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14.10: Generalized Impedance

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  • We have dealt in Chapter 13 with a sinusoidally varying voltage applied to an inductance, a resistance and a capacitance in series. The equation that governs the relation between voltage and current is

    \[V=L\dot I + RI + Q/C.\label{14.10.1}\]

    If we multiply by \(C\), differentiate with respect to time, and write \(I\) for \(\dot Q\), this becomes just

    \[C \dot V = LC \ddot I + RC \dot I + I.\label{14.10.2}\]

    If we suppose that the applied voltage \(V\) is varying sinusoidally (that is, \(V=\hat{V}e^{j\omega t}\), or, if you prefer, \(V=\hat{V}\sin \omega t\)), then the operator \(d^2/dt^2\), or "double dot", is equivalent to multiplying by \(-\omega ^2\), and the operator \(d/dt\), or "dot", is equivalent to multiplying by \(j\omega\). Thus Equation \ref{14.10.2} is equivalent to

    \[j\omega CV = -LC\omega^2I+ jRC\omega I + I.\label{14.10.3}\]

    That is, \[V=[R+jL\omega + 1/jC\omega]I.\label{14.10.4}\]

    The complex expression inside the brackets is the now familiar impedance Z, and we can write


    But what if \(V\) is not varying sinusoidally? Suppose that \(V\) is varying in some other manner, perhaps not even periodically? This might include, as one possible example, the situation where \(V\) is constant and not varying with time at all. But whether or not \(V\) varying with time, Equation \ref{14.10.2} is still valid – except that, unless the time variation is sinusoidally, we cannot substitute \(j\omega\) for \(d/dt\). We are faced with having to solve the differential equation \ref{14.10.2}.

    But we have just learned a neat new way of solving differential equations of this type. We can take the Laplace transform of each side of the equation. Thus

    \[C\bar{\dot V} = LC \bar{\ddot I} + RC \bar{\dot I} + \bar{I}.\label{14.10.6}\]

    Now we are going to make use of the differentiation theorem, equations 14.7.2 and 14.7.3.

    \[C(s\bar{V}-V_0) = LC(s^2\bar{I} - sI_0 - \dot I_0) + RC(s\bar{I} - I_0) + \bar{I}.\label{14.10.7}\]

    Let us suppose that, at \(t=0\), \(V_0\) and \(I_0\) are both zero – i.e. before \(t=0\) a switch was open, and we close the switch at \(t=0\). Furthermore, since the circuit contains inductance, the current cannot change instantaneously, and, since it contains capacitance, the voltage cannot change instantaneously, so the equation becomes

    \[\bar{V} = (R+Ls+1/Cs)\bar{I}.\label{14.10.8}\]

    This is so regardless of the form of the variation of \(V\): it could be sinusoidal, it could be constant, or it could be something quite different. This is a generalized Ohm's law. The generalized impedance of the circuit is \(R+Ls+\frac{1}{Cs}\). Recall that in the complex number treatment of a steady-state sinusoidal voltage, the complex impedance was \(R+jL\omega+\frac{1}{jCw}\).

    To find out how the current varies, all we have to do is to take the inverse Laplace transform of


    We look at a couple of examples in the next sections.