## 2 Theoretical Analysis
### Decoupling the Cladding-Induced Interference
So, Equation (8) is firstly substituted into the first \(R^{\prime}{}_{4,3}(\alpha)\) in Equation (7), then \(R^{\prime}{}_{5,4}(\alpha)\) becomes:  
\[R^{\prime}{}_{5,4}(\alpha)=R_{5,4}(\alpha)+\frac{R_{4,3}(\alpha)e^{-2\beta_{4 }(d_{3}-d_{4})}[1-R_{5,4}(\alpha)R_{5,4}(\alpha)]}{1+R_{5,4}(\alpha)R^{\prime} {}_{4,3}(\alpha)e^{-2\beta_{4}(d_{3}-d_{4})}}+\frac{R^{\prime}{}_{3,2}(\alpha) e^{-2\beta_{3}(d_{2}-d_{3})}e^{-2\beta_{4}(d_{3}-d_{4})}[1-R_{4,3}(\alpha)R_{5,4}( \alpha)][1-R_{5,4}(\alpha)R_{5,4}(\alpha)]}{[1+R_{4,3}(\alpha)R^{\prime}{}_{4, 3}(\alpha)e^{-2\beta_{4}(d_{3}-d_{4})}][1+R_{5,4}(\alpha)R^{\prime}{}_{4,3}( \alpha)e^{-2\beta_{4}(d_{3}-d_{4})}]} \tag{9}\]  
where, (\(d_{3}-d_{4}\)) is the thickness of the cladding, \(R_{5,4}(\alpha)\) and \(R_{4,3}(\alpha)\) are only related to cladding material. The \(R^{\prime}{}_{4,3}(\alpha)\) that still remains in Equation (9) can be approximated by \(R_{4,3}(\alpha)\), since the first reflection was much stronger than that of the other reflections, and \(R_{4,3}(\alpha)\) significantly influences the value of \(R^{\prime}{}_{4,3}(\alpha)\). Thus, the equations in the first line of Equation (9) is only related to the material and thickness of the cladding. Substituting Equation (9) into Equation (4), \(\Delta U\) could be divided into two terms:  
\[\begin{array}{l}\Delta U=\Delta U_{1}+\Delta U_{2}\\ =j\pi\omega\mu_{0}I(\omega)\times\int_{0}^{\infty}e^{-2\alpha l_{1}}S(\alpha)[R _{5,4}(\alpha)+\frac{R_{4,3}(\alpha)e^{-2\beta_{4}(d_{3}-d_{4})}[1-R_{5,4}( \alpha)R_{5,4}(\alpha)]}{1+R_{5,4}(\alpha)R^{\prime}{}_{4,3}(\alpha)e^{-2\beta _{4}(d_{3}-d_{4})}}]d\alpha\\ +j\pi\omega\mu_{0 As \(\beta_{3}=\alpha\), \(R_{4,3}(\alpha)=-R_{5,4}(\alpha)\), \(\Delta U_{2}\) in Equation (10) could be rewritten as follows:  
\[\Delta U_{2}=j\pi\omega\mu_{0}I(\omega)\int_{0}^{\infty}e^{-2\alpha(l_{1}+d_{2}- d_{3})-2\beta_{4}(d_{3}-d_{4})}S(\alpha)R^{\prime}_{3,2}(\alpha)\times Coef(\alpha)d\alpha \tag{11}\]  
\[Coef(\alpha)=\frac{1-R_{5,4}(\alpha)R_{5,4}(\alpha)}{1-R_{5,4}(\alpha)R^{ \prime}_{3,2}(\alpha)e^{-2\beta_{3}(d_{2}-d_{3})}}\frac{1-R_{5,4}(\alpha)R_{5,4 }(\alpha)}{1+R_{5,4}(\alpha)R^{\prime}_{4,3}(\alpha)e^{-2\beta_{4}(d_{3}-d_{4})}} \tag{12}\]  
where the parameter \(Coef(\alpha)\) is used for simplicity. As the commonly used cladding is 0.3\(-\)0.7 mm thick, the commonly used insulation is 40 mm to a few hundred mm thick, and the sensor lift-off is a few mm [24]. Thus, the cladding thickness, \((d_{3}-d_{4})\), is much smaller than the sum of the sensor lift-off and the insulation thickness, \((l_{1}+d_{2}-d_{3})\). Therefore, \(e^{-2\alpha(l1+d2-d3)-2\beta 4(d3-d4)}\) could be approximated by \(e^{-2\alpha(l1+d2-d3)}\). Then, Equation (11) could be rewritten as follows:  
\[\Delta U_{2}=j\pi\omega\mu_{0}I(\omega)\int_{0}^{\infty}e^{-2\alpha(l_{1}+d_{ 2}-d_{3})}S(\alpha)R^{\prime}_{3,2}(\alpha)\times Coef(\alpha)d\alpha \tag{13}\]  
In Equation (13), as \(S(\alpha)R^{\prime}_{3,2}(\alpha)\times Coef(\alpha)\) is continuous in the range [0, \(\infty\)], and \(e^{-2\alpha(l1+d2-d3)}\) is continuous with its values always greater than zero. Thus, the first integral mean value theorem [25] could be used, and Equation (13) becomes:  
\[\Delta U_{2}=j\pi\omega\mu_{0}I(\omega)\times S(\varepsilon)R^{\prime}_{3,2}( \varepsilon)Coef(\varepsilon)\times\int_{0}^{\infty}e^{-2\alpha(l_{1}+d_{2}- d_{3})}d\alpha \tag{14}\]  
where \(\varepsilon\) is a value between 0 and \(\infty\). Calculating \(\int_{0}^{\infty}e^{-2\alpha(l1+d2-d3)}\) d\(\alpha\) gives \(\int_{0}^{\infty}e^{-2\alpha(l1+d2-d3)}\) d\(\alpha=\)1/2(\(l_{1}+d_{2}-d_{3}\)), thus (14) could be rewritten as follows:  
\[\Delta U_{2}=\frac{j\pi\omega\mu_{0}I(\omega)\times S(\varepsilon)R^{\prime} _{3,2}(\varepsilon)Coef(\varepsilon)}{2(l_{1}+d_{2}-d_{3})} \tag{15}\]  
Because \(j\pi\omega\mu_{0}I(\omega)\) and \(S(\varepsilon)R^{\prime}_{3,2}(\varepsilon)\times Coef(\varepsilon)\) are independent of \((d_{2}-d_{3})\) and \(l_{1}\), \(\Delta U_{2}\) is inversely proportional to \((l_{1}+d_{2}-d_{3})\).