## 3 Design of SIRAC  
### The SI Algorithm  
While considering the NMP linear system:  
\[ \begin{array}{l}\dot{x}(t)=Ax(t)+Bu(t)\\ y(t)=Cx(t)+Du(t)\\ z(t)=C_{m}x(t)+D_{m}u(t)\end{array}  \tag{13}\]  
where \(x\in\mathbb{R}^{n}\), \(u\in\mathbb{R}^{p}\), \(y\in\mathbb{R}^{m}\), \(z(t)\) is the measurement output, and \(A\), \(B\), \(C\), \(D\), \(C_{m}\), and \(D_{m}\) are suitable dimension matrices. If \(y_{d}(t)\) is the desired trajectory, the desired input \(u_{d}(t)\) and state \(x_{d}(t)\) meet the following equations:  
\[ \begin{array}{l}\dot{x}_{d}(t)=Ax_{d}(t)+Bu_{d}(t)\\ y_{d}(t)=Cx_{d}(t)+Du_{d}(t)\\ z_{d}(t)=C_{m}x_{d}(t)+D_{m}u_{d}(t)\end{array}  \tag{14}\]  
then the \(H_{\infty}\) controller can stabilize the output trajectory. A transform of coordinates \(T\) is used in order to find the inverse input-state of \([u_{d}(t),x_{d}(t)]\), such that:  
\[\left[\xi(t),\eta(t)\right]^{T}=Tx(t) \tag{15}\]  
where \(\xi(t)\) includes the output \(y(t)\) and its time derivatives, that is,  
\[\xi(t)\equiv\left[y_{1},\dot{y}_{1},\cdots,\frac{d^{r_{1}-1}y_{1}}{dt^{r_{1}- 1}},y_{2},\dot{y}_{2},\cdots,\frac{d^{r_{2}-1}y_{2}}{dt^{r_{2}-1}},\cdots,y_{ m},\dot{y}_{m},\cdots,\frac{d^{r_{m}-1}y_{m}}{dt^{r_{m}-1}}\right]^{T} \tag{16}\]  
\[\frac{d^{r_{k}}y_{k}}{dt^{r_{k}}}=C_{p}A^{\eta_{k}}x(t)+C_{p}A^{\eta_{k}-1}Bu(t) \tag{17}\]  
\[D=0 \tag{18}\]  
Figure 2: The desired trajectory of the altitude. where \(r=[r_{1},r_{2},\cdots,r_{m}]\) is the relative degree vector and \(C_{p}\) is the pth row of \(C\). For the aircraft, the state \(x(t)=[\delta_{e}(t),\delta_{t}(t),V(t),\gamma(t),q(t),\theta(t),h(t)]^{T}\) and the output \(y(t)=[h(t),V(t)]^{T}\). According to [29], the relative degree \(r=[3,2]\), so the \(\xi(t)\) can be selected as:  
\[\xi(t)\equiv\left[h(t),\dot{h}(t),\ddot{h}(t),V(t),\dot{V}(t)\right]^{T} \tag{19}\]  
Additionally, the system internal state \(\eta(t)=[q(t),\theta(t)]^{T}\). Now, Equation (15) can be rewritten as:  
\[\left[h(t),\dot{h}(t),\ddot{h}(t),V(t),\dot{V}(t),q(t),\theta(t)\right]^{T}=Tx (t) \tag{20}\]  
In order to find \(T\), assume  
\[u(t)=[\delta_{\alpha c},\delta_{tc}]^{T}=[\delta_{e},\delta_{t}]^{T}=\left[ \begin{array}{cccccc}1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0\end{array} \right]\!\!x(t) \tag{21}\]  
Substitute Equation (21) into Equation (17) yields  
\[\frac{d^{n}y_{k}}{dt^{n}}=\left\{C_{p}A^{n_{k}}+C_{p}A^{n_{k}-1}B\!\left[ \begin{array}{cccccc}1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0\end{array} \right]\!\!\right\}x(t) \tag{22}\]  
According to Equations (20) and (22), we get  
\[T=\left[ \begin{array}{cccccc}0&0&0&0&0&1\\ \left\{C_{1}A^{1}+C_{1}A^{0}B\!\left[\begin{array}{cccccc}1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0\end{array}\right]\!\!\right\}\\ \left\{C_{1}A^{2}+C_{1}A^{1}B\!\left[\begin{array}{cccccc}1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0\end{array}\right]\!\!\right\}\\ 0&0&1&0&0&0&0\\ \left\{C_{2}A^{1}+C_{2}A^{0}B\!\left[\begin{array}{cccccc}1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0\end{array}\right]\!\!\right\}\\ 0&0&0&0&1&0&0\\ \end{array} \right] \tag{23}\]  
Currently, the system Equation (13) can be rewritten into the new coordinates, as:  
\[\dot{\xi}(t)=\hat{A}_{1}\xi(t)+\hat{A}_{2}\eta(t)+\hat{B}_{1}u(t) \tag{24}\]  
\[\dot{\eta}(t)=\hat{A}_{3}\xi(t)+\hat{A}_{4}\eta(t)+\hat{B}_{2}u(t) \tag{25}\]  
where  
\[\left[ \begin{array}{cc}\hat{A}_{1}&\hat{A}_{2}\\ \hat{A}_{3}&\hat{A}_{4}\end{array} \right]=TAT^{-1}\!\left[ \begin{array}{c} \hat{B}_{1}\\ \hat{B}_{2}\end{array} \right]=TB \tag{26}\]  
Without the loss of general, assume that the desired output trajectory \(y_{d}(t)\) and its time derivatives are specified, that is, \(y^{(r)}(t)=y^{(r)}_{d}(t)\). Subsequently, the following steps are used to find the desired input \(u_{d}(t)\) and desired state \(x_{d}(t)\). **Step 1:** Find the inverse system. Equation (17) can be rewritten as:  
\[y^{(r)}_{d}(t)=\left[\ddot{h}_{d},\dot{V}_{d}\right]^{T}=A_{x}x_{d}(t)+B_{y}u_ {d}(t) \tag{27}\]where  
\[A_{x}=\left[ \begin{array}{c}C_{1}A^{3}\\ C_{2}A^{2}\end{array} \right],B_{y}=\left[ \begin{array}{c}C_{1}A^{3-1}B\\ C_{2}A^{2-1}B\end{array}