待解决问题
LTE层映射中层数为3的映射
离问题结束还有0天0小时  |  提问者:wumisty   |  提问时间:2015-1-12 21:53
LTE层映射中层数为3码字为2的映射中 协议上面是如此描述的:
x_0(i)=d_0(i)
x_1(i)=d_1(2i)
x_2(i)=d_1(2i+1)
M_symb_layer = M_symb_0=M_symb_1/2;
最后调制完了应该是
x_0(0) x_0(1) ....x_0(M_symb_layer )
X= x_1(0) x_1(1) .... x_1(M_symb_layer ) 吧
x_2(0) x_2(1) ... x_1(M_symb_layer )
但是看了很多代码都是如下一样将2个码字的数据分成3份,x0为第一份,x1和x2一次交替取后面两份,这样的话就不应该满足M_symb_layer = M_symb_0=M_symb_1/2这个式子了??
代码1:
x1 = d(1:(M_symb/TxLayer_Num));
for i = 1:(M_symb/TxLayer_Num)
x2(i) = d((M_symb/TxLayer_Num)+2*(i-1)+1);
x3(i) = d((M_symb/TxLayer_Num)+2*(i-1)+2);
end
x = [x1; x2; x3];
代码2:
if(layers==3 && num_codewords==2)
{    
for(i=0; i<inlen/3; i++)
  {
   outbuf[i]= inbuf[i];
   outbuf[i+(inlen/3)] = inbuf[inlen/3 + 2*i];
   outbuf[i+(inlen/3)*2] = inbuf[inlen/3 + 2*i + 1];
  }
              
  outlen = inlen;
}
 
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