## 6 State Observer  
For this case, the incremental state \(\Delta x(k)\) is not accessible. For this reason, a state observer is required. The incremental state can be estimated using the following expression [35]:  
\[\Delta x_{e}(k)=x_{e}(k)-x_{e}(k-1) \tag{27}\]  
Furthermore, it is possible to use an incremental state observer, which is described below [15]:  
\[\Delta x_{e}(k+1)=A\Delta x_{e}(k)+B\Delta u(k)+H(y(k)-y(k-1)-\text{C}\Delta x _{e}(k)) \tag{28}\]  
where \(H\in\Re^{(n\times q)}\). The error estimate would be:  
\[\begin{split} e(k+1)&=\Delta x(k+1)-\Delta x_{e}(k+ 1)\\ &=(A\Delta x(k)+B\Delta u(k))-(A\Delta x_{e}(k)+B\Delta u(k)+\\ &\qquad H(y(k)-y(k-1)-\text{C}\Delta x_{e}(k)))\end{split} \tag{29}\]  
From Equations (5), (6) and (8), we obtain:  
\[\begin{split} y(k)-y(k-1)&=a_{y}+\text{C}x(k)-a_{y} -\text{C}x(k-1)\\ &=C(x(k)-x(k-1))=\text{C}\Delta x(k)\end{split}\]  
Then, Equation (29) would be:  
\[\epsilon(k+1)=(A-HC)\epsilon(k) \tag{30}\]  
The method to calculate H has been described in [15] and is obtained as follows:  
\[H=AC^{t} \tag{31}\]  
Finally, the observer of the incremental state extended vector is obtained:  
\[x_{ae}(k)=\left[\begin{array}{c}y(k)\\ \Delta x_{e}(k)\end{array}\right] \tag{32}\]  
and the controller is described as follows:  
\[\begin{split}\Delta^{2}u(k)=K(x_{x^{2}r}-x_{a^{2}}(k))& =\left[\begin{array}{cc}K_{y}&K_{\Delta y}&K_{x}\end{array}\right] \left(\left[\begin{array}{c}y_{r}\\ \Delta y_{r}\\ 0\end{array}\right]-\left[\begin{array}{c}y(k)\\ \Delta y(k)\\ \Delta^{2}x(k)\end{array}\right]\right)\\ &=\left[\begin{array}{cc}K_{y}&K_{\Delta y}&K_{x}\end{array}\right] \left[\begin{array}{c}y_{r}-y(k)\\ \Delta y_{r}-\Delta y(k)\\ -\Delta^{2}x(k)\end{array}\right] \tag{39}\]  
where \(K\in\Re^{(p\times(2q+n))}\), \(K_{y}\in\Re^{(p\times q)}\), \(K_{\Delta y}\in\Re^{(p\times q)}\) and \(K_{x}\in\Re^{(p\times n)}\), and the control action is  
\[\begin{split}\Delta u(k)&=\Delta u(k-1)+\Delta^{2}u(k)\\ u(k)&=u(k-1)+\Delta u(k)\end{split} \tag{40}\]  
If the feedback system is stable and steady state is approached, the following is true:  
\[\lim_{k\rightarrow\infty}\Delta^{2}u(k)=0,\lim_{k\rightarrow\infty}\Delta^{2 }x(k)=0\]  
and, thus,  
\[\lim_{k\rightarrow\infty}K_{y}(y_{r}-y(k))=0,\lim_{k\rightarrow\infty}K_{ \Delta y}(\Delta y_{r}-\Delta y(k))=0\]  
So, if \(K_{y}\) and \(K_{\Delta y}\) are of rank \(q\), it is verified that  
\[\lim_{k\rightarrow\infty}y(k)=y_{r},\lim_{k\rightarrow\infty}\Delta y(k)= \Delta y_{r}\]  
Therefore, the controlled system has a steady-state error equal to zero in response to step-type inputs and disturbances, and even to ramp-type inputs. It is important to highlight that the control action (40) has been calculated in an incremental-2 form, which is equivalent to applying a double integral control action. Similarly to the incremental model, for the incremental-2 model, it is necessary to use a state observer, since \(\Delta^{2}x(k)\) is not accessible. Based on Equation (27), the incremental-2 state can be estimated as follows:  
\[\Delta^{2}x_{e}(k)\Delta x_{e}(k)-\Delta x_{e}(k-1) \tag{41}\]  
With Equations (41) and (28), the observer can be described as  
\[\Delta^{2}x_{e}(k+1)=A\Delta^{2}x_{e}(k)+B\Delta^{2}u(k)+H\Big{(}\Delta y(k)- \Delta y(k-1)-C\Delta^{2}x_{e}(k)\Big{)} \tag{42}\]  
where \(H\in\Re^{(n\times q)}\) is calculated in the same way as in Equation (31).Thus, the observer of the extended vector of the incremental-2 state would be  
\[x_{a^{2}e}(k)=\left[\begin{array}{c}y(k)\\ \Delta x_{ac}(k)\end{array}\right]=\left[\begin{array}{c}y(k)\\ \Delta y(k)\\ \Delta^{2}x_{e}(k)\end{array}\right] \tag{43}\]