Question 6.3.1: Coupled RC Loops Determine the transfer function Vo(s)/Vs (s...

The energy in this circuit is stored in the two capacitors. Because the energy stored in a capacitor is expressed by Cv^{2}/2 , appropriate choices for the state variables are the voltages v_{1} and v_{o}. The capacitance relations are
\frac{dv_{o}}{dt} = \frac{i_{3}}{C}      \frac{dv_{1}}{dt} = \frac{i_{2}}{C}         (1)
For the right-hand loop,
i_{3} = \frac{v_{1}  −  v_{o}}{R}             (2)
From equation (1)
\frac{dv_{o}}{dt} = \frac{1}{RC} (v_{1}  −  v_{o})                (3)
For the left-hand loop,
i_{1} = \frac{v_{s}  −  v_{1}}{R}               (4)
From conservation of charge and equations (2) and (4),
i_{2} = i_{1}  −  i_{3} = \frac{v_{s}  −  v_{1}}{R}  −  \frac{v_{1}  −  v_{o}}{R} = \frac{1}{R} (v_{s}  −  2v_{1} + v_{o})          (5)
Using this with equation (2) gives
\frac{dv_{1}}{dt} = \frac{1}{RC} (v_{s}  −  2v_{1} + v_{o})        (6)
Equations (3) and (6) are the state equations. To obtain the transfer function V_{o}(s)/V_{s} (s) , transform these equations for zero initial conditions to obtain
s V_{o}(s) = \frac{1}{RC} [V_{1}(s)  −  V_{o}(s)]
sV_{1}(s) = \frac{1}{RC} [V_{s} (s)  −  2V_{1}(s) + V_{o}(s)]
Eliminating V_{1}(s) from these two equations gives the transfer function
\frac{V_{o}(s)}{V_{s} (s)} = \frac{1}{R^{2}C^{2}s^{2}  +  3 RCs  +  1}          (7)