## 4 Gearshift Control Problem
### Inertia Phase Control
#### 4.3.2 Feedback Control  
To derive the feedback control law, the forward kinetic model should be used. However, as has been discussed before, \(T_{out}\) is the load on the wheel and is hard to measure. It should be substituted by \(\dot{\omega}_{S1S2}\), which can be attained indirectly. As before, substituting \(T_{BK2}\) with \(T_{dir}\), \(\dot{\omega}_{R1}\) can be expressed as Equation (38).  
\[\dot{\omega}_{R1}= \frac{1}{\rho}\Bigg{\{}(1+k_{1}+k_{2})k_{1}k_{2}T_{dirv}-(1+k_{1}) (1+k_{2})k_{1}k_{2}T_{in} \tag{38}\] \[+\Bigg{[}(1+k_{1}+k_{2})k_{1}k_{2}J_{S1S2}-(1+k_{2})k_{1}J_{C1R2} \Bigg{]}\dot{\omega}_{S1S2}\Bigg{\}},\]  
where  
\[\rho=(1+k_{1}+k_{2})(1+k_{1})(1+k_{2})J_{R1}+(1+k_{2})k_{1}^{2}J_{C1R2}. \tag{39}\]  
From Equation (38), it is obvious that \(\dot{\omega}_{R1}\) is related to three quantities, namely, \(T_{dir}\), \(T_{in}\), and \(\dot{\omega}_{S1S2}\). If the term regarding \(\dot{\omega}_{S1S2}\) is omitted, Equation (38) turns into  
\[\dot{\omega}_{R1}=\frac{1}{\rho}\Bigg{[}(1+k_{1}+k_{2})k_{1}k_{2}T_{dirv}-(1+k _{1})(1+k_{2})k_{1}k_{2}T_{in}.\Bigg{]} \tag{40}\]  
The term regarding \(T_{dirv}\) can be treated as known disturbances and compensated in a feedforward way. To make it clear, rewrite Equation (38) in the state-space form:  
\[\dot{x}=ax+bu+b_{k}d_{k}+b_{u}d_{u} \tag{41}\]  
with  
\[\left\{\begin{aligned} x&=\omega_{R1}\\ a&=0\\ u&=T_{in}\\ b&=-\frac{(1+k_{1})(1+k_{2})k_{1}k_{2}}{\rho}\\ b_{k}d_{k}&=\frac{(1+k_{1}+k_{2})k_{1}k_{2}}{\rho} T_{dirv},\end{aligned}\right. \tag{42}\]  
where \(d_{k}\) and \(d_{u}\) denote known and unknown disturbances, respectively. With \(b_{u}=b\), a time-domain disturbance observer (DO) can be used to estimate \(d_{u}\):  
\[\left\{\begin{aligned} \dot{z}&=-Lb(z+Lx)-L(ax+bu+b_{k}d_{k})\\ \dot{d}_{u}&=z+Lx.\end{aligned}\right. \tag{43}\]  
Note that \(L\) should satisfy \(-Lb<0\). The component for disturbance compensation in \(u\) is as follows:  
\[u_{du}=-\hat{d}_{u}. \tag{44}\]  
Note that \(u_{feedforward}\) in Equation (37) can be written as follows:  
\[u_{feedforward}=u_{du}+u_{inv}, \tag{45}\]  
where  
\[u_{dk} =-\frac{b_{k}d_{k}}{b}=\frac{1+k_{1}+k_{2}}{1+k_{1}+k_{2}+k_{1}k _{2}}T_{dirv} \tag{46}\] \[u_{inv} =-\left(\frac{1+k_{1}+k_{2}}{k_{1}k_{2}}J_{R1}+\frac{k_{1}}{k_{2} +k_{1}k_{2}}J_{C1R2}\right)\dot{\omega}_{R1}, \tag{47}\]where \(u_{dk}\) is used to compensate the known disturbances \(b_{k}d_{k}\), while \(u_{inv}\) is derived from the nominal model. Assuming that both \(b_{k}d_{k}\) and \(b_{u}d_{u}\) are completely compensated, the state-space model in Equation (41) turns into the nominal one and can be expressed as follows:  
\[\dot{x}_{n}=ax_{n}+bu_{n}. \tag{48}\]  
The feedback control law for the nominal model is  
\[u_{feedback}=-F(x-r), \tag{49}\]  
where \(F\) is the feedback gain and can be roughly determined as  
\[F=\frac{\ln|\omega_{R1}(t_{1})|-\ln E}{b(t_{f}-t_{1})}, \tag{50}\]  
where \(E\) is the level of residual tracking error when \(t_{f}\) is reached. Note that the sign of \(F\) should be determined in accordance with \(a\) and \(b\) to ensure convergence. This is obvious by substituting Equation (49) into (48), whose solution is  
\[x_{n}(t)=\left(x_{n}(t_{1})+\frac{bF}{a-bF}\right)e^{(a-bF)t}-\frac{bF}{a-bF}r, \tag{51}\]  
where the reference signal \(r\) is from the fifth-degree polynomial expressed by Equation (34).  
For the exponential convergence of \(x_{n}(t)\) to \(x_{n}(\infty)=-\frac{bF}{a-bF}r\), \(F\) should satisfy that  
\[a-bF<0. \tag{52}\]  
Finally, the control law for \(T_{in}\) in the inertia phase can be summarized as  
\[u=u_{dk}+u_{inv}+u_{du}+u_{feedback}. \tag{53}\]  
Block diagram of the slip speed controller is shown in Figure 9.