This post assumes a working knowledge of elementary circuit theory as well as Fourier Analysis.

It seems to me that often in an introductory circuits course complex impedance is a major concept but its mathematical basis is never taught. Thus I'd like to take a moment to discuss some of the mathematics surrounding this concept.

The resistor is a purely resistive elementary circuit element possessing a time independent current voltage characteristic described by Ohm's Law.

\[ V(t) = I(t) R \]The capacitor is a purely reactive elementary circuit element that stores energy in its electric field.

\[ I(t) = C \frac{dV(t)}{dt} \]The inductor is a purely reactive elementary circuit element that stores energy in its magnetic field.

\[ V(t) = L \frac{dI(t)}{dt} \]What we aim to do is to derive a linear current voltage characteristic for the two reactive circuit elements by reducing the differential equations to algebraic equations. Using the Fourier Transform we can compute a complex quantity in the frequency domain that is analogous to resistance in the time domain.

The resistance is both time independent and frequency independent so its corresponding complex impedance is rather boring.

\begin{aligned} \newcommand{\FT}[1]{\mathcal F \left\{#1\right\}} \newcommand{\e}[1]{e^{#1}} \FT{V(t)} &= \FT{I(t) R} \\ V(j\omega) &= I(j\omega) R \\ \frac{V(j\omega)}{I(j\omega)} &= Z_R = R \end{aligned}The capacitance is a reactive quantity so its corresponding complex impedance is a phase shift.

\begin{aligned} \FT{i(t)} &= \FT{C \frac{dV(t)}{dt}} \\ I(j \omega) &= C j \omega V(j \omega) \\ \frac{V(j \omega)}{I(j \omega)} &= Z_C = \frac{1}{j \omega C} \end{aligned}The inductance is a reactive quantity so its corresponding complex impedance is a phase shift.

\begin{aligned} \FT{V(t)} &= \FT{L \frac{di(t)}{dt}} \\ V(j \omega) &= L j \omega I(j \omega) \\ \frac{V(j \omega)} {I(j \omega)} &= Z_L = j \omega L \end{aligned}For this concept of complex impedance to be useful it is necessary to find appropriate expressions for the current and voltage in the frequency domain. It is common for both the input and output to a system be sinusoidal which results in their Fourier Transforms being less than ideal expressions to work it. In this case we determine the coefficients of the Fourier Series and use those as our input and output.

Let us assume that we have a simple series LC circuit driven by a voltage source and we are asked to find the current passing through the resistor. In this case our input voltage is a sinusoid.

\[ V(t) = A \cos(j \omega t + \phi) \]We can express this input as a complex exponential using Euler's Formula. This representation is the phasor or phase vector of the input.

\[ V(t) = \Re\{A \e{j \omega t} \e{\phi}\} \]Because we wish to work with the input in the frequency domain by using the coefficients of the Fourier Series we transform the input by taking the coefficients of the time dependent portion of the signal.

\[ \tilde{V} = \Re\{A \e{\phi}\} \]We can calculate the total impedance of the circuit by combining the impedance of each individual element in the same manner that we use to calculate the total resistance.

\[ \tilde{Z} = R + j \omega L \]We can express this impedance as a complex exponential using Euler's Formula. This representation is the phasor or phase vector of the impedance.

\[ \tilde{Z} = \left|\tilde{Z}\right| \e{\varphi \tilde{Z}} = \sqrt{R^2 + \omega^2 L^2} \e{\arctan\frac{\omega L}{R}} \]To calculate the current passing through the resistor we then apply Ohm's Law in the frequency domain.

\begin{aligned} \tilde{I} &= \frac{\tilde{V}}{\tilde{Z}} = \Re\left\{ \frac{A \e{\phi}}{ \sqrt{R^2 + \omega^2 L^2} \e{\arctan \frac{\omega L}{R}}} \right\} \\ &= \frac{A}{\sqrt{R^2 + \omega^2 L^2}} \Re\left\{ \e{\phi - \arctan \frac{\omega L}{R}} \right\} \end{aligned}At this point we have the frequency domain representation of the output. Usually we wish to convert this back to the time domain representation. In this case the time domain representation is a sinusoid.

\[ i(t) = \frac{A}{\sqrt{R^2 + \omega^2 L^2}} \cos \left(j \omega t + \phi - \arctan \frac{\omega L}{R} \right) \]This is the general procedure involved in solving a reactive circuit in the frequency domain. Note that while the example shows a cosinusoidal input the input can be any function. A periodic function can be decomposed into a Fourier Series and solved by taking the coefficients of the terms as the input phasors while a nonperiodic function can be solved by using an input phasor equal to its Fourier Transform. Of course a periodic function can also be solved by using an input phasor equal to its Fourier Transform but this is usually more work because the Fourier Transform of a periodic function usually involves Dirac Delta functions which are not easily manipulated.

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