This is small example showing how to use Mason rule to ﬁnd the transfer function of an RLC circuit.
Solving the circuit loops () applied to each loop gives (all in done in Laplace domain)
The variables are . 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 to each side of the ﬁrst equation, and add to each side of the second equation, this gives
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 for the above using Mason rule.