2 sinusoidal circuits

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Circuits Sinusoidal Excitation 1 M H Miller

Sinusoidal Circuit Analysis

There are two particularly interesting aspects of sinusoidal circuit analysis. First anoften-disregarded property of a sinusoid is that is a periodic function, and as such hasneither beginning nor end. Thus designating a reference origin as t = 0 (or otherwise)can have no physical consequence, since such a designation is arbitrary. In contrastspecifying an origin in general does affect the specific description of events. The otherinteresting aspect is that much of the process of analyzing sinusoidal circuits hasalready been studied. The essential effort to make here is recognizing this is so, and

understanding why it is so.

IntroductionAny (non-pathological) function can be represented by an appropriate superposition of sine waves of different frequencies, either as a Fourier series or as a Fourier integral. The response of a linear network toa composite sinusoidal signal can be determined for each sinusoidal term separately, and the overallresponse then obtained by superposition. This is in part the basis of the fundamental importance of theanalysis of the response of a linear circuit to a single sinusoid of arbitrary frequency. Circuit behavior forcomposite signals can be inferred from knowledge of this sinusoidal response. Hence what is consideredin this course primarily is the analysis of a linear circuit with an arbitrary single-frequency sinusoidalexcitation.

General Remarks on SinusoidsThere are several useful relationships between the sine and cosine functions, and uniform circular rotation.Neither the sine nor the cosine has a beginning or an end; both are periodic functions. A kind of 'start'/'finish' relationship is a description of one period of the sinusoid; the remainder of the function is adisplaced repetition of that period. A geometrical description of various sinusoid relationships is drawnbelow. It is formally premature to characterize the figure as representing either a sine or a cosine; it couldbe either. However because the sine is zero when its argument is zero (or an odd multiple of π) it is thesine that would ordinarily be drawn as shown. The cosine is 1 when its argument is zero (or an evenmultiple of π), and ordinarily it would be drawn with that value at the origin. The difference between a sineand cosine drawing is a displacement of 1/4 the period. Thus the drawing could represent cos(ωt - 4t/T),although as noted ordinarily the axis would be shifted to cancel the constant in the argument.

However it is quite usual to have to compare several sinusoids concurrently, and a single axis shift will not

cancel all the constant arguments, e.g., sin(ωt) and cos(ωt+4t/T).




As noted elsewhere there is a close relationship between a sinusoid and uniform circular rotation, asillustrated in the figure to the right. The figure captures asingle instant of the sinusoid history. As time progresses theangle changes, but the relationships remain the same.

The period of the sinusoid corresponds to a single rotation of the radius around the circumference. This leads to a commonmethod of describing the sinusoid period independent of the

actual period; one uses the angular measure corresponding toa single rotation, i.e., 360˚ or 2π in radian measure.

A half-period (or half-cycle) is 180˚ or π radians. Angularmeasure is used commonly to describe the difference between corresponding points on two relatedsinusoids. This is illustrated below. Generally, unless otherwise stated a counterclockwise rotation aroundthe unit circle corresponds to an evolution of the sinusoid from left to right. In the figure the displacementbetween the peaks of the two sinusoids is θ whereas the displacement between the peaks of the samesinusoid is 360˚; 0 ≤ θ ≤ 360˚ .

The waveforms commonly are considered to evolve in time left to right, i.e. for a given sinusoid the furtherright a point is located the later in time it occurs. A similar interpretation is applied on comparing twosinusoids of the same period. The sinusoid whose peak is on the right side of the relative displacement

marking is said to 'lag' the other sinusoid by a phaseangle θ; it reaches its peak θ degrees later then theother sinusoid Conversely the other sinusoid is saidto ‘lead’; the relationship is a mutual one. Notecarefully that 'lead' and 'lag' are comparative measures;one sinusoid leads or lags another sinusoid. Thisphase angle concept applies just to a description of sinusoids of the same period, but then sinusoids playan exceptional role in circuit theory.Incidentally, although it is convenient to use peaks (+or -) or zero crossings as distinctive points on asinusoid clearly if the peaks are displaced by θ then all

pairs of corresponding points also are displaced by the same angle θ.

Suppose a voltage Asin ωt is applied across a capacitor; the current through the capacitor then is Aωcos ωt.If either the voltage or current is sinusoidal then the other also is sinusoidal. There is a phase shift of 90˚between the voltage and current. The cosine peaks at t=0, while the sine peaks 90˚ later, i.e., further to theright. Hence the current leads the voltage in a capacitor. This is a reflection of the physical requirementthat charge flow before the voltage is changed. Note also that the amplitude of the current is large the largerω is, i.e., the faster the voltage is changed. Since a given voltage change requires a given charge change (fora given capacitance; q = Cv) one should expect that the faster the voltage changes the faster the current hasto change.

Similar relationship apply to an inductor except, as should be expected, the voltage leads the current




Circuit Equations; sinusoidal excitationThe starting point for the analysis procedure is again the recognition that for a linear circuit the applicationof KVL, KCL, and the V-A relations ultimately ends up with a linear differential equation with constantcoefficients. In the present case, i.e. for a circuit containing only sinusoidal sources, the 'driving function' isinusoidal since differentiation of a sine or cosine source strength is sinusoidal, and sines and cosines of the same frequency can be combined into a single sinusoid, e.g.

Hence quite generally the ultimate formal result of the analysis process is an equation for a circuit variablewith the general form

As before the complete solution of this equation is a sum of a general (transient) partial solution and apartial solution particular to the specific driving function on the right side of the equation. In virtually allcases however the transient solution is not of specific interest, and it is simply assumed that at the time thecircuit is being considered enough time has passed for inevitably decaying transients to become negligible.Thus only the particular (i.e. sinusoidal) solution is considered.

The particular solution is has been shown by mathematicians to have the general formΑsin(ωt) + Βcos(ωt),

and substitution into the equation leads to a trigonometric equation which can be solved for A and B in astraightforward way. Although we actually will use a modified procedure a relatively uncomplicatedillustration of direct substitution is provided below first.

The analysis procedure used starts with a loop equation for the circuit considered; this is item a) in the

figure. Simply as a matter of convenience the equation as a whole is differentiated term by term to convertit to a pure differential equation shown as item b); the solution is of course the same, and has the generalform shown in item c), where A and B are constant coefficients (of integration) to be determined. Thissolution is substituted into the equation, resulting in item d). The equation must be satisfied at all times,and in particular at the conveniently selected times ωt = 0 and ωt = 90º. The convenience lies in the factthat the sine is zero and the cosine 1 in the one case, and vice versa in the other. On this basis we obtaintwo independent equations from which to determine A and B, resulting in item f). The second form isobtained using the trigonometric identity sin(θ – φ) = sinθ cosφ - sinφ cosθ and defining θ = ωt,

φ = tan–1ωRC.




This brute force approach is not the preferred method to analyze the circuit. It is much simpler to beindirect, and for that reason indirection is an almost universal solution procedure. Indeed the illustrativecircuit very nearly can be analyzed by inspection if, of course, the underlying process is appreciated!Underlying the indirection is use of Euler's Theorem, the separation property of the real and imaginaryparts of a complex number, and superposition. The process is illustrated formally at first, to make clearwhat is a much simpler application in practice.

The first equation (below) repeats the original equation, the one we actually want to solve. A subscript 'r' isadded to identify the solution to this equation explicitly. The second equation duplicates the first, exceptfor the driving function. Its driving function is chosen to accommodate the transformation into the thirdequation. A subscript 'i' is added to identify this solution.

The third equation is obtained by multiplying the second by the (constant) complex operator j, and addingthe result to the first equation. The driving function for the second equation was chosen specifically so thatadding it to the first provides the complex exponential driving function via Euler's theorem. The solution tothis third equation is the sum of the solutions to the other two equations, y = yr + jyi. It is to make this sothat the driving functions have been configured. The ‘r’ solution corresponds to the real part of thecomplex exponential, and the ‘i’ part corresponds to the imaginary part. This relationship makes the twosolutions separately identifiable if the solution to the third equation can be obtained.

What makes this apparent added complexity worthwhile is the fact that the particular solution for the third

equation has the form Ke jωt

, where K is a constant multiplier. This is a most important consequence,because it means that differentiation of a variable is equivalent to algebraic multiplication by jω, andintegration is equivalent to division by jω. Substitution into the differential equation then leads to analgebraic expression, much easier to deal with than a differential equation. And as stated knowing thesolution for the 'complex' equation it is easy enough to extract the desired solution, say Re[y] = yr .Actually almost always in practice, as we will see, the circuit information sought is available directly fromthe complex solution by itself.

Creating the 'complex' equation is described as a 'transformation' to the ‘frequency’ domain because it isthe frequency variable ω (i.e. the coefficient of t) that appears explicitly in the algebraic expression. Incontrast the original equation is said to be in the 'time' domain.




Illustrative AnalysisAs is done for the transient case we can take considerable advantage of foreknowledge of the nature of thetransformation, i.e., simply knowing in advance that it can be done enables us to avoid most of the effort of actually doing it.

a) Suppose for simplicity (and by invoking superposition to assert no loss of generality) the circuit tobe analyzed contains just one source; e.g. the circuit drawn to the right.. The first decision iswhether the time domain solution is to be the real or theimaginary part of the complex solution. Since the two partsof the complex solution are obtained concurrently, and it is

more or less as easy to extract the real part, as it is to extractthe imaginary part; pick whatever seems best. (As always thegeneral mathematical arbitrariness of the choice should notreally be used to make an arbitrary choice; take someadvantage of the freedom even if only an esthetic preference).

b)

Insert an additional source, another voltage source in series with a voltage source, another currentsource in parallel with a current source. Choose the added source strength so that thesuperposition of the sources has an complex exponential source strength. In the circuit illustratedthe source is Ecos(ωt); hence add jEsin(ωt) in series. If the source had been Esin(ωt) one could,

for example, add –jEcos(ωt) to obtain the combined source strength -jEe jωt. Of course in generalyou should remember your choice, i.e. remember whether ultimately the solution you want toextract is the real or the imaginary part of the complex solution.

c)

The capacitor and inductor volt-ampere relations in the time domain are differential equations, butin the frequency domain, i.e., after the source transformation, they simplify to the algebraicexpressions I =jωCV and V=jωLI. These have the same form as Ohm's Law, i.e., the voltage andcurrent are related by a (in this case, imaginary) constant. All the derivations involving Ohm's lawdo not depend for the validity on the name of the proportionality constant, i.e., how it is written, butonly on it being constant. Hence all the circuit analysis techniques studied for resistive circuitsapply when capacitors and inductors are included, with the qualification of course that the properproportionality constant is used. The arithmetic is more complex (!), but in exchange for a quitemodest complexity the range of circuits readily amenable to analysis and design is expandedenormously.

d) Consider the illustrative circuit as before, redrawn below on the left. Transform to the frequencydomain by adding a voltage source jAsin(ωt) in series with the cosine source, to form the complex

source Ee jωt as indicated on the right..

To analyze the circuit write the loop equation Ee jωt = Ie jωtR + Ie jωt /jωC, where the loop current (for the

particular solution) is known to be proportional to e jωt. Actually, in usual practice, the equation would be

written simply as E = IR + I/jωC, with the common e jωt factor suppressed –all variables contain this factorand it reduces clutter not to carry it along as a common factor of every term. (Remember however that it isnecessary to restore the exponential factor as part of the variable when done.)

Solve for I (and restoring the exponential factor) the loop current is,




The solution to the original equation was chosen to be the real part of the complex solution, so we wantRe(I). There are several variations on obtaining the real part, all producing the same result. Perhaps thesimplest is first to convert the denominator from rectangular to polar form, combine that with the polar formof the numerator, and use Euler's theorem to take the real part:

Compare this with the solution obtained before, or better, compare the procedures used.The alternative trigonometric form shown before may be obtained directly (if desired) from the identity

where both the numerator and denominator are multiplied by the imaginary operator j.

To calculate, say, the voltage across the capacitor, we have Vc = I/jωC,

and




Sinusoidal Analysis Examples

Example 1: The linearity of the circuit is the basis for the assertion that if the voltage source strength is,say doubled, all voltages and currents are doubled. If the sourcestrength is multiplied by 1.7456 (arbitrary number) all voltages andcurrents are multiplied by this same number. This is the basis for thefollowing technique for analyzing the circuit shown. Suppose thesource strength (unknown value) is changed so as to make V(2,3) =1∠0˚. The current through the 2.5Ω then is 0.4 (ampere). The

current through the inductor is 1/j = -j, and the source current is 0.4-j.The source voltage then is (0.4-j)(5 -j2.5) +1 = 0.5 - j6 (volts). Butthe actual source voltage is 1 volt. Hence scale all the circuit voltagesand currents calculated for V(2,3) = 1 by a factor 1/(0.5 - j6) = 0.166∠ 85.23˚. This then provides thevalue of V(2,3) for the 1 volt source.

Note: PSPICE can be used to compute the circuit voltages and currents. Although it does not provide for adirect analysis of the circuit it can do so indirectly, helped by a little mathematical slight of hand. Forsinusoidal excitation L and C always appear with ω as a multiplying factor, i.e., either as ωL or ωC. Since aspecific frequency is not involved in the analysis we can simply choose a frequency such that ω=1. In thiscase ωL = L, and ωC = C. Hence the inductive reactance of j1 corresponds to a 1H inductance, and thecapacitive reactance of -j2.5 corresponds to a capacitance of 1/2.5 = 0.4F. The netlist for the example circuitis shown below. PSPICE wants to know the start and stop frequencies (which are the same since we needcompute at only a single frequency), i.e., 1/ ω = 0.159155 for ω =1. Note that PSpice uses frequency, notradian frequency.

AC AnalysisVS 1 0 AC 1R12 1 2 5L23 2 3 1R23 2 3 2.5C30 3 0 .4

* Perform an AC analysis for just one frequency, 1 Hz..AC LIN 1 .159155 .159155

*Save all branch voltages (magnitude and phase , and real and imaginary) in the .out file.PRINT AC VM(R12), VP(R12),VR(R12), VI(R12),+VM(L23), VP(L23),VR(L23), VI(L23),+VM(R23), VP(R23),VR(R23), VI(R23),+ VM(C30),VP(C30),VR(C30),VI(C30)

.END

The branch voltages (extracted from the .out file) are:FREQ VM(R12) VP(R12) VR(R12) VI(R12)1.592E-01 8.944E-01 1.704E+01 8.552E-01 2.621E-01

VM(L23) VP(L23) VR(L23) VI(L23)1.592E-01 1.661E-01 8.524E+01 1.379E-02 1.655E-01

VM(R23) VP(R23) VR(R23) VI(R23)

1.592E-01 1.661E-01 8.524E+01 1.379E-02 1.655E-01VM(C30) VP(C30) VR(C30) VI(C30)

1.592E-01 4.472E-01 -7.296E+01 1.310E-01 -4.276E-01




Example 2: While it is generally a good idea to solvesome of these problems by hand (in part to acquire basicfamiliarity with complex number arithmetic) a point of diminishing returns is reached where it can become justtedious and pointless. This example uses PSPICE tocalculate the node voltages for the circuit shown. Thenode-to-datum matrix equations are shown below,followed by data from the .out file. See the preceding

example for the 'trick' in writing the netlist in terms of reactance.

Note particularly that procedurally this is the same as for a resistive circuit, except of course that theappropriate proportionality constant is used. Compare this analysis with that in which theintegro–differential equations are used directly.

AC AnalysisR10 1 0 6R12 1 2 2R20 2 0 2

R23 2 3 2I30 0 3 AC 12C34 3 4 1R40 4 0 1L13 1 3 3.AC LIN 1 .159155 .159155.PRINT AC VM(1), VP(1), VM(2), VP(2),

+VM(3), VP(3), VM(4),VP(4).END

FREQ 1.592E-01

VM(1) VP(1)

7.961E+00 -5.490E+01VM(2) VP(2)6.713E+00 -3.647E+01VM(3) VP(3)1.284E+01 -2.516E+01VM(4) VP(4)9.077E+00 1.984E+01




Example 3: For this example we calculate Vo in thecircuit below using Thevenin's theorem. In the first box(just below) the Thevenin (open-circuit) voltage iscalculated. In the second box the Thevenin impedanceis calculated by setting all fixed source strengths to zeroand calculating the impedance 'seen' by the 1Ω. Analternative calculation of the Thevenin impedance isshown in the third box; this is the 'short-circuit' currentcalculation. Finally the fourth box shows the

calculation of Vo.



Sinusoidal Notes 10 ECE 210 MHM W'96

Example 4: For this example a mesh analysis is shown. However it is instructive to consider alternativemethods as well. A nodal analysis is an obvious alternative. Consider also converting the seriescombination of a voltage source and impedance to a Norton equivalent (both sources), then combining thecurrent sources and calculating the node (2) voltage. Or consider applying 'superposition'; evaluate thecontribution of each voltage source separately and then combine the contributions.

Example 5: In preceding illustrations we have effectively 'tricked' PSpice into performing a complex(frequency domain) analysis, but there is something to be wary of. The preceding problems started withthe circuit already described in the frequency plane, and nothing required determining explicitly whether thesolution desired was the real or the imaginary part of the complex solution. When and if that questionbecomes pertinent some care is needed.

The reason lies with the specification of an AC source, for example a voltage source, in PSpice:Vname +node -node AC amplitude phase

If the phase is omitted it is assumed implicitly to be zero. What is not directly evident is that PSpiceinterprets this to refer to a cosine wave, i.e.

Vname +node -node AC K θis interpreted as Kcos(ωt + θ). If you want interpretation as Ksin(ωt) make θ = 90˚,

Consider a simple circuit consisting of a source VS in series with a resistor in series with a capacitor; thenetlist is given below.

AC Analysis

VS 1 0 AC 1 0 ; COS source*VS 1 0 AC 1 -90 ;SIN sourceR12 1 2 3C23 2 0 0.25.AC LIN 1 .159155 .159155.PRINT AC IM(R12) IP(R12) IR(R12), II(R12),VR(1), VI(1), VM(1), VP(1).END

First the computation is performed with zero phase shift, i.e., for a cos source. The source transforms intothe frequency domain as 1∠0˚, and the loop current as 1/(3-j4) = 0.12+j0.16. The PSpice computationproduces:

FREQ IM(R12) IP(R12) IR(R12) II(R12)

1.592E-01 2.000E-01 5.313E+01 1.200E-01 1.600E-01VM(1) VP(1) VR(1) VI(1)1.000E+00 0.000E+00 1.000E+00 0.000E+00

Note that the source is the real part of the complex expression. The time domain solution for the loopcurrent is the real part of the complex current, i.e., 0.12 cos(2πt).

For the second computation a 90˚ phase angle is added to make the source waveform a sine. Note that thesource is the imaginary part of the complex frequency domain source strength. The time domain solutionfor the loop current is the imaginary part of the complex current, i.e., -0.12sin(2πt).



Sinusoidal Notes 11 ECE 210 MHM W'96

FREQ IM(R12) IP(R12) IR(R12) II(R12) 1.592E-01 2.000E-01 -3.687E+01 1.600E-01 -1.200E-01

VM(1) VP(1) VR(1) VI(1)1.000E+00 -9.000E+01 5.451E-17 -1.000E+00

To sum up: while in general one can choose to make the time domain solution either the real or theimaginary part of the frequency domain solution only one choice may be made at any one time. SincePSpice makes a default choice users of the program must take that into account.

Example 6: What to do if a circuit contains both sine and a cosine sources, e.g. the circuit drawn to theleft. There is nothing is basically different to do.The important point to note is that you have just onechoice as to whether to use the real or the imaginarypart of the complex solution. Use either one, butonly one. Thus to use the real part use

sin(θ) = Re[ -je jθ, i.e., add a source -jcos(θ). Orequivalently recognize that sin(θ) = cos(θ-90˚). To use the imaginary part note that

cos(θ) = Im[je jθ] = sin( θ +90˚). Incidentally, the illustrative circuit is simple enough so that theseconsiderations can be avoided by an application of superposition. But don't take the easy road here; write asingle node equation and calculate I.

And then there is PSpice. The cosine source is entered with zero phase angle, the sine source is enteredwith a -90˚ phase angle, and it is the real part of the complex solution that is to be used in the time domain1.

AC AnalysisVSin 1 0 AC 2 -90L12 1 2 4R20 2 0 10C23 2 3 0.25VCos 3 0 AC 6.AC LIN 1 .159155

.159155.PRINT AC VR(1), VI(1),VR(2), VI(2),+ VR(3), VI(3)

.END

FREQ 1.592E-01VR(1) VI(1)1.090E-16 -2.000E+00VR(2) VI(2)-5.000E+00 1.500E+01VR(3) VI(3)6.000E+00 0.000E+00

2 sinusoidal circuits

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