# Performance of Dynamic Time Division Broadcast Protocol with Rateless Coding
## 3 Performance Analysis
### Diversity-Multiplexing Trade-off
#### 3.2.2. DMT of Sub-DTDBC-RC and DTDBC-RC  
When \(\rho\rightarrow+\infty\), it is impossible to obtain the limit value of the expected rate of DTDBC-RC from Equation (9), and the DMT of DTDBC-RC cannot be obtained with the conventional outage and expected rate analysis like in Section 3.2.1. Hence, we first derive the DMT of sub-DTDBC-RC and then the DMT of DTDBC-RC by setting the number of subslots \(N\rightarrow+\infty\). As for the DMT of sub-DTDBC-RC, it is given in the following theorem. Theorem 3.2: _The DMT of sub-DTDBC-RC is_  
\[d_{sub-DTDBC-RC}(r)=\left\{ \begin{array}{ll}2(1-\frac{r}{N}),&\text{if }0< \alpha\leq\frac{2}{5}\\ 2-\frac{\alpha r}{N(1-2\alpha)},&\text{if }\frac{2}{5}<\alpha\leq\frac{1}{2+r/N}\\ \frac{1}{\alpha}-1-\frac{r}{N}.&\text{if }\frac{1}{2+r/N}<\alpha<\frac{1}{2}, \end{array} \right. \tag{17}\]  
_where \(N\) is the number of subslots of sub-DTDBC-RC._  
Figure 3: The calculation of \(d_{1}(r)\). **Proof. ** When \(\rho\to+\infty\), we have \(C_{sub-DTDBC-RC}^{\inf}=\lim_{\rho\to+\infty}C_{sub-DTDBC-RC}=\frac{N(1-\alpha) R}{\alpha}\) from Equation (11). Let \(C_{sub-DTDBC-RC}^{\inf}=r\log\rho\), where \(r\) (\(0\leq r\leq 1\)) is the multiplexing gain of sub-DTDBC-RC, \(C=\frac{N(\alpha r)}{N(1-\alpha)}\log\rho\). Then, \(\Sigma_{1}\) and \(\Sigma_{2}\) of \(P_{out}\) in Equation (5) are calculated as follows. _Calculation of \(\Sigma_{1}\)._ The calculation of \(\Sigma_{1}\) follows the same way as in the proof of Theorem 2, only with the exception that \(C\) is now replaced by \(C=\frac{\alpha r}{N(1-\alpha)}\log\rho\). Substituting \(C\) into Equation (14), \(\Sigma_{1}\doteq\rho^{-2(1-r/N)^{+}}\)  
_Calculation of \(\Sigma_{2}\)._ Following the same way as in the proof of Theorem 2, only \(T_{1}\) is needed for calculation of \(\Sigma_{2}\) and then \(P_{out}\) (since \(\Sigma_{2}\doteq T_{1}\) ). First, when \(\alpha\leq\frac{1}{3}\), \(\Sigma_{2}\doteq\rho^{-\infty}\); when \(\alpha>\frac{1}{3}\), substituting \(C=\frac{\alpha r}{N(1-\alpha)}\log\rho\) into \(T_{1}\) in Equation (8), we get \(T_{1}\doteq P_{r}\left((1-w)^{+}>\frac{r}{N}\text{ and }\alpha\left(1-u\right)^{+}+(1-2 \alpha)\left(1-w\right)^{+}<\frac{\alpha r}{N}\right)\doteq\rho^{-d_{2}(r)}\), where \(d_{2}(r)\) is the solution of the following optimization problem based on Lemma 2. \[d_{2}(r)=\min(u+w),\]  
\[s.t.,\ 0\leq w\leq 1-\frac{r}{N},\ 0\leq u\leq 1\ and\ 1-\alpha-\frac{\alpha r}{N}\leq\alpha u +(1-2\alpha)\,w.\]  
Moreover, \(\frac{1-\alpha-\alpha r/N}{\alpha}<\frac{1-\alpha-\alpha r/N}{1-2\alpha}\) (since \(\alpha>\frac{1}{3}\)). Thus, to calculate \(d_{2}(r)\), we only need to compare \(\frac{1}{\alpha}-1-\frac{r}{N}\) with \(1\). _Case I_: when \(\frac{1}{3}<\alpha\leq\frac{1}{2+r/N}\), \(\frac{1}{\alpha}-1-\frac{r}{N}\geq 1\), and \(\Sigma_{2}\doteq\rho^{-d_{2}(r)}\) with \(d_{2}(r)=2-\frac{\alpha r}{N(1-2\alpha)}\). _Case II_: when \(\frac{1}{2+r/N}<\alpha<\frac{1}{2}\), and \(\Sigma_{2}\doteq\rho^{-d_{2}(r)}\) with \(d_{2}(r)=\frac{1}{\alpha}-1-\frac{r}{N}\). The calculation of \(d_{2}(r)\), which follows the same way as for calculating \(d_{1}(r)\) in the proof of Theorem 2, is simple and thus omitted here. Concluding the above process, we have (i) when \(0<\alpha\leq\frac{1}{3}\), \(P_{out}\doteq\rho^{-2(1-r/N)^{+}}\); (ii) when \(\frac{1}{3}<\alpha\leq\frac{1}{2+r/N}\),  
\[P_{out}\doteq\left\{ \begin{array}{ll}\rho^{-2(1-r/N)^{+}},&\text{if }\frac{1}{3}<\alpha\leq\frac{2}{5}\\ \rho^{-(2-\frac{\alpha r}{N(1-2\alpha)})^{+}}.&\text{if }\frac{2}{5}<\alpha\leq \frac{1}{2+r/N};\end{array} \right.\]  
(iii) and when \(\frac{1}{2+r/N}<\alpha<\frac{1}{2}\), \(P_{out}\doteq\rho^{-\left(\frac{1}{\alpha}-1-\frac{r}{N}\right)^{+}}\). With the above results, the DMT of sub-DTDBC-RC, i.e., \(d_{sub-DTDBC-RC}=-\lim_{\rho\to+\infty}\frac{\log P_{out}}{\log\rho}\), is readily obtained by comparing \(\Sigma_{1}\) and \(\Sigma_{2}\). With Theorem 3, the DMT of DTDBC-RC is readily obtained by setting \(N\to+\infty\) as follows.