6.002 Notes: Sinusoidal Steady State and the ImpedanceMethodProf. Karl K. Berggren, Dept. of EECS

October 26, 2021

Sinusoidal Steady State and the Impedance Method

You have probably seen impedance in some context before, perhapsin a physics class, or in some discussions in an introductory lab class.If so, you may remember the key formulas and perhaps how to applythem, but here I’d like to not only explain how to use impedance asa concept, but more importantly exactly why it works. That way, youwill know when you can apply this method (namely, when consider-ing sinusoidal steady state signals), and when it is not appropriate. In more advanced classes, more general

use of the impedance method is oftentaught by using transform methodsto map non sinusoidal signals intotheir sinusoidal components. However,for now you should only use theimpedance method when dealing withsinusoidal signals.

Superposition of Sinusoidal Sources

Remember that this simple mathematical relation between sinusoidswas only true for linear transformations, i.e. for transformationsthat satisfy superposition and homogeneity. The question is, arecapacitors and inductors linear?

Well, let’s see first if superposition applies. If we find two solu-tions that satisfy the element relation of a capacitor or inductor, willtheir sum also satisfy the relation? Let’s try it out for two pairs ofvariables, i1, v1 and i2, v2 that both satisfy the constitutive relation fora capacitor:

i1 = Cdv1

dt,

i2 = Cdv2

dt

⇒ i1 + i2 = Cd(v1 + v2)

dt.

(1)

So yes, superposition applies to capacitors (and similarly, to induc-tors).

Now let’s check if homogeneity applies:

αi1 = Cd(αv)

dt. (2)

So yes, homogeneity also applies. So, by the same arguments weused to claim that resistors were linear, these devices are also linear.This realization can be surprising until one fully understands themeaning of linearity as it is used in circuit theory.

What is perhaps even more surprising is that the principle ofhomogeneity applies even if the scaling factor α is imaginary. For

6.002 notes: sinusoidal steady state and the impedance method 2

a linear function, then if f (x) = y then f ( j x) = j y and alsof (a + j b) = ay + j by. That means that mathematically (althoughof course not physically), you can work with circuits with complexsource strengths and parameter values, and they can be analyzed justas you are accustomed to for real source strengths and parametervalues.

The remarkable consequence of this fact is that if we have a realcircuit that we need to analyze, we can add any imaginary sourceswe want to it (obviously without breaking the existing topology), andthen simply take the real part of the answer to solve for the original(real) circuit. I.e. in linear circuits, there is no "mixing" of real andimaginary components. Solving a problem with complex sources islike solving two unrelated problems, one with real sources and onewith imaginary sources. The real and imaginary parts of the solutiongive the answers to each of these problems respectively.

The process of adding imaginary sources basically just looks liketransforming the original source into one with a complex drive (be-cause current sources in parallel and voltage sources in series simplyadd). This process is illustrated in the sketch below for a currentsource: a complex voltage source can be similarly constructed byadding adding an imaginary source in series with a real one.

Ie j ωt I cos ωt j I sin ωt

i = Ie j ωt = I cos(ωt) + j I sin(ωt) (3)

This equation basically means that we can model a cosinusoidalcurrent source as the real part of a complex current source. It is im-portant to remember that the actual current source has no complexcharacter, we’re just taking advantage of the mathematical trick thatbecause of linearity, we can add any complex value to a source and itcan’t change the real component of the system (by superposition). Soat the end of the calculation, we just take the real part of the solutionand we have our problem solved.

Just being able to add imaginary sources to a problem withoutbreaking it does not imply that would be useful. So why would onewant to use an imaginary source? Why go to the trouble of addingstill more sources to an already complicated problem? Surprisingly,

6.002 notes: sinusoidal steady state and the impedance method 3

adding these imaginary sources makes the problem much easier tosolve, as long as it is done correctly. The trick is to add sources inseries with the voltage sources (and in parallel with the currentsources) so that the combined sources looks like a single complexsource. When this is done properly for sinusoidal signals, this singlesource appears to have a strength that rotates continuously in a circu-lar path on the complex plane. We will see that this makes life mucheasier for us down the road—it permits us to use the impedancemethod.

The Impedance Method

The reason complex drives are useful is illustrated immediately byconsidering a network consisting of an inductor in series with a cur-rent source.

iL=Ie j ωt L

+

−

vL

iL

Calculating the complex voltage vL and its relation to the complexcurrent through the inductor iL, we can save a lot of trouble andeffort relative to solving a full differential equation. Instead, we startwith the constitutive relation for an inductor,

vL = LdiLdt

(4)

and plug in our expression for iL. From this we see that

vL = LdIe j ωt

dt= L j ωIe j ωt

= j ωLiL= ZL iL.

(5)

where we’ve introduced a new symbol here ZL = j ωL which standsfor the impedance of the inductor. Notice the relation we just derivedrelates a voltage and a current, just like Ohm’s law! So ZL must haveunits of resistance (Ω).

We can simplify (5) even further by realizing that both sides ofthe equation vL = ZL iL include a time dependence that can becanceled out. We need to separate out this time dependence so wedefine complex oscillation amplitudes V` and I`, so that one can writevL = V`e j ωt and iL = I`e j ωt, in which case V`e j ωt = ZL I`e j ωt. The

6.002 notes: sinusoidal steady state and the impedance method 4

resulting simplified equation just relates complex values with no timedependence:

V` = ZL I`. (6)

That was the key step. The time dependence cancels out, resultingin a problem with no time dependence. But we already know howto solve problems that don’t include a time dependence, so we’vereduced the problem to one that we’ve solved before.

Because this looks so much like Ohm’s law, we can now redrawour circuit, replacing the sinusoidal drive and the inductor whichmuch simpler DC components (admittedly with complex values forparameter strengths now).

I j ωL

where we have used the symbol to represent an impedance,real, imaginary, or complex. Basically, these elements act as resistors,but we don’t draw resistors in their spot so that we remember thatthey have complex parameter values. The disadvantage is that nowwhen you solve the circuit for the real voltage (across the inductor,say), you need to remember to multiply by the time-dependent termand take the real part of the result, so the voltage vL is not I` j ωL asyou might expect, but rather

vL(t) = <[ j ωLI`e j ωt]

= | j ωLI`| cos (ωt +∠( j ωLI`)))

= ωLI` cos (ωt + π/2) .

(7)

A very similar derivation (with a voltage source and a capacitorrather than a current source and an inductor) can be used to showthat the impedance of a capacitor is ZC = 1/( j ωC).

Filter Example

So let’s now think about a simple circuit used as a filter. A filter is acomponent with an input and output that provides a different scalingof the input sinusoidal signal (in amplitude and phase) dependingon the frequency. They are used extensively to reduce noise, enhancesignal strength, or modify a circuit’s phase. The circuit schematic is:

6.002 notes: sinusoidal steady state and the impedance method 5

vIN

R

L+

−vOUT

although it is not typically drawn this way. I drew it like this to em-phasize the similarity to a voltage divider. It is more typically drawnas shown in the margin, which is topologically identical to the figureabove.

+

−

vIN

R

L

+

−

vOUT

In the past, we typically have asked students to determine anoutput value from a circuit. But for a filter, we want the user to beable to vary the input, which will vary the output. So instead wenow want only to find a way to relate the output to the input, i.e. todetermine how the output depends on the input.

Because the circuit is linear, both input and output will be sinu-soidal waves with the same frequency. And because any two sinu-soidal waves can be related by the ratio of their amplitudes and thedifference of their phases, all we need to do is determine which com-plex number relates the two.

This complex number depends on frequency, and thus is called afunction (because its value is a function of ω). When plotted vs. ω ithelps provide an intuitive picture of the performance of the filter. Wecall this function the transfer function because it characterizes thetransfer of a signal from the input to the output.

Now that we understand better how and why the impedancemethod works, we don’t need to go through the full justification forit each time we use it. We can more or less just follow a recipe of asequence of steps:

1. Redraw the circuit as one that consists only of complex impedancesand DC complex sources;

−+ Vin

ZR=R

ZL= j ωL+

−Vout

2. Analyze the circuit using or DC analysis methods (in this case it is

6.002 notes: sinusoidal steady state and the impedance method 6

a simple voltage divider).

Vout =ZL

ZL + ZRVin

=j ωL

(R + j ωL)Vin

= H( j ω)Vin,

(8)

where we’ve introduces the symbol H( j ω) = Vout/Vin for thetransfer function of the system.

3. Analyze the transfer function’s frequency dependence. In thiscase, we see that as ω → 0, H( j ω) → j ωL/R so as ω gets smaller,the voltage will drop. This means low frequency signals will befiltered. Conversely, as ω → ∞, H( j ω) → 1, thus the outputjust reflects the input, with no change in phase or amplitude. Thismeans that high frequency signals will pass through... we call thisa High Pass Filter.

Some exercises for yourself to see if you know this material

Don’t refer to notes when working these exercises–try to answerthese questions just with paper and pencil in front of you.

1. Starting with the constitutive law of an inductor, derive its impedance.

2. Starting with the constitutive law of a capacitor, derive its impedance.

3. Create a random topology of one resistor, one capacitor, and onesinusoidal source and solve for the transfer function using thesource as the input, and any other circuit variable as the output.

Some Useful Videos to Watch

If you’re looking for some weekend YouTube to watch that can alsoeducate you about this subject, check out a few videos I’ve made:

1. https://youtu.be/y5OePLS_R8I Impedance Explained

2. https://youtu.be/GwXenS7BUHo Solving for a voltage usingimpedance analysis. High-pass inductive filter example.

Glossary

Complex Amplitude Amplitude of a sinusoidal signal (current orvoltage) including an assumed imaginary term. The magnitude ofthe complex amplitude will equal the magnitude of the real signalamplitude, and the argument of the complex amplitude will be

6.002 notes: sinusoidal steady state and the impedance method 7

the phase of the real signal relative to assumed zero phase for anunshifted pure cosine.

Filter Circuit designed to block or attenuate certain frequencies froma signal while allowing other frequencies to pass through.

High-Pass Filter Circuit designed to block or attenuate low fre-quencies from a signal while allowing higher frequencies to passthrough.

Impedance Ratio of complex voltage to complex current going througha device.

Transfer Function Gain of a circuit with sinusoidal drive as a functionof j ω.

Acknowledgements

Thanks to the many students, TAs, and faculty who have contributedto the formulation of the ideas in these notes over many years ofteaching, and to Rinske Wijtmans for help formatting the final docu-ment.

SSS SummaryI think it is worth succinctly summarizingthe main logical flow and results of theimpedance method:Consider an arbitrary circuit element,NS5 -entIAist Iacos (wt+ PA)(Ug = Vg cos (wt +46)then we apply the real-world constitutiverelation to relateis to voiAeigiiattroi-CduBdt+ NB-dt=[Voed-> IA cos (wt+fx) = c d Viscos (wt+43))j (wt + 4B)=IA> Re[ Fge det rass] Re (ca va edat o uredjest rela)>> Re[In edilen e dict] = Re (C d. Vo editsjust[]= Re [cused yo jwe dit]dt-doRe duedt

-redwo[j CV/=> Re [(52) sedut j = Re [lowVs) -edutId ] =Ia - IA eIf we also notice that I'm ( I ne jout): Inljuc je dut)jelawhere=and Te = Vz ejtowe canthen writejust= jwc Vo e justIae- Ia = jw CVGis a solution to the original problem.>b1.e. if this relation is true, the deviceconstitutive relation will be satisfied.At this point we observe that thisequation looks exactly like Ohm'slaw, if Ia and toDCbranch variables. The correspondingresistance would be Yjwc.wereYjwc. To be

clear that this is in no way an actualresistance, we call it an impedance and useIt V6 --Z="/jwcthe symbol Z to denote it.We can visualize how the impedancepicture relates to the original circuit byreversing the steps we took to get toit, i.e. by multiplying by ejust andthen taking the real part.I+ V-VetejutRe+ VBwsxe: HoZZ = juciI at Ve aretime independentcomplex numbersia & VB are&measureablesinusoidal circuitvariables.

Topof PageThis same process can be repeated forall the components we haveused so farIt=le- )tegutxVocos(wt+Q)VAwhere V = Vse jeVoRe[_xejutIs = Is eilIs cos(wt+cp)where we now represent the transformatto the impedance with just =.MoR=>Z=RHIZ=VjwcC008oZ=j w L

Because of the analogy to resistance,impedances can be combined just likeresistances,III=Z,ZzZ, + ZzZ,11ZzZ, IIZz= Z, ZzZ,+ ZzThe basic approach one should thus takewhen solving such problems is this:1. Verify that all the source are indeedin the sinusoidal state, centered aroundzero.2. Convert to a DC problem withcomplex sources a impedances.

3. Analyze for the complex Dc branchvariables of interest.4. Compute the actual branch variablesof interest by multiplying the results)from 3. by edwl and taking thereal part.5. Perform additional analysis ofnon-linear expressions (die power)as needed.Power and ImpedanceWhen calculating power in the sinusoidal steadystate, one has to remember that the complexamplitudes are not actual currents and voltagesand power is not linear relative to i and v.Thus one has to return to first principles.

Consider the sinusoidal steady state element:+ VadIZajutizRecall that pain and is Re[Ia e dist]and vs Re[va e just= [pz Rellimle ilet+25a)].Re [valedlote cva),= p = 1 Ialcos (wt+LIa)-/Valcos(wt + Lva)+j.Because the system is in sinusoidal steadystate, we can calculate the average bytemporalTintegrating across a periodIa Valcos(wt +LI a)cos Cwt + LVa) at<p>- tl'II. ??>Xaverage withrespect totimeAt this point, we can use the fact thatVa = Z Ia→ Lva = LZ + LIato write"T<p>" 17. Valcos(wt+) cos( ut +23+LZ?t

We can ignore the LIa term because weare integrating over a full period, and are freeto shift the integration limits by a constantand for periodic functions, the phase at which youT=begin and end cannot matter.=><p><•+S'Ita Valcos (wit) cos (wt + [2) dt.Noting that cas (wtr LZ) = cos wtcos Lz - suutsin LZand also noting so cos wut som ut dt =0 weSTcan show thatTT<pt +Silaval cos'ut cos LZ it(=lIa Val cos (LZ).21=Notice that for capacitors and inductors LZ* 1 l =) <p>2=0 as expected. However< p+<p>p = Ia Val because LZ-0.I=for resistors2