# Robust Transceiver Design for IRS-Assisted Cascaded MIMO Communication Systems
## 3 Achievable Rate (Full-Duplex Relay)
### Mathematical Preliminaries
One simple example of such a mapping can be found in \[R^{\text{FD}}_{\text{sr}} =\log_{2}\frac{\left|\sigma_{t}^{2}\mathbf{I}_{K_{r}}+\mathbf{ \hat{H}}_{tot_{2}}\mathbf{Q}_{s}\mathbf{\hat{H}}_{tot_{2}}^{H}\!+\!\mathbf{ \hat{H}}_{tot_{2}}\mathbf{Q}_{s}\mathbf{\hat{H}}_{tot_{2}}^{H}+\mathbf{\hat{H} }_{r}\mathbf{Q}_{r}\mathbf{\hat{H}}_{r}^{H}\right|}{\left|\sigma_{t}^{2} \mathbf{I}_{K_{r}}+\mathbf{\hat{H}}_{tot_{2}}\mathbf{Q}_{s}\mathbf{\hat{H}}_{ tot_{2}}^{H}+\mathbf{\hat{H}}_{r}\mathbf{Q}_{r}\mathbf{\hat{H}}_{r}^{H}\right|}, \tag{9}\]7 Here, an example of mapping from a three-dimensional to a two-dimensional space is shown. The left shape shows the feasible set for the IRS with three elements in a real valued space. The cube belongs to the case of \(T^{\prime}\), i.e., constraints \(-1\leq\theta_{m}\leq 1,\forall m\), while the sphere shows the constraint \(\theta_{1}^{2}+\theta_{2}^{2}+\theta_{3}^{2}\leq 1\) which belongs to \(T^{\prime\prime}\). On the right, the feasible sets belonging to the two aforementioned regions after performing mapping \(f\) are presented as an example. It can be seen that the mapping of the first set of constraints (the hexagon) covers the whole area of that of the second one (the ellipse). One important key is, as mapping is a linear function, we have \(\mathcal{A}\subset\mathcal{B}\to f(\mathcal{A})\subset f(\mathcal{B})\), where \(\mathcal{A}\) and \(\mathcal{B}\) are two arbitrary sets and \(f\) is the mapping. In general, as the number of IRS elements or the dimensions of \(\bar{\mathbf{h}}_{r}\) increase, the mapping of the hypercube becomes more and more complicated and finding the optimal distance becomes more difficult. However, there is an upper bound for this distance. As shown in Figure 3, if instead of the cube, we limit the feasible set of IRS elements to the sphere inside the cube, i.e., replacing (18b) with (19b), the solution to the problem becomes \(GE\geq GF\). It turns out that finding \(GE\) is very simple as by the definition we have \(\sigma_{\min}(\mathbf{H}_{IR}diag(\mathbf{H}_{RI}^{T}))=OE\), and also we know that \(\sqrt{T_{r}}=GO\). Therefore, we can conclude \(GE=\sqrt{T_{r}}-\sigma_{\min}(\mathbf{H}_{IR}diag(\mathbf{H}_{RI}^{T}))\). Finally, we use one last upper bound to make the original problem even easier to solve. Note that if instead of the ellipse, we consider the circle inscribed in it, we will have \(\max_{\bar{\mathbf{h}}_{r}}\ \min_{\mathbf{\Theta}}||\bar{\mathbf{h}}_{r}+( \mathbf{H}_{IR}*\mathbf{H}_{RI}^{T})\theta||_{2}=\sqrt{T_{r}}-\sigma_{\min}( \mathbf{H}_{IR}*\mathbf{H}_{RI}^{T}),\forall\bar{\mathbf{h}}_{r}\). As a result, we have  
\[R^{\text{FD}}_{\text{sr}} =\log_{2}\frac{\left|\sigma_{t}^{2}\mathbf{I}_{K_{r}}+\mathbf{ \hat{H}}_{tot_{2}}\mathbf{Q}_{s}\mathbf{\hat{H}}_{tot_{2}}^{H}\!+\!\mathbf{ \hat{H}}_{tot_{2}}\mathbf{Q}_{s}\mathbf{\hat{H}}_{tot_{2}}^{H}+\mathbf{\hat{H} }_{r}\mathbf{Q}_{r}\mathbf{\hat{H}}_{r}^{H}\right|}{\left|\sigma_{t}^{2} \mathbf{I}_{K_{r}}+\mathbf{\hat{H}}_{tot_{2}}\mathbf{Q}_{s}\mathbf{\hat{H}}_{ tot_{2}}^{H}+\mathbf{\hat{H}}_{r}\mathbf{Q}_{r}\mathbf{\hat{H}}_{r}^{H}\right|}, \tag{9}\]8  
where \(T^{\prime}=\big{(}\sqrt{T_{r}}-\sigma_{\min}(\mathbf{H}_{IR}*\mathbf{H}_{RI}^{ T})\big{)}^{2}\). It is worth mentioning that the geometrical representation for the optimization problem (13) is different because there, considering that the RSI wants to bring the worst representation against the IRS configuration and covariance matrices simultaneously, the RSI cannot freely span the whole circle. This is due to the fact that some regions in the circle might not be a good choice when it comes to RSI design against covariance matrices. However, if the best representation of RSI against the covariance matrices also provides the best RSI against the IRS configuration, the solution to (14) and (13) will be the same. Eventually, instead of optimization problem (13), one can solve optimization problem (14). The solution to the new problem is guaranteed to be achievable by the original problem as well.