Question 6.1: A Series RLC Circuit Consider the series RLC circuit shown i...

a. Applying Kirchhoff’s voltage law to the single loop along the clockwise direction gives

v_{\mathrm{R}}+v_{\mathrm{L}}+v_{\mathrm{C}}-v_{\mathrm{a}}=0

For the series loop, the same current flows through each element. The expressions for v_{R^{\prime}} v_{\mathrm{L}^{\prime}} and v_{\mathrm{C}} are

\begin{aligned} & v_{\mathrm{R}}=R i, \\ & v_{\mathrm{L}}=L \frac{\mathrm{d} i}{\mathrm{~d} t}, \\ & v_{\mathrm{C}}=\frac{1}{C} \int i \mathrm{~d} t . \end{aligned}

We then have

R i+L \frac{\mathrm{d} i}{\mathrm{~d} t}+\frac{1}{C} \int i \mathrm{~d} t=v_{\mathrm{a}}

Note that the above equation is an integral-differential equation, not a differential equation. To eliminate the integral term, we take the time derivative of both sides of the equation. Rearranging the terms results in

L \frac{\mathrm{d}^{2} i}{\mathrm{~d} t^{2}}+R \frac{\mathrm{d} i}{\mathrm{~d} t}+\frac{1}{C} i=\frac{\mathrm{d} v_{\mathrm{a}}}{\mathrm{d} t}

which is a second-order differential equation for current i with the time derivative of the applied voltage as the forcing function.

b. Taking the Laplace transform of the above differential equation yields

L s^{2} I(s)+R s I(s)+\frac{1}{C} I(s)=s V_{\mathrm{a}}(s)

The transfer function relating the input voltage v_a(t) and the output current i(t) is

\frac{I(s)}{V_{\mathrm{a}}(s)}=\frac{s}{L s^{2}+R s+(1 / C)}

c. Note that the capacitor voltage v_{\mathrm{C}} does not appear explicitly in the differential equation. To determine the transfer function V_{\mathrm{C}}(s) / V_{\mathrm{a}}(s), we use the result of the transfer function I(s) / V_{\mathrm{a}}(s) and apply the voltage-current relation for a capacitor

v_{\mathrm{C}}=\frac{1}{C} \int i \mathrm{~d} t

which gives

V_{\mathrm{C}}(s)=\frac{1}{C s} I(s)

Thus, the transfer function relating the input voltage v_{\mathrm{a}}(t) and the capacitor voltage v_{\mathrm{C}}(t) is

\frac{V_{\mathrm{C}}(s)}{V_{\mathrm{a}}(s)}=\frac{1}{C s} \frac{I(s)}{V_{\mathrm{a}}(s)}=\frac{1}{L C s^{2}+R C s+1} .

The circuit in Figure 6.18 can also be modeled using a differential equation for charge q. Recall that current is the time rate of change of charge, i=\mathrm{d} q / \mathrm{d} t or q=\int i \mathrm{~d} t. Rewriting the current-related terms in the equation R i+L(\mathrm{~d} i / \mathrm{d} t)+(1 / C) \int i \mathrm{~d} t=v_{\mathrm{a}} in terms of q gives

L \frac{\mathrm{d}^{2} q}{\mathrm{~d} t^{2}}+R \frac{\mathrm{d} q}{\mathrm{~d} t}+\frac{1}{C} q=\nu_{\mathrm{a}}

which is a second-order differential equation for the charge q(t) with the applied voltage as the forcing function.