Sinusoidal Circuit Analysis
There are two particularly interesting aspects of sinusoidal circuit analysis. First an
often-disregarded property of a sinusoid is that is a periodic function, and as such has
neither 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 contrast
specifying an origin in general does affect the specific description of events. The other
interesting aspect is that much of the process of analyzing sinusoidal circuits has
already been studied. The essential effort to make here is recognizing this is so, and
understanding why it is so.
Introduction
Any (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 to
a composite sinusoidal signal can be determined for each sinusoidal term separately, and the overall
response then obtained by superposition. This is in part the basis of the fundamental importance of the
analysis of the response of a linear circuit to a single sinusoid of arbitrary frequency. Circuit behavior for
composite signals can be inferred from knowledge of this sinusoidal response. Hence what is considered
in this course primarily is the analysis of a linear circuit with an arbitrary single-frequency sinusoidal
excitation.
General Remarks on Sinusoids
There 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 a
displaced repetition of that period. A geometrical description of various sinusoid relationships is drawn
below. It is formally premature to characterize the figure as representing either a sine or a cosine; it could
be either. However because the sine is zero when its argument is zero (or an odd multiple of π) it is the
sine that would ordinarily be drawn as shown. The cosine is 1 when its argument is zero (or an even
multiple of π), and ordinarily it would be drawn with that value at the origin. The difference between a sine
and 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).
Circuits Sinusoidal Excitation
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M H Miller
As noted elsewhere there is a close relationship between a sinusoid and uniform circular rotation, as
illustrated in the figure to the right. The figure captures a
single instant of the sinusoid history. As time progresses the
angle 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 common
method of describing the sinusoid period independent of the
actual period; one uses the angular measure corresponding to
a single rotation, i.e., 360˚ or 2π in radian measure.
A half-period (or half-cycle) is 180˚ or π radians. Angular
measure is used commonly to describe the difference between corresponding points on two related
sinusoids. This is illustrated below. Generally, unless otherwise stated a counterclockwise rotation around
the unit circle corresponds to an evolution of the sinusoid from left to right. In the figure the displacement
between the peaks of the two sinusoids is θ whereas the displacement between the peaks of the same
sinusoid is 360˚; 0 ≤ θ ≤ 360˚ .
The waveforms commonly are considered to evolve in time left to right, i.e. for a given sinusoid the further
right a point is located the later in time it occurs. A similar interpretation is applied on comparing two
sinusoids 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 phase
angle θ; it reaches its peak θ degrees later then the
other sinusoid Conversely the other sinusoid is said
to ‘lead’; the relationship is a mutual one. Note
carefully that 'lead' and 'lag' are comparative measures;
one sinusoid leads or lags another sinusoid. This
phase angle concept applies just to a description of
sinusoids of the same period, but then sinusoids play
an exceptional role in circuit theory.
Incidentally, although it is convenient to use peaks (+
or -) or zero crossings as distinctive points on a
sinusoid 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 the
right. Hence the current leads the voltage in a capacitor. This is a reflection of the physical requirement
that charge flow before the voltage is changed. Note also that the amplitude of the current is large the larger
ω is, i.e., the faster the voltage is changed. Since a given voltage change requires a given charge change (for
a given capacitance; q = Cv) one should expect that the faster the voltage changes the faster the current has
to change.
Similar relationship apply to an inductor except, as should be expected, the voltage leads the current
Circuits Sinusoidal Excitation
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M H Miller
Circuit Equations; sinusoidal excitation
The starting point for the analysis procedure is again the recognition that for a linear circuit the application
of KVL, KCL, and the V-A relations ultimately ends up with a linear differential equation with constant
coefficients. In the present case, i.e. for a circuit containing only sinusoidal sources, the 'driving function' is
sinusoidal 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 variable
with the general form
As before the complete solution of this equation is a sum of a general (transient) partial solution and a
partial solution particular to the specific driving function on the right side of the equation. In virtually all
cases however the transient solution is not of specific interest, and it is simply assumed that at the time the
circuit 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 form
Αsin(ωt) + Βcos(ωt),
and substitution into the equation leads to a trigonometric equation which can be solved for A and B in a
straightforward way. Although we actually will use a modified procedure a relatively uncomplicated
illustration 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 convert
it to a pure differential equation shown as item b); the solution is of course the same, and has the general
form shown in item c), where A and B are constant coefficients (of integration) to be determined. This
solution 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 fact
that the sine is zero and the cosine 1 in the one case, and vice versa in the other. On this basis we obtain
two independent equations from which to determine A and B, resulting in item f). The second form is
obtained using the trigonometric identity sin(θ – φ) = sinθ cosφ - sinφ cosθ and defining θ = ωt,
φ = tan–1ωRC.
Circuits Sinusoidal Excitation
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M H Miller
This brute force approach is not the preferred method to analyze the circuit. It is much simpler to be
indirect, and for that reason indirection is an almost universal solution procedure. Indeed the illustrative
circuit 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 imaginary
parts of a complex number, and superposition. The process is illustrated formally at first, to make clear
what 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' is
added to identify the solution to this equation explicitly. The second equation duplicates the first, except
for the driving function. Its driving function is chosen to accommodate the transformation into the third
equation. 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 adding
the result to the first equation. The driving function for the second equation was chosen specifically so that
adding it to the first provides the complex exponential driving function via Euler's theorem. The solution to
this third equation is the sum of the solutions to the other two equations, y = yr + jy i. It is to make this so
that the driving functions have been configured. The ‘r’ solution corresponds to the real part of the
complex exponential, and the ‘i’ part corresponds to the imaginary part. This relationship makes the two
solutions 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 Kejωt, 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, and
integration is equivalent to division by jω. Substitution into the differential equation then leads to an
algebraic expression, much easier to deal with than a differential equation. And as stated knowing the
solution 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 from
the complex solution by itself.
Creating the 'complex' equation is described as a 'transformation' to the ‘frequency’ domain because it is
the frequency variable ω (i.e. the coefficient of t) that appears explicitly in the algebraic expression. In
contrast the original equation is said to be in the 'time' domain.
Circuits Sinusoidal Excitation
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M H Miller
Illustrative Analysis
As is done for the transient case we can take considerable advantage of foreknowledge of the nature of the
transformation, 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 to
be analyzed contains just one source; e.g. the circuit drawn to the right.. The first decision is
whether the time domain solution is to be the real or the
imaginary part of the complex solution. Since the two parts
of the complex solution are obtained concurrently, and it is
more or less as easy to extract the real part, as it is to extract
the imaginary part; pick whatever seems best. (As always the
general mathematical arbitrariness of the choice should not
really be used to make an arbitrary choice; take some
advantage of the freedom even if only an esthetic preference).
b) Insert an additional source, another voltage source in series with a voltage source, another current
source in parallel with a current source. Choose the added source strength so that the
superposition of the sources has an complex exponential source strength. In the circuit illustrated
the 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 -jEejωt. Of course in general
you should remember your choice, i.e. remember whether ultimately the solution you want to
extract 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, but
in the frequency domain, i.e., after the source transformation, they simplify to the algebraic
expressions I =jωCV and V=jωLI. These have the same form as Ohm's Law, i.e., the voltage and
current are related by a (in this case, imaginary) constant. All the derivations involving Ohm's law
do not depend for the validity on the name of the proportionality constant, i.e., how it is written, but
only on it being constant. Hence all the circuit analysis techniques studied for resistive circuits
apply when capacitors and inductors are included, with the qualification of course that the proper
proportionality constant is used. The arithmetic is more complex (!), but in exchange for a quite
modest complexity the range of circuits readily amenable to analysis and design is expanded
enormously.
d) Consider the illustrative circuit as before, redrawn below on the left. Transform to the frequency
domain by adding a voltage source jAsin(ωt) in series with the cosine source, to form the complex
source Eejωt as indicated on the right..
To analyze the circuit write the loop equation Eejωt = IejωtR + Ie jωt/jωC, where the loop current (for the
particular solution) is known to be proportional to ejωt. Actually, in usual practice, the equation would be
written simply as E = IR + I/jωC, with the common ejωt factor suppressed –all variables contain this factor
and it reduces clutter not to carry it along as a common factor of every term. (Remember however that it is
necessary to restore the exponential factor as part of the variable when done.)
Solve for I (and restoring the exponential factor) the loop current is,
Circuits Sinusoidal Excitation
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M H Miller
The solution to the original equation was chosen to be the real part of the complex solution, so we want
Re(I). There are several variations on obtaining the real part, all producing the same result. Perhaps the
simplest is first to convert the denominator from rectangular to polar form, combine that with the polar form
of 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
Circuits Sinusoidal Excitation
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M H Miller
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 source
strength is multiplied by 1.7456 (arbitrary number) all voltages and
currents are multiplied by this same number. This is the basis for the
following technique for analyzing the circuit shown. Suppose the
source 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). But
the actual source voltage is 1 volt. Hence scale all the circuit voltages
and currents calculated for V(2,3) = 1 by a factor 1/(0.5 - j6) = 0.166∠ 85.23˚. This then provides the
value 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 a
direct analysis of the circuit it can do so indirectly, helped by a little mathematical slight of hand. For
sinusoidal excitation L and C always appear with ω as a multiplying factor, i.e., either as ωL or ωC. Since a
specific frequency is not involved in the analysis we can simply choose a frequency such that ω=1. In this
case ωL = L, and ωC = C. Hence the inductive reactance of j1 corresponds to a 1H inductance, and the
capacitive reactance of -j2.5 corresponds to a capacitance of 1/2.5 = 0.4F. The netlist for the example circuit
is shown below. PSPICE wants to know the start and stop frequencies (which are the same since we need
compute at only a single frequency), i.e., 1/ω = 0.159155 for ω =1. Note that PSpice uses frequency, not
radian frequency.
AC Analysis
VS
1
R12 1
L23 2
R23 2
C30 3
0
2
3
3
0
AC
1
5
1
2.5
.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)
1.592E-01
8.944E-01
1.704E+01
8.552E-01
VM(L23)
VP(L23)
VR(L23)
1.592E-01
1.661E-01
8.524E+01
1.379E-02
VM(R23)
VP(R23)
VR(R23)
1.592E-01
1.661E-01
8.524E+01
1.379E-02
VM(C30)
VP(C30)
VR(C30)
1.592E-01
4.472E-01
-7.296E+01 1.310E-01
Circuits Sinusoidal Excitation
7
VI(R12)
2.621E-01
VI(L23)
1.655E-01
VI(R23)
1.655E-01
VI(C30)
-4.276E-01
M H Miller
Example 2: While it is generally a good idea to solve
some of these problems by hand (in part to acquire basic
familiarity with complex number arithmetic) a point of
diminishing returns is reached where it can become just
tedious and pointless. This example uses PSPICE to
calculate the node voltages for the circuit shown. The
node-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 the
appropriate proportionality constant is used. Compare this analysis with that in which the
integro–differential equations are used directly.
AC Analysis
R10
1
0
6
R12
1
2
2
R20
2
0
2
R23
2
3
2
I30
0
3
AC
12
C34
3
4
1
R40
4
0
1
L13
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
Circuits Sinusoidal Excitation
FREQ 1.592E-01
VM(1)
7.961E+00
VM(2)
6.713E+00
VM(3)
1.284E+01
VM(4)
9.077E+00
8
VP(1)
-5.490E+01
VP(2)
-3.647E+01
VP(3)
-2.516E+01
VP(4)
1.984E+01
M H Miller
Example 3: For this example we calculate Vo in the
circuit below using Thevenin's theorem. In the first box
(just below) the Thevenin (open-circuit) voltage is
calculated. In the second box the Thevenin impedance
is calculated by setting all fixed source strengths to zero
and calculating the impedance 'seen' by the 1Ω. An
alternative calculation of the Thevenin impedance is
shown in the third box; this is the 'short-circuit' current
calculation. Finally the fourth box shows the
calculation of Vo.
Circuits Sinusoidal Excitation
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M H Miller
Example 4: For this example a mesh analysis is shown. However it is instructive to consider alternative
methods as well. A nodal analysis is an obvious alternative. Consider also converting the series
combination of a voltage source and impedance to a Norton equivalent (both sources), then combining the
current sources and calculating the node (2) voltage. Or consider applying 'superposition'; evaluate the
contribution 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 with
the circuit already described in the frequency plane, and nothing required determining explicitly whether the
solution desired was the real or the imaginary part of the complex solution. When and if that question
becomes 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 PSpice
interprets this to refer to a cosine wave, i.e.
Vname
+node -node AC
K
θ
is 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; the
netlist is given below.
AC Analysis
VS
1
0
*VS
1
0
R12
1
2
C23
2
0
.AC
LIN 1
.PRINT AC IM(R12)
.END
AC
AC
1
0
; COS source
1
-90 ;SIN source
3
0.25
.159155
.159155
IP(R12) IR(R12), II(R12),VR(1), VI(1), VM(1), VP(1)
First the computation is performed with zero phase shift, i.e., for a cos source. The source transforms into
the frequency domain as 1∠0˚, and the loop current as 1/(3-j4) = 0.12+j0.16. The PSpice computation
produces:
FREQ
1.592E-01
IM(R12)
2.000E-01
VM(1)
1.000E+00
IP(R12)
5.313E+01
VP(1)
0.000E+00
IR(R12)
1.200E-01
VR(1)
1.000E+00
II(R12)
1.600E-01
VI(1)
0.000E+00
Note that the source is the real part of the complex expression. The time domain solution for the loop
current 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 the
source is the imaginary part of the complex frequency domain source strength. The time domain solution
for the loop current is the imaginary part of the complex current, i.e., -0.12sin(2πt).
Sinusoidal Notes
10
ECE 210 MHM W'96
FREQ
1.592E-01
IM(R12)
2.000E-01
VM(1)
1.000E+00
IP(R12)
-3.687E+01
VP(1)
-9.000E+01
IR(R12)
1.600E-01
VR(1)
5.451E-17
II(R12)
-1.200E-01
VI(1)
-1.000E+00
To sum up: while in general one can choose to make the time domain solution either the real or the
imaginary part of the frequency domain solution only one choice may be made at any one time. Since
PSpice 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 the
left. There is nothing is basically different to do.
The important point to note is that you have just one
choice as to whether to use the real or the imaginary
part of the complex solution. Use either one, but
only one. Thus to use the real part use
sin(θ) = Re[ -jejθ, i.e., add a source -jcos(θ). Or
equivalently recognize that sin(θ) = cos(θ-90˚). To use the imaginary part note that
cos(θ) = Im[jejθ] = sin( θ +90˚). Incidentally, the illustrative circuit is simple enough so that these
considerations can be avoided by an application of superposition. But don't take the easy road here; write a
single node equation and calculate I.
And then there is PSpice. The cosine source is entered with zero phase angle, the sine source is entered
with a -90˚ phase angle, and it is the real part of the complex solution that is to be used in the time domain1.
AC Analysis
VSin
1
0
AC
2
-90
L12
1
2
4
R20
2
0
10
C23
2
3
0.25
VCos 3
0
AC
6
.AC
LIN 1
.159155
.159155
.PRINT AC VR(1), VI(1),VR(2), VI(2),
+ VR(3), VI(3)
.END
Sinusoidal Notes
FREQ
1.592E-01
VR(1)
1.090E-16
VR(2)
-5.000E+00
VR(3)
6.000E+00
11
VI(1)
-2.000E+00
VI(2)
1.500E+01
VI(3)
0.000E+00
ECE 210 MHM W'96