## 5 Loss with an Increasing Number of Satellites  
From the operator's perspective, the priority is to maximize the number of satellites for a given transmitted/consumed power while meeting a certain target error rate. The analysis conducted in the previous section using Theorem 1, reported in Figure 13 (\(M=63\)), Figure 14 (\(M=100\)), and Figure 15 (\(M=150\)), leads to the following observations:  
* The case \(N_{S}=2\) suffers from a 3 dB loss because with a probability of 0.5 the signal is zero. At low \(E_{b}/N_{0}\), the error probability for \(N_{S}=3\) is, therefore, smaller than that for \(N_{S}=2\), even if the interference is obviously higher. At \(P_{b}(e)=10^{-2}\) and \(M=150\)--\(N_{S}=4\) is better than \(N_{S}=2\). * Similar phenomena occur with \(N_{S}=4\) and \(N_{S}=5\): \(N_{S}=5\) is better than \(N_{S}=4\) at higher values of \(P_{b}(e)\) (e.g., \(P_{b}(e)=10^{-2}\)), whereas \(N_{S}=5\) is worse than \(N_{S}=4\) at smaller values of \(P_{b}(e)\) (e.g., \(P_{b}(e)=10^{-4}\)). * The error floor for \(N_{S}=8\) is smaller than the error floor for \(N_{S}=7\); using the approximation for \(P_{b}(e)\), we see that when \(N_{S}=8\), there is a term \((1-p_{0}-p_{0}^{2})\) at the denominator, whereas the term at the denominator is \(1-p_{0}^{2}\) for \(N_{S}=7\); note that the two values of \(p_{0}\) are not equal for \(N_{S}=7\) and \(8\), but this is the only difference in the formula. These considerations suggest that it would be useful to have an analytical result to calculate the loss as a function of the number of satellites. This is the result of Theorem 2 presented in this section. Given the value \(\left(\frac{E_{b}}{N_{0}}\right)^{*}\) [dB] necessary to achieve an error rate \(P_{b}(e)^{*}\) for the uncoded 2-PAM, for a single satellite, the value \(\left(\frac{E_{b}}{N_{0}}\right)^{\prime}=\left(\frac{E_{b}}{N_{0}}\right)^{ *}+Loss\) [dB] necessary to achieve the same error rate in \(N_{S}\) satellites and the spreading factor \(M\) is given by the following results:  
**Theorem 2**. : _For an uplink CDM system serving \(N_{S}\) satellites with a saturated amplifier, the approximate loss in dB required to achieve a given \(P_{b}(e)^{*}\) in the case of a spreading factor \(M\) is given by the following:_  
Figure 15: Comparison between exact theoretical (lines) and approximate (dashed lines) CDM bit error probabilities for a saturated HPA with random spreading codes, with spreading factor \(M=150\), for \(N_{S}=2,3,4,5,6,7,8,15,20\) satellites. * \(N_{S}\) _odd:_ \[Loss_{dB}=-10\log_{10}N_{S}-10\log_{10}\left(p_{0}^{2}-\frac{2(1-p_{0}^{2})}{M} \left(\frac{E_{b}}{N_{o}}\right)^{*}\right)\] _where we have the following:_ \[p_{0}=\binom{N_{S}-1}{(N_{S}-1)/2}2^{-(N_{S}-1)}\,.\]
* \(N_{S}\) _even:_ \[Loss_{dB}=-10\log_{10}N_{s}-10\log_{10}\left(p_{0}^{2}-\frac{2(1-p_{0}-p_{0}^{ 2})}{M}\left(\frac{E_{b}}{N_{o}}\right)^{*}\right)\] _where we have the following:_ \[p_{0}=\binom{N_{S}-1}{N_{S}/2-1}2^{-(N_{S}-1)}\,.\]  
The proof of this theorem is available in Appendix B. To fully understand the impact of the HPA, it is useful to compare it with an ideal, fully linear system. Theorem 2 allows us to obtain the following results:  
* Figure 16 shows the overall loss with respect to the ideal 2-PAM of the CDM system with saturated HPA at \((E_{b}/N_{0})^{*}=\eta=9.58\) dB (i.e., at \(P_{b}(e)^{*}=10^{-5}\)): note that, for \(M\to\infty\), an odd number of satellites gives rise to a smaller loss (less than 2 dB), whereas an even number gives a higher loss (more than 2 dB). * Figure 17 shows the extra loss (with respect to the same CDM system and an ideal perfectly linear amplifier) due to the saturated HPA at \((E_{b}/N_{0})^{*}=\eta=9.58\) dB (i.e., at \(P_{b}(e)^{*}=10^{-5}\)).