## Appendix A Proof of Theorem 1
The exact error probability is as follows: \[r(m)=\frac{1}{\sqrt{T_{c}}}\int_{mT_{c}}^{(n+1)T_{c}}r(t)dt=\sqrt{P_{max}T_{c} }\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]+n(m),\]5 The random variable \(z(m)\) has a mean value, mean square value, and variance \[r(m)=\frac{1}{\sqrt{T_{c}}}\int_{mT_{c}}^{(n+1)T_{c}}r(t)dt=\sqrt{P_{max}T_{c} }\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]+n(m),\]6 respectively. The approximate error probability given in (A3) applies also to the case \(N_{S}=2\). * \(N_{S}\) odd The random variable \(d\) is the sum of an even number of terms \(\pm 1\) and is zero with a probability of the following: \[r(m)=\frac{1}{\sqrt{T_{c}}}\int_{mT_{c}}^{(n+1)T_{c}}r(t)dt=\sqrt{P_{max}T_{c} }\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]+n(m),\]7it is positive (i.e., \(\geq 2\)) with a probability of the following:  
\[r(m)=\frac{1}{\sqrt{T_{c}}}\int_{mT_{c}}^{(n+1)T_{c}}r(t)dt=\sqrt{P_{max}T_{c} }\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]+n(m),\]8  
it is negative (i.e., \(\leq 2\)) with a probability of the following:  
\[r(m)=\frac{1}{\sqrt{T_{c}}}\int_{mT_{c}}^{(n+1)T_{c}}r(t)dt=\sqrt{P_{max}T_{c} }\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]+n(m),\]9  
Therefore  
* \(z(m)=1\) with a probability of \(p_{0}\). * \(z(m)=c_{i}(m)\) with a probability of \(p_{p}\). * \(z(m)=-c_{i}\) with a probability of \(p_{n}=p_{p}\). Overall, \(z\) is the sum of \(M\) random variables \(z(m)\) such that we have the following:  
* \(z(m)=1\) with a probability of \(\pi_{p}=p_{0}+p_{p}=(1+p_{0})/2\). * \(z(m)=-1\) with a probability of \(\pi_{m}=p_{p}=(1-p_{0})/2\). Therefore, the error probability for \(N_{s}\) odd is as follows:  
\[r=\sum_{m=0}^{M-1}r(m)c_{i}(m)=\sqrt{P_{max}T_{c}}\left(\sum_{m=0}^{M-1}c_{i}( m)\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]\right)+n=\sqrt{P_{max}T_{c} }z+n,\]0 (A4)  
Note the following:  
\[r=\sum_{m=0}^{M-1}r(m)c_{i}(m)=\sqrt{P_{max}T_{c}}\left(\sum_{m=0}^{M-1}c_{i}( m)\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]\right)+n=\sqrt{P_{max}T_{c} }z+n,\]1  
\[r=\sum_{m=0}^{M-1}r(m)c_{i}(m)=\sqrt{P_{max}T_{c}}\left(\sum_{m=0}^{M-1}c_{i}( m)\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]\right)+n=\sqrt{P_{max}T_{c} }z+n,\]2  
and \(z\sim\mathcal{N}(Mp_{0},M(1-p_{0}^{2}))\) and  
\[r=\sum_{m=0}^{M-1}r(m)c_{i}(m)=\sqrt{P_{max}T_{c}}\left(\sum_{m=0}^{M-1}c_{i}( m)\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]\right)+n=\sqrt{P_{max}T_{c} }z+n,\]3  
The error probability is reduced to the following:  
\[r=\sum_{m=0}^{M-1}r(m)c_{i}(m)=\sqrt{P_{max}T_{c}}\left(\sum_{m=0}^{M-1}c_{i}( m)\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]\right)+n=\sqrt{P_{max}T_{c} }z+n,\]4 (A5)  
Note that when \(N_{S}\) is odd, the average power at the output of the saturated amplifier is equal to the maximum power:  
\[r=\sum_{m=0}^{M-1}r(m)c_{i}(m)=\sqrt{P_{max}T_{c}}\left(\sum_{m=0}^{M-1}c_{i}( m)\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]\right)+n=\sqrt{P_{max}T_{c} }z+n,\]5  
We now introduce \(E_{b}\) (energy per bit) in (A2)-(A5). Note that during a time interval \(T_{b}=MT_{c}\), the number of transmitted bits, \(N_{S}\), corresponds to the number of satellites to be served. Two cases should be considered:  
* The amplifier is nonlinear and at saturation, in which case, \(E_{b}/N_{0}\) is set according to the maximum power: \(N_{S}E_{b}=P_{max}MT_{c}\). * The amplifier is ideal and linear, in which case, \(E_{b}/N_{0}\) is set according to the average power: \(N_{s}E_{b}=P_{av}MT_{c}=P_{max}MT_{c}(1-\tilde{\pi}_{0})\). Recall that \(\tilde{\pi}_{0}=0\) for \(N_{s}\) odd. The difference in the definition of the signal-to-noise ratio in the two cases is  
\[r=\sum_{m=0}^{M-1}r(m)c_{i}(m)=\sqrt{P_{max}T_{c}}\left(\sum_{m=0}^{M-1}c_{i}( m)\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]\right)+n=\sqrt{P_{max}T_{c} }z+n,\]6  
which is negligible when \(N_{S}\) is large, but is equal to 3 dB when \(N_{S}=2\).