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How to use Mason rule to obtain transfer function of simple RLC electric circuit

Nasser M. Abbasi

November 9, 2015   Compiled on January 30, 2024 at 11:22pm


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 {align*} \left ( R_{1}+sL\right ) I_{1}-I_{2}Ls-V_{in}\left ( s\right ) & =0\\ \left ( R_{2}+\frac {1}{Cs}\right ) I_{2}+LsI_{2}-I_{1}Ls & =0\\ V_{out}\left ( s\right ) & =R_{2}I_{2} \end {align*}

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 {align*} I_{1} & =\left ( R_{1}+sL\right ) I_{1}-I_{2}Ls-V_{in}\left ( s\right ) +I_{1}\\ I_{2} & =I_{2}+\left ( R_{2}+\frac {1}{Cs}\right ) I_{2}+LsI_{2}-I_{1}Ls \end {align*}

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 {align*} \frac {V_{out}}{V_{in}} & =\frac {\sum _{i=1}^{1}M_{i}\Delta _{i}}{1-\sum \text {one at time}+\sum \text {2 at times}}\\ & =\frac {\left ( -1\right ) \left ( -Ls\right ) \left ( R_{2}\right ) }{1-\sum \left ( R_{1}+Ls+1\right ) +\left ( \frac {1}{Cs}+R_{2}+Ls+1\right ) +\sum \left ( R_{1}+Ls+1\right ) \left ( \frac {1}{Cs}+R_{2}+Ls+1\right ) }\\ & =\frac {LsR_{2}}{1-\left ( R_{1}+R_{2}+\frac {1}{Cs}+2Ls+2\right ) +\left ( R_{1}+Ls+1\right ) \left ( R_{2}+\frac {1}{Cs}+Ls+1\right ) }\\ & =\frac {LsR_{2}}{\frac {1}{Cs}\left ( R_{1}+Ls\right ) \left ( CLs^{2}+CR_{2}s+1\right ) } \end {align*}