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

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. 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*}