## Appendix A Proof of Lemma 4
Indeed, the breaking point of interest is one of the roots of the polynomial (A3). These roots are analyzed by considering that the parameter \(g_{1}\) is variable. This allows studying the behavior of the roots of (A3) through a root locus diagram. Note that this root locus diagram does not depict the movement of the roots of the closed-loop injection scheme but the movement of the breaking points related to the system. Then, it is possible to associate a fictitious closed-loop system with Equation (A3). Let us consider an open loop system given by  
\[\frac{Y_{3}(z)}{U_{3}(z)}=\frac{\alpha_{1}bb_{1}e^{b_{1}T}}{b_{1}\big{(}e^{b_{ 1}T}z-1\big{)}(az^{n}+anz^{n-1}(z-e^{aT}))+ab_{1}e^{b_{1}T}z^{n}(z-e^{aT})}.\] (A4)  
With proportional action, \(R_{3}(z)=U_{3}(z)-g_{1}Y_{3}(z)\), it is possible to obtain the closed-loop system given by  
\[\frac{Y_{3}(z)}{R_{3}(z)}=\frac{\alpha_{1}bb_{1}e^{b_{1}T}}{b_{1}\big{(}e^{b_{ 1}T}z-1\big{)}(az^{n}+anz^{n-1}(z-e^{aT}))+ab_{1}e^{b_{1}T}z^{n}(z-e^{aT})+ \alpha_{1}bb_{1}e^{b_{1}T}g_{1}}.\] (A5)  
Note that the characteristic equation of the fictitious system (A5) is equivalent to Equation (A3). Then, in this new fictitious system (A5), it is possible to analyze the behavior of the breaking points. Taking into consideration an analysis of the locus of the roots for the system (A4) with \(g_{1}\) as a variable parameter, Figure A5 shows that there exists a breaking point between the pole located within the unit circle and the pole located outside. The location of the breaking point is obtained by solving for \(g_{1}\) from the characteristic polynomial Equation (A5) and its corresponding derivative, provided by  
\[\frac{dg_{1}}{dz}=-a\mathrm{e}^{-b_{1}T}z^{n-2}\Big{(}(e^{b_{1}T}(n+1)(n+2)z^ {2}-\Big{(}e^{T(b_{1}+a)}+1\Big{)}n(n+1)z+e^{aT}n^{2}-e^{aT}n\Big{)}.\] (A6)One of the roots of (A6), defining the behavior of the breaking point of interest, is given by  
\[P(z)=\frac{\mathrm{e}^{-b_{1}\tau/n}\Bigg{(}\sqrt{\frac{n\Big{(}\Big{(}n(n+1) \Big{(}\mathrm{e}^{(b_{1}+a)\tau/n}-2\Big{)}+8\Big{)}\mathrm{e}^{(b_{1}+a)\tau/n }+n(n+1)\Big{)}}{n+1}+n\Big{(}\mathrm{e}^{(b_{1}+a)\tau/n}+1\Big{)}}}{2(n+2)}.\] (A7)  
In order to get closed-loop stability in the scheme shown in Figure 1, the breakpoints derived from the fictitious system (A1) should be within the unit circle. Now, this issue is analyzed and ensured. It is important to note that the approach used in this proof guarantee that the breakpoint falls within the unit circle when \(n\to\infty\), or equivalently \(T\to 0\). This is with the intention of obtaining the stability conditions of the continuous time injection scheme shown in Figure 1. Therefore, we can obtain  
\[\lim_{n\to\infty}P(z)=1.\] (A8)  
From (A8), the breaking point is at the limit of stability. The key point is to determine if the breakpoint (A8) tends to one from the left or from the right. Note that when the limit of the Equation (A8) tends to 1 from the left, then the system has the breaking point within the unit circle, and therefore there is a stability region. Let us assume that the condition \(\tau<\frac{1}{a}-\frac{1}{b_{1}}+\sqrt{\frac{1}{a^{2}}+\frac{1}{b_{1}^{2}}}\) in the limit (A8). This implies that the breaking point of interest is within the unit circle, and there exists a region of stability. Finally, to end the proof, it is important to demonstrate that \(n-2\) roots of (A2) lie inside the unit circle assuming the condition \(\tau<\frac{1}{a}-\frac{1}{b_{1}}+\sqrt{\frac{1}{a^{2}}+\frac{1}{b_{1}^{2}}}\). Then, considering the continuous case (\(n\to\infty\), or equivalently, when \(T\to 0\)) for the characteristic Equation (A2), one obtains,  
\[\lim_{n\to\infty}\Big{[}\alpha_{1}\Big{(}g_{1}z-g_{1}\mathrm{e}^{-b_{1}T}+ \alpha_{2}g_{2}\Big{)}+z^{n}\Big{(}z-\mathrm{e}^{aT}\Big{)}\Big{(}z-\mathrm{e }^{-b_{1}T}\Big{)}\Big{]}=(z-1)^{2}\lim_{n\to\infty}(z^{n}+g_{1})=0.\] (A9)  
It is important to note that while two poles are close to \(z=1\), \(n-2\) poles lie close to the point \((-g_{1})^{1/n}\) inside the unit circle.