Question 6.12: Impedance Method: An RLC Circuit For the electric circuit in...
Impedance Method: An RLC Circuit
For the electric circuit in Example 6.4, use the impedance method to derive the differential equation relating the output voltage v_{o}(t) to the input voltage v_{\mathrm{a}}(t). Assume zero initial conditions.
The original electric circuit is shown in Figure 6.55a. We can replace the passive elements with their impedance representations and redraw the circuit in the s domain as shown in Figure 6.55b. Note that the resistor R is in parallel connection with the capacitor C. The corresponding equivalent impedances are
\begin{aligned} & Z_{1}(s)=L s \\ & \frac{1}{Z_{2}(s)}=\frac{1}{R}+\frac{1}{1 / C s}, \end{aligned}
or
Z_{2}(s)=\frac{R}{R C s+1} .
For the equivalent impedance circuit in Figure 6.55c, we apply Kirchhoff’s voltage law,
Z_{1}(s) /(s)+Z_{2}(s) I(s)-V_{\mathrm{a}}(s)=0,
where the current is related to the output voltage by
V_{\mathrm{o}}(s)=Z_{2}(s) I(s).
Thus, we have
Z_{1}(s) \frac{V_{\mathrm{o}}(s)}{Z_{2}(s)}+V_{\mathrm{o}}(s)=V_{\mathrm{a}}(s),
which gives the transfer function relating the input voltage v_{a} and the output voltage v_{o},
\frac{V_{\mathrm{o}}(s)}{V_{\mathrm{a}}(s)}=\frac{Z_{2}(s)}{Z_{1}(s)+Z_{2}(s)}=\frac{R}{R L C s^{2}+L s+R} .
By transforming V_{\mathrm{o}}(s) / V_{\mathrm{a}}(s) from the s domain to the time-domain with the assumption of zero initial conditions, we obtain the differential equation of the system
R L C \ddot{v}_{\mathrm{o}}+L \dot{v}_{\mathrm{o}}+R v_{\mathrm{o}}=R v_{\mathrm{a}^{\prime}},
which is the same as the one obtained in Example 6.4.
