和路径损耗指数：

Taps=4;                                         % Number of Taps
p1=0.5/2.3;                                     % Power of Tap1
p2=0.9/2.3;                                     % Power of Tap2
p3=0.7/2.3;                                     % Power of Tap3
p4=0.2/2.3;
gain1=sqrt(p1/2)*[randn(1,N) + j*randn(1,N)];   % Gain for Tap1
gain2=sqrt(p2/2)*[randn(1,N) + j*randn(1,N)];   % Gain for Tap2
gain3=sqrt(p3/2)*[randn(1,N) + j*randn(1,N)];   % Gain for Tap3
gain4=sqrt(p4/2)*[randn(1,N) + j*randn(1,N)];   % Gain for Tap4

% dist = distance between sender and transmitter ; -pathLossExp = path loss exponent

gain1=gain1*dist^(-pathLossExp/2);  % Path Loss for gain1
gain2=gain2*dist^(-pathLossExp/2);  % Path Loss for gain2
gain3=gain3*dist^(-pathLossExp/2);  
gain4=gain4*dist^(-pathLossExp/2);  

x11=x(:);
x12=reshape(x11,1,length(x11));
i=1:length(x12);        
delay1=1; 
for i=delay1+1:length(x12) % Producing one sample delay in Tap2 w.r.t. Tap1
   x13(i)=x(i-delay1);
end
delay2=2;
for i=delay2+1:length(x12) % Producing two sample delay in Tap2 w.r.t. Tap1
   x14(i)=x(i-delay2);
end
delay3=3;
for i=delay3+1:length(x12) % Producing three sample delay in Tap2 w.r.t. Tap1
   x15(i)=x(i-delay3);
end
x1=reshape(x13,(n+3),length(x13)/(n+3));
x2=reshape(x14,(n+3),length(x14)/(n+3));
x3=reshape(x15,(n+3),length(x15)/(n+3));
ch1=repmat(gain1,(n+3),1);     
ch2=repmat(gain2,(n+3),1);
ch3=repmat(gain3,(n+3),1);
ch4=repmat(gain4,(n+3),1);
data_channel=x.*ch1+x1.*ch2+x2.*ch3+x3.*ch4;  % Passing data through channel

但是，我需要一些帮助来验证包含的路径丢失是否正确。

如何将路径损耗包含在瑞利信道中？

0 个答案: