# Robust Transceiver Design for IRS-Assisted Cascaded MIMO Communication Systems
## 3 Achievable Rate (Full-Duplex Relay)
### Algorithm Description
We have

\[\mathbf{H}_{R\uparrow}\mathbf{\Theta}\mathbf{H}_{IR}= \left(\sum_{r_{1}=1}^{K_{r}}\sum_{r_{2}=1}^{M}h_{RIr_{1},r_{2}} \left|r_{1}\right\rangle\!\!\left\langle r_{2}\right|\right)\left(\sum_{r_{3} =1}^{M}\theta_{r_{3},r_{3}}\left|r_{3}\right\rangle\!\!\left\langle r_{3} \right|\right)\left(\sum_{r_{4}=1}^{M}\!\!\sum_{r_{5}=1}^{K_{t}}h_{IRr_{4},r_ {5}}\left|r_{4}\right\rangle\!\!\left\langle r_{5}\right|\right) \tag{17}\]8 \[\mathbf{H}_{R\uparrow}\mathbf{\Theta}\mathbf{H}_{IR}= \left(\sum_{r_{1}=1}^{K_{r}}\sum_{r_{2}=1}^{M}h_{RIr_{1},r_{2}} \left|r_{1}\right\rangle\!\!\left\langle r_{2}\right|\right)\left(\sum_{r_{3} =1}^{M}\theta_{r_{3},r_{3}}\left|r_{3}\right\rangle\!\!\left\langle r_{3} \right|\right)\left(\sum_{r_{4}=1}^{M}\!\!\sum_{r_{5}=1}^{K_{t}}h_{IRr_{4},r_ {5}}\left|r_{4}\right\rangle\!\!\left\langle r_{5}\right|\right) \tag{17}\]9 (107) \[=\sum_{r_{1}=r_{2}=r_{3}=r_{4}}^{K_{r}}\!\!\sum_{r_{3}=r_{4}}^{M }\!\!\sum_{r_{5}=1}^{K_{t}}h_{RIr_{1},r_{2}}\theta_{r_{3},r_{3}}h_{IRr_{4},r_ {5}}\!\!\left|r_{1}\right\rangle\!\!\left\langle r_{2}\right|\!\!\left\langle r _{3}\right|\!\!\left\langle r_{4}\right\rangle\!\!\left\langle r_{5}\right|\]0Now we define

\[=\sum_{r_{1}=r_{2}=r_{3}=r_{4}}^{K_{r}}\!\!\sum_{r_{3}=r_{4}}^{M }\!\!\sum_{r_{5}=1}^{K_{t}}h_{RIr_{1},r_{2}}\theta_{r_{3},r_{3}}h_{IRr_{4},r_ {5}}\!\!\left|r_{1}\right\rangle\!\!\left\langle r_{2}\right|\!\!\left\langle r _{3}\right|\!\!\left\langle r_{4}\right\rangle\!\!\left\langle r_{5}\right|\]1 (A47) \[=\sum_{r_{1}=r_{2}=r_{3}=r_{4}}^{K_{r}}\!\!\sum_{r_{3}=r_{4}}^{M }\!\!\sum_{r_{5}=1}^{K_{t}}h_{RIr_{1},r_{2}}\theta_{r_{3},r_{3}}h_{IRr_{4},r_ {5}}\!\!\left|r_{1}\right\rangle\!\!\left\langle r_{2}\right|\!\!\left\langle r _{3}\right|\!\!\left\langle r_{4}\right\rangle\!\!\left\langle r_{5}\right|\]2 (A48)

It is obvious that \(\psi_{1}>0\) and \(\psi_{2}\geq 0\). Now we have

\[=\sum_{r_{1}=r_{2}=r_{3}=r_{4}}^{K_{r}}\!\!\sum_{r_{3}=r_{4}}^{M }\!\!\sum_{r_{5}=1}^{K_{t}}h_{RIr_{1},r_{2}}\theta_{r_{3},r_{3}}h_{IRr_{4},r_ {5}}\!\!\left|r_{1}\right\rangle\!\!\left\langle r_{2}\right|\!\!\left\langle r _{3}\right|\!\!\left\langle r_{4}\right\rangle\!\!\left\langle r_{5}\right|\]3 (A49)

and

\[=\sum_{r_{1}=r_{2}=r_{3}=r_{4}}^{K_{r}}\!\!\sum_{r_{3}=r_{4}}^{M }\!\!\sum_{r_{5}=1}^{K_{t}}h_{RIr_{1},r_{2}}\theta_{r_{3},r_{3}}h_{IRr_{4},r_ {5}}\!\!\left|r_{1}\right\rangle\!\!\left\langle r_{2}\right|\!\!\left\langle r _{3}\right|\!\!\left\langle r_{4}\right\rangle\!\!\left\langle r_{5}\right|\]4 (A50)

Finally, one can conclude

\[=\sum_{r_{1}=r_{2}=r_{3}=r_{4}}^{K_{r}}\!\!\sum_{r_{3}=r_{4}}^{M }\!\!\sum_{r_{5}=1}^{K_{t}}h_{RIr_{1},r_{2}}\theta_{r_{3},r_{3}}h_{IRr_{4},r_ {5}}\!\!\left|r_{1}\right\rangle\!\!\left\langle r_{2}\right|\!\!\left\langle r _{3}\right|\!\!\left\langle r_{4}\right\rangle\!\!\left\langle r_{5}\right|\]5 (A51)

## Appendix E

Here we show that if \(\gamma_{r_{i}}\geq\gamma_{r_{i+1}}\) and \(\sigma_{1_{i}}^{2}\gamma_{s_{\rho(i)}}\geq\sigma_{1_{i+1}}^{2}\gamma_{s_{\rho (i+1)}}\) then \(\gamma_{r_{i}}\sigma_{s_{\rho(i)}}^{2}\geq\gamma_{r_{i+1}}\sigma_{r_{\rho(i+1)}} ^{2}\). First we define \(f(x,y)=\sqrt{x^{2}+axy}-x-b,\;x\geq 0,\;y\geq 0\) in which \(a\) and \(b\) are positive constants. Now we have,

\[=\sum_{r_{1}=r_{2}=r_{3}=r_{4}}^{K_{r}}\!\!\sum_{r_{3}=r_{4}}^{M }\!\!\sum_{r_{5}=1}^{K_{t}}h_{RIr_{1},r_{2}}\theta_{r_{3},r_{3}}h_{IRr_{4},r_ {5}}\!\!\left|r_{1}\right\rangle\!\!\left\langle r_{2}\right|\!\!\left\langle r _{3}\right|\!\!\left\langle r_{4}\right\rangle\!\!\left\langle r_{5}\right|\]6 (A52)

In addition, for \(\frac{\partial f}{\partial x}\) we have

\[=\sum_{r_{1}=r_{2}=r_{3}=r_{4}}^{K_{r}}\!\!\sum_{r_{3}=r_{4}}^{M }\!\!\sum_{r_{5}=1}^{K_{t}}h_{RIr_{1},r_{2}}\theta_{r_{3},r_{3}}h_{IRr_{4},r_ {5}}\!\!\left|r_{1}\right\rangle\!\!\left\langle r_{2}\right|\!\!\left\langle r _{3}\right|\!\!\left\langle r_{4}\right\rangle\!\!\left\langle r_{5}\right|\]7 (A53)

One can check that for positive values \(x,\;y\) and \(a\), we always have \(\frac{2x+ay}{2\sqrt{x^{2}+axy}}\geq 1\).