返回信息流谁能解释一下,TSE今天下午讲的,INR/SNR--〉无穷的时候,为什么两用户的容量/速率 能达到1。 问了半天,还是不明白,很丢人的。
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Re: 清华办的无线通信讲席教授组系列报告通知:牛人如织
heyjude
2009/3/24镜像同步7 回复
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I guess it should like this:
Since INR/SNR -> infinite, the crosstalk dominates. So t1 -> r2 and t2 -> r1 are both error-free. Then r1 and r2 can exchange their received data, which leads to an 1 bit/s/Hz capacity.
【 在 heyjude 的大作中提到: 】
: 谁能解释一下,TSE今天下午讲的,INR/SNR--〉无穷的时候,为什么两用户的容量/速率 能达到1。 问了半天,还是不明白,很丢人的。
It should be like this:
For Two User SISO case, when log INR/log SNR >= 2, the channel t1->r2 is much stronger than the channel t1->r1. When r2 treat its desired signal as noise, the achievable rate t1->r2 is still bigger than t1->r1 such that r2 can decode the message of t1. Since we are talking about a reliable decoding scheme, r2 of course can decode its desired message, therefore t1,t2->r2 form a MAC. Similar analysis can be obained at r1, thus each receiver can decode both messages at no rate loss.
【 在 ismydoom 的大作中提到: 】
: I guess it should like this:
: Since INR/SNR -> infinite, the crosstalk dominates. So t1 -> r2 and t2 -> r1 are both error-free. Then r1 and r2 can exchange their received data, which leads to an 1 bit/s/Hz capacity.
嗯,受教了。想明白了。
关键是干扰和噪声本质上的信号结构不同,噪声是个实心儿的平台,而干扰只是一个空心儿的隔板。因此低于噪声平台的话死活也是过不去,但是低于干扰平台则可以钻过去。
so fancy!