## 2 Materials and Methods
### Floquet Exponents
\begin{array}{cc}-\gamma-\mu&1\\ -1&-\mu\end{array} \right)\left( \begin{array}{c}a\\ b\end{array} \right). \tag{94}\]  
It is convenient to express the term \(C\) as in Equation (95) to find the solution to this system. \[\mathbf{C}=\left( \begin{array}{cc}-\gamma-\mu&1\\ -1&-\mu\end{array} \right). \tag{95}\]  
The mathematical expression in Equation (95) can be simplified by considering that it is a function of \(A_{1}\) and \(A_{2}\), as shown in Equation (96). \[\mathbf{C}=\mathbf{A}_{1}+\mathbf{A}_{2}, \tag{96}\]  
where both terms \(A_{1}\) and \(A_{2}\) are expressed according to the parameter \(\gamma\) and \(\mu\), as shown in Equations (97) and (98):  
\[\mathbf{A}_{1}=\left( \begin{array}{cc}-\frac{\gamma}{2}-\mu&0\\ 0&-\frac{\gamma}{2}-\mu\end{array} \right), \tag{97}\]\[\mathbf{A}_{2}=\bigg{(} \begin{array}{cc}-\frac{\gamma}{2}&1\\ -1&\frac{\gamma}{2}\end{array} \bigg{)}. \tag{98}\]  
The advantage of using \(A_{1}\) and \(A_{2}\) is that the expression can be simply expressed as shown in Equation (99). \[e^{\mathbf{C}t}=e^{\mathbf{A}_{1}t}e^{\mathbf{A}_{2}t}, \tag{99}\]  
where  
\[e^{\mathbf{A}_{1}t}=e^{-\left(\frac{\gamma}{2}+\mu\right)t}\mathbf{I}_{2}, \tag{100}\]  
and  
\[e^{\mathbf{A}_{2}t}=\left( \begin{array}{cc}-\frac{\gamma}{\sqrt{\alpha}} \sin\left(\frac{1}{2}\sqrt{\alpha}t\right)+\cos\left(\frac{1}{2}\sqrt{\alpha} t\right)&\frac{2}{\sqrt{\alpha}}\sin\left(\frac{1}{2}\sqrt{\alpha}t\right)\\ -\frac{2}{\sqrt{\alpha}}\sin\left(\frac{1}{2}\sqrt{\alpha}t\right)&\frac{ \gamma}{\sqrt{\alpha}}\sin\left(\frac{1}{2}\sqrt{\alpha}t\right)+\cos\left( \frac{1}{2}\sqrt{\alpha}t\right)\end{array} \right), \tag{101}\]  
with \(\alpha=4-\gamma^{2}\). In consequence,  
\[\left( \begin{array}{c}a\left(T-\frac{d}{2}\right)_{-}\\ b\left(T-\frac{d}{2}\right)_{-}\end{array} \right)=e^{-\left(\frac{\gamma}{2}+ \mu\right)\left(T-\frac{d}{2}\right)}e^{\mathbf{A}_{2}\left(T-\frac{d}{2} \right)}\left( \begin{array}{c}a\left(\frac{d}{2}\right)_{+}\\ b\left(\frac{d}{2}\right)_{+}\end{array} \right). \tag{102}\]  
By replacing the matrix \(\left( \begin{array}{c}a\left(\frac{d}{2}\right)_{+}\\ b\left(\frac{d}{2}\right)_{+}\end{array} \right)\) the expression obtained is that presented in (103). \[\left( \begin{array}{c}a\left(T-\frac{d}{2}\right)_{-}\\ b\left(T-\frac{d}{2}\right)_{-}\end{array} \right)=e^{-\left(\frac{\gamma}{2}+ \mu\right)\left(T-\frac{d}{2}\right)}e^{\mathbf{A}_{2}\left(T-\frac{d}{2} \right)}\left(e^{-\mu\frac{d}{2}}\left( \begin{array}{cc}e^{-\gamma\frac{d}{ 2}}&0\\ 0&1\end{array} \right)+\mathbf{A}\right)\left( \begin{array}{c}a\left(0\right) \\ b\left(0\right)\end{array} \right). \tag{103}\]  
Now, we integrate over the second discontinuity to find the state at \(t=(T-\frac{d}{2})_{+}\):  
\[\left( \begin{array}{c}a\left(T-\frac{d}{2}\right)_{+}\\ b\left(T-\frac{d}{2}\right)_{+}\end{array} \right) =e^{-\left(\frac{\gamma}{2}+\mu\right)\left(T-\frac{d}{2}\right)} e^{\mathbf{A}_{2}\left(T-\frac{d}{2}\right)}\left(e^{-\mu\frac{d}{2}} \left( \begin{array}{cc}e^{-\gamma\frac{d}{2}}&0\\ 0&1\end{array} \right)+\mathbf{A}\right)\left( \begin{array}{c}a\left(0 \right)\\ b\left(0\right)\end{array} \right)\] \[+\mathbf{B}\left( \begin{array}{c}a\left(0\right)\\ b\left(0\right)\end{array} \right). \tag{104}\]  
In order to simplify the expression:  
\[\mathbf{H}=e^{-\left(\frac{\gamma}{2}+\mu\right)\left(T-\frac{d}{2}\right)}e ^{\mathbf{A}_{2}\left(T-\frac{d}{2}\right)}\left(e^{-\mu\frac{d}{2}}\left( \begin{array}{cc}e^{-\gamma\frac{d}{2}}&0\\ 0&1\end{array} \right)+\mathbf{A}\right). \tag{105}\]  
Now, \(a(T)\) and \(b(T)\) are calculated, integrating over the last section:  
\[\left( \begin{array}{c}a\left(T\right)\\ b\left(T\right)\end{array} \right)=e^{-\mu\frac{d}{2}}\left( \begin{array}{cc}e^{ -\gamma\frac{d}{2}}&0\\ 0&1\end{array} \right)\left( \begin{array}{c}a\left(T-\frac{d}{2}\right)_{+}\\ b\left(T-\frac{d}{2}\right)_{+}\end{array} \right). \tag{106}\]  
Therefore,  
\[\left( \begin{array}{c}a\left(T\right)\\ b\left(T\right)\end{array} \right)=e^{-\mu\frac{d}{2}}\left(