## 2 Powertrain Model
Since a two-speed transmission is considered, each operation mode can be driven in two speeds \((j\in\{1,2\})\)  
Figure 4: Models of the powertrain components and the most important parameters. Furthermore, the control variable for the electric drive \(u_{\text{ED}}\) is defined according to current operation state \(s_{\text{m}}\) and the corresponding degree of freedom:  
\[u_{\text{ED}}= \begin{cases}\begin{array}{ll}\omega_{\text{ED}}&\text{for $s_{ \text{m}}=1\ldots 2$,}\\ T_{\text{ED}}&\text{for $s_{\text{m}}=3\ldots 4$,}\\ f(T_{\text{FD}},\omega_{\text{FD}})&\text{for $s_{\text{m}}=5\ldots 6$.}\end{array}\end{cases}  \tag{3}\]  
In the case of the EM mode, no degree of freedom exists, and consequently, \(T_{\text{ED}}\) and \(\omega_{\text{ED}}\) are completely defined by means of the torque and angular velocity at the final drive. The model of the DHT comprises a simple planetary gear model with the stationary gear ratio \(i_{0}\). Its sun gear is connected to the ICE, its ring gear to the electric drive, and its carrier to the two-speed transmission. Considering this configuration as well as the transmission ratios of the two-speed transmission \(i_{j}\) and the electric drive \(i_{\text{ED}}\), the torques and angular velocities in the CVT mode are determined according to the following equation:  
\[ \begin{array}{l}T_{\text{ED}}=-i_{0}\cdot\left(\left(1+i_{0}\right)\cdot i _{\text{ED}}\cdot i_{j}\right)^{-1}\cdot T_{\text{FD}},\\ T_{\text{ICE}}=\left(\left(1+i_{0}\right)\cdot i_{j}\right)^{-1}\cdot T_{ \text{FD}},\\ \omega_{\text{ICE}}=\left(1+i_{0}\right)\cdot i_{j}\cdot\omega_{\text{FD}}+i_{ 0}\cdot i_{\text{ED}}^{-1}\cdot\omega_{\text{ED}},\end{array}  \tag{4}\]  
with the angular velocity of the electric drive \(\omega_{\text{ED}}\) as the degree of freedom (\(u_{\text{ED}}=\omega_{\text{ED}}\)). Analogously, the angular speeds and torques in the PAR mode can be derived, yielding the following:  
\[ \begin{array}{l}\omega_{\text{ED}}=-i_{\text{ED}}\cdot i_{j}\cdot\omega_{ \text{FD}},\\ \omega_{\text{ICE}}=i_{j}\cdot\omega_{\text{FD}},\\ T_{\text{ICE}}=i_{j}^{-1}\cdot T_{\text{FD}}+i_{\text{ED}}\cdot T_{\text{ED}}, \end{array}  \tag{5}\]  
where the torque of the electric drive \(T_{\text{ED}}\) needs to be chosen (\(u_{\text{ED}}=T_{\text{ED}}\)). For the EM mode, only the torque and angular velocity of the electric drive must be determined:  
\[ \begin{array}{l}T_{\text{ED}}=-\left(i_{\text{ED}}\cdot i_{j}\right)^{-1} \cdot T_{\text{FD}},\\ \omega_{\text{ED}}=-i_{\text{ED}}\cdot i_{j}\cdot\omega_{\text{FD}}.\end{array}  \tag{6}\]  
The ICE is represented by a map, which describes the specific fuel consumption \(b_{\text{e}}\) as a function of the torque \(T_{\text{ICE}}\) and rotational speed \(\omega_{\text{ICE}}\). The mass flow rate of the fuel is determined according to the following equation:  
\[\dot{m}_{\text{f}}=\frac{b_{\text{e}}(T_{\text{ICE}},\omega_{\text{ICE}})\cdot T _{\text{ICE}}\cdot\omega_{\text{ICE}}}{3.6\cdot 10^{6}}, \tag{7}\]  
with \(\dot{m}_{\text{f}}\) in g\(\cdot\)s\({}^{-1}\). Similarly, the required electric power \(P_{\text{el}}\) of the electric drive is determined by  
\[P_{\text{el}}=T_{\text{ED}}\cdot\omega_{\text{ED}}+P_{\text{ED,loss}}(|T_{ \text{ED}}|,|\omega_{\text{ED}}|), \tag{8}\]  
where the power loss \(P_{\text{ED,loss}}\) represents the result of the corresponding map, shown in Figure 4. This map also includes the power losses of the inverter. Since the electric power \(P_{\text{el}}\) must be provided by the battery, the battery current is expressed as  
\[\dot{q}=-i_{\text{Batt}}=-\frac{V_{0}(SoC)}{2\cdot R_{\text{i}}(SoC)}+\sqrt{ \left(\frac{V_{0}(SoC)}{2\cdot R_{\text{i}}(SoC)}\right)^{2}-\frac{P_{\text{el }}}{z_{\text{cell}}\cdot R_{\text{i}}(SoC)}}. \tag{9}\]  
This follows from the simplified equivalent circuit in Figure 4, where \(z_{\text{cell}}\) is the number of cells, \(V_{0}\) is the internal cell voltage, and \(R_{\text{i}}\) is the internal cell resistance.