## 4 Gearshift Control Problem
### Torque Phase Control  
In the torque phase, the load on the motor shaft should be gradually transferred from the off-going brake to the oncoming one. Requirements for torque phase control are as follows:  
* The off-going brake should be kept engaged during the whole process and no slip occurs. * The load and the pressure on the off-going brake should be reduced to zero at the same time. Under the first gear, BK1 is locked, which results in  
\[\dot{\omega}_{S1S2}=(1+k_{2})\dot{\omega}_{C2}, \tag{21}\]  
where \(\dot{\omega}_{C2}\) can be expressed with \(T_{out}\), \(T_{BK2}\), and \(T_{in}\) by solving Equations (19)-(21):  
\[\dot{\omega}_{C2}=\frac{1}{(1+k_{2})^{2}J_{S1S2}+J_{C2}}\left[(1+k_{2})T_{in}- T_{out}-\frac{1+k_{2}}{k_{1}}T_{BK2}\right]. \tag{22}\]  
The inertia of the motor is rather small compared to that of the engine, and therefore \(J_{S1S2}\) can be neglected. Based on this assumption, Equation (22) is simplified as  
\[J_{C2}\dot{\omega}_{C2}=(1+k_{2})T_{in}-\frac{1+k_{2}}{k_{1}}T_{BK2}-T_{out}. \tag{23}\]  
Thus, the driving torque on the output shaft is  
\[T_{driv}=(1+k_{2})T_{in}-\frac{1+k_{2}}{k_{1}}T_{BK2}. \tag{24}\]  
Note that \(T_{driv}\) consists of two terms regarding \(T_{in}\) and \(T_{BK2}\). The former is the motor torque; and the latter is determined by the pressure on BK2 (since it is slipping during the torque phase). Thus, \(T_{driv}\) can be manipulated either through controlling the motor or BK2. As an internal torque, \(T_{BK1}\) is determined by \(T_{in}\) and \(T_{BK2}\) as well:  
\[T_{BK1}=k_{2}T_{in}-\frac{1+k_{1}+k_{2}}{k_{1}}T_{BK2}. \tag{25}\]  
During the torque phase, there are totally three components to be controlled, namely, the motor and two brakes. However, according to Equations (24) and (25), only one variable can be freely determined. Recalling the requirement on controllability of the shift time, \(T_{BK1}\) (transmitted torque of the off-going brake) should better be selected as the one being manipulated. For example, as the simplest case, \(T_{BK1}\) can be commanded to decrease to zero linearly within a specified duration. Since BK1 is required to be kept engaged along the whole torque phase, its transmitted torque should be no larger than the torque capacity \(T_{cap,BK1}\). \[T_{BK1}\leq T_{cap,BK1}. \tag{26}\]Additionally, recall the second requirement that \(T_{BK1}\) and \(T_{cap,BK1}\) be reduced to zero at the same time, which suggests the following control law for \(T_{BK1}\):  
\[T_{BK1}=\lambda T_{cap,BK1}, \tag{27}\]  
where \(\lambda\) is a coefficient that satisfies \(\lambda\leq 1\). For practical use, \(\lambda\) is advised to fall between 0.90 and 0.99. As has been discussed before, the trajectory of \(T_{cap,BK1}\) in torque phase can be designed as a linear curve:  
\[T_{cap,BK1}=T_{cap,BK1}(t_{0})-\frac{T_{cap,BK1}(t_{0})}{t_{1}-t_{0}}(t-t_{0}),\quad t_{0}\leq t\leq t_{1}, \tag{28}\]  
where \(t_{0}\) and \(t_{1}\) are the start and end point of the torque phase, respectively. Duration of the torque phase can thus be arbitrarily manipulated (within the capacity and constraint of components) through the parameter \(t_{1}\). \(T_{driv}\) in Equation (24) is specified by the vehicle controller unit (VCU) and related to position of the accelerator pedal. Therefore, it can be treated as known. \(T_{in}\) and \(T_{B2}\) can be derived by solving Equations (24) and (25):  
\[T_{BK2} =\frac{k_{1}k_{2}}{1+k_{1}+k_{2}+k_{1}k_{2}}T_{driv}-\frac{k_{1} }{1+k_{1}}T_{BK1} \tag{29}\] \[T_{in} =\frac{1+k_{1}+k_{2}}{1+k_{1}+k_{2}+k_{1}k_{2}}T_{driv}-\frac{1}{ 1+k_{1}}T_{BK1}. \tag{30}\]  
Equations (27)-(30) are control laws for the torque phase. It should be noted that the \(T_{cap,BK1}(t_{0})\) in Equation (28) should be determined carefully, so as not to cause sudden change of \(T_{in}\) and \(T_{BK2}\).