Complex Impedance

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.