home
PDF (letter size)
PDF (legal size)

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

Nasser M. Abbasi

November 9, 2015   Compiled on May 20, 2020 at 10:23pm

Contents

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.

pict

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.

pict

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*}