接着爲了確定計算小區速率在空分上出了什麼問題,我們將這種計算小區速率的思想運用到時間同步的方法中來和paper的結果作對比,以便找出其中的問題。
clear;
SNR=0;
result=zeros(1,10);
for i=1:10000
result1=zeros(1,10);
pu=10;
%pr=10db
pr=10;
r=-3.8;
K=15;
z=0.01;
d_in=rand(1,K)*0.99+0.01;
theta_in=rand(1,K)*(2*pi);
d_out=rand(6,K)*0.99+0.01;
theta_out=rand(6,K)*(pi);
for j=1:K
x=1;
result2=zeros(1,10);
for M=50:50:500
%產生alpha
beta_of_alpha=zeros(1,7);%總共三個區域
d1=sqrt(2^2+d_out(1,j)^2-2*2*d_out(1,j)*cos(theta_out(1,j)));
d2=sqrt(2^2+d_out(2,j)^2-2*2*d_out(2,j)*cos(theta_out(2,j)));
d3=sqrt(2^2+d_out(3,j)^2-2*2*d_out(3,j)*cos(theta_out(3,j)));
d4=sqrt(2^2+d_out(4,j)^2-2*2*d_out(4,j)*cos(theta_out(4,j)));
d5=sqrt(2^2+d_out(5,j)^2-2*2*d_out(5,j)*cos(theta_out(5,j)));
d6=sqrt(2^2+d_out(6,j)^2-2*2*d_out(6,j)*cos(theta_out(6,j)));
beta_of_alpha(1)=z*d_in(j)^r;
beta_of_alpha(2)=z*d1^r;
beta_of_alpha(3)=z*d2^r;
beta_of_alpha(4)=z*d3^r;
beta_of_alpha(5)=z*d4^r;
beta_of_alpha(6)=z*d5^r;
beta_of_alpha(7)=z*d6^r;
alpha_jk=sum(pr*beta_of_alpha*K)+1;
%產生first項
beta_target_user=z*d_in(j)^r;
first=pu*pr*beta_target_user^2*K*(M-1)/alpha_jk;
%產生的second項
second=pu*beta_target_user*(1-pr*beta_target_user*K/alpha_jk);
%產生third項
third=0;
for m=2:7
third1=pu*beta_of_alpha(m)*(1+pr*beta_of_alpha(m)*K*(M-2)/alpha_jk);
third=third+third1;
end
% third=pu*beta_of_alpha(2)*(1+pr*beta_of_alpha(2)*K*(M-2)/alpha_jk)+pu*beta_of_alpha(3)*(1+pr*beta_of_alpha(3)*K*(M-2)/alpha_jk);
%產生forth項,forth項共分爲小區內的干擾和小區間的干擾
forth_1=sum(pu*z*d_in.^r)-pu*z*d_in(j);
%注意相鄰小區到本小區的距離的計算
forth_2=sum(sum(pu*z*d_out_ture.^r))-sum(pu*z*d_out_ture(:,j).^r);
% forth_2=sum(pu*z*d_out_ture(1,:).^r)-pu*z*d_out_ture(1,j)^r;
% forth_3=sum(pu*z*d_out_ture(2,:).^r)-pu*z*d_out_ture(2,j)^r;
forth=forth_1+forth_2;
result2(x)=first/(second+third+forth+1);
x=x+1;
end
result1=result1+result2;
end
result=result+result1;
end
result=result/10000;
result0=6/15*log2(1+result);
M=[50:50:500];
plot(M,result0,'x-');
xlabel('Numbers of antennas at BS');
ylabel('Sum rate(bits/symbol)');
hold on;
運行結果表明:確實小於paper中的結果。
新的問題出現了:查找代碼中的問題項。
依然是forth項的問題嗎?
- 看了代碼,沒有感覺到forth項的代碼有什麼問題
- 基於第一點,我們可以得到一個結論:在那些距離基站較近的用戶可以有更大的速率
d的分佈如下:
在d=0.8148時,我們有result
在d=00651的時候,我們有result
這個值是合理的嗎???
更多謎底等待明天探索。