## 3 Impact of Power Amplifier Nonlinearities on the System Performance  
Our first goal is to study the impact of input power on system performance when HPA non-linearity is taken into consideration. The signal \(s_{T}(t)\) from Equation (1) is used to drive the HPA with different nominal input powers \(P_{in}\), and the corresponding output power \(P_{out}\) is evaluated. The results are presented in Figure 4.  
Observing Figure 4, we note the following:  
* When \(N_{S}=2\), the maximum output power is limited to 28 dBW (instead of \(P_{max}=31\) dBW), and this maximum is reached when \(P_{in}=-68\) dBW;
* When \(N_{S}=6\), max \(P_{out}=29.36\) dBW at \(P_{in}=-62\) dBW;
* When \(N_{S}=11\), and max \(P_{out}=P_{max}=31\) dBW at \(P_{in}=-54\) dBW. There is clearly a gap between the output power when the input is a CDM signal and the output power when the input is a pure sinusoid ('No modulation' curve in Figure 4). The  
Figure 4: HPA output power versus input power for various numbers of satellites. Figure 3: AM/AM curve for the considered high-power amplifier. reason is that, when the input power is such that the HPA works completely at saturation, the input signal is clipped, i.e., it is as if the input signal were as follows: \[s_{T}^{\prime}(t)=\sqrt{\frac{P_{in}}{N_{S}}}\text{ sign}\left[\sum_{i=0}^{N_{S}-1}\sum_{n}b_{i}(n)P_{T_{b}}(t-nT_{b})\sum_{m}c_{i}(m)P _{T_{c}}(t-mT_{c})\right]\] (2) where \(\text{ sign}\ (x)\) is the sign (or signum) function, equal to 1 if \(x>0\), to \(-1\) if \(x<0\), to 0 if \(x=0\). We can observe the following: * When there are \(N_{S}=2\) satellites, \(s_{T}(t)\) is either equal to 0 (with a probability of 0.5) or to \(\sqrt{\frac{P_{in}}{2}}\) (with a probability of 0.25) or to \(-\sqrt{\frac{P_{in}}{2}}\) (with a probability of 0.25); therefore, the signal at the HPA output has either instantaneous power \(P_{max}\) (with a probability of 0.5) or 0 (with a probability of 0.5); the average power at the HPA output is \(P_{out,av}=P_{max}/2\) (\(31-3=28\) dBW, as in Figure 4). * When there are \(N_{S}\) satellites, \(s_{T}(t)\) is equal to zero with a probability of \[\tilde{\pi}_{0}=\binom{N_{S}}{N_{S}/2}\bigg{[}\frac{1}{2}\bigg{]}^{N_{S}}\] if \(N_{S}\) is even (at a given instant \(t\) half of the chips are \(+1\), half are \(-1\)), whereas, \(\tilde{\pi}_{0}=0\) when \(N_{s}\) is odd. If the input power \(P_{in}\) is large enough, the output instantaneous power is \(P_{max}\) with a probability of \(\tilde{\pi}_{1}=1-\tilde{\pi}_{0}\), and it is zero with a probability of \(\tilde{\pi}_{0}\). The average output power is, therefore, \[P_{out,av}=(1-\tilde{\pi}_{0})P_{max}\] and the ratio between \(P_{max}\) and \(P_{out,av}\) is \[10\log_{10}\frac{P_{max}}{P_{out,av}}=-10\log_{10}(1-\tilde{\pi}_{0}).\] This loss cannot be recovered, it is, at most, 3 dB when \(N_{S}=2\), and it decreases when \(N_{S}\) increases. * When there are 6 satellites, \(x(t)\) is zero with a probability of \[\tilde{\pi}_{0}=\binom{6}{3}\bigg{[}\frac{1}{2}\bigg{]}^{6}=\frac{6\times 5 \times 4}{3\times 2}\ \frac{1}{64}=\frac{20}{64}=0.3125\,.\] Assuming that, when \(x(t)\
eq 0\) (probability 0.6875), the HPA output is \(\pm\sqrt{P_{max}}\), then the average output power is \(P_{out,av}=0.6875\,P_{max}\) (\(31-1.627\)=29.37 dBW, against the measured 29.36 dBW in Figure 4). * When there are 11 satellites, the probability that \(x(t)=0\) is zero (there are 11 chips that add together, so the result can never be 0). The output average power \(P_{out,av}\) can reach the maximum value \(P_{max}\) provided that \(P_{in}\) is large enough. In general, an odd number of satellites allows us to obtain the following: \(P_{out,av}=P_{max}\). However, when \(P_{out,av}=P_{max}\), it is as if we were transmitting the sign of the sum of the chips, instead of the sum of the chips. Consequently, in the following, we assume that the sign of the sum of the transmitted signals is sent, as shown in Figure 5. This approach was first proposed in [25].