Transient Analysis
In this course we consider (with a few exceptions) linear circuits, or more explicitly circuits containing
linear independent and controlled sources, R and now L and C circuit elements. The LC elements introduce
volt-ampere constitutive relations that involve differentiation (or integration) with respect to time of voltages
and currents. KVL and KCL do not involve specific circuit elements and are applied as described before.
Hence the application of KVL and KCL provide a set of linear equations, the novelty now being the
inclusion of terms involving derivatives of variables. (If integrals turn up in an equation differentiate the
equation as a whole to obtain only derivatives.) As before the independent equations so obtained are used
to solve for one or another variable. To do so the equations are multiplied by constants and added to
eliminate variables in favor of a specific variable. An equation can be differentiated repeatedly to provide a
variable term in a form suitable for elimination by combination with another equation. The net result of
these manipulations is to obtain linear differential equations with constant coefficients in one variable
having the general form:
The Am are constant coefficients, and the term F(t) on the right represents the combined effect of any
source terms in the circuit. The initial discussion of capacitance and inductance the circuit equations
obtained are of this form. A more involved illustration is provided later.
A linear differential equation with constant coefficients has a unique solution, and a necessary and
sufficient condition for a valid solution is:
a) The proposed solution, on substitution into the equation, leads to an identity, i.e., the proposed
solution 'satisfies’ the equation.
b) The proposed solution must include as many unspecified constants as the order of the
differential equation; these correspond to constants of integration.
By and large a differential equation is 'solved' by guessing at the solution and then verifying that the
existence conditions for the unique solution are satisfied. Of course one need (and should not) not make
the same guess repeatedly; the solution is unique so that a solution established earlier (even if by someone
else) can be reused. A solution is commonly expressed as a superposition of two partial solutions; one
partial solution satisfies the equation with F(t) set to zero, and includes the necessary constants of
integration. This is often called the 'general' solution. The second partial solution, the 'particular' solution,
when substituted in the equation on the left produces an identify with the right side term F(t). The general
solution corresponds to the circuit response to energy stored in capacitor charges and inductor currents,
and is dependent on the specific circuit elements and circuit topology involved. The particular solution is
that part of the circuit behavior 'forced' by sources if any in the circuit.
There are three cases of special importance, which have much broader implications than we deal with in this
course. We consider two of these cases at present, and the third (sinusoidal sources) elsewhere.
The first case considered is that where the circuit contains no sources, i.e. F(t) = 0, so that only the general
solution is involved. If the circuit includes the slightest resistance, an inevitable occurrence in nature, then
current flow involves energy dissipation. From this two general conclusions may be drawn. First, absent
sources the circuit must at the outset have energy stored to support current (and associated dissipation).
This storage takes one or the other (or both) forms, energy initially stored in a capacitor of energy initially
stored in an inductor. The second inference is that since the stored energy is inevitably finite it will
eventually be fully dissipated if the circuit contains resistance (inevitable in practice), and current flow
eventually will cease. This latter conclusion is the basis for the common description of the general solution
to the equation as the ‘transient’ solution. The circuit invariably reaches a steady state sooner or later. The
transient solution is of general significance in evaluating circuit behavior in response to a sudden change.
Odd things, sometimes useful,, sometimes destructive, may happen during the transient interval, and
understanding their nature is important.
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The second case we consider here is that for which F(t) is a constant. Unfortunately the real importance of
this case is not easily presented in an introductory case. Suffice to say that a mathematical procedure
(‘convolution’) exists which enables circuit behavior to more or less arbitrary F(t) to be determined in
terms of behavior with F(t) constant. The innermost secrets of general circuit behavior are contained in the
special case, and can be revealed by application of appropriate methods. Relatively simple illustrations of
this are provided subsequently; more advanced study is needed for a fuller appreciation of convolution.
The solution of the case F(t) = constant is developed in two steps. First the general (transient) solution is
obtained as it would be for F(t) = 0. Then the ‘particular’ solution is obtained; this is the part of the
solution that is ‘particular’ to the specific F(t) involved. The ‘complete’ solution is obtained by adding the
transient to the particular solution.
The third case, the one deferred for latter consideration, involves sources with sinusoidally varying voltages
and currents.
Source-Free Circuits; Transient Solution
The first case taken up is that where there are no sources in the circuit, so that F(t) = 0. Only the 'natural'
circuit behavior is involved. It follows on the basic of energy conservation that circuit voltages and currents
generally must continually decrease. For if the circuit contains the slightest resistance there is an inevitable
dissipation of the finite energy initially stored in the circuit capacitances and inductors. The linear
differential equation with constant coefficients has been extensively studied. As noted before it has a
unique solution.
In particular the general solution has the form est, where s is a constant whose value is to be determined.
Note the interesting property of the exponential function that differentiation is equivalent to multiplication
by s, i.e. d(est,)/dt = s est.
Substitution into the differential equation leads to the requirement
y ( Amsm + A m-1sm-1 + … + A1s + A 0 ) = 0
where the polynomial in parentheses must be zero to avoid the physically trivial mathematical solution y=0.
There are two cases to consider. Assume for the moment that all the roots of the polynomial are distinct.
Then a solution with the requisite number of undetermined constants is
y = Bm esmt + B m-1 esm-1t + … + B1 es1t + B 0 es0t
where sm, sm-1, …, s1, s0 are the polynomial roots, and Bm, Bm-1 , … ,B1, B0 are constant coefficients
whose value is to be determined. Based on the energy conservation requirement, i.e., the eventual
dissipation of all of the initial energy stored in the circuit, we may anticipate that all the roots will have
negative real parts.
It may happen that the polynomial has repeated roots, i.e., one or more roots of the form (s - sr)r. In the
solution form as given above this would lead to several of the terms condensing into a single term, and the
requirement that the number of unspecified independent constants equal the order of the equation would
not be satisfied. It can be verified that a root of multiplicity r corresponds to a contribution to the solution
of the form
(Br0 + Br1t + B r2t2 + … + Br-1tr-1)esrt
Note that the circuit determines the form of the equation, including the root values. In this respect note that
the energy in a capacitor, for example, depends only on the charge stored at the time the energy is evaluated.
That charge is the cumulative consequence of current to the capacitor in the past, but the details of that
earlier current do not matter. It is only the net charge resident at a given time that determines the stored
energy. A similar observation applies to inductors. Often the constants are determined from 'initial
conditions', meaning simply those conditions at t=0 are specified. Occasionally conditions specified at
t–> ∞ are useful. But in fact independent conditions at any time may be used. After all a constant by its
nature has the same value whenever evaluated.
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One final comment to make before presenting some illustrations. To change the energy stored in a
capacitor, i.e. to change the voltage across the capacitor, a net change in capacitor charge is necessary. But
absent an impractical infinite current a current requires a finite time to transport a finite charge. Hence the
voltage across a capacitor cannot change instantaneously in practice. A similar constraint prevents the
current flowing in an inductor from changing instantaneously, since that would require a infinite rate of
change of voltage.
¶ Source-Free Circuit; Transient Illustration
Hindsight plays an essential role in simplifying the
process of obtaining the transient solution. We need
not actually derive the appropriate differential equation
to solve, but instead deal with relatively much simpler
algebraic equations. The circuit drawn to the right is
used to illustrate the difference in effort. The voltage
pulse is applied long enough (for t ≤ 0) so that the
circuit comes to a steady state, and then at t=0 the
pulse is removed. Note that the circuit must come to a
steady state eventually in order to avoid conflicts with KVL; the capacitors cannot continue to charge
indefinitely since this means a continual increase in voltage across a capacitor. The circuit performance, in
particular the node voltage e(t) is to be determined for t ≥ 0.
Simply for pedagogical illustration, and not in any way to suggest it as the general approach to be adopted
we initially set up a solution by a straightforward application of KVL, KCL, and the constitutive relations of
the circuit elements. (This might be considered an illustration of how not to go about analyzing the circuit.)
The circuit as it is effectively after the source voltage goes to
zero is redrawn to the left. The elements R1 and C1 are
considered each as separate branches, and a node voltage ex is
assigned as shown.
Note that in the steady state at the start of the transient (t ≤ 0)
there is no current through either capacitance since that would
imply a change of voltage with time. Absent current through the
R2 the initial node voltage e(t=0) = 0. However the initial voltage across C1, which must be the same as
just before the source voltage is removed, is not zero; in fact ex(t=0+ ) – e(t=0+ ) = Vs. Hence C2 stores no
energy initially, while C1 stores C1 Vs2/2. All this energy eventually is dissipated and as t->∞ all currents
approach zero.
Formal equations to affect a solution are:
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At this point we ‘guess’ at a solution of the form est substitute into the differential equation to obtain an
algebraic equation, and proceed as described before. We are assured there is a solution, and that the
solution is unique. But even for a relatively simple case such as the one considered the mathematical
manipulations while straightforward are nevertheless contorted. Fortunately there is a much simpler
procedure.
The simplification derives from prior knowledge of important aspects of the circuit. A formal analysis
always leads to a linear differential equation with constant coefficients and a zero driving function. There is
no need actually to make the analysis to support this conclusion. Thus anticipate directly a solution of the
form est. Differentiation and integration of an exponential is a well-known operation, and the volt-ampere
expressions for a capacitor, for example, becomes I = sCV; the volt-ampere expression for an inductor
becomes V = sLI. Thus operations of differentiation and integration are equivalent (for the transient
calculation) to multiplication or division by s.
These expressions are formally identical to Ohm's Law, i.e. current and voltage are linearly related. The
particular proportionality constant is different but this does not affect any of the formal derivations for
various techniques of analyzing resistive circuits studied before. The relationships are not changed because
we substitute a symbol 1/sC or sL in place of R; the significant thing is that the proportionality factor is a
constant. All the resistive circuit procedures developed before may be applied, of course using the
appropriate proportionality constant for each of the circuit elements involved.
The circuit is redrawn to the left. This time the capacitors are
labeled in terms of the proportionality constant simply as a
convenient reminder of the algebraic terminal volt-ampere relation
that will be used in place of the differential relationship. KVL
requires that the voltage drop across the series connection of R1
and sC1 is the sum of the voltage drops across the individual
elements. Hence
E = iR1 + i/sC1
Now write by inspection (after a bit of familiarization) a single node equation
and simplify algebraically to
This is the same quadratic equation as is determined by substitution in the differential equation, but
obtained with considerably less effort. For a specific numerical illustration assume VS = 1v, R1 = R2 =
1Ω, and C1 = C2 = 1F. The two distinct roots of the polynomial expression are readily calculated and the
solution written as
e(t) = Ae-0.382t + Be -2.618t
where A and B are as yet undetermined constants. Note that roots are negative, as they must be to avoid a
conflict with KVL and KC (since otherwise currents and voltages would increase indefinitely). The
undetermined constants A and B correspond to two constants of integration for the second order equation
involved.
Of course to complete the solution we must evaluate A and B. The constants of integration reflect the
accumulated history of the circuit up to the time at which the analysis begins. At that time, i.e., at t=0 the
circuit is in a steady-state condition, i.e., currents and voltages are no longer changing. Hence the current
through either capacitor is zero, and consequently there is no voltage drop across either resistor. Hence the
voltage across C2 is zero, and the voltage across C1 is Vs. Note that capacitor voltages cannot change
instantaneously; a finite current must flow for a finite time to change capacitor charge.
To complete the solution we first require e(t=0) = 0, i.e., A+B =0. For a second independent condition we
can write a loop equation around the R1-C1-R2 loop. Although the voltage across a capacitor cannot
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change instantaneously the current through a capacitor can so change.. Capacitor voltage cannot change
until stored charge has changed, but charge can start to change immediately. Hence the current I through
R1 (into the datum) must be VS/R1 on the basis of the initial capacitor voltage and Ohm’s Law. The
current through R2 is initially zero, since e(t=0) = 0. Therefore
VS/R1 = -C2de/dt
= C2(0.382A + 2.618B)
Hence
A = -B = -0.447VS/R1C2 = -0.447.
Incidentally note again that the current through R2, which initially is zero, changes abruptly to Vs/R1 at t =
0. Such a current spike can be quite large, depending on circuit details, and may be destructive.
Source-Free Transient; Another Illustration
The switch in the circuit drawn to the right is opened at t = 0,
after a steady state condition has been reached. In that steady
state condition there is no voltage across the inductor, but
there is a current through the inductor iL(t=0) = 10/5 = 2a.
The current cannot change instantaneously, and so when the
switch opens this initial current flows through the shunt 5Ω
resistor creating an instantaneous 10v drop across that 5Ω. A
KVL loop equation (for example) requires iL(2s + 5) = 0; this
anticipates a solution of the form est. Hence require s = -2.5, and iL = Be -2.5t. To meet the initial
condition on the current B = 2a.
The voltage v across the 5Ω is 10 e-2.5t.
Yet Another Illustration
In the circuit drawn to the right the switch S is opened at t = 0,
after the circuit has reached an equilibrium state. In steady
state all transients have passed. Hence there is no voltage
across the inductor L, and so no current through R1 (Ohm's
Law). The steady-state current through L is the same as the
current through R, and applying KVL and Ohm's Law
determines to be V/R.
With switch S open the current in L must flow immediately
into R1 since the inductor current cannot change instantaneously.(KCL). This produces a voltage drop
across R1 and so a change in current as the stored inductive energy is dissipated in R1.
Anticipate a solution in the form of the form est, and write a KVL equation about the R1-L loop;
sLi + iR1 =0
-R1t/L
and so require s = -R/L, i = Ke
where K is a constant. To calculate the constant K requires a
knowledge of at least one point on the I vs t curve; the initial condition I(t=0) = V/R will do, Hence
I(t) = (V/R) e-R1t/L.
Similarly
v(t) = Ldi/dt = L (-R1/L)(V/R)( e-R1t/L) = -R1(V/R)( e-R1t/L)
Old Problem (but with significance)
Continually using initial conditions or conditions as t-> ∞ to characterize
circuit energy tends to leave an erroneous impression that there is nothing
else to use. This example is intended to dispel that impression. In this
example we suppose that (by means not shown explicitly) the capacitor
was charged to 5 volts and connected at t=0 to the combination of a 2Ω
and 3Ω resistors. However the switch shown remains closed until t=1
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second, at which time it opens. In other words the circuit electrical configuration changes at t=1 second.
The circuit may be analyzed separately in the two intervals, 0 ≤ t ≤ 1, and t ≥ 1. For 0 ≤ t ≤ 1 the circuit is
simply a series combination of a 1F capacitor and a 2Ω resistor, and it is not difficult to determine
i(t) = 2.5e-2t, and v(t) = 5e-2t.
For t ≥ 1 determining the form of the solution also is not difficult. The switch is open, and the circuit
becomes equivalent to a series combination of a 1F capacitor and a 2 + 3 = 5Ω resistor. Hence i(t) = Ae-5t
where the constants of integration A and B are to be determined. What is needed is some information
which relates the circuit state just before the switch opens to the state just after the switch opens. There is
nothing really new here; the voltage across the capacitor cannot change instantaneously, and so
v(t =1 ) = v(t =1+ ). Now v(t=1) = 5e-2, and v(=1+) = Be–5.. Hence v(t) = 5e-2e-5(t-1) for t ≥ 1. From
Ohm's law we have i(t) = e-2e-5t for t ≥ 1. It is convenient to use t -1 in the exponent both as a reminder
that this part of the solution is valid only for t ≥ 1 and because t-1=0 at this starting point. In this way it is
easier to see that that the solution is the expected exponential decay starting from t' = t - 1 = 0.
Incidentally note that i(t=1-) ≠ i(t=1+ ), i.e., the current changes abruptly. Since there is no inductor in this
circuit there is no theoretical problem.
Stored Magnetic Energy ‘Gotcha’
In the inductor circuit shown the steady-state condition with the switch closed is with a
current E/R2 flowing through the inductor, no current flows through R1 since the
inductor voltage drop is zero. The voltage at node 'a' then is E. Suppose the switch is
opened at t = 0; this reduces the current through R2 to zero. Since the inductor
current must remain unchanged initially that current immediately flows through R1,
raising the voltage at 'a' to E + IR1 = E + ER1/R2 immediately. Thereafter the voltage
at 'a' decays exponentially back to E with tile constant L/R1. The larger R1 the faster
the decay, but also the higher the voltage 'spike' at 'a'. Suddenly opening an inductive
circuit carrying current can be hazardous; the stored magnetic energy does not
dissipate spontaneously.
Capacitor Current ‘Spike’
Initially one capacitor in the circuit shown is charged to an initial voltage of Vo volts; the other is fully
discharged. The discharged capacitor stores no energy of course. The other
capacitor stores an amount of energy equal to CVo2/2. When the switch
closes the initially discharged capacitor charges at the expense of the other
capacitor. Eventually a steady state is reached in which each capacitor is
charged to a voltage V/2. (No charge is lost, and symmetry dictates the two
capacitors have equal charges, i.e., half the charge each.). The energy stored
in each capacitor is therefore C (Vo/2)2/2, so that the energy remaining in the circuit is CV2/4. Since
energy is conserved current flowing in the resistor must have dissipated the difference between the initial
and final stored energy,
Calculate the current I for t ≥ 0 as
And then show
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############################Series Resonant Circuit (important special case)
The transient response (general solution) of an RC or an RL circuit is in general a sum of exponential
terms with a negative real coefficient in the exponent. This is a reflection of energy conservation- a
resistance always consumes energy and the source-free circuit has only the finite initial energy stored in a
capacitor or in an inductor with which to maintain a current. Hence eventually circuit activity must end.
In general the same conclusion may be drawn for the transient behavior of an RLC circuit, i.e., when both
inductive and capacitive energy storage is available. However some interesting new behavior is possible,
under appropriate conditions, involving the transfer of energy from one type of storage to the other. For
example capacitive energy can support a current flow, although of course the capacitive energy stored (as
measured by the capacitor voltage) decreases in doing so. But this current can be associated with inductive
energy storage, and it can in turn support a return of energy to the capacitor. This energy exchange is the
novel behavior examined in the circuit illustrated below, left. .
As usual we anticipate the exponential form of the particular (transient)
solution est, and use this information to transform the differential voltampere relations to algebraic expressions. A loop equation provides
I(R + sL + 1/sC) = 0, and for a non-trivial solution the coefficient of I
must be zero: s2 + 2αs + ωo2 = 0. Here ωo2 = 1/LC and = 2α = R/L.
The roots of this quadratic are
There are three cases to consider. If the discriminant of the quadratic is positive both roots are real but
unequal, and the circuit voltages and currents decay monotonically; this is called the 'overdamped' case.
The 'underdamped' case is that for which the discriminant is negative, and the roots are a complex conjugate
pair with equal negative real parts. The imaginary part of the root then corresponds to a sinusoidal
variation, although the negative real part assures that current flow eventually decays to zero. The third case
is the 'critically damped' case and is so to speak the boundary between the other two cases.
For an illustrative calculation we take L= 1H, C= 1F, and assume an initial capacitor voltage of 1 volt and
zero loop current. Then ωo = 1. To produce the overdamped case we take R=4Ω, corresponding to
α=2 > ωo. The roots then are -2 ± √3, and the solution for (say) the voltage across the capacitor is
The coefficients are calculated from the two initial conditions Vcap = 0, and d(Vcap)/dt =loop current = 0.
A computer computation is compared to the theoretical calculation below.
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To illustrate the underdamped case we use R = 1Ω. Then α = 1/2 and the roots of the quadratic are
(1± j√3)/2. Hence
where Euler's theorem is used to convert the imaginary exponentials a more convenient trigonometric form.
The initial conditions again are used to evaluate the constants of integration, and a PSpice computation is
compared to the theoretical calculation below.
Finally for the critically damped case we make R = 2Ω, so that there is a double root s = -1. The
solution in this case takes the special form (A+Bt) exp-t, and application of the initial conditions finds
vcap = (1+t)e -t. The computed and theoretical plots are compared below.
For ease of comparison all three cases are re-plotted below on the same axes.
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There is one special case of some interest, and that is for a theoretically lossless circuit, i.e., R=0Ω. In this
case the initial energy of the circuit is not diminished by current flow, but simply is exchanged continually
between the inductor and capacitor. (A voltage across the capacitor also is across the inductor, requiring a
current change. The capacitor discharges in increasing the current; the current increases at a decreasing rate
eventually reaching a maximum when the capacitor is fully discharged. The current then decreases as it
recharges the capacitor.) The roots in this case are pure imaginary, and the circuit voltages and currents are
sinusoidal. For the initial conditions as specified and ωo = 1, Vcap = cos(t). Note: The loop current and
the capacitor voltage actually are computed using R = 1µΩ since PSpice as most such programs require a
non-zero resistance. The period of the sinusoid (see plot below) is 2π as expected. Note also that the
stored capacitor energy (CV2/2) is an extremum when the stored inductive energy (LI2/2)is zero, and viceversa. While a lossless LC circuit is a theoretical abstraction a very practical equivalent is one in which the
inevitable losses are continually replaced by appropriate means; this is an advanced topic for another place
and time.
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¶ Fixed-Source Circuit; Transient Solution
In the case now considered fixed (DC) sources are assumed to be among the circuit elements, and so the
right side of the circuit differential equation may be a constant (including zero), i.e., F(t) = K. The complete
solution then includes in addition to the general solution discussed above an added part, the particular
solution. The particular solution is simply y = K/A0 (see above for the differential equation). The
complete solution then is this particular solution added to the general solution. It is often easy to determine
the particular solution for a constant-source circuit directly on noting that it corresponds to the circuit
performance as t -> ∞, i.e., when the transient particular solution has dissipated all the initial stored energy.
In the steady state all derivatives with respect to time are zero as t -> ∞; there is then no current through a
capacitor, nor is there voltage across an inductor.
The solution procedure is easier to illustrate than to describe.
Consider the circuit drawn to the right, with R1 = R2 = 1Ω, C1 =
C2 = 1F, and Vs = 1 volt. Suppose the capacitors are initially
uncharged. The general solution is determined with the source
strength VS set to zero, assuming the exponential form of the
solution.. A node equation (for example) obtains
Note the description of the 'resistance' of the series combination of R1 and C1 as the sum of R1 and 1/sC1;
the branch current is the node voltage divided by this 'resistance'. The roots of the polynomial expression
are readily determined and the general solution written as
Ae-0.382t + Be -2.618t
The particular solution (with VS = 1v) corresponds to d/dt = 0 (i.e. t -> ∞). There is no current through
either capacitor, the current in R2 must then be zero (KCL), and therefore the voltage across C2 in
particular is zero. The voltage across C1 is 1 volt (KVL).
The solution for, say the voltage across C2, has the form
VC2(t) = Ae-0.382t + Be -2.618t + 0
This expression 'satisfies' the circuit equation and has the necessary two undetermined constants. The '0'
term is included here explicitly simply to emphasize that the particular solution is accounted for.
Two independent conditions characterizing the circuit are needed to determine the two constants of
integration A and B. Since the capacitor is (specified to be) initially uncharged require VC2(t=0) = 0 as
one condition. For the other condition we may use the fact that the initial current through C2 is 1A. This
follows on noting that VC1(t=0) = 0, for then KVL requires a 1 volt drop across the R1 and Ohm's Law
requires the 1A current. Since the voltage across R2 is zero initially it conducts no current, and KCL
requires the 1A to flow through C2. Apply these conditions to determine A and B:
VC2(t) = 0.4472(e-0.382t - e-2.618t )
Similarly the solution for, say, the voltage across the C1 has the form
VC1(t) = Ce-0.382t + De-2.618t +1
with the initial conditions VC1(t=0) = 0, and IC1(t=0) = 1A. Note that the voltage across C1 as t -> ∞ is 1
volt; this is the particular solution for this voltage. Evaluate the constants of integration to find
VC1(t) = -0.7236e-0.382t - 0. 2769e-2.618t +1
Finally the current through C1= 1F is found simply by differentiating VC1(t):
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IC1(t) = 0.2764e-0.382t - 0. 7236e-2.618t
Computer computations (PSpice) are compared to the theoretical calculations below.
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Simplified Convolution Illustration
It was suggested above that the special case of a fixed-source particular solution actually has general
applicability. The particular solution embodies information about the circuit to which it applies, and this
can be used to evaluate response when other types of sources are present. Although a detailed discussion
of the general methodology involved is outside the scope of the course it is instructive to consider two
special cases by way of illustrating the ideas involved.
In the single time-constant circuit to the right, for example,
the driving source is a voltage pulse; the objective is as
usual to determine the branch voltages and currents, say
Vc(t) to be specific. To do so we apply superposition. We
suppose the pulse voltage source to be formed by two
voltage sources in series, one providing a step function
voltage rise at t=0 of amplitude Vo and the other a step
function voltage drop of -Vo at t=To. The combined effect is that of the actual pulse source. Since the
circuit is a linear one we may consider each source separately, combining the separate contributions to, say,
Vc(t) to obtain the solution for the pulse source.
The solution for the t=0 step input is V1c(t) = Vo(1-e-t/RC). Obtaining this result is left(actually done
earlier) as an exercise, although it is easy to verify that KVL, KCL, and the branch volt-ampere relations are
satisfied. Moreover Vc(t=0) = 0, the presumed initial state of the capacitor. In addition the initial capacitor
current is CdVc/dt is Vo/R as required. This establishes the expression as being the unique solution for the
initial step.
We do not need to make a major effort to obtain the solution for the negative step, but it is necessary to be
careful in how it is expressed. Circuit response to a step input relative to the time the step is applied is
formally the same whenever the step is applied. The negative step here is applied at t=To; branch voltages
and currents (for the partial contribution of the negative step) are zero for t < To. Thus write
V2c(t) = –Vo( 1-e-(t-To)/RC) valid for t ≥ To. The time origin remains at t=0; the exponent for the
negative step is zero at t=To.
The total solution is written in two parts:
For 0 ≤ t < To
Vc(t) =V1c(t) = Vo( 1-e-t/RC)
For To ≤ t
Vc(t) =V1c(t) + V2c(t)
= Vo (1-e-t/RC) –Vo (1-e-(t-To)/RC)
= Vo (1-e-To/RC)e-(t-To)/RC = constant e-(t-To)/RC
The last form can be interpreted usefully in the following manner. The constant coefficient of the
expression is the voltage across the capacitor at t=To; this is seen from the expression valid for t ≤ To and
is the result of charging of the capacitance by the initial step function.. At t=To the driving voltage is
removed. For what happens next (t ≥ To) we need not be concerned with earlier times; as far as the
capacitor is concerned its entire earlier experience is summarized in the capacitor charge (or voltage) at
t=To. Hence we can analyze the circuit for t ≥ To as a source-free circuit with the initial capacitor voltage
as given at t =To. This is a simple exponential decay, except that it should be noted that time is to be
measured from an origin t' = t -To.
A PSPICE computer calculation of the capacitor voltage follows for several normalized values of the pulse
period. Note the voltage rises exponentially starting from t=0 towards the (normalized) pulse amplitude as
expected. On the trailing edge of the pulse the voltage decays exponentially from whatever voltage it
reached at that time.
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The (normalized) current through the resistor is plotted next. For the initial step the current decays
exponentially from its initial value as the capacitor charges. On the trailing edge of the pulse a current
equal to the initial current but of the opposite polarity is generated, causing the resistor current to reverse
and discharge the capacitor. Note that the capacitor voltage does not change at t = To. Hence if the source
voltage abruptly drops by Vo, the resistor voltage IR must drop by an equal amount, in accord with KVL
In more advanced courses this idea of superposition will be developed so as to enable expressing the
response to any (mathematically non-pathological) source function as an integral (superposition) of the
step function response. But, just for fun of a sort, consider the following more involved illustration.
Suppose R=1Ω and C=1F in the preceding circuit, and instead of a pulse the source function is 'ramp', i.e.,
V(t) = 0 for t ≤ 0, V(t) = t for 0 ≤ t ≤ 1, and V(t) = 1 for t ≥ 1. This function has a derivative which is 0 for
t ≤ 0, 1 for 0 ≤ t ≤ 1, and 0 for t > 1; in other words the derivative is a pulse. Since the circuit is linear any
linear operation in time can be performed on the circuit variables (provided it is performed on all of them of
course) and the results correspond to the sources having the same operation performed on them.
Integration is a linear operation. Hence take the pulse response computed above (particularized for
simplicity to R=1Ω, C = 1F, To = 1 second, Vo = 1 volt) and integrate it to obtain the ramp response!
The pulse solution (particularized) is:
For 0 ≤ t < 1
Vc(t) = 1-e-t
For 1 ≤ t
Vc(t) = (1-e-1)e-(t-1)
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and integrating to get the ramp solution:
Note that the integration always is from 0 to t, and for 1≤ t this involves a different integrand from 0 to 1
than from 1 to t.
The PSPICE plot below compares the computed capacitor voltage against the theoretical calculations.
Consider This (but just for more fun)
The results of the computer computation for the previous discussion are repeated below as V(ramp) and
V(r–out). For an approximate computation a superposition of four step functions is used (V(step). The
computed response of the circuit is plotted as V(s-out). The point here is not the fairly good agreement of
the approximation with the actual ramp response. Rather it is note that the convolution process mentioned
before consists basically of approximating a voltage (or current) as a superposition of step functions. Of
course the approximation becomes better if more steps each with smaller amplitude are used for the
approximation. Ultimately differential steps are used, and instead of s sum of pulse response it is an
integral that is used. It is this integral procedure that is ‘convolution’. While we do not consider
convolution further here it is useful to emphasize that circuit response to a constant source has a wider
application than might be anticipated.
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M H Miller
Series Resonant Circuit (important special case)
The transient response (general solution) of an RC or an RL circuit is in general a sum of exponential
terms with a negative real coefficient in the exponent. This is a reflection of energy conservation- a
resistance always consumes energy and the source-free circuit has only the finite initial energy stored in a
capacitor or in an inductor with which to maintain a current. Hence eventually circuit activity must end.
In general the same conclusion may be drawn for the transient behavior of an RLC circuit, i.e., when both
inductive and capacitive energy storage is available. However some interesting new behavior is possible,
under appropriate conditions, involving the transfer of energy from one type of storage to the other. For
example capacitive energy can support a current flow, although of course the capacitive energy stored (as
measured by the capacitor voltage) decreases in doing so. But this current can be associated with inductive
energy storage, and it can in turn support a return of energy to the capacitor. This energy exchange is the
novel behavior examined in the circuit illustrated below, left. .
As usual we anticipate the exponential form of the particular (transient)
solution est, and use this information to transform the differential voltampere relations to algebraic expressions. A loop equation provides
I(R + sL + 1/sC) = 0, and for a non-trivial solution the coefficient of I
must be zero; s2 + 2αs + ωo2 = 0. Here ωo2 = 1/LC and = 2α = R/L.
The roots of this quadratic are
There are three cases to consider. If the discriminant of the quadratic is positive both roots are real and
unequal, and the circuit voltages and currents decay monotonically; this is called the 'overdamped' case.
The 'underdamped' case is that for which the discriminant is negative, and the roots are a complex conjugate
pair with equal negative real parts. The imaginary part of the root then corresponds to a sinusoidal
variation, although the negative real part assures that current flow eventually decays to zero. The third case
is the 'critically damped' case and is so to speak the boundary between the other two cases.
For an illustrative calculation we take L= 1H, C= 1F, and assume an initial capacitor voltage of 1 volt and
zero loop current. Then ωo = 1. To produce the overdamped case we take R=4Ω, corresponding to
α=2 > ωo. The roots then are -2 ± √3, and the solution for (say) the voltage across the capacitor is
The coefficients are calculated from the two initial conditions Vcap = 0, and d(Vcap)/dt =loop current = 0.
A computer computation is compared to the theoretical calculation below.
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To illustrate the underdamped case we use R = 1Ω. Then α = 1/2 and the roots of the quadratic are
(1± j√3)/2. Hence
where Euler's theorem is used to convert the imaginary exponentials a more convenient trigonometric form.
The initial conditions again are used to evaluate the constants of integration, and a PSPICE computation is
compared to the theoretical calculation below.
Finally for the critically damped case we make R = 2Ω, so that there is a double root s = -1. The
solution in this case takes the special form (A+Bt) exp-t, and application of the initial conditions finds
vcap = (1+t)e -t. The computed and theoretical plots are compared below.
For ease of comparison all three cases are re-plotted below on the same axes.
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There is one special case of some interest, and that is for a theoretically lossless circuit, i.e., R=0Ω. In this
case the initial energy of the circuit is not diminished by current flow, but simply is exchanged continually
between the inductor and capacitor. (A voltage across the capacitor also is across the inductor, requiring a
current change. The capacitor discharges in increasing the current; the current increases at a decreasing rate
eventually reaching a maximum when the capacitor is fully discharged. The current then decreases as it
recharges the capacitor.) The roots in this case are pure imaginary, and the circuit voltages and currents are
sinusoidal. For the initial conditions as specified and ωo = 1 , Vcap = cos(t). Note: The loop current and
the capacitor voltage as computed using R = 1µΩ since PSpice as most such programs require a non-zero
resistance. The period of the sinusoid is 2π as expected in the plot below. Note also that the stored
capacitor energy (CV2/2) is an extremum when the stored inductive energy (LI2/2)is zero, and vice-versa.
While a lossless LC circuit is a theoretical abstraction a very practical equivalent is one in which the
inevitable losses are replaced by appropriate means; this is an advanced topic for another place and time.
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