search

aprilbian / deepjscc-lplusplus

Source code for paper deepjscc-l++
20 stars 1 forks source link

about the pl #1

Open Luckwjf opened 3 months ago

Luckwjf commented 3 months ago

“to matrix ̃ Z with dimension NF × NT , where NF NT = 2ρLN with NT = hI wI .”其中NF NT = 2ρLN，NT = hI wI。 Why is there a 2 here? Can you help me answer this question?

aprilbian commented 3 months ago

thanks for the question, it is because of both the real and imaginary part of the signal.

Luckwjf commented 3 weeks ago

感谢老师的回答呀，还想要请教一下代码的问题。sig_s = self.pwr_normalize(sig_s)*np.sqrt(bw/self.max_trans_feat) 这行代码中为什么要在乘np.sqrt(bw/self.max_trans_feat)，有什么含义嘛

aprilbian commented 3 weeks ago

因为那个self.pwr_normalize(sig_s)之后，整体的sig_s的power其实比真实值要大了，

我们需要的是bw unit_trans_feat这么大的power，但是实际上给的是max_trans_feat unit_trans_feat这么大的power，所以要乘上一个np.sqrt(bw/self.max_trans_feat)归一化一下

你可以仔细看一下pwr_normalize的代码，或者自己设个断点查一下

Luckwjf commented 3 weeks ago

感谢老师的回答，是不是经过掩码之后如果不乘上一个np.sqrt(bw/self.max_trans_feat)进行归一化的话，那么这个信号的功率其实是相当于发送了max_trans_feat unit_trans_feat这么信号的功率，但实际上我们要进行带宽压缩实际上传输的是max_trans_feat unit_trans_feat的一部分，也就是所占总的bw/self.max_trans_feat，所以要乘上一个np.sqrt(bw/self.max_trans_feat)进行缩小归一化。老师我这样的理解是正确的嘛

aprilbian commented 3 weeks ago

感谢老师的回答，是不是经过掩码之后如果不乘上一个np.sqrt(bw/self.max_trans_feat)进行归一化的话，那么这个信号的功率其实是相当于发送了max_trans_feat unit_trans_feat这么信号的功率，但实际上我们要进行带宽压缩实际上传输的是max_trans_feat unit_trans_feat的一部分，也就是所占总的bw/self.max_trans_feat，所以要乘上一个np.sqrt(bw/self.max_trans_feat)进行缩小归一化。老师我这样的理解是正确的嘛 对的