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This is small example showing how to use Mason rule to find the transfer function $\frac{V_{o}ut(s)}{V_{i}n(s)}$ of an RLC circuit.

Solving the circuit loops ($V=Ri$) applied to each loop gives (all in done in Laplace domain)$$\begin{array}{rl}(R_1+sL)I_1-I_{2}Ls-V_{in}\left(s\right)& =0\\ (R_2+\frac{1}{Cs})I_2+Ls{I}_2-I_{1}Ls& =0\\ V_{out}\left(s\right)& =R_2{I}_2\end{array}$$

The variables are $I_1,I_2$. In Mason, each variable goes to a node. Hence so we need to have each variable by on its own on the the LHS. To do this, do this trick: Add $I_1$ to each side of the first equation, and add $I_2$ to each side of the second equation, this gives$$\begin{array}{rl}{I}_1& =(R_1+sL)I_1-I_{2}Ls-V_{in}\left(s\right)+I_1\\ I_2& =I_2+(R_2+\frac{1}{Cs})I_2+Ls{I}_2-I_{1}Ls\end{array}$$

Now set up the signal graph, assign a node to each variable. The input and output go a node also. This is the result.

Now we Find $\frac{V_{out}}{V_{in}}$ for the above using Mason rule.$$\begin{array}{rl}\frac{V_{out}}{V_{in}}& =\frac{\sum _{i=1}^1{M}_i{\mathrm{\Delta }}_i}{1-\sum \text{one at time}+\sum \text{2 at times}}\\ & =\frac{(-1)(-Ls)\left(R_2\right)}{1-\sum (R_1+Ls+1)+(\frac{1}{Cs}+R_2+Ls+1)+\sum (R_1+Ls+1)(\frac{1}{Cs}+R_2+Ls+1)}\\ & =\frac{Ls{R}_2}{1-(R_1+R_2+\frac{1}{Cs}+2Ls+2)+(R_1+Ls+1)(R_2+\frac{1}{Cs}+Ls+1)}\\ & =\frac{Ls{R}_2}{\frac{1}{Cs}(R_1+Ls)(CL{s}^2+C{R}_{2}s+1)}\end{array}$$