Question 6.3.2: Circuit Analysis Using Impedance For the circuit shown in Fi...

Note that R and C are in parallel. Therefore their equivalent impedance Z(s) is found from
\frac{1}{Z(s)} = \frac{1}{1/Cs} + \frac{1}{R}
or
Z(s) = \frac{R}{RCs  +  1}
An impedance representation of the equivalent circuit is shown in Figure 6.3.3b. In this representation we may think of the impedance as a simple resistance, provided we express the relations in Laplace transform notation. Kirchhoff’s voltage law gives
V_{s}(s)  −  R_{1} I (s)  −  Z(s) I (s) = 0
The output voltage is related to the current by V_{o}(s) = Z(s) I (s) . Eliminating I (s) from these two relations gives
V_{s} (s)  −  R_{1} \frac{V_{o}(s)}{Z(s)}  −  V_{o}(s) = 0

which yields the desired transfer function:
\frac{V_{o}(s)}{V_{s} (s)} = \frac{Z(s)}{R_{1}  +  Z(s)} = \frac{R}{RR_{1} Cs  +  R  +  R_{1}}

This network is a first-order system whose time constant is
τ = \frac{R R_{1}}{R  +  R_{1}}