A Series RLC Circuit Consider the series RLC circuit shown in Figure 6.18. a. Derive the differential equation in terms of the loop current i. b. Determine the transfer function I(s)/Va(s), which relates the source voltage va(t) to the loop current i(t). Assume that all the initial conditions are

Question

A Series RLC Circuit

Consider the series RLC circuit shown in Figure 6.18.

a. Derive the differential equation in terms of the loop current [latex]i[/latex].

b. Determine the transfer function [latex]I(s) / V_{\mathrm{a}}(s)[/latex], which relates the source voltage [latex]v_{\mathrm{a}}(t)[/latex] to the loop current [latex]i(t)[/latex]. Assume that all the initial conditions are zero.

c. Determine the transfer function [latex]V_{\mathrm{C}}(s) / V_{\mathrm{a}}(s)[/latex], which relates the source voltage [latex]v_{\mathrm{a}}(t)[/latex] to the capacitor voltage [latex]v_{\mathrm{C}}(t)[/latex]. Assume that all the initial conditions are zero.

Accepted Answer

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

[latex]v_{\mathrm{R}}+v_{\mathrm{L}}+v_{\mathrm{C}}-v_{\mathrm{a}}=0[/latex]

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

[latex] \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} [/latex]

We then have

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

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

[latex]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}[/latex]

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

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

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

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

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

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

[latex]v_{\mathrm{C}}=\frac{1}{C} \int i \mathrm{~d} t[/latex]

which gives

[latex]V_{\mathrm{C}}(s)=\frac{1}{C s} I(s)[/latex]

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

[latex]\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} .[/latex]

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

[latex]L \frac{\mathrm{d}^{2} q}{\mathrm{~d} t^{2}}+R \frac{\mathrm{d} q}{\mathrm{~d} t}+\frac{1}{C} q= u_{\mathrm{a}}[/latex]

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

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

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