## Appendix A Proof of Theorem 1  
Proof of Theorem 1.: We assume that the receiver correctly acquired the received signal frequency and phase and that the phase jitter is small. This hypothesis allows for a simplification of the analysis, where only the in-phase component of the 2-PSK received signal is relevant. \[s_{T}^{\prime}(t)=\sqrt{P_{max}}\text{ sign}\left[\sum_{i=0}^{N_{S}-1}\sum_{n}b_{i}(n)P_{T_{b}}(t-nT_{b}) \sum_{m}c_{i}(m)P_{T_{c}}(t-mT_{c})\right]\]  
The received signal is \(r(t)=s_{T}(t)+n(t)\) where \(n(t)\) is white Gaussian noise with a power spectral density of \(N_{0}/2\). During the \(m\)-th chip interval in the first-bit period, the receiver of satellite \(i\) evaluates the following:  
\[r(m)=\frac{1}{\sqrt{T_{c}}}\int_{mT_{c}}^{(n+1)T_{c}}r(t)dt=\sqrt{P_{max}T_{c} }\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]+n(m),\]  
where \(n(m)\sim\mathcal{N}(0,N_{0}/2)\), and then we have the following:  
\[r=\sum_{m=0}^{M-1}r(m)c_{i}(m)=\sqrt{P_{max}T_{c}}\left(\sum_{m=0}^{M-1}c_{i}( m)\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]\right)+n=\sqrt{P_{max}T_{c} }z+n,\]  
where \(n\sim\mathcal{N}(0,MN_{0}/2)\). Term \(z\) can be written as follows:  
\[z=\sum_{m=0}^{M-1}c_{i}(m)\text{ sign}\left[b_{i}(0)c_{i}(m)+\sum_{j=0,j\
eq i }^{N_{S}-1}d_{j}(m)\right]\]  
where \(d_{j}(m)=b_{j}(0)c_{j}(m)\in\{\pm 1\}\) can be assumed statistically independent of \(c_{i}(m)\) for \(i\
eq j\). Note that \(z\) is the sum of terms \(z(m)\) with three values: \(c_{i}(m)\) (if the argument of the sign function is strictly positive), \(-c_{i}(m)\) (if the argument is strictly negative), and 0 (if the argument is zero); therefore, \(z\) can take integer values from \(-M\) to \(M\). Without loss of generality, we consider the transmitted symbol \(b_{i}(0)=1\). Then, the error probability can be written as follows:  
\[P_{b}(e) =P(r<0|b_{i}(0)=1)=P\Big{(}\sqrt{P_{max}T_{c}}z+n<0\Big{|}b_{i}(0)=1 \Big{)}\] \[=P\Big{(}n>\sqrt{P_{max}T_{c}}z\Big{|}b_{i}(0)=1\Big{)}\] \[=\sum_{\xi=-M}^{M}\frac{1}{2}\ \text{erfc}\left(\sqrt{\frac{P_{max}T_{c}}{ MN_{0}}}\ \zeta\right)P(z=\zeta)\,.\]  
Having assumed \(b_{i}(0)=1\), \(  
The exact expression above can be approximated using the central limit theorem on \(z\): it represents the sum of \(M\) (typically large) i.i.d. random variables \(z(m)\) with the mean and variance as follows:  
\[\mu_{z(m)}=\pi_{0}\times 0+\pi_{p}\times 1+\pi_{n}\times(-1)=p_{0}\]  
\[\mathbb{E}\{z(m)^{2}\}=1-\pi_{0}=1-p_{0};\ \ \sigma_{z(m)}^{2}=1-p_{0}-p_{0}^{2}.\]  
Therefore, \(z\sim\mathcal{N}(Mp_{0},M(1-p_{0}-p_{0}^{2}))\) and  
\[r=\sqrt{P_{max}T_{c}}z+n\sim\mathcal{N}\Big{(}\sqrt{P_{max}T_{c}}Mp_{0},\ M(1-p_{0}-p_{0}^{2})P_{max}T_{c}+MN_{0}/ 2\Big{)}.\]  
The error probability is reduced to the following:  
\[r(m)=\frac{1}{\sqrt{T_{c}}}\int_{mT_{c}}^{(n+1)T_{c}}r(t)dt=\sqrt{P_{max}T_{c} }\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]+n(m),\]0 (A3)  
Note that when \(N_{S}\) is even, the average power at the output of the saturated amplifier is  
\[r(m)=\frac{1}{\sqrt{T_{c}}}\int_{mT_{c}}^{(n+1)T_{c}}r(t)dt=\sqrt{P_{max}T_{c} }\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]+n(m),\]1  
where  
\[r(m)=\frac{1}{\sqrt{T_{c}}}\int_{mT_{c}}^{(n+1)T_{c}}r(t)dt=\sqrt{P_{max}T_{c} }\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]+n(m),\]2  
* \(N_{S}=2\) In this case, having set \(i=0\), without loss of generality, we have the following: \[r(m)=\frac{1}{\sqrt{T_{c}}}\int_{mT_{c}}^{(n+1)T_{c}}r(t)dt=\sqrt{P_{max}T_{c} }\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]+n(m),\]3 The term \(z(m)=0\) if \(d_{1}(m)\
eq c_{0}(m)\), which occurs with a probability of \[r(m)=\frac{1}{\sqrt{T_{c}}}\int_{mT_{c}}^{(n+1)T_{c}}r(t)dt=\sqrt{P_{max}T_{c} }\text{ sign}\left[\sum_{j=0}^{N_{S}-1}b_{j}(0)c_{j}(m)\right]+n(m),\]4 If \(d_{1}(m)=c_{0}(m)\), which occurs with a probability of \(\pi_{p}=1/2\), then \(z(m)=1\). Therefore, \(z\) is the sum of \(M\) terms that can take values \(1\) or \(0\) with the same probability.