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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

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