In this paper:https://arxiv.org/pdf/1805.08898.pdf

enter image description here enter image description here

I can understand why C4 means $P_E > P^*$,I just need to times $1-\rho_k$ in both two side,the C4 will become $(1-\rho_k)[\sum\limits_{k=1}^{K}\mathbf h_k^H \mathbf F_j \mathbf h_k+\sigma^2_{\alpha_k}]\ge P$,and it is the same as $P_E > P^*$,but why can C5 also mean $\sum\limits_{k=1}^{K}tr(\mathbf F_{K4E})>P_T$?In here,you can see $F_{K4E}$ as $F_K$.

$P_T$ is the power budget,it means that i have only $P_T$ power to send the message out

$\sum\limits_{k=1}^{K}tr(\mathbf F_{K4E})$ means that the total power i need to send the power

C5 comes from this formula: enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.