# A Theoretical Framework to Determine RHP Zero Dynamics in Sequential Interacting Sub-Systems
## 2 The Polynomial Form to Determine Transmission Zeros
Proposition 1: _Let \(S(s)\in\mathbb{R}[s]^{n\times n}\) be the Smith matrix, and given two disjoint sets \(\mathcal{Q}\) and \(\mathcal{R}\), the determinant has the following structure:_  
\[det(S)=\prod_{i=1}^{n}S_{i,i}+(-1)^{n+1}\prod_{j=2}^{n}S_{j,j-1}S_{1,n} \tag{4}\]  
_where the elements \(S_{i,j}\in\mathcal{Q}\) and the elements \(S_{i,j}\in\mathcal{R}\) with \(i\
eq j\)._  
The proof for Proposition 1 is determined straightforwardly through induction, starting from \(n=2\). Proposition 2: _Let \(S(s)\in\mathbb{R}[s]^{n\times n}\) be the Smith matrix, and given two disjoint sets \(\mathcal{Q}\) and \(\mathcal{R}\), the determinant of the minor \(n-1\) of \(S\) can have three mutually exclusive structures:_  
\[det([S]_{i,j}^{n-1})= \begin{cases}\prod\limits_{i=1}^{n-1}S_{i,j}&A\\ (-1)^{n}\prod\limits_{j=2}^{n-1}S_{j,j-1}S_{1,n}&B\\ \prod\limits_{i,j=1}^{n-1}S_{i,j}&C\end{cases}  \tag{5}\]  
_where Cases \(A\) and \(B\) are evaluated only with elements from sets \(\mathcal{Q}\) and \(\mathcal{R}\), respectively, and Case \(C\) has elements from both sets._  
The proof of Proposition 2 is determined by investigating the three possibilities to compute an \(n-1\) minor by deleting a row \(i\) and column \(j\) (i.e., \(i=j\) for Case A, \(i<j\) for B and \(i>j\) for C). **Proposition 3**. : _Let \(G(s)\in\mathbb{R}[s]^{k\times k}\) be a transfer function matrix with poles of multiplicity 1, \(S(s)\in\mathbb{R}[s]^{k\times k}\) a Smith matrix associated with \(G(s)\), and \(k\in\mathbb{N}\), with \(k>2\), then the product of \(k\) different elements in \(S(s)\) has the form:_  
\[\prod_{i,j=1}^{k}S_{i,j}=h^{k-1}(s)p_{k}(s)\]  
_where \(h(s)\) is the LCM of the denominators in \(G(s)\) and \(p_{k}(s)\) is a polynomial that depends on the elements \(S_{i,j}\)._  
Proposition 3 allows one to separate the product of \(k\) different non-zero elements of the matrix \(S(s)\) into two terms:  
* \(h(s)\), a fix polynomial whose roots do not depend on the product's elements and
* \(p_{k}(s)\), a polynomial whose roots depend on the product's elements. As such, using Proposition 3 and (5), the following results for \(\Delta_{n-1}\) and \(\Delta_{n}\) are obtained:  
\[\Delta_{n-1}(s) = GCD\{[S]_{i,j}^{n-1}(s)\}=GCD\{det([S]_{i,j}^{n-1}(s))\} \tag{6}\] \[= GCD\{h^{n-2}(s)p_{i,j}(s)\}=h^{n-2}(s)\] \[\forall i,j=\overline{1,n}.\] \[\Delta_{n}(s) = GCD\{[S]_{i,j}^{n}(s)\}=GCD\{S(s)\}=det(S(s))\] (7) \[= \prod_{i=1}^{n}S_{i,j}(s)+(-1)^{n+1}\prod_{j=2}^{n}S_{i,j-1}(s)S_ {1,n}(s).\]  
The equality in (6) holds for the polynomial \(h(s)\), as it is independent of \(i\) and \(j\). After some algebraic manipulation (7) can be written for \(\Delta_{n}(s)\) as:  
\[\prod_{i=1}^{n}c_{i}h^{n-1}(s)[\prod_{i=1}^{n}\gamma_{i}(sT_{2}+1 )\cdots(sT_{3n-1}+1)+\] \[+(-1)^{n+1}\prod_{i=1}^{n}(1-\gamma_{i})(sT_{1}+1)\cdots(sT_{3n}+ 1)] \tag{8}\]  
Having (8) and (6), one can compute (3), and after the division of \(\epsilon_{n}\) by \(h(s)\), accordingly with the Smith-McMillan form, the final form for the transmission zero polynomial is:  
\[\epsilon_{n}(s)=\prod_{i=1}^{n}c_{i}[\prod_{i=1}^{n}\gamma_{i}(sT _{2}+1))\cdots(sT_{3n-1}+1)\over\epsilon_{\epsilon_{A}(s)}}+\] \[+(-1)^{n+1}\underbrace{\prod_{i=1}^{n}(1-\gamma_{i})(sT_{1}+1) \cdots(sT_{3n}+1)}_{\epsilon_{B}(s)}] \tag{9}\]  
where \(\epsilon_{A}(s)\) is a \(2n\) degree polynomial and \(\epsilon_{B}(s)\) is a \(n\) degree polynomial, both with all the coefficients strictly positive (Under the condition that \(\gamma_{i}\) belongs to: \(0<\gamma_{i}<1\;\;\forall i=\overline{1,n}\)), defined as follows:  
\[\Delta_{n-1}(s) = GCD\{[S]_{i,j}^{n-1}(s)\}=GCD\{det([S]_{i,j}^{n-1}(s))\} \tag{6}\]0 \[\Delta_{n-1}(s) = GCD\{[S]_{i,j}^{n-1}(s)\}=GCD\{det([S]_{i,j}^{n-1}(s))\} \tag{6}\]1  
**3. Evaluation of the Transmission Zero Polynomial**  
The roots of the zero polynomial depend on the number of modules \(n\) present in the system (i.e., whether \(n\) takes odd or even values, it influences the sign of the second term in (9)).