# Parallel Hybrid Electric Vehicle Modelling and Model Predictive Control
## 2 Modelling of the Parallel HEV
At a low speed of less than 40 km/h, the clutch is open, and only the main electric motor EM1 propels the HEV. The contribution of some other exponential coefficients is small and can be ignored. The vehicle rolling resistance torque at a low speed can be simplified as:  
\[M_{v}=M_{v0}+k_{v}\omega_{3} \tag{2}\]  
where \(M_{v0}\) is the initial resistance constant of air drag and rolling friction. \(k_{v}\) is a linear coefficient that depends on the gear ratio. On the first part, the torque applied is:  
\[M_{1o}=J_{1}\dot{\omega}_{1} \tag{3}\]  
This torque can be calculated as:  
\[M_{1o}=M_{ICE}+M_{M2}-M_{C} \tag{4}\]  
where \(M_{ICE}\) is the torque from ICE; \(M_{M2}\) is the torque from motor ME2; and \(M_{C}\) is the torque from the clutch. When the clutch is locked, the clutch torque \(M_{C}\) is the maximum static clutch friction:  
\[M_{C}=\frac{2}{3}r_{C}F_{NC}\mu_{S}\text{ when }\left(M_{C}=M_{fmax}^{Static}\right) \tag{5}\]  
where \(r_{C}\) is the clutch radius; \(F_{NC}\) is the normal force; and \(\mu_{S}\) is the clutch friction coefficient. When the clutch is in the transitional period, \(M_{C}<M_{fmax}^{Static}\), the clutch torque is:  
\[M_{C}=r_{C}F_{NC}sign(\omega_{1}-\omega_{2})\mu_{K}\text{ when }\left(M_{C}<M_{ fmax}^{Static}\right) \tag{6}\]  
where \(\mu_{K}\) is the clutch slipping coefficient. Figure 2: Simplified structure of the parallel HEV. On the second part, the torque applied on the main motor ME1 is:  
Figure 2.0  
The sum of inertias is calculated as:  
Figure 2.1  
The torque changing is calculated as:  
Figure 2.2  
The balance of torque \(M_{2o}\) is calculated as:  
Figure 2.3  
where \(\eta\) is the transmission efficiency of the gearbox and the differential gear. The above torque equations can be transformed to the following dynamic equations:  
Figure 2.4  
The angular acceleration of the shaft 1 is calculated as:  
Figure 2.5  
where \(k_{\beta 1}\) is the shaft 1 friction coefficient. The angular acceleration of the shaft 2 is calculated as:  
Figure 2.6  
where \(k_{\beta 2}\) is the shaft 2 friction coefficient. Finally, the angular acceleration of the shaft 3 is calculated as:  
Figure 2.7  
where \(k_{\beta 3}\) is the shaft 3 friction coefficient. The jerk on the drivetrain is calculated as:  
Figure 2.8  
The torque generated on the main motor is calculated as:  
Figure 2.9  
where \(M_{DC\_MOTORR}\) is the main motor torque; \(k_{T}\) is the motor constant, \(k_{T}=\frac{M_{T\text{\tiny{T}}}}{T_{\text{\tiny{T}}}}\) (Nm/A); \(k_{E}\) is the electromotive force (EMF) constant, \(k_{E}=k_{T}\); \(R_{I}\) is the resistance; \(V\) is the voltage supply; and \(\omega\) is the angular velocity. Now we proceed and transform all the above equations into a first order linear system as:\[\dot{\theta}_{1}=\left[ \begin{array}{cccccccc}0+&\omega_{1}+&0+&0+&0+&0+&0\\ \end{array} \right]+\left[ \begin{array}{cccccccc}0+&0+&0+&0+&0\\ \end{array} \right] \tag{17}\]  
\[\dot{\omega}_{1}=\left[ \begin{array}{cccccccc}0+&\frac{-\left(k_{\mu_{1}}+ \frac{k_{\mu_{2}}k_{\mu_{3}}}{k_{1}}\right)\omega_{1}}{I_{1}}+&0+&0+&0+&0\\ \end{array} \right]+\left[ \begin{array}{cccccccc}\frac{M_{ICE}}{I_{1}}+&\frac {k_{\mu_{1}}V_{1}}{R_{1}h_{1}}+&0+&\frac{-M_{\rm C}}{I_{1}}+&0\\ \end{array} \right] \tag{18}\]  
\[\dot{\theta}_{2}=\left[ \begin{array}{cccccccc}0+&0+&0+&\omega_{2}+&0+&0+&0\\ \end{array} \right]+\left[ \begin{array}{cccccccc}0+&0+&0+&0+&0+&0\\ \end{array} \right] \tag{19}\]  
\[\dot{\omega}_{2}=\left[ \begin{array}{cccccccc}0+&0+&0+&0+&\frac{-\left(k_{ \mu_{2}}+\frac{k_{\mu_{2}}k_{\mu_{3}}}{k_{1}^{2}}\right)\omega_{2}}{I_{2}^{3 }}+&-\frac{k_{\mu_{3}}\omega_{3}}{I_{1}^{2}}\end{array} \right]+\left[ \begin{array} []{cccccccc}0+&0+&0+&0+&0\\ \end{array} \right] \tag{20}\]  
\[\dot{\theta}_{3}=\left[ \begin{array}{cccccccc}0+&0+&0+&0+&\omega_{3}+&0\\ \end{array} \right]+\left[ \begin{array}{cccccccc}0+&0+&0+&0+&0\\ \end{array} \right] \tag{21}\]  
\[\dot{\omega}_{3}=\left[ \begin{array}{cccccccc}0+&0+&0+&0+&\frac{k_{\mu_{3} }\omega_{3}}{I_{3}}+&0\\ \end{array} \right]+\left[