## 4 Dynamic Inverse Controller Design Based on the DNN Model  
In order to realize the tracking control of FAHV velocity and altitude, a controller is designed based on the DNN model (5).  
Assume that the reference signal is \(z_{r}\), the output tracking error is  
\[e_{m}=z_{r}-z. \tag{19}\]  
Taking time derivative of \(e_{m}\) and using (5), we have  
\[\begin{array}{l}\dot{e}_{m}=\dot{z}_{r}-C\hat{x}\\ =\dot{z}_{r}-CA\hat{x}-CB\hat{W}\Phi(\hat{x})-CB\hat{\beta}\Psi(\hat{x})u\end{array}. \tag{20}\]  
Design the linearizing feedback control law as  
\[u=\left(CB\hat{\beta}\Psi(\hat{x})\right)^{-1}\left(\dot{z}_{r}-CA\hat{x}-CB \hat{W}\Phi(\hat{x})+Ke_{m}\right), \tag{21}\]  
where \(K=diag\{k_{i}\},k_{i}>0,i=1,2\).  
Then, substituting (21) into (20), we obtain  
\[\dot{e}_{m}=-Ke_{m}. \tag{22}\]  
Therefore, the error \(e_{m}\) will converge to the origin exponentially.  
**Remark 2**.: _To apply the control law (21), we have to assure the existence of \(\left(CB\hat{\beta}\Psi(\hat{x})\right)^{-1}\)._  
To analyze the existence of \(\left(CB\hat{\beta}\Psi(\hat{x})\right)^{-1}\), the learning laws (15) and (17) can be written in matrix form as  
\[\begin{array}{l}\dot{\hat{W}}=-\dot{\widetilde{W}}=\eta EPBS\\ \dot{\hat{\beta}}=-\dot{\widetilde{\beta}}=\lambda EPBS_{1}U\end{array}, \tag{23}\]\[\text{where}\quad\eta = \quad\begin{bmatrix}\eta_{1}&&&\\ &\eta_{2}&\\ &&\ddots&\\ &&&\eta_{11}\end{bmatrix},E = \quad\begin{bmatrix}e_{1}&&&\\ &e_{2}&\\ &&\ddots&\\ &&&e_{11}\end{bmatrix},S = \quad\begin{bmatrix}\Phi^{T}(\hat{x})\\ \Phi^{T}(\hat{x})\\ \vdots\\ \Phi^{T}(\hat{x})\end{bmatrix},\] \[\lambda = \begin{bmatrix}\lambda_{1}&&&\\ &\lambda_{2}&&\\ &&\ddots&\\ &&&\lambda_{11}\end{bmatrix},S_{1} = \quad\begin{bmatrix}\Psi_{1}(\hat{x})&&&\\ &\Psi_{2}(\hat{x})&&\\ &&\ddots&\\ &&&\Psi_{3}(\hat{x})\end{bmatrix},U = \begin{bmatrix}u&\\ u&&\\ &&\ddots&\\ &&&u\end{bmatrix},B,\] and \(P\) are as listed in Equations (6) and (14), respectively.  
For our problem, expand matrix \(CB\hat{\beta}\Psi(\hat{x})\), and we have  
\[CB\hat{\beta}\Psi(\hat{x})=\left[\begin{array}{cccccc}1&0&0&0&\cdots&0\\ 0&0&1&0&\cdots&0\end{array}\right]\begin{bmatrix}b_{1}&&&\\ &b_{2}&&\\ &&\ddots&\\ &&&b_{11}\end{bmatrix}\begin{bmatrix}\hat{\beta}_{11}&0&\cdots&0\\ 0&\hat{\beta}_{22}&&\vdots\\ \vdots&\ddots&0\\ 0&\cdots&0&\hat{\beta}_{11,11}\end{bmatrix}\begin{bmatrix}\psi_{11}&\psi_{12 }\\ \psi_{21}&\psi_{22}\\ \vdots&\vdots\\ \psi_{11,1}&\psi_{11,2}\end{bmatrix}\,. \tag{24}\]  
Since \(B\) and \(\hat{\beta}\) are diagonal matrices and \(b_{i}\
eq 0,\forall i=1,2,\cdots,n\), and because the outputs of the FAHV are the velocity and the altitude, we only need to meet the condition \(\hat{\beta}_{11}\
eq 0\) and \(\hat{\beta}_{33}\
eq 0\). A projection algorithm was proposed to assure the condition [27, 28, 40], but it needs to know the optimal value of \(\beta^{*}\), and this is difficult to obtain.  
As can be seen from (24), to guarantee that \(CB\hat{\beta}\Psi(\hat{x})\) is reversible, one can make the elements \(\psi_{12}\) and \(\psi_{31}\) equal to zero when generating matrix \(\Psi(\hat{x})\). Additionally, it is seen from (23) that the condition \(\hat{\beta}_{11}\
eq 0\) and \(\hat{\beta}_{33}\
eq 0\) can be satisfied by adjusting the parameters \(\lambda\) and \(B\).