## 2 Materials and Methods
### Floquet Exponents
\begin{array}{c}x_{1}^{*}\left(T-\frac{d^{*}}{2} \right)\\ x_{2}^{*}\left(T-\frac{d^{*}}{2}\right)\end{array} \right)\left( \begin{array}{ cc}\zeta_{2}&\zeta_{3}\end{array} \right). \tag{81}\]  
The expression is obtained:  
\[\left( \begin{array}{c}x_{1}^{*}\left(\frac{d^{*}}{2}\right)\\ x_{2}^{*}\left(\frac{d^{*}}{2}\right)\end{array} \right)=e^{A_{1}\frac{C}{2}} \left( \begin{array}{c}x_{1}^{*}\left(0\right)\\ x_{2}^{*}\left(0\right)\end{array} \right)+\frac{d^{*}}{2}B. \tag{82}\]  
and  
\[\left( \begin{array}{c}x_{1}^{*}\left(T-\frac{d^{*}}{2}\right)\\ x_{2}^{*}\left(T-\frac{d^{*}}{2}\right)\end{array} \right)=e^{A_{2}\left(T-d^ {*}\right)}\left( \begin{array}{c}x_{1}^{*}\left(\frac{d^{*}}{2}\right)\\ x_{2}^{*}\left(\frac{d^{*}}{2}\right)\end{array} \right)+A_{2}^{-1}\left(e^{A_ {2}\left(T-d^{*}\right)}-\mathbf{I}_{2}\right)B. \tag{83}\]  
It is finally arrived at that the variational equation that governs the evolution of the disturbance is given by:  
\[\left( \begin{array}{c}\dot{a}\\ b\end{array} \right) =\left( \begin{array}{cc}-\gamma-\mu&0\\ 0&-\mu\end{array} \right)\left( \begin{array}{c}a\\ b\end{array} \right)+\mathbf{A}\delta\left(t-\frac{d}{2}\right)\left( \begin{array}{c}a(0)\\ b(0)\end{array} \right) \tag{84}\] \[+\left( \begin{array}{cc}0&1\\ -1&0\end{array} \right)\left( \begin{array}{c}a\\ b\end{array} \right)\theta\left(t-\frac{d}{2}\right)\] (85) \[+\mathbf{B}\delta\left(t-\left(T-\frac{d}{2}\right)\right)\left( \begin{array}{c}a(0)\\ b(0)\end{array} \right)\] (86) \[+\left( \begin{array}{cc}0&-1\\ 1&0\end{array} \right)\left( \begin{array}{c}a\\ b\end{array} \right)\theta\left(t-\left(T-\frac{d}{2}\right)\right). \tag{87}\]  
where \(d^{*}=d\) is obtained simply for convenience. The next step is to solve this variational equation. The integrals of the perturbation equations must be done piecewise, and due to the presence of the function Dirac delta, integration over discontinuities is also required. Initially, a solution between \(t=0\) and \(t=\frac{d}{2}\) is obtained to calculate \(a\left(\frac{d}{2}\right)_{-}\) and \(b\left(\frac{d}{2}\right)_{-}\). Now, in \(0<t<\frac{d}{2}\), the variational equation is converted to:  
\[\left( \begin{array}{c}\dot{a}\\ \dot{b}\end{array} \right)=\left( \begin{array}{cc}-\gamma-\mu&0\\ 0&-\mu\end{array} \right)\left( \begin{array}{c}a\\ b\end{array} \right) \tag{88}\]  
Then, the solution of this system is given by Equation (89). \[\left( \begin{array}{c}a\\ b\end{array} \right)=e^{-\mu t}\left( \begin{array}{cc}e^{-\gamma t}&0\\ 0&1\end{array} \right)\left( \begin{array}{c}a(0)\\ b(0)\end{array} \right), \tag{89}\]  
and consequently,  
\[\left( \begin{array}{c}a\left(\frac{d}{2}\right)_{-}\\ b\left(\frac{d}{2}\right)_{-}\end{array} \right)=e^{-\mu\frac{d}{2}}\left( \begin{array}{cc}e^{-\gamma\frac{d}{2}}&0\\ 0&1\end{array} \right)\left( \begin{array}{c}a(0)\\ b(0)\end{array} \right). \tag{90}\]  
After integrating over the discontinuity to find \(a\left(\frac{d}{2}\right)_{+}\) and \(b\left(\frac{d}{2}\right)_{+}\), we finally obtain Equation (91). \[\left( \begin{array}{c}a\left(\frac{d}{2}\right)_{+}\\ b\left(\frac{d}{2}\right)_{+}\end{array} \right)=\left( \begin{array}{c}a \left(\frac{d}{2}\right)_{-}\\ b\left(\frac{d}{2}\right)_{-}\end{array} \right)+\mathbf{A}\left( \begin{array} []{c}a(0)\\ b(0)\end{array} \right). \tag{91}\]  
Therefore,  
\[\left( \begin{array}{c}a\left(\frac{d}{2}\right)_{+}\\ b\left(\frac{d}{2}\right)_{+}\end{array} \right)=\left(e^{-\mu\frac{d}{2}} \left( \begin{array}{cc}e^{-\gamma\frac{d}{2}}&0\\ 0&1\end{array} \right)+\mathbf{A}\right)\left( \begin{array}{c}a(0)\\ b(0)\end{array} \right). \tag{92}\]  
Now, after integrating between \(t=\frac{d}{2}\) and \(t=T-\frac{d}{2}\) with this new initial condition, the expression found is:  
\[\left( \begin{array}{cc}a\left(T-\frac{d}{2}\right)_{-}&b\left(T-\frac{d}{2} \right)_{-}\end{array} \right)^{T}. \tag{93}\]  
First, in this interval, the disturbed system is as in Equation (94):  
\[\left( \begin{array}{c}\dot{a}\\ \dot{b}\end{array} \right)=\left(