# Continuous-Time Perfect Control Algorithm--A State Feedback Approach
## 5 Stability Properties of the New Perfect Control Law  
The stability of perfect control algorithm has already been discussed widely for a class of discrete-time systems [26]. In the literature, it can be found that the stability property should be understood in terms of the BIBS approach. It is caused mainly by the fact that the perfect control output remains stable even for plants with closed-loop poles located outside the stable region. Thus, the input-to-state characteristics are the main concern here. Of course, according to Equation (13), the CTPC algorithm can be treated as the inverse model control-oriented state-feedback system occupied by  
\[\mathbf{u}(t)=-\mathbf{K}\mathbf{x}(t), \tag{18}\]  
where \(\mathbf{K}=(\mathbf{C}\mathbf{B})^{\mathrm{R}}\mathbf{C}(\mathbf{A}+\mathbf{ I}\frac{1}{\Delta t})\), whose closed-loop control poles can be calculated in accordance with the following formula  
\[det(\mathbf{A}-\mathbf{B}\mathbf{K})=0. \tag{19}\]  
It is clear that the discussed stability depends on the part connecting vectors \(\mathbf{x}(t)\) and \(\dot{\mathbf{x}}(t)\); thus, the closed-loop poles can be obtained. The static non-autoregressive part disappears as soon as the output reaches its reference value, which essentially occurs after a single calculation step \(\Delta t\). In conclusion, the stability of the closed-loop CTPC law is determined by the roots of characteristic equation as follows:  
\[det(s\mathbf{I}-\mathbf{A}+\mathbf{B}(\mathbf{C}\mathbf{B})^{\mathrm{R}} \mathbf{C}(\mathbf{A}+\mathbf{I}\frac{1}{\Delta t}))=0. \tag{20}\]  
Here again, we need to emphasize that the numerical solution of stability problem can be considered in the context of the inverse matrix calculation, with an infinite number of possible outcomes--even with the ability to almost arbitrary allocation of the number of zero and nonzero closed-loop poles. **Remark 5**. : _The pole-placement feature is connected with the rank of the closed-loop perfect control system matrix. The mentioned rank seems to be dependent on the size differences derived from \(\mathbf{A}\),\(\mathbf{B}\),\(\mathbf{C}\) matrices. This intriguing phenomenon is still under investigation._  
Interestingly, in comparison with the discrete-time instances, the zero closed-loop poles are not obtained here. This is caused by the fact that the additional part of state feedback (\(\mathbf{I}\frac{1}{\Delta t}\)) entails a drift of the CTPC algorithm poles that were originally equal to zero. In this case, the zero eigenvalues of the closed-loop system matrix are transformed to those connected with the solver-derived \(\Delta t\) parameter. The influence of solver time seems to be an obvious consideration when aiming for proper output behavior. If the goal is to obtain the zero control error in a single solver step, it is clear that if the solver step takes a shorter time \(\Delta t\), then the dynamics applied to the output and state variables should be faster to overcome the shorter calculation interval. In order to explain the crucial role of the step time \(\Delta t\), the closing statements are formulated below. **Theorem 2**. : _The continuous-time perfect control law as in Equation (12) should be understood in terms of the single step time \(\Delta t\) tending to zero._  
**Proof. ** Immediately after substitution of formula (12) to right-invertible nondelayed plants (1), we obtain the expression  
\[\dot{\mathbf{y}}(t)=\frac{1}{\Delta t}(\mathbf{y_{ref}}(t+\Delta t)-\mathbf{y}(t )), \tag{21}\]  
which can be transformed, according to the relation (7), to  
\[\left(\frac{\Delta t-1}{\Delta t}\right)[\mathbf{y_{ref}}(t+\Delta t)-\mathbf{ y}(t)]=\mathbf{0}. \tag{22}\]  
For \(\Delta t\to 0\), we receive \(\mathbf{y}(t)\rightarrow\mathbf{y_{ref}}(t+\Delta t)\), which ends the proof. **Corollary 1**. : _The perfect regulation (13) having \(\mathbf{y_{ref}}(t)=\mathbf{0}\) occurs iff the relation_  
\[\Delta t\mathbf{y}(t)=\mathbf{y}(t), \tag{23}\]  
_determined by \(\Delta t\to 0\), \(\mathbf{y}(t)\to 0\), holds._  
**Remark 6**.