Download Sinusoidal Circuit Analysis: Understanding Linear Response to Excitation and more Study notes History in PDF only on Docsity!
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).
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
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 + jyi. 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.
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 + Iejωt/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,
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 0 AC 1 R12 1 2 5 L23 2 3 1 R23 2 3 2. C30 3 0.
- Perform an AC analysis for just one frequency, 1 Hz. .AC LIN 1 .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- VM(L23) VP(L23) VR(L23) VI(L23) 1.592E-01 1.661E-01 8.524E+01 1.379E-02 1.655E- VM(R23) VP(R23) VR(R23) VI(R23) 1.592E-01 1.661E-01 8.524E+01 1.379E-02 1.655E- VM(C30) VP(C30) VR(C30) VI(C30) 1.592E-01 4.472E-01 -7.296E+01 1.310E-01 -4.276E-
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. .PRINT AC VM(1), VP(1), VM(2), VP(2), +VM(3), VP(3), VM(4),VP(4) .END
FREQ 1.592E-
VM(1) VP(1)
7.961E+00 -5.490E+
VM(2) VP(2)
6.713E+00 -3.647E+
VM(3) VP(3)
1.284E+01 -2.516E+
VM(4) VP(4)
9.077E+00 1.984E+
Sinusoidal Notes 10 ECE 210 MHM W'
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 AC 1 0 ; COS source *VS 1 0 AC 1 -90 ;SIN source R12 1 2 3 C23 2 0 0. .AC LIN 1 .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 into the frequency domain as 1∠0˚, and the loop current as 1/(3-j4) = 0.12+j0.16. The PSpice computation produces:
FREQ IM(R12) IP(R12) IR(R12) II(R12) 1.592E-01 2.000E-01 5.313E+01 1.200E-01 1.600E- VM(1) VP(1) VR(1) VI(1) 1.000E+00 0.000E+00 1.000E+00 0.000E+
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 11 ECE 210 MHM W'
FREQ IM(R12) IP(R12) IR(R12) II(R12)
1.592E-01 2.000E-01 -3.687E+01 1.600E-01 -1.200E-
VM(1) VP(1) VR(1) VI(1)
1.000E+00 -9.000E+01 5.451E-17 -1.000E+
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 - L12 1 2 4 R20 2 0 10 C23 2 3 0. VCos 3 0 AC 6 .AC LIN 1. . .PRINT AC VR(1), VI(1),VR(2), VI(2),
FREQ 1.592E-
VR(1) VI(1)
1.090E-16 -2.000E+
VR(2) VI(2)
-5.000E+00 1.500E+
VR(3) VI(3)
6.000E+00 0.000E+
