# Performance of Dynamic Time Division Broadcast Protocol with Rateless Coding
## 3 Performance Analysis
### Diversity-Multiplexing Trade-off
#### 3.2.1 DMT of TDBC-IR  
For TDBC-IR, its DMT is presented in the following theorem for comparison. **Theorem 2**. : _The DMT of TDBC-IR is given as: (i) when \(0\leq r\leq 0.5\),_  
\[d_{TDBC-IR}(r)=\left\{ \begin{array}{ll}2(1-r),&\text{if }0<\alpha\leq\frac{2}{3} \\ 2-\frac{\alpha r}{(1-2\alpha)},&\text{if }\frac{2}{3}<\alpha\leq\frac{1}{2+r}\\ \frac{1}{\alpha}-1-r.&\text{if }\frac{1}{2+r}<\alpha<\frac{1}{2};\end{array} \right. \tag{12}\]  
_(ii) when \(0.5<r\leq 1\),_  
\[d_{TDBC-IR}(r)=\left\{ \begin{array}{ll}2(1-r),&\text{if }0<\alpha\leq\frac{1}{3-r} \\ \frac{1}{\alpha}-1-r.&\text{if }\frac{1}{3-r}<\alpha<\frac{1}{2}.\end{array} \right. \tag{13}\]  
Proof. : Since \(P_{out}\) in Equation (5) is decided by \(\Sigma_{1}\) and \(\Sigma_{2}\), to prove the theorem, we first calculate \(\Sigma_{1}\) and then \(\Sigma_{2}\) with \(C\) expressed as the function of the multiplexing gain and when \(\rho\to+\infty\). _Calculation of \(\Sigma_{1}\)_. When \(\rho\to+\infty\), \(C_{TDBC-IR}^{\inf}=\lim\limits_{\rho\to+\infty}C_{TDBC-IR}=\frac{(1-\alpha)C} {\alpha}\), based on Equation (10). Let \(X_{S_{1}S_{2}}\doteq\rho^{-u}\), \(X_{S_{1}R}\doteq\rho^{-v}\) and \(X_{S_{2}R}\doteq\rho^{-u}\), then according to Lemma 1, \(u\), \(v\), and \(w\) are i.i.d. random variables with identical probability density function \(p_{z}=\rho^{-z}\) with \(z\in\{u,v,w\}\). Let \(C_{TDBC-IR}^{\inf}=r\log\rho\), where \(r\) (\(0\leq r\leq 1\)) is the multiplexing gain of TDBC-IR, then \(C=\frac{\alpha r}{1-\alpha}\log\rho\) and we have (based on Equations (5) and (6))  
\[\Sigma_{1}\doteq P_{r}\left(\Theta_{S_{1}S_{2}}^{1}\cap\Theta_{S_{1}R}^{1} \right)+P_{r}\left(\Theta_{S_{1}S_{2}}^{1}\cap\Theta_{S_{2}R}^{2}\right), \tag{14}\]  
where the exponential equality holds because \(\min\left(P_{r}\left(\Theta_{S_{1}S_{2}}^{1}\cap\Theta_{S_{2}R}^{2}\right),P_ {r}\left(\Theta_{S_{1}S_{2}}^{1}\cap\Theta_{S_{1}R}^{1}\right)\right)\geq P_{r }\left(\Theta_{S_{1}S_{2}}^{1}\cap\Theta_{S_{1}R}^{2}\right)\). Then, by substituting \(X_{S_{1}S_{2}}\doteq\rho^{-u}\), \(X_{S_{1}R}\doteq\rho^{-v}\) and \(X_{S_{2}R}\doteq\rho^{-w}\) into Equation (14), we have  
\[ \begin{array}{l}\Sigma_{1}\doteq P_{r}\left(X_{S_{1}S_{2}}<z_{0}\text{ and }X_{S_{2}R}<z_{0}\right)+P_{r}\left(X_{S_{1}S_{2}}<z_{0}\text{ and }X_{S_{1}R}<z_{0}\right)\\ \doteq P_{r}\left(u\geq 1-r\right)P_{r}\left(v>1-r\right)+P_{r}\left(u\geq 1-r \right)P_{r}\left(w\geq 1-r\right)\\ \doteq\rho^{-2(1-r)^{+}}+\rho^{-2(1-r)^{+}}\doteq\rho^{-2(1-r)^{+}},\end{array}  \tag{15}\]  
where the second exponential equality is derived based on Lemma 2. _Calculation of \(\Sigma_{2}\)_. Based on Equations (14) and (8), we have \(T_{2}\doteq T_{1}\
ot>T_{3}\) and then  
\[P_{out}=\Sigma_{1}+e^{-z_{0}/\Omega_{1}}T_{1}+e^{-z_{0}/\Omega_{2}}T_{2}-T_{3} \doteq\rho^{-2(1-r)^{+}}+T_{1}. \tag{16}\]  
With this equation, \(\Sigma_{2}\doteq T_{1}\), we only need to calculate \(T_{1}\) to obtain \(\Sigma_{2}\), and then \(P_{out}\). The following process calculates \(T_{1}\). First, when \(\alpha\leq\frac{1}{3}\), \(T_{1}=T_{2}=T_{3}=0\) (see from Equations (2)-(4)) and then \(\Sigma_{2}\doteq T_{1}\doteq\rho^{-\infty}\); second, when \(\alpha>\frac{1}{3}\),  
\[ \begin{array}{l}T_{1}=P_{r}\left(X_{S_{1}S_{2}}<z_{0},\,X_{S_{2}R}>z_{0}\text{ and }\left(1+\rho X_{S_{1}S_{2}}\right)^{\alpha}\left(1+\rho X_{S_{2}R}\right)^{1-2 \alpha}<2^{(1-\alpha)C}\right)\\ \doteq P_{r}\left(\left(1-w\right)^{+}>r\text{ and }\alpha\left(1-u\right)^{+}+ \left(1-2\alpha\right)\left(1-w\right)^{+}<\alpha r\right).\end{array} \]  
Let \(T_{1}\doteq\rho^{-d_{1}(r)}\), then \(d_{1}(r)\) is the solution of the following optimization problem based on Lemma 2. \[d_{1}(r)=\min(u+w),\]  
\[\text{s.t., }0\leq w\leq 1-r,\,0\leq u\leq 1\text{ and }1-\alpha-\alpha r\leq\alpha u+\left(1-2\alpha\right)w.\]  
Since \(\alpha>\frac{1}{3}\), \(\frac{1-\alpha-\alpha r}{\alpha}<\frac{1-\alpha-\alpha r}{1-2\alpha}\). To obtain \(d_{1}(r)\), we only need to compare \(\frac{1}{\alpha}-1-r\) with \(1\).