## 4 PCE Dependency on PAPR and Power Level  
In this section, the impact of PAPR and received signal power level on the PCE of the rectifier is presented by both simulations and measurements. We derive the harvested DC power depending on the received signal using a circuit-based analysis [17], considering the basic rectifier scheme as in Figure 3. The basic rectifier consists of a rectifier matching network avoiding RF signal reflection, a rectifying diode as the main RF-DC transformer, and a low-pass RC filter to filter out the RF response and to control the ripple. Since symbol rate and channel fading are slow compared to the circuit's transient time needed for reaching a constant DC output, we consider constant channel coefficients and symbol during the considered time block. Knowing the received signal with power splitting coefficient \(\rho\) and the matched rectifier resistance \(R_{match}\), we can compute the instantaneous voltage into the rectifier \(v_{in}\left(t\right)\) as [17]  
\[v_{in}\left(t\right)=\Re\left[\sum_{n=1}^{N_{t}}\left(h_{n}w_{n}x_{n}+n_{n} \right)e^{2\pi f_{n}t}\right]\sqrt{\rho R_{match}}. \tag{11}\]  
Following the matching network of the rectifier, the voltage seen from the diode to the end of the circuit is \(v_{D_{in}}\left(t\right)\). Subsequently, we can derive the system model as:  
\[v_{D}(t) =v_{D_{in}}\left(t\right)-v_{o}(t), \tag{12a}\] \[i_{D}(t) =i_{C}(t)+i_{R}(t),\] (12b) \[i_{D}(t) =I_{s}\left(e^{\frac{v_{D}\left(t\right)}{nv_{T}}}-1\right),\] (12c) \[i_{C}(t) =C\frac{dv_{o}(t)}{dt},\ i_{R}(t)=\frac{v_{o}(t)}{R}, \tag{12d}\]  
where \(v_{D}(t)\), \(v_{D_{in}}(t)\), \(v_{o}(t)\), \(i_{D}(t)\), \(i_{C}(t)\), \(i_{R}(t)\), \(I_{s}\), \(n\), \(V_{T}\) represent the voltage over the diode, instantaneous voltage into the diode, instantaneous voltage at the output of the diode, current flowing through the diode, current through the capacitor, current through the resistor, diode's characteristic saturation current, ideality factor, and thermal voltage, respectively, as can be seen in Figure 3.  
After proper transformation, the instantaneous output voltage is derived by solving the ordinary differential equation (ODE) function as below  
\[\frac{dv_{o}(t)}{dt}=\frac{I_{s}}{C}\left(e^{\frac{v_{D_{in}}\left(t\right)-v _{o}(t)}{nv_{T}}}-1\right)-\frac{v_{o}(t)}{RC}. \tag{13}\]  
Since the right hand side of Equation (13) is a complicated non-linear function, a unique analytical solution is impossible or very hard to achieve. Thus we propose to use an ODE solver to achieve the numerical solution. Figure 3: Basic rectifier scheme. Having obtained the instantaneous voltage in time domain, the output DC voltage \(V_{DC}\) is the mean value of the diode's output voltage \(v_{o}\) over a steady period as [17]  
\[V_{DC}=\frac{1}{T-t_{\text{steady}}}\int_{t_{\text{steady}}}^{T}v_{o}\left(t \right)dt, \tag{14}\]  
where \(t_{\text{steady}}\) is the starting point of the charging steady state and \(T\) is the total examined time of the output signal. The output DC power \(P_{DC}\) equals \(\frac{V_{DC}^{2}}{R}\). To compare the performance of different waveforms and equalization strategies, we define the normalized PCE as the harvested DC power divided by the transmitted power:  
\[v_{in}\left(t\right)=\Re\left[\sum_{n=1}^{N_{t}}\left(h_{n}w_{n}x_{n}+n_{n} \right)e^{2\pi f_{n}t}\right]\sqrt{\rho R_{match}}. \tag{11}\]0  
with \(P_{n}\) the power level of \(n\)-th tone. To show how the PAPR and power level of the received signal influence the PCE of the rectifier, we simulate the DC output power of the half-wave rectifier using signals of varying PAPR at different power levels. We consider that the rectifier consists of a HSMS2850 Schottky diode, 150 pF load capacitor, 10 k\(\Omega\) load resistor, and an optimized input matching network ensuring the full received RF signal goes through the rectifier. As the received signal, we consider two-tone multisine signals at different power levels and varying PAPR.