Chapter 10: First-Order RL Circuits – Electric Circuit Analysis

Chapter 10

First-Order RL Circuits

CHAPTER OBJECTIVES
  • To develop the differential equations for series and parallel RL circuits
  • Initial condition, its need and interpretation
  • Complementary function, particular integral and their interpretation
  • Natural response, transient response and forced response in an RL circuit
  • Interpretation of various response components
  • Nature and details of step response of RL circuit and time-domain specifications based on it
  • Time constant and various interpretations for it
  • Steady-state response versus forced response and various kinds of steady-state response
  • Zero-input response and zero-state response and their interpretation
  • Linearity and superposition principle as applied to various response components
  • Impulse response and its importance
  • Equivalence between impulse forcing function and non-zero initial condition
  • Zero-state response for other inputs from impulse response
  • Relations between δ(t), u(t) and r(t) and corresponding responses
  • Zero-state response of RL circuit for exponential and sinusoidal inputs
  • Frequency response of RL circuit
  • General analysis procedure for single time constant RL circuits
INTRODUCTION

Energy storage elements – inductance and capacitance – lend memory to the electrical circuits and make interesting dynamic behaviour in them possible. The simplest of such dynamic circuits will contain one such energy storage element along with passive resistors and linear dependent sources. Further simplification is possible by ruling out dependent sources and limiting the resistors to one. We consider one such two-element circuit containing one energy storage element in detail in this chapter. Specifically, we study series and parallel RL circuits with voltage source and current source excitations of different waveshape with and without initial energy storage.

These circuits are simple in structure. However, that does not imply that their dynamic behaviour is trivial. The simple nature of these circuits does not imply that they are unimportant. In fact, simple RL and RC circuits find applications in almost all areas of Electrical and Electronics Engineering. Moreover, they display almost all the important features of dynamics displayed by the general class of ‘Linear Time-Invariant (LTI) Circuits’. They form very simple and effective circuit examples to convey important concepts in the analysis of LTI circuits.

Circuits containing a large number of inductors, capacitors and dependent sources will require simulation tools for quantitative analysis. However, a thorough understanding of the behaviour of simple circuits like the ones covered in this chapter will help the engineer develop qualitative insight into such circuits and arrive at approximate quantitative results. No complex and costly circuit simulation package is a satisfactory substitute for this kind of analytic capability in an engineer.

We use simple RL circuits to bring out many basic concepts in linear dynamic circuits in this chapter.

10.1 THE SERIES RL CIRCUIT

The series RL circuit with all circuit variables identified as per passive sign convention is shown in Fig. 10.1-1. Unit step voltage applied to the circuit appears in the same figure. The circuit contains three two-terminal elements. Each two-terminal element is associated with two variables – a voltage variable and a current variable. We use passive sign convention and in passive sign convention the positive current enters the element at the positive of voltage variable assigned to the element. Thus, vL and iL are the voltage and current variables of the inductor as per passive sign convention. The corresponding variables for the remaining two elements are shown in Fig. 10.1-1. All these six variables are functions of time though the dependence on time is not explicitly marked.

Fig. 10.1-1 (a) The series RL circuit (b) Unit step voltage waveform

10.1.1 The Series RL Circuit Equations

The analysis of this circuit aims at solving for all these six variables as functions of time. The circuit has three nodes. The three elements are in series and they will have a common current through them and hence the following equation is valid at all t:

 

iL(t) = iR(t) = – iS(t)

 

The circuit has one mesh and it has to satisfy KVL at all t in this mesh. Applying KVL in this mesh will give us an equation tying up the three voltage variables – vS, vR and vL – and thereby reducing the number of independent variables among the three to two. The relevant KVL equation is

 

Thus, we have eliminated vR by expressing it as the difference between the other two voltage variables vS and vL. vS is a source function – it is the time function representing the voltage waveform of an independent source and hence it is a known function. Therefore, there is only one voltage variable and we choose that to be vL.

Now, we have two variables – vL and iL. We also have the element equations of inductor and resistor to tie them up and get a single equation in iL as follows.

Thus, a first-order linear differential equation with constant coefficients describes the behaviour of current in a series RL circuit for all time t. The order of a differential equation is the degree of highest derivative appearing in the equation and it is one in this equation. The differential equation is linear since all derivatives of the unknown function iL and the function iL itself appear in the equation with an exponent of unity, the three coefficients in the three terms in the equation are not functions of iL and the forcing function on the right-hand side of the equation is a function of time only and is not dependent on the unknown function iL in any way. R and L are circuit parameters that are assumed to be constants. Therefore, the coefficients of differential equation turn out to be constants.

10.1.2 Need for Initial Condition Specification

The differential equation derived above describes the variation of the inductor current for all t in the time-domain. It is so because we have used KCL and KVL in deriving it and these two laws hold at all time instants. Does it mean that we will be able to solve for iL(t) for all t – from infinite past to infinite future? Yes, provided we know the function vS(t) for the entire time range –∞ < t < ∞. That is going to be a problem!

Do we really have to get back as far as –∞ in the time axis? Not really. We need to get back to the time instant at which our inductor was manufactured. It is indeed difficult to manufacture an inductor such that it takes birth with some initial energy trapped in it! Therefore, we may safely assume that the inductor had zero initial energy and hence zero initial current (energy storage in an inductor is proportional to square of current flowing through it) when it was manufactured. In fact, when we talk of infinite past, we have this time instant in mind. Therefore, we can be sure that the inductor had zero current in it in the infinite past i.e., at t = –∞. Now, if we know vS(t) from that instant onwards up to the present instant, we will be able to solve for iL(t) from t = –∞ to the present instant by integrating the governing differential equation.But didn’t we assume that the inductor was connected along with the resistor and source right at the instant it was manufactured and it remained connected so from then on? We cannot assume any such thing. Therefore, we must know whatever that happened to the inductor from the instant it was manufactured to the instant it was connected in this circuit we are trying to solve, if we are to solve this circuit at all. This is due to the fact that the function vS (t) can at best be known only from the instant at which this circuit came into existence and the inductor may have been subjected to various voltages in various other circuits before this circuit was wired up. In that case, the inductor will be carrying a current

where –∞ refers to the instant of manufacturing of the inductor and tcreate refers to the instant at which the RL circuit under discussion came into being. The voltage vL(t) in the above integral refers to all the voltage that was applied to the inductor during this time interval. Thus, the inductor carries its accumulated past in the form of an initial current I0 given by the above integral when it enters the RL circuit we are trying to analyse. Notice that the voltage appearing in the integrand has no relation with the source that is applied subsequently to the RL circuit.

Briefly, we need to know the past of inductor. However, fortunately we need not know everything about its past – we need only the value of the above integral. We will be able to solve for iL(t) for all instants after tcreate if we know the value of I0 along with vS(t) from the instant it was applied, i.e., from tcreate. Obviously, we need to know everything about its past if we want to solve for iL(t) from t = –∞ onwards. That is too much of a past to carry. Therefore, we would want to solve for iL(t) only for ttcreate usually. We need the value of I0, the initial current in the inductor at an instant just prior to tcreate for that. This single number condenses all the past history of inductor as far as the effect of voltages applied to it in the past on the evolution of its current in future is concerned. This number is called the initial condition for the inductor.

The time instant tcreate is to be understood as the instant from which we know the data required for solving the differential equation. That is the time instant at which the initial condition is specified for the inductor and it is the time instant from which we have complete knowledge about the input source voltage. Moreover, it is the time instant from which the inductor should act as an element in this circuit and only in this circuit. It is customary in Circuit Theory to set this instant as the time-zero instant, i.e., tcreate = 0 unless there is some specific reason for making it different.

Usually, some kind of switching action takes place in the circuit at this instant. It could be a switching that applies a specific voltage waveform at its input. Or, it could be a switching operation which changes the structure of the circuit – for example, one element in the circuit may have been kept shorted by closing a switch across it and now at t = 0 that switch is opened. Such switching action usually brings in jump discontinuities in circuit variables. Jump discontinuities in variables involved in differential equations are difficult to handle mathematically unless singularity functions are brought in. We do not want to do that in this book. Our way of handling such discontinuities in circuit variables will be circuit-theoretic and we need to define two more time instants to facilitate our circuit-theoretic reasoning in such situations. These two time instants are t = 0 and t = 0 +. t = 0 is a time instant which is to the left of t = 0 in the time axis. However, the time interval [0, 0] is of infinitesimal width, i.e., 0 is arbitrarily close to 0, but always less than 0. Similarly, 0 + is on the right of 0 and is arbitrarily close to 0. Thus, 0 < 0 < 0 + while 0 – 0 ≈ 0 and 0 + – 0 ≈ 0.

10.1.3 Sufficiency of Initial Condition

We defined three important time instants – t = 0, 0 and 0 +. We also concluded that, with the initial condition for iL specified and with the knowledge of vS from the instant at which the initial condition is specified, we will be able to solve the differential equation for iL. But, which instant among the three instants above should be used for specifying initial condition?

The three time instants have been defined only to make it easy to handle possible discontinuities in the input function vS. If the source function vS is a continuous function of t we do not need t = 0 and t = 0 +. In that case, initial condition is specified at t = 0. However, if vS has a discontinuity or singularity (for example, a jump discontinuity as in step function or an impulse located at t = 0) at t = 0, it is possible that either iL or its derivative will undergo step changes at t = 0. Therefore, the time instant at which the initial condition is applicable must be clearly specified. That instant has to be t = 0. Since it will not make any difference in the case where vS is a continuous function, we will stipulate that initial condition be always be specified at t = 0.

The initial current in the inductor at t = 0+ can be different from its value specified at t = 0for certain kinds of input functions. We will illustrate how the initial condition value at 0 + can be calculated from initial condition value at 0 and nature of vS later in this chapter. We have to address another issue at present.

We now assert that iL(t) for t ≥ 0+ can be obtained if initial condition I0 at t = 0 is known and vS(t) for t ≥ 0+ is known. This is equivalent to asserting that the net effect of all the voltage applied across the inductor in the time range –∞ < t ≤ 0 on the evolution of its current in the time range t ≥ 0+ is encoded in a single number I0, the initial condition at t = 0. This in effect says that the past which the inductor remembers is contained in I0.

They are strong assertions and they need to be proved. We offer a plausibility reasoning to convince ourselves that these assertions are true. We assume that the function vS(t) is continuous at t = 0 in the reasoning that follows.

We recast the circuit equations of the series RL circuit in Fig. 10.1-1 in the following manner:

Further, we rewrite the last equation by splitting the range of integration into two sub-ranges as follows. Notice the change in time-range of applicability.

We recognise the second term on the right-hand side as (R/L)I0 where I0 is the initial condition for the inductor.Therefore, we write

It is obvious from Eqn. 10.1-1 that the solution for vL(t) for t ≥ 0 will depend only on I0 and the values of vS for t ≥ 0. A simple method to integrate the equation numerically is outlined below.

Divide the time interval [0, t] into many small intervals, each of width ∆t. Let N be the number of such intervals. Then

gives N + 1 values of vL(t) in the interval [0, t] at equally spaced sub-intervals. The accuracy of calculation can be improved by increasing the number of sub intervals (i.e., by decreasing ∆t). iL(t) can be found by evaluating the first derivative of vL(t) and multiplying it by L once vL(t) is calculated with sufficient accuracy.

Thus, I0 specified at some time point and vS(t) from that time point will be sufficient data needed to solve for inductor current in an RL circuit from that time point onwards.

10.2 SERIES RL CIRCUIT WITH UNIT STEP INPUT – QUALITATIVE ANALYSIS

We study the behaviour of inductor current in a series RL circuit excited by a unit step voltage source qualitatively in this section. The initial current in the inductor at t = 0 is assumed to be zero for the purpose of this analysis. The relevant circuit with the manner in which the unit step voltage is realised in practice shown clearly appears in Fig. 10.2-1.

Unit step voltage is defined the following manner. It has a step discontinuity at t = 0 and strictly speaking the time instant t = 0 is excluded from the domain of the function. However, it is customary in Circuit Theory to represent this function with a solid line jumping from 0 to 1 at t = 0 in its plot as shown in Fig. 10.1-1

The switch S2 was closed at t = –∞ and is opened at t = 0 and remains open up to t = + ∞. The switch S1 is open from t = –∞, is closed at t = 0 and remains closed up to t = + ∞. We assume that the switches open and close in zero time intervals. The switch S2 applies zero V to the circuit for all t 0 and the switch S1 applies the 1V available from the DC battery for all t ≥ 0 +, thereby effecting unit step voltage application to the circuit.

Fig. 10.2-1 Series RL circuit with unit step voltage input

10.2.1 From t = 0 to t = 0 +

The voltage applied to inductor in the interval t ∈ (–∞, 0] is zero. Hence, the current in the inductor at t = 0 is 0 A. The voltage applied by the unit step input source has a jump discontinuity from 0V to 1V at t = 0. Can this jump discontinuity result in a change in inductor current at t = 0 + ? We relate the possible change in inductor current over the time interval [0, 0 + ] to the voltage across inductor during that interval by employing the element equation of inductor.

The resistor and the inductor always share the input voltage. Thus, the resistor too may absorb part of the input voltage during the interval [0, 0 + ] if there is a current in the circuit. Therefore, the voltage available to the inductor during this interval can only be less than what is available in the source. More importantly, this voltage can assume only finite (though not determinate in view of the discontinuity in the input function) values in this interval. Further, the interval is infinitesimal in width. The area under a finite-valued function over an infinitesimal interval is zero. Therefore, iL(0) = iL(0+) i.e., the inductor current is continuous across the interval [0,0 +]. It takes an impulse function at t = 0 to generate finite area over infinitesimal time interval. Thus, we conclude that, the current in an inductor at t = 0 and t = 0+ in any circuit will be same unless the circuit can support impulse voltage across that inductor at t = 0.

There is a difference between applying impulse voltage and supporting impulse voltage. If we connect an independent voltage source with its source function = δ(t) to a circuit, then, we are applying an impulse voltage to the circuit. However, suppose we are connecting an independent current source which suddenly changes its current from 0A to 1A at t = 0 to an inductor. Now, the inductor has to change its current from 0A to 1A instantaneously at t = 0 and it will produce a back e.m.f of (t) V across it in that process. The current source has to absorb that voltage in order to satisfy Kirchhoff’s Voltage Law. But ideal current sources do not complain even if they are called upon to support infinite voltage. Hence, in this instance, the circuit supports impulse voltage, but does not apply it.

10.2.2 Inductor Current Growth Process

The initial current in the inductor at t = 0 is zero in the circuit under consideration. This implies that the initial energy storage in the inductor is zero. This state of zero initial energy is also indicated often by phrases ‘initially at rest’ and ‘initially relaxed’. The input voltage is a unit step function and it remains within finite limits at all instants. Thus, the current in the inductor cannot change over the small time interval between t = 0 and t = 0 +. Therefore, current in the circuit at t = 0 + is zero.

However, the input voltage which was at 0 V at t = 0 has become 1V at t = 0 + while the current remained at zero value. The resistor can absorb only 0V with a zero current flowing through it. This implies that the voltage across inductor will undergo a sudden jump at t = 0 from 0 V to 1V. In fact, all sudden changes in the input voltage of a series RL circuit will have to appear across the inductor i.e., the resistor will refuse to have anything to do with the sudden jumps in input voltage and will dump all such jumps onto the inductor. This is so because if the resistor absorbs even a small portion of the jump in input voltage, the circuit current too will have to have a jump discontinuity. But the inductor does not allow that unless the circuit can apply or support impulse voltage. Therefore, the inductor insists that the current through it remain continuous (unless impulse voltage can be supported) and it is willing to pay the price for that by absorbing the discontinuities in the input source voltage. Since the inductor will maintain the current continuous, the voltage across the resistor too will remain continuous. In fact, this is one of the applications of RL circuit – to make the voltage across a load resistance smooth when the input voltage is choppy.

The current in an inductor is related to voltage across it through the relation

This shows that, with a voltage of 1V at t = 0 + across it, the current in the inductor cannot remain at its current value of zero forever. The current has to grow since its derivative is + ve and hence it starts growing at the rate of 1/L A/s initially. This in turn implies that the slope of current Vs. time curve at t = 0 + will be 1/L A/s.

However, as the current in the inductor grows under the compulsion of voltage appearing across it, the resistor starts absorbing voltage. This results in a decrease in the inductor voltage since vR and vL will add up to 1V by KVL at all t ≥0 +. Decrease in vL results in a decreasing rate of change of iL since rate of change of iL is directly proportional to vL. This means that the tangent drawn on the iL vs. t curve will start with a slope of 1/L A/s at t = 0 + and will bend progressively towards the time axis as t increases. The current in the inductor which starts out with a certain initial momentum loses its momentum with time and grows at slower and slower rates as time increases due to the resistor robbing an increasing portion of source voltage from the inductor. Therefore, we expect the current function to be of convex shape.

Will the growth process ever end and will the inductor current reach a steady value i.e., a constant value? Let us assume that it does so. Then, what is the value of current that can remain constant in the circuit? If the current is constant, the inductor will demand only zero voltage for allowing that current. That will mean that all the source voltage, i.e., 1V will have to be absorbed by the resistor. The resistor will do that only if the current through it is 1/R A. Therefore, if the circuit can reach the end of growth process and become steady, it will do so only at this unique value of current – 1/R A. Thus, iL = 1/R A is a possible steady end to the growth process we described above. However, detailed mathematical solution will tell us that inductor current never attains this value; but approaches this value asymptotically with time.

10.3 STEP RESPONSE OF RL CIRCUIT BY SOLVING DIFFERENTIAL EQUATION

We have already derived the first-order linear differential equation with constant coefficients that describe the behaviour of inductor current in a series RL circuit for all t. This governing differential equation is

where τ = L/R, iL(t) is the current in the inductor (as well as the circuit current) and vS(t) is the input voltage source function (also called excitation function, forcing function, input function etc.) which must be defined for all t if Eqn. 10.3-1 is to be used.

Notice that the domain of the functions appearing in the governing differential equation is the entire time axis from –∞ to + ∞ since the differential equation is obtained by a systematic application of KCL and KVL. They are basic conservation laws in essence and hence must remain true at all instants of time. Therefore, we must know the input function for all t if we are to solve for iL(t) for all t. But, neither can we know input source function for all t in a practical circuit problem nor do we want to solve for current for all t. Also, the input function may be ill-defined at certain instants of time. This makes a detailed discussion of the way input functions are specified in circuit problems necessary.

10.3.1 Interpreting the Input Forcing Functions in Circuit Differential Equations

We can identify a sequence of time instants at which certain ‘switching events’ will take place in any circuit problem in general. Quite often, this sequence of time points may contain only one entry – as in the case of an RL circuit with battery connected to it by a switch that closes at some specified time instant. However, it is quite possible that a sequence of switching events will take place in our circuit in a more complex setting. Among these various switching instants there will be one that is the earliest instant after which we have complete information about all source functions applied to the circuit – this instant is customarily (though not necessarily) marked as t = 0 in circuit problems. The circuit problem involves solving for all the circuit variables as functions of time from the first switching instant onwards.

Fig. 10.3-1 (a) Applying unit step voltage to RL circuit (b) 1V DC switching in RL circuit

The switch S1 in the circuit in Fig. 10.3-1 (a) is closed at t = 0 and the switch S2 is open at t = 0 to bring about an abrupt change in the applied voltage from 0 to 1 V. If S2 was closed from t = –∞, the current in inductor at t = 0 can only be zero. Whatever energy was possibly trapped in the inductor at t = –∞ would have got dissipated in the resistor by current flow through closed circuit. Hence, there is no need for specifying initial current in the inductor since the applied voltage is known for all t. The differential equation describing the circuit in Fig. 10.3-1 (a) is

The solution of the equation will give the so-called unit step response of the RL circuit. Unit Step function is defined from –∞. Hence, the initial instant in this circuit is t = –∞ and not t = 0. All circuits are assumed to be relaxed at t = –∞ by default.

We just do not know anything about applied voltage for t < 0 in the circuit in Fig. 10.3-1 (b). This is so because voltage across an open–circuit is not decided by the open–circuit; but it is decided by the elements connected on the right side of open–circuit. The initial condition at t = 0 for inductor current in this circuit can be zero or non-zero. Zero initial condition does not imply that no voltage was ever applied to inductor in the past. Rather, it means that whatever be the voltage waveform applied to the inductor in the past, the area under that waveform from –∞ to 0in the time axis was zero. Similarly, a non-zero initial condition will imply that, some extra circuit arrangement that is not shown in the circuit diagram was employed to apply suitable voltage to inductor in the past in order to take its current to the specified value at t = 0. In any case, it does not matter if we are interested in the circuit solution only for t ≥0 +. The differential equation describing the circuit is as in the following equation:

Note that Eqn. 10.3.1 cannot describe the circuit in Fig. 10.3-1 (b) even if the initial condition is specified as 0 A. This is so since solution of Eqn. 10.3.1 will show that the circuit current is = 0 for all t ≤ 0, whereas zero current specified at t = 0 in circuit (b) does not imply that the current was zero for all t ≤ 0. However, the circuit solution for t ≥ 0 + will be the same in both circuits if I0 = 0 A is specified for the circuit in Fig. 10.3-1 (b).

10.3.2 Complementary Function and Particular Integral

The governing equation for inductor current in a series RL circuit with unit step voltage input is

where α = R/L and β = 1/L. There is a jump discontinuity from 0 to 1V at t = 0 in the applied voltage. This jump in applied voltage travels straight to the inductor and appears as a jump in voltage across it at t = 0.

Particular Integral of a differential equation is the solution term due to the forcing function. The domain of particular integral is same as the domain of the input function. In this case, the input is 1V in t ∈ (0,∞). Hence, the particular integral will be valid for t ∈ (0,∞). The particular integral in the present case is obtained by solving the following equation in the interval t ∈ (0,∞):

The differential equation in Eqn. 10.3-4 is to be true for all t ∈ (0,∞). This is possible only if iL is such a function that its first derivative along with its own copy becomes a constant for all t. The only such function is iL = c, a constant. Substituting this trial solution in Eqn. 10.3-4, we get that the particular integral is β/α A for t ∈ (0,∞).

What about particular integral for t ∈ (–∞,0)? The voltage applied in this interval was 0 V. Therefore, the particular integral is 0 for t ∈ (–∞,0).

Therefore, the particular integral of the circuit changes from 0 to β/α A at t = 0. That is, there is a step discontinuity in the particular solution at t = 0. However, such an instantaneous change in inductor current is not permitted in an RL circuit unless impulse voltage is applied across the inductor. The inductor current has to be continuous at t = 0.

Hence, we need another term in the circuit solution which will force the total circuit solution to satisfy the inductor current continuity requirement at t = 0.

This additional function has to satisfy a constraint. The solution iL(t) satisfies the differential equation in Eqn. 10.3-4 at all t∈ 0. Therefore, the additional function we are to going to add to this solution to enforce compliance with continuity requirement at t = 0 should not add anything to the right side of differential equation. This implies that it has to be a function that will satisfy the following differential equation for all t excluding t = 0.

A trivial solution to this equation is iL = 0. We are not interested in that. We recast the above equation as

and note the fact that if iL is a function of time, then both sides of this equation will be functions of time. Two functions of time can be equal to each other over their entire domain if and only if they are same kind of functions – they must have same shape when plotted. Hence, we look for functions which produce a copy (probably scaled versions) of themselves on differentiation.

Sinusoidal function comes to our mind first. But then, we remember that sinusoidal function can be covered by exponential function with imaginary exponent by Euler’s formula. Hence, eγ t where γ can be a complex number is a function with the desired property. So we try out Aeγt, where A is an arbitrary constant, as a possible solution in Eqn. 10.3-5. We get

A = 0 is a trivial solution. eγ t = 0 cannot be a solution since the equation has to be true for all t. Therefore, γ has to be equal to –α. Thus, A eα t is the solution for the Eqn. 10.3-5 in any given interval. The differential equation cannot help us in deciding the value of A. It decides only the value of exponent in the exponential function. This solution we have arrived at i.e., A eα t is called the complementary function of the differential equation in Eqn. 10.3-5 and the differential equation with zero forcing function is called the homogeneous differential equation.

Now that we have got the function required to enforce compliance with the current continuity requirement in the circuit, let us proceed to form the total solution for iL in series RL circuit with step input. There are two intervals over which the particular integral is known. They are (–∞,0] and [0 +, ∞). Let the complementary solution that we accept for t∈(–∞,0] be A_ eαt and the complementary solution that we accept for t∈ (–∞,0-] be A + e–αt .

The voltage applied to the circuit was zero right from t = –∞ up to t = 0. Hence, the circuit current has to remain zero over (–∞,0]. Therefore, the amplitude of complementary solution for t ∈ (–∞, 0] has to be chosen as zero, i.e., A_ = 0. However, this will mean that there will be a step discontinuity at t = 0 in the complementary solution for any non-zero A +. That is precisely what we want. We want the discontinuity in complementary solution to cancel the discontinuity in particular integral at t = 0 and thereby make inductor current continuous at t = 0. Obviously, A + must be chosen as negative of the size of discontinuous jump in a particular integral. This gives us A + = –1/R. Hence, the final solution is

Equation 10.3-7 gives the unit step response (abbreviated as step response) of current in a series RL circuit.

10.3.3 Series RL Circuit Response in DC Voltage Switching Problem

We have seen that switching on a 1V DC source to a series RL circuit is not the same as applying a unit step voltage source across it. Let us see how the difference affects the circuit solution. The relevant circuit is given as circuit in Fig. 10.3-1 (b). We do not know what was done to the inductor before t = 0. All we know is that the voltage applied across the circuit is 1V for all t ≥ 0 + with switching taking place at t = 0. Hence, we need to know the inductor current at t = 0 if we are to solve for inductor current for t ≥ 0 +. Let this current iL (0) = I0 be given as data.

We divide the time-axis into two non-overlapping semi-infinite intervals (–∞,0] and [0 +,∞). The time-point t = 0 is excluded from both.

We do not know the particular integral for t ∈ (–∞,0] since we do not know what was the voltage function applied to the circuit during that interval. We know that the complementary solution part for t ∈ (–∞,0] can be A_ e-αt ; but we cannot fix the value of A_ since the nature of input during the relevant interval is unknown. Therefore, we cannot solve for inductor current in this circuit for t ∈ (–∞,0].

We know the particular solution t ∈ [0 +,∞). It is 1/R A as in the case of unit step response. We also know that the complementary solution for this interval can be A + e–αt. We are given that iL (0) = I0. No impulse voltage is applied across the circuit. Hence, the inductor current has to remain continuous in the interval [0, 0 + ] and iL(0 +)has to be equal to iL (0). Therefore, the total solution for t ∈ [0 +,∞) has to start at I0.

Substituting initial condition,

The complete solution for the current can be written as

For the particular case with I0 = 0 A, the circuit solution will be

This is not the same as unit step response given in Eqn. 10.3-7.

10.4 FEATURES OF RL CIRCUIT STEP RESPONSE

Step response in electrical circuit analysis context implies application of the unit step function, u(t), as the input. The response to this unit step application can be described in terms of a chosen circuit variable, which may be a voltage variable, a current variable or a linear combination of voltage and current variables. We had chosen the current through inductor as the response variable in the case of series RL circuit. The current waveform was shown to be

The primary objective in applying an input function to a circuit is to make some chosen output variable in the circuit behave in a desired manner. This is why the input function is called a forcing function. Input function is a command to the circuit to vary its response variable in a manner similar to its own time variation. Application of unit step input is equivalent to a command to the circuit to change its response variable in a step-wise manner in this sense. Similarly, when we switch on a voltage vS(t) = Vm sin ωt V at t = 0 to any circuit, we are, in effect, commanding the circuit to make the chosen response variable follow this function in shape. A purely memory less circuit will follow the input command with no delay. However, circuits with memory elements will not.

Inductor constitutes electrical inertia. It does not like to change its current and resists any such current change by producing a back e.m.f across it – the magnitude of this e.m.f is directly proportional to the rate at which the inductor current changes. Other elements in the circuit (usually voltage sources, switches etc.) will have to supply the voltage demanded by the inductor if the desired current change is to take place. This is the price the other elements in the circuit have to pay for demanding the lazy inductor to change its current. The price is heavier if the required change in current is to be accomplished faster.

It is still more instructive to look at the ‘inertia’ aspect of inductor from its energy storage capability. An inductor stores energy in its magnetic field. The energy stored in the field is proportional to square of current through inductor. Thus, if we want to change the current through inductor, we have to supply energy to the inductor or absorb energy from the inductor. Note that we do not have to do any such thing if the current through inductor is at a constant level. Let us assume that we want to change the inductor current from I1 to I2 (I2 > I1). By the time we have done it, we would have given the inductor 0.5L(I22 I12) Joules of energy. We can pump energy into the inductor only by pumping power into it. Therefore, a voltage has to appear across the inductor whenever its current tries to change. Further, energy has to be pumped into inductor at a fast rate if the current in inductor is to change fast. That means that power flow into inductor has to be increased if the inductor current is to change fast. And that is why voltage across inductor becomes higher and higher when a given amount of current change is sought to be attained in shorter and shorter time intervals.

Consider a similar situation in translational mechanics. A mass M is forced to move against friction. Assume that the frictional force is proportional to velocity of the mass and that there is no sticking friction. Now, if we apply a constant force to the mass we know that (i) the mass reaches a final speed at which the applied force is met exactly by the frictional force acting against motion and (ii) it takes some time to reach this situation. Mass M does not like to move due to its inertia – it is in the nature of objects in this world to stay put. They prefer it that way. Similarly, it is in the nature of inductor to stay put as far as its current is concerned. However, objects in this world do yield to forces eventually. In the above case, since the mass M shows a tendency to stay put even after the force has come into action, it has to absorb the entire force initially. In that process, it gets accelerated. Hence, for a brief period initially, a major portion of the applied force is used to accelerate the mass and only a minor portion is used for meeting friction. This proportion will change with time and finally no force will be spent on accelerating the mass and entire force will be spent on countering friction. Hence, initially the ‘inertial nature’ of mass dominates the situation and puts up a stiff fight with the force that is a command to the mass to move at a constant speed. Slowly the resistance from the mass weakens and inexorably the force subjugates the inertial nature of mass. And after sufficient time has elapsed, the applied force wins the situation; the mass yields almost completely to the force command and moves at an almost constant speed commensurate with the level of friction present in the system.

This tussle between the inherent inertial nature of systems and the compelling nature of forcing functions is a common feature in dynamic systems involving memory elements and is present in electrical circuits too. Thus, the response immediately after the application of a forcing function in a circuit will be a compromise between the inherent natural laziness of the system and the commanding nature of forcing function. The circuit expresses its dislike to change by spewing out a time function, which quantitatively describes its unwillingness to change. The forcing function wears down this natural cry from the circuit gradually and establishes its supremacy in the circuit in the long run – by forcing all circuit variables to vary as per its dictate in the long run.

The total response in the circuit is always a mixture of these two with the component from forcing function dominating almost entirely in the long run and the natural component from the circuit’s inherent inertia ruling in the beginning. It should be noted at this point that it is quite possible that neither component will succeed in overpowering the other in some circuits. Such circuits are called marginally stable circuits. Further, there are circuits in which the natural component will not only refuse to yield but grow without limit as time increases; thereby overpowering the forcing function with time. Such circuits are called unstable circuits. We will take up such circuits in later chapters. However, at present, we deal with circuits that yield to the forcing function in the long run – called stable circuits.

The time function that the circuit employs to protest against change is called the natural response of the circuit and the time function that the forcing function establishes in the response variable is called the forced response. The natural response means precisely that – it encodes the basic nature of the circuit and has nothing to do with the nature of forcing function. Its shape and other features (except amplitude) are decided by the nature and number of energy storage elements in the circuit, the way these energy storage elements are connected along with resistive elements to form the circuit, etc. Thus, its shape depends only on the nature of elements and the topology of the circuit and does not depend on the particular shape and value of forcing function – it is natural to the circuit. But its magnitude will depend on initial condition and forcing function too.

The series RL circuit with voltage source excitation howls ‘exponentially’ when forcing function commands its current to change. In fact, all stable dynamic systems described by a ‘linear first-order ordinary differential equation with constant coefficients’ will cry out exponentially when they are asked to change. They all have a natural response of the type Ae–αt where α which decides the shape of response, is decided by system parameters (R and L in the present instance) and A is decided by initial condition and the initial value of forced response. The forcing function along with initial condition will decide the magnitude of natural response, but not the shape.

The shape of natural response does not depend on forcing function and hence must be the same for a zero forcing function and a non-zero forcing function. A non-zero response with a zero forcing function can exist if the circuit starts out with initial energy at t = 0. This is similar to a mass, which has been accelerated to some velocity before t = 0, slowing down to zero speed after t = 0 under the effect of friction with no other force applied to it. Thus, it follows that we can find out the shape of natural response by solving the differential equation describing the circuit response with forcing function set to zero. But that will be the homogeneous differential equation and we know that its solution is the complementary function of the equation. The complementary solution of the differential equation describing the current in the inductor in our RL circuit was shown to be an exponential function with negative real index earlier. Thus, we conclude that the complementary solution of the describing differential equation of a circuit yields the natural response of the circuit, whereas the particular integral corresponding to the applied forcing function yields the forced response.

10.4.1 Step Response Waveforms in Series RL Circuit

There are only three circuit variables in a series RL circuit and they are iL(t),vR(t) and vL(t) as marked in Fig. 10.1-1. The expressions for vR(t) and vL(t) may be worked out from the solution for iL(t).

We introduce normalisation in these expressions before we plot them by dividing the expressions by the final steady value or the maximum value as applicable. Similarly, we define normalised time variable tn as t/τ. This results in

where the second subscript ‘n’ indicates normalised variables.

Fig. 10.4-1 Series RL circuit step response – normalised waveforms

These waveforms appear in Fig. 10.4-1. The inductor current rises from zero level at t = 0 + and approaches the normalised final value of 1 as tn approaches ∞. It never touches the final value of 1 since the exponential function never becomes zero. The growth of inductor current gradually loses momentum resulting in the convex shape of current waveform. Simultaneously, the voltage across inductor decays exponentially and tends to go to zero as tn approaches ∞.

Most of the rise in iL and fall in vL take place within first three units of normalised time, i.e., within 3τ s of actual time. The values of e–1, e–2 and e–3 are 0.368, 0.135 and 0.05, respectively. Hence, the step response current in a RL series circuit rises to 63.2% of its final value in the first τ seconds. The voltage across inductor in the same circuit falls by 63.2% from its value at t = 0 + to reach 36.8% of its initial value in the first τ seconds. During the second τ s period, inductor current rises by another 23.3% to reach 86.5% of its final value. Correspondingly, the voltage across inductor falls to 13.5% of its initial value at the end of 2τ s. And at the end of 3τ s, 95% of the transient is over and the inductor current is only 5% away from its final value.

The value of inductor current is 99% and 99.33% of final value at 4.6τ s and 5τ s respectively. Hence, we can consider the natural response of a series RL circuit to be practically over within 5τ s where τ = L/R. During the first five τ periods, the response in the circuit is undergoing a transient phase before reaching a practically steady situation. This period – i.e., the period during which the natural response component is not negligible– is termed as the transient period and a value of 5τ is usually assigned to it in the case of first-order circuits. This leads to another name for natural response – it is also called the transient response. However, we have to be careful about this name since it gives us an impression that this response component is only transient and hence it will vanish with time invariably. This is not always so. There are circuits in which natural response either persists indefinitely or increases with time. Therefore, do not expect transients to vanish with time in all cases.

In the case of series RL circuit with unit step voltage input, the natural response (or transient response) term in inductor current is – e–t/τ/R A and the forced response is 1/R A.

10.4.2 The Time Constant ‘τ’ of a Series RL Circuit

The quantity L/R symbolically represented by τ has turned out to be an important one for RL circuit by now. The unit of L is V-s/A and the unit of R is V/A resulting in a dimension of time for this quantity with Seconds as its unit. Hence, this quantity is defined as the Time Constant of series RL circuit.

One of the interpretations assigned to the time constant is that it is the time taken by an initially relaxed RL circuit to reach 63.2% of its final current value. That is not very satisfactory. After all, 63.2% is not a neat round number or a particularly significant one. However, 50% would have been a neat measure. In fact, the time taken by a first-order system step response to reach 50% of its final value is termed its half-life and this turns out to be ≈0.693 τ in the case of an RL circuit.

Speaking qualitatively, we can appreciate the fact that τ is a measure of the duration taken by the circuit to reach the final current value. RL circuit causes delay in step response current due to the memory capability of inductor. The inductor remembers its initial state and its memory prevents it from allowing sudden changes in current through it. But how deep is its memory in time? How persistent is its memory? Time constant provides answers to these questions.

A large τ implies deeper memory and therefore increased duration will be required for the forcing function to compel the inductor to go to final current value. A larger τ implies a more persistent memory and a heightened tendency on the part of the circuit to keep its current smooth in time-domain.

Time constant has another interesting interpretation. We had noted earlier that the current in the initially relaxed RL circuit starts rising at the rate of 1/L A/s at t = 0 +. Refer to Fig. 10.4-2. Tangents to the normalised inductor current plot and the normalised inductor voltage plot are drawn at normalised time instants of 1, 2 and 3. It is clear from this figure that if the inductor current had continued to rise at the same rate of rise it had at t = 0 +, it would have reached the final value at tn = 1 unit i.e., at t = τ. The slope of ILn at t = 0 + is 1 normalised unit of current per unit of normalised time. Therefore, the normalised current would have reached 1 unit at tn = 1 if the rate of rise had remained unchanged. One unit of normalised time amounts to one τ of real time. Thus, time constant is the time the current in an initially relaxed RL circuit would have taken to reach the final steady value had the initial rate of rise been maintained throughout. Equivalently, it is the time the voltage across inductor would have taken to reach zero if the initial rate of fall could be maintained throughout.

Fig. 10.4-2 Current slope based interpretation of time constant

However, time constant is even more than that. Consider the remaining two tangents at tn = 2 and tn = 3 in Fig. 10.4-2. Moving along either tangent line will take us to the final value of inductor current in one unit of normalised time away from the instant at which the tangent is drawn. If this is true about these two time instants, then, it must be true for all time instants since there is nothing special about these two. Let us examine the slope of inductor current in detail.

Let Δtn be the time required from tn to reach the final value of 1 with rate of change of iLn(t) held at this value. Then,

Thus, time constant of RL circuit is the additional time required from the current instant for the step response in the circuit to reach the final steady value assuming that the rate of rise of response is held constant at its current value from that instant onwards.

10.4.3 Rise Time and Fall Time in First-Order Circuits

‘Rise time (tr)’ and ‘Fall time (tf)’ are two measures of time delay defined in the context of unit step response for linear dynamic systems. These two measures are quite general in definition in order to accommodate wide variety of systems having many terms in their transient response. However, in the case of simple first-order systems there is a direct relationship between the time constant and rise and fall times.

Rise time is defined as the time interval between the first 10% point and first 90% point in the rising step response of a system where the percentages are to the base of the final step response value. Similarly, fall time is defined as the time interval between the first 90% point and the first 10% point in the step response of a system where the response variable is such that it starts at a non-zero initial value and decays to zero value in the long run. The percentages are to the base of the initial response value in this case.

These two definitions are illustrated in the case of series RL circuit step response in Fig. 10.4-3. Normalised variables are used in this figure and the corresponding normalised time points at which the 10% and 90% crossover takes place are also marked in the figure. From the figure, it is clear that rise time and fall time of this circuit are equal to ≈ 2.2τ s, where τ is the time constant of the circuit. This result is valid for any circuit described by a first-order linear differential equation with constant coefficients and has nothing to do with the inductive nature of the circuit under consideration.

Fig. 10.4-3 Rise time and fall time in series RL circuit

These two measures are defined for a general circuit of any order and therefore serve as measures of delay in response and depth of memory in the circuit in situations where a single time constant cannot be identified as the major delaying factor in the circuit.

10.4.4 Effect of Non-Zero Initial Condition on DC Switching Response of RL Circuit

We have been dealing with the step response of an RL circuit till now. We will deepen our understanding in this sub section by taking up the DC switching response of a series RL circuit with non-zero initial current in the inductor at t = 0.

We have already derived the expression for inductor current in this case as

Normalising this equation by using 1/R A as the current base and τ s as the time base, we get the normalised form with I0n as the normalised initial current at t = 0.

This expression can also be written in a form that shows the natural response (transient response) and forced response components clearly separated out as below.

This equation is plotted in Fig. 10.4-4 with solid curves showing the total response and dotted curves showing the natural response or transient response. Curves are shown for four values of initial condition at t = 0. They are –0.5, 0, 0.5 and 1.5. All values are normalised ones. A negative initial condition value indicates that the initial current in the inductor at t = 0 was in a direction opposite to that of forced response. The forced response in all cases is represented by a horizontal line with intercept of unity in the vertical axis.

Fig. 10.4-4 Total response and transient response in series RL circuit step response for various initial currents

The waveforms in Fig. 10.4-4 and Eqn. 10.4-6 bring out the following aspects of RL circuit DC switching response with non-zero initial condition.

  1. The transient response (natural response) of an RL circuit contains two contributions – one from the initial condition specification and another from the value of forced response at t = 0 +. The magnitude of the transient term is decided by these two quantities. Thus, the role of transient response is seen clearly to be one of bridging the gap between the initial current in the inductor and the final current in the inductor. Transient response thus enforces compliance with the initial condition specification in the circuit.
  2. The total response is a rising response if the initial current at t = 0 is less than the final current value. It is a falling response if initial current is more than final current.
  3. There will be no transient response in the circuit if the initial current specified at t = 0 is equal to the final current value in magnitude and direction.
  4. Consider a new current variable defined as ∆iLn(t) = iLn(t) – I0n, i.e., the change in inductor current from its initial value. Substituting Eqn. 10.4-6 in this definition, we get

     

    ΔiLn(t) = (1 – I0n)(1 – e–tn); for tn
    ≥ 0+.

     

  5. Compare this expression for the change in inductor current with the inductor current expression for initially relaxed circuit. We can see that whatever that has been said about time constant becomes applicable in relation to the change in inductor current rather than to the total current when initial current is non zero. The final value of this change is 1–I0n and the change in inductor current rises to 63.2% of its final value in one time constant, 86.5% of its final value in 2 τ s etc. Similarly, the change in inductor current covers the 10% to 90% range in 2.21τ s where the percentages are to the base of 1–I0n.

10.4.5 Free Response of Series RL Circuit

We consider a special case of an RL circuit with zero forcing function in this sub section. Obviously, the solution for inductor current in this source-free RL circuit will contain only complementary solution. The particular integral is zero since forcing function is zero. The complementary solution is of the form Ae–αt where α = R/L. Applying initial condition to this solution makes it clear that A = 0 unless the initial condition specified at t = 0 is non-zero. Thus, a source-free RL circuit can have a non-zero solution only if the inductor has some energy trapped in it at t = 0. This energy storage must have been created by some source prior to t = 0.

Fig. 10.4-5 Source-free RL circuit and waveforms

Consider the circuit in Fig. 10.4-5 (a). The switch S1 was closed long back and the circuit has attained the final inductor current value of 1/R A by the time t = 0 is reached. At t = 0, the switch S1 is opened and the switch S2 is closed simultaneously. Thus, a source-free series RL circuit with an initial current of I0 (which is equal to 1/R A in the circuit in Fig. 10.4-5) is set up at t = 0. The circuit in Fig. 10.4-5 (b) is equivalent to the circuit in Fig. 10.4-5 (a) for t = 0 +.

The expressions for inductor current and circuit voltages are derived as below follows:

The current in the circuit decays exponentially from I0 to zero with a time constant equal to L/R seconds. This is shown in Fig. 10.4-5 (c). The corresponding voltage across inductor is negative valued and decays with the same time constant. The circuit voltages are shown in Fig. 10.4-5 (d).

Example: 10.4-1

Obtain an expression for voltage across the resistor in an initially relaxed series RL circuit for rectangular pulse voltage input defined as vS(t) = 1 V for 0 ≤ t T and 0 V elsewhere. Plot the response for (i) T = 0.2τ, (ii) T = τ and (iii) T = 2τ.

Solution

The differential equation for iL(t) for the interval [0 +, T] is

Particular integral that is valid in the interval [0 +, T] is β/α = 1/R. The complementary function is of the form A e t / τ. The circuit is initially relaxed. Applying initial condition to total solution and solving for A, we get the total solution as

The inductor current would have followed this expression till there is a change in input source function or circuit structure. There is a change in the applied voltage at t = T in the present example. The voltage applied for all t ≥ T + is zero. Thus, the circuit is described by the following differential equation for T + t < ∞.

The particular integral for this equation is zero. The complementary function is again A e (tT) / τ (but valid only for t ≥ T+) with the value of A to be decided. The value of A is found out from the value of current at t = T +. However, since there was no impulse voltage involved in the circuit at t = T, the value of current at t = T+ and t = T will be same. This value can be obtained by substituting t = T in Eqn. 10.4-7.

Therefore, the expression for vR(t) is

The subscript ‘n’ indicates normalisation with respect to τ. The plots of resistor voltage with normalised time for various T/τ ratios are shown in Fig. 10.4-6.

Fig. 10.4-6 Single pulse response of RL circuit in Example: 10.4-1

Example: 10.4-2

Solve for i and v as functions of time in the circuit in Fig. 10.4-7.

Fig. 10.4-7 Circuit for Example: 10.4-2

Solution

This circuit was already in DC steady-state at t = 0. At t = 0, the switch closes, thereby forming a source-free RL circuit on the right side and a simple resistive circuit on the left side. These two circuits do not interact after t = 0 except that the current through the switch will be a combination of the currents from these two circuits.

Inductor is a short for DC steady-state. Therefore, the initial current in the inductor at t = 0 was 10V/20Ω = 0.5A from top to bottom. Since the switching at t = 0 does not involve impulse voltage, the inductor current remains at 0.5A at t = 0 + too.

Thus, a source-free RL circuit with initial current of 0.5A is set up at t = 0. The various current components in the circuit after t = 0 are marked in Fig. 10.4-8.

Fig. 10.4-8 Circuit for solving Example: 10.4-2

Example: 10.4-3

Solve the circuit in Fig. 10.4-9 for the current through the switch as a function of time.

Fig. 10.4-9 Circuits for Example: 10.4-3

Solution

In this example, the two meshes in the circuit interact after t = 0. We can solve this circuit in many ways – branch-current method, mesh analysis, Thevenin’s equivalent, etc. are some possibilities. First, we solve it by branch-current method.

Various branch currents in the circuit are identified in Fig. 10.4-9 (b). We have to get a differential equation in the current variable i.

Applying KCL at switch node gives us i2 = i1i.

Applying KVL in the first mesh gives us 15 = 10(i1 + i) ⇒ i1 = 1.5 – i

i2 = 1.5 – i i = 1.5 – 2i

Applying KVL in the second mesh gives us 10 i = 10 i2 + 0.015 (di2 /dt) = 15–20i –0.03(di/dt)

Therefore, the differential equation governing i is di/dt + 1000 i = 500 for t ≥ 0 +.

Initial condition for i, i.e., its value at t = 0 +, is needed. Initial value of i2 is 15V/20Ω = 0.75A since the circuit was in DC steady-state prior to switching. Since i2 = 1.5 – 2i, value of i at t = 0 + will be (1.5 – 0.75)/2 = 0.375A. The particular integral of the differential equation for i is 500/1000 = 0.5A. Time constant is 1/1000.

Therefore, i = (C e–1000 t + 0.5) A. Evaluating C from initial condition for i at t = 0 +, we get C = 0.375 – 0.5 = – 0.125.

Therefore, the switch current i = (0.5 – 0.125 e–1000 t) A for t ≥ 0+.

Let us solve the same problem by mesh analysis. The relevant circuit with two mesh currents – I1 and I2 – identified is shown in Fig. 10.4-10 (a).

Fig. 10.4-10 Circuits for mesh analysis and Thevenin’s equivalent analysis in Example: 10.4-3

The two mesh equations are

Eliminating I1 from second equation using the first equation and simplifying, we get

The initial condition for I2 at t = 0 + is same as the initial condition for inductor current at that instant. This value is 0.75A.Particular integral is 0.5A. Time constant is 1/1000 s.

Using this solution in the first mesh equation we can get

∴Current through the switch = I1 I2 = 0.5 – 0.125e–1000 t A for t≥ 0 +

This circuit problem can also be solved by using Thevenin’s theorem. The circuit portion to the left of inductor may be replaced by its Thevenin’s equivalent as shown in Fig. 10.4-10 (b). Inductor current can be obtained from this circuit. Once inductor current is available, we will be able to get back to the switch current using KCL or KVL. This is illustrated now in the following.

The initial current in the inductor is 0.75A. The circuit in Fig. 10.4-10 (b) is a simple series RL circuit and its particular integral is 7.5/15 = 0.5A. Its time constant is 15mH/15Ω = 1ms. Therefore, its solution is = C e–1000 t + 0.5. Evaluating the initial condition constant C and completing the solution, we get, inductor current = 0.5 + 0.25 e–1000 t A.

We derived the differential equations governing three variables in the circuit – the branch current in the central limb, second mesh current and the inductor current in the process of solving this circuit. The left-hand side of all the three differential equations had the same coefficients. (Why?)

We also notice that the time constant of the circuit can be easily found as L/Rth where Rth is the Thevenin’s equivalent resistance appearing across the inductor. However, Thevenin’s equivalent is found by deactivating all independent sources. Therefore, the time constant of a single-inductor circuit can be found by replacing all independent voltage sources by short-circuits and all independent current sources by open circuits and finding the equivalent resistance connected across the inductor. We illustrate this procedure further in the next example.

Example: 10.4-4

Show that the current in 18mH inductor in the circuit in Fig. 10.4-11 (a) will go to zero as t → ∞. Also, find the inductor current and currents delivered by the voltage sources as functions of time. Find out how long we have to wait for the inductor current to fall below 100mA.

Fig. 10.4-11 Circuits for Example: 10.4-4 (a) Circuit for the problem (b) Circuit for finding Thevenin’s equivalent (c) Thevenin’s equivalent

Solution

First, we find the time constant effective after t = 0 +. The Thevenin’s equivalent of the circuit connected across the inductor is evaluated by using the circuit in Fig. 10.4-11 (b) and the resulting equivalent is shown in Fig. 10.4-11 (c). Since the voltage source in Thevenin’s equivalent is zero-valued, the inductor current will have a zero steady-state value. The time constant of the circuit is 18mH/12Ω = 1.5ms.

We find the initial condition for inductor current next. The circuit was in steady-state prior to switching at t = 0. Inductor is replaced by a short-circuit DC steady-state. Therefore, the inductor current at t = 0 must have been 7/14 = 0.5A and it will be 0.5A at t = 0 + since there is no impulse voltage involved in the switching.

Now, the circuit is a series RL circuit with a known initial condition and DC sources. We know the solution for such a circuit. It is of general format – A e –t/τ + C – where C is the particular integral (therefore, the DC steady-state value) and A is the arbitrary constant to be found out from initial condition. This is the general format of solution for any circuit variable in a first-order circuit with DC excitation.

We have two ways to find the currents delivered by the voltage sources – i1 and i2. In the first method, we find the voltage across M and N in circuit Fig. (a) as vMN = 8i + 0.018 di/dt and then find i2 as (7– vMN)/6 and i1 as (vMN– 14)/12.

In the second method, we realise that all variables in this circuit will have a A et/τ term and a steady-state term and that we can find the arbitrary constant A if we know the value of the particular variable at 0 +. Therefore, we set out to find the initial and final (i.e., steady-state) value of the source currents. The inductor current was at 0.5A at t = 0 +. The voltage across inductor at that instant is –6V from Eqn. 10.4-9. Therefore, vMN = 8x0.5 – 6 = –2 V at t = 0 +. Therefore, initial value of i1 at t = 0 + = (–2 – (–14))/12 = 1A. Final current in inductor is zero. Hence, final current in both sources will be same and equal to (7 + 14)V/(6 + 12)Ω = 7/6 = 1.167A.

Therefore, our required currents are

Fig. 10.4-12 Waveforms of (a) voltages and (b) source and inductor currents in Example: 10.4-4

These are plotted in Fig. 10.4-12. The time required for inductor current to go below 100mA is found as follows:

 

0.1= 0.5e–t/1.5 A; t in ms , ∴ = ln 0.2 = –1.6094 and t = 2.414 ms
10.5 STEADY-STATE RESPONSE AND FORCED RESPONSE

The total response in an RL circuit with any forcing function will consist of two terms – the transient response (or natural response) and the forced response. The transition from initial state of the circuit (which is encoded in a single number in the form of initial inductor current specification at t = 0) to the final state (in which only forced response will be present) is accomplished with the help of the transient response. Now, we introduce a new term called steady-state response and relate it to the response terms we are already familiar with.

Our study of the solution of differential equation describing the RL circuit has shown us that the total response will always contain two components – the transient response and the forced response. Of course, forced response will be zero if forcing function is zero i.e., in a source-free circuit. Similarly, the transient response term may become zero under certain suitable initial condition values. But these are special situations and, in general, there will be two terms in the total response. This is true not only for an RL circuit but also for any linear circuit described by linear ordinary differential equations with constant coefficients. Such a circuit of higher order will have two groups of terms in its total solution – first group constituting transient response containing one or more terms and the second group constituting forced response containing one or more terms depending on the type of forcing function. Thus, forced response is a response component which is always present in the total response of a circuit except when the forcing function itself is zero.

We have seen that the transient response of an RL circuit contains exponential function of the form e–αtwhere α is a positive number decided by R and L. Such an exponential function with negative real index will taper down towards zero as t approaches ∞. Hence, we expect the transient response in an RL circuit to vanish with time quite irrespective of the forced response component. Therefore, we expect that there will only be the forced response component active in the circuit in the long run i.e., after sufficient time had been allowed for the transient response to die down. When all the transient response terms in all the circuit variables in the circuit have died down to negligible levels (they never die down to zero) and the only response component in all the circuit variables is the forced response component, we say the circuit has reached the steady-state with respect to the particular forcing function that was applied to the circuit. Notice that under steady-state conditions the transient response terms should not be present in any circuit variable at all. Or, in other words, there cannot be a circuit which attains steady-state in some of its variables and does not attain steady-state in yet others.

Therefore, a circuit will reach steady-state if and only if all its transient response terms are of decreasing type. Moreover, the only response that will continue in the circuit after it has reached steady-state is the forced response component. Therefore, steady-state response is same as forced response with the condition that the steady-state will exist only if all the transient response terms are of damped nature – i.e., decreasing functions of time. Thus, steady-state response is another name for forced response when transient response is assured to die down to negligible levels. Forced response will always be present; but steady-state response need not be.

Consider the circuit in Fig. 10.5-1. The 1V source on the left side has set up an initial current of 1A in the inductor of 1H at t = 0. The switch S1 is opened and switch S2 is closed at t = 0 to apply a 1V source right across the inductor. The current in the inductor is shown in Fig. 10.5-1 (b). We note that with a bounded input (1V DC source is a bounded input), we get a current in the inductor which is not bounded. Also, the transient response (1e–0t) does not decrease with time. Therefore, there is no steady-state in this circuit though there is a forced response.

Fig. 10.5-1 A circuit with no steady-state and its step response

10.5.1 The DC Steady-State

It is to be noted is that the steady-state attained by a circuit is intimately connected with the type of forcing function applied to the circuit. We cannot expect the generalisations arrived at based on the steady-state behaviour for a particular forcing function to hold in the case of steady-state behaviour for another forcing function.

For example, consider the steady-state in series RL circuit when input is a unit step voltage. The forced response in this case is a constant current of value 1/R A. The voltage across inductor with a constant current through it can only be zero. But zero is indeed a constant. Thus, we see that, under steady-state condition with step input, all the circuit variables become constants. The only constant voltage an inductor can have across it is zero if the current through it is also constrained to remain constant since v = L di/dt for an inductor. Noting the ‘constant’ nature of input voltage and ‘constant’ nature of all circuit variables under steady-state, we name this kind of steady-state as the DC steady-state.

Under DC steady-state (provided the circuit can reach such a steady-state – transient response has to die down for that), inductors in a circuit can have any constant valued current; however, their voltages will be constrained to remain at zero. But that is similar to the definition of a short-circuit – except that the voltage across a short-circuit is zero for any current. Thus, we can solve for DC steady-state response in RL circuits (containing one or more inductors) by replacing all inductors by short-circuits – provided all independent sources in the circuit are switched DC sources or step functions and a DC steady-state can exist in the circuit. Subject to the above conditions, we can state that ‘an inductor is a short-circuit under DC steady-state’.

While we are on the topic of steady-state, we might as well look at the remaining two kinds of steady-state we will need in the analysis of dynamic circuits.

10.5.2 The Sinusoidal Steady-State

Sinusoidal steady-state refers to the steady-state that gets established in a circuit when all the independent sources in the circuit are sinusoids of same angular frequency. Like DC steady-state, this steady-state too can exist only if all the terms in transient response die down to negligible levels with time.

The word ‘steady’ has the literal meaning of ‘unchanging’. This unfortunately gives an impression that steady-state is that state in a circuit in which all circuit variables are unchanging in time. This is an error that a beginner in Circuit Analysis has to guard against. Steady-state does not necessarily mean that circuit response is unchanging in time. It is so only in the case of a DC steady-state.

To understand the meaning of the word ‘steady’ in the Circuit Analysis context, we have to look at the input forcing function and find out those features of the forcing function which remain unchanging with time. In the case of DC or step inputs, the input value itself is unchanging in [0 +, ∞). In the case of sinusoidal input forcing function the amplitude of the sinusoid, its angular and cyclic frequencies, its phase and its shape in one period (i.e., sinusoidal shape) remain constant in time. Therefore, if a steady-state exists in a circuit under the action of such a forcing function, we can expect all circuit variables to have sinusoidal shape, fixed amplitudes, fixed frequency which is same as that of the forcing function and fixed phase with respect to forcing function. This is what is meant by sinusoidal steady state.

 

Thus, a circuit excited by one or more sinusoidal forcing functions of same frequency is said to have reached sinusoidal steady-state if all its transient response components have died down and all its circuit variables have sinusoidal waveshape with same frequency as that of forcing functions and fixed amplitudes and phase angles.

The waveforms in Fig. 10.5-2 show the applied voltage and inductor current in an initially relaxed RL circuit with R = 0.33 Ω and L = 0.33 H. A sinusoidal voltage = 1 sin (5t) V was switched on to the circuit at t = 0. The current waveform shows the exponential transient response in the first few seconds clearly. After about 10 s or so, the transient response has decayed to negligible level and the response contains only a sinusoidal waveform that is of same frequency as that of applied voltage. It has fixed amplitude and a fixed phase with respect to the input sine wave. Thus, the circuit has reached sinusoidal steady-state within few time constants (τ = 1s).

Fig. 10.5-2 Waveforms illustrating sinusoidal steady state

Sinusoidal steady-state is also referred to as AC steady-state in Circuit Analysis literature.

10.5.3 The Periodic Steady-State

This is the third kind of steady-state that can come up in linear circuits. Here, the input forcing function is periodic but not sinusoidal. Therefore, a single number like amplitude in the case of a sinusoid is not available for this input. The only aspect that is steady about it is its frequency. Thus, we expect the circuit to reach a steady-state (if it can) in which all circuit variables will be periodic with same frequency as that of input. But the waveshape of response variables will not be the same as the waveshape of input forcing function. The sinusoidal steady-state we discussed in the previous sub section is indeed a periodic steady-state; but it is more than that – in sinusoidal steady-state, the response waveform is same as that of forcing function. In a periodic steady-state, such a constraint may not be satisfied.

Fig. 10.5-3 Waveforms illustrating periodic steady-state

The waveforms shown in Fig. 10.5-3 illustrate the attainment of periodic steady-state in an RL circuit (R = 0.33Ω, L = 0.33 H, I0 = 0, τ= 1s) driven by a voltage = (1 sin 5t + 0.4 sin 15t) V from t = 0 + onwards. Notice that the waveshape of current is not the same as that of applied voltage. However, the current is periodic with same period and its waveshape remains the same in successive periods after it has reached steady-state.

We saw that inductors can be replaced by short-circuits for DC steady-state analysis. However, we notice that the currents in the inductors in circuits are time-varying currents in sinusoidal steady-state and periodic steady-state conditions. Hence inductors cannot be replaced by short-circuits for all kinds of steady-state analysis – that works only for DC steady-state analysis.

Example: 10.5-1

Periodic steady-state can come up in RL circuits when one or more sources in the circuit are periodic functions of time. But even when all sources are DC sources there can be a periodic steady-state in the circuit if some parameter in the circuit is varying periodically with time. The circuit for this example appears in Fig. 10.5-4. The switch S in the circuit is operated periodically at a switching frequency of 1kHz. In every switching cycle, it is kept closed for half the time and then kept open for remaining half of cycle time i.e., 0.5ms. After a large number of switching cycles a steady repeating pattern of current gets established in the circuit. We describe this steady pattern in this example.

Fig. 10.5-4 Circuit for Example: 10.5-1

Assume that the circuit is in periodic steady-state. Let the current start with a value I1 when the switch is closed in the beginning of a steady-state cycle. Then, the current increases along an exponential with τ of 2ms towards 2A (because 2 is the particular integral under this condition). But the current is not allowed to reach 2A since the switch goes open after 0.5ms when the current is I2. Now, a new transient starts, trying to take the current from this value to 1A (because 1 is the particular integral now) exponentially with a time constant of 1ms. But this exponential is terminated after another 0.5ms and then the cycle starts all over again. At the end of the second 0.5ms, the current in the circuit will be I1 if the circuit is in periodic steady-state. We convert the above reasoning into equations and solve for I1 and I2.

The corresponding values for 500Hz switching also were worked out and they are I1 = 1.186 A and I2 = 1.506 A. The plot of inductor current for 500Hz switching and 1kHz switching are shown in Fig. 10.5-5 (a) and (b).

Fig. 10.5-5 Periodic steady-state in circuit in Example: 10.5-1 for (a) 500Hz and (b) 1kHz switching

The total resistance in the circuit was changing abruptly between 10 and 20 Ω. But the current in the circuit does not show any discontinuity. This once again illustrates the fact that inductor smoothes a circuit current. Moreover, we see from Fig. 10.5-5 that, the smoothing effect is more when circuit time constants are larger than the switching cycle period. In fact, the current tends to become almost constant at 1.33A as inductance value in this circuit is increased. (The reader is encouraged to ponder over why it should it be 1.33A and why the average value of waveforms in Fig. 10.5-5 also should be 1.33A.)

10.6 LINEARITY AND SUPERPOSITION PRINCIPLE IN DYNAMIC CIRCUITS

We have dealt with DC switching response of an RL circuit with 1V input. How do we get the solution if it is not 1V but V V that is switched onto the circuit? Can we just multiply the unit switching response by V to get the solution for this input?

We know that memory less circuits containing linear passive resistors, linear dependent sources and independent sources will be linear and will obey superposition principle. We examine the issue of linearity of circuits containing one or more energy storage elements along with resistors, linear dependent sources and independent sources in this section. We are already familiar with a particular decomposition of total response in such a circuit in terms of transient response and forced response. We will see in this section that yet another decomposition of total response into the so-called zero-input response and zero-state response is needed in view of linearity considerations in the circuit.

An electrical element is called linear if its element equation obeys superposition principle. Superposition principle involves two sub-principles – principle of additivity and principle of homogeneity. We have seen earlier that an inductor with an element relation v = L di/dt and a capacitor with an element relation i = C dv/dt are linear elements. But will an interconnection of such linear elements (and independent sources) into a circuit result in a linear system? A linear system is one in which every response variable in the system obeys superposition principle.

Intuitively we expect an interconnection of linear elements and independent sources to yield a linear circuit; but mathematically it is not that simple. It requires to be proved. We accept the result that a circuit formed by interconnecting linear passive elements, linear dependent sources and independent sources will be a linear circuit. Such a linear circuit has to obey superposition principle.

Therefore, we must be able to get iL(t) in a series RL circuit with V u(t) V as its input source function by scaling the unit step response by V. Assuming an initial condition of I0 at t = 0, this scaling results in Eqn. 10.3-8 getting multiplied by a dimensionless scalar V to yield

as the solution. But this solution is incorrect because the current at t = 0 + is VI0 according to this equation rather than the correct value of I0.It looks as if the principle of homogeneity is not valid here. Let us try to get the solution without resorting to linearity.

The complementary solution is A eαt with α= 1/τ= R/L and the particular integral is V/R. Therefore, the total solution is iL(t) = A eαt + V/R. Substituting the initial condition at t = 0 + and solving for A, we get the final solution as iL(t) = (I0 V/R)e–αt + V/R = I0 e–αt + V/R (1 – e–αt) for t ≥ 0 +. Thus, the correct solution is,

The first term in Eqn. 10.3-8 does not get multiplied by V when the DC voltage magnitude is scaled by V. The second term in the same equation gets scaled by V. This means that the forced response (the constant term in the solution) obeys the principle of homogeneity.

Now, let us solve the circuit for three situations (a) with V1 (b) V2 and (c) V1 + V2 as the DC voltages with the same initial condition for all the three cases. The expression for iL(t) in the three cases can be derived as

These expressions show that the forced response part obeys the principle of additivity also. In all cases we see that neither the total response nor the transient response (or natural response) obeys superposition principle.

We have arranged the terms in the expression for inductor current in Eqn. 10.6-1 and Eqn. 10.6-2 in a special manner – the dependence on initial condition is contained in the first term and the dependence on step magnitude is contained in the second term. We notice that it is not only the forced response component which satisfies the superposition principle, but the entire second term which depends on forcing function satisfies the superposition principle. However, both terms contribute to transient response and transient response does not satisfy superposition principle.

We notice further that the first term depends only on initial condition and will be the total response if there is no forcing function, i.e., the first term is the response in a source-free circuit. Similarly, the second term depends on forcing function and does not depend on initial condition. The second term will be the total response if the circuit is initially relaxed. These observations will remain valid for any forcing function. The nature of the second term will change with the nature of forcing function. However, the resolution of total response into two components –one that depends entirely on initial condition and another that depends entirely on forcing function – will be possible for any forcing function in any linear circuit.

Here too we accept a result proved in the general theory of linear systems without worrying about its proof.

 

The response for t ≥ 0 + that results from initial condition alone (that is, with zero input for t ≥ 0 +) is called ‘zero-input response’. The response for t ≥ 0 + that results from application of input for t ≥ 0 + with zero initial condition is called ‘zero-state response’.

The total response in a linear time-invariant circuit containing energy storage elements can be found by adding the zero-input response and zero-state response together. Zero-input response will depend only on the initial state of the circuit as encoded in its initial condition specifications. Zero-state response will depend only on forcing function.

Now, we focus on the zero-input response of an RL circuit. This is the response in a source-free circuit due to its initial energy alone. It is I0eαt A with the usual interpretations for all the symbols. It must be obvious that the zero-input response will scale with I0, i.e., when the initial condition value is multiplied by a real constant the zero-input response also gets multiplied by the same constant. Similarly, when two different values of initial condition I01 and I02 result in two different zero-input responses, the zero-input response with the initial condition value at I01 + I02 will be the sum of the two zero-input responses observed in the first two cases. Thus, zero-input response of RL circuit (and, of all linear time-invariant circuits) obeys superposition principle with respect to initial condition values.

Thus, we see that both zero-state response and zero-input response obey superposition principle individually.

 

Zero-input response follows superposition principle with respect to initial condition values and zero-state response obeys superposition principle with respect to input source functions.

Therefore, total response will not follow superposition principle with respect to forcing function or initial condition – only its components will obey superposition principle.

Figure 10.6-1 shows the zero-input response and zero-state response components along with the total current response in an RL circuit under DC voltage switching condition for various values of normalised initial condition values. Decomposition of total response into transient response and forced response for the same circuit was shown in Fig. 10.4-4. Compare these two decompositions.

Fig. 10.6-1 Decomposition of total response into zero-input response and zero-state response

Both zero-input response and zero-state response will contain natural response terms. However, the natural response component in zero-state response has amplitude that depends on forcing function value and does not depend on initial condition value. Zero-input response and part of zero-state response together will form transient response. The remaining part of zero-state response will be the forced response.

10.7 UNIT IMPULSE RESPONSE OF SERIES RL CIRCUIT

We look at the response of series RL circuit in Fig. 10.1-1 with vS(t) = δ(t), the unit impulse voltage. We have discussed the unit impulse function in detail in Chapter 3. We saw that we can view unit impulse function as a limiting case of a rectangular pulse waveform of amplitude 1/Δt V located between t = 0 and t = Δt as Δt → 0. The pulse always maintains unit area under it by increasing the amplitude as its duration decreases. We use this interpretation of unit impulse function to analyse impulse response of an RL circuit first. The circuit is assumed to be initially relaxed.

Fig. 10.7-1 Pulse response of series RL circuit

The applied pulse and circuit current responses are shown in Fig. 10.7-1. The response in the interval [0 +, Δt] will be the same as the unit step response scaled by 1/Δt, At Δt the applied voltage at input goes to zero, i.e., the input gets shorted and a source-free RL circuit starts executing its zero-input response from that point onwards. The amplitude of this response has to be the initial current value just prior to Δt. This value is marked as IP in the Fig. 10.7-1. An expression for IP can be obtained by evaluating the first part of response at Δt as shown in Eqn. 10.7-1.

As Δt is decreased, the value of IP increases and moves towards left in time axis. Eventually, it attains a limit as Δt → 0. This limiting value can be found by using series expansion for exponential function as in Eqn. 10.7-2 the following equation:

Therefore, as Δt → 0 the first part of response tends to become a jump by 1/L A and second part becomes an exponential from t = 0 + onwards with 1/L as the starting value. And in the limit, the impulse response becomes

The only steady feature the unit impulse function possesses after t = 0 + is constancy at zero value. Therefore, we expect the forced response to be zero. The only response component that can be there in the impulse response is the transient response component. And, since the initial condition at t = 0 is zero (because the voltage applied to the circuit in the interval (∞, 0] is zero) it follows that zero-input response is zero. Therefore, the impulse response appearing in Eqn. 10.7-3 is the zero-state response for unit impulse application.

Now, consider the series RL circuit with an initial condition of 1/L A at t = 0 and vS(t) = 0 for t ≥ 0 +. This is a source-free RL circuit with non-zero initial energy. It will have only zero-input response component in its response and its total response will be

which is exactly the same as the unit impulse response. Thus, the effect of applying δ(t) is to change the initial condition of inductor between t = 0 and t = 0 +. Remember, we have always pointed out that initial condition at t = 0 and t = 0 + will be the same only if no impulse voltage is applied in the circuit or supported in the circuit. Applying δ(t) amounts to keeping the circuit shorted for t ≤ 0 and t ≥ 0 + and applying an undefined voltage at t = 0 such that a finite V-s area of 1V-s is dumped into the circuit at t = 0. This results in changing the inductor current by 1V-s/L henry = 1/L A between t = 0 and t = 0 +.

The equivalence between non-zero initial condition at t = 0 and application of suitably sized impulse voltage at t = 0 is further clarified by the relations in Eqn. 10.7-4 where vL and iL are voltage across an inductor and current through the inductor, respectively.

Circuit solution for t ≥ 0 + requires only the value of initial current at t = 0 +. This value is the sum of the first two definite integrals in Eqn. 10.7-4. It does not matter which integral contributes how much as long as the sum of their contributions remain constant as far as iL(t) after t = 0 + is concerned. First integral gives the initial current in inductor due to all the voltages applied to it in the past. Second integral will be non-zero only if impulse voltage is applied to inductor at t = 0. As far as iL(t) after t = 0 + is concerned these two terms are interchangeable. Therefore, an initial current of I0 in the inductor at t = 0 may be replaced by zero initial current at t = 0 and an impulse voltage LI0 δ(t) with correct polarity in series with the inductor.

But the voltage variable that appeared in Eqn. 10.7-4 was the voltage across inductor. If we want to replace a non-zero initial current at t = 0 by an impulse voltage source, we must ensure that impulse source appears fully across inductor and does not lose itself across other elements in the circuit. Therefore, we have to argue that the δ(t) we applied at the input travels through R and appears fully across L. The definition of unit impulse function avoids defining it at t = 0 and makes up for that by providing its area content in an infinitesimally small interval around t = 0.

Now, consider applying this to an RL circuit. Refer to Fig. 10.1-1. Applying KVL in the mesh, we get vR(t) + vL(t) = vS(t) for all t. Since this relation is true for all t, we can integrate both sides of the equation between same two limits to get the following relation.

Area under vR(t) between two instants t1 and t2 + Area under vL(t) between two instants t1 and t2 = Area under vS(t) between two instants t1 and t2. Taking t1 = 0 and t2 = 0 + this relation results in

We need to show that the first integral on the left side is zero. Let us assume that it is zero. Then, second integral is 1. The second integral value divided by L gives the change in inductor current over the interval [0, 0 +]. Therefore, the inductor current change is 1/L A and the inductor current is 1/L A since circuit was initially relaxed. This is the maximum value that the current can have at t = 0 + since we had assumed that the first integral value is zero. Thus, the current through the circuit during [0, 0 +] is confined between 0 and a maximum of 1/L A. This implies that though the current during this infinitesimal interval is indeterminate in value (because there is a jump discontinuity in it) it remains upper-bounded and lower-bounded and hence is finite-valued in the interval. A finite-valued function integrated over infinitesimal interval results in zero integral value. After all, there is no area under a rectangle if the rectangle is of finite length in one direction and of infinitesimal length in the other direction. Therefore the first integral, i.e., the portion of area content of δ(t) that gets lost across resistor, is zero and our assumption to that effect is correct. Now, we may similarly assume that this integral is non-zero and prove that the assumption leads to a contradiction. Therefore, all the area content of δ(t) appears across inductor itself.

 

Non-zero initial current of I0 in an inductor L at t = 0 in a circuit can be replaced by zero initial current at t = 0 along with an impulse voltage source LI0δ(t) in series with the inductor for solving the circuit in the domain [0 +, ∞).

Moreover, we have resolved the small problem we had with linearity and superposition principle in dynamic circuits. In fact, there is no problem. Superposition principle is fully obeyed by linear dynamic circuits. Only that we have to apply it carefully when there are non-zero initial condition values specified. In that case, we have to remember that each initial condition represents a source and that it becomes a multi-source problem. When changes are effected in a source, superposition principle has to be applied to that component of total response contributed by the particular source.

10.7.1 Zero-State Response for Other Inputs from Impulse Response

The unit impulse response (hereafter this will be referred to as ‘the impulse response’ – that it is a zero-state response and that unit impulse is applied will be implicit) is the most important response for a linear circuit. We will see in a later chapter that zero-state response for any other well-behaved forcing function can be found from it. In this sub section, we will show that it is an easy matter to get zero-state response for step input, pulse input and ramp input functions from the impulse response by integrating it.

First, we establish the relation between unit impulse function and unit step function. It is a simple one. Running integral of unit impulse function results in unit step function. Unit impulse function is defined as

Consider the following integral.

Obviously, the value of f(t) is zero for –∞ < t ≤ 0 since δ(t) is zero in that range. But area under impulse function undergoes a rapid change by 1 when t goes from 0 to 0 +. Therefore, f(0) = 0, whereas f(0 +) = 1. Therefore, t = 0 is a point of discontinuity in f(t) and has to be excluded from its domain. f(t) remains at 1 for t ≥ 0 + since δ(t) is zero for t ≥ 0 +. Therefore, the description of f(t) coincides with the definition of u(t),

Now, we turn our attention to the differential equation describing series RL circuit for zero-state response with some input function vS(t) switched onto the circuit at t = 0. Here, vS(t) is a function which is defined for all t and may not be zero for t < 0. We make the applied voltage equal to zero in the negative time axis and equal to this function in the positive time axis by multiplying it with u(t) to find the zero-state response.

Since this equation is true for all t (it is a KVL equation), we can integrate both sides from –∞ to t and write

Changing the order of integration and differentiation

The last equation is the differential equation describing the zero-state response for integrated input function.

 

Therefore, we can state that the zero-state response in a linear circuit for an integrated input function is equal to integral of the zero-state response for the input function.

Strictly speaking, we showed this only for a first-order differential equation; but there is nothing in the proof which limits it to first-order differential equation alone. This result is valid for any linear lumped circuit described by linear differential equation of any order. The integration has to be performed from –∞ in theory. However, since we know that the zero-state response is zero from –∞ to t = 0, we need to integrate from t = 0 to t only.

Integral of δ(t) gives u(t). Therefore, integral of impulse response should give unit step response (usually referred to as the step response). This is verified by carrying out the integration below.

Zero-state unit impulse response of an RL circuit

It can be easily verified that unit ramp function is the integral of unit step function. These three basic input functions and their relations are shown in Fig. 10.7-2.

Fig. 10.7-2 Impulse, step and ramp functions and their relations

Therefore, the ramp response can be found by integrating the step response as below.

Ramp response and its components are shown in Fig. 10.7-3. The voltage across resistor is plotted instead of inductor current.

Fig. 10.7-3 Unit ramp response of RL circuit

The unit ramp function has a kink at t = 0 and hence it is not differentiable at t = 0. However, it is differentiable at all other time instants. Hence, first derivative of r(t) will be a function defined as 0 for t ≤ 0, 1 for t ≥ 0 + and undefined at t = 0. But that is the unit step function. Therefore, unit step function is the first derivative of unit ramp function.

Unit step function is not even continuous at t = 0. Obviously, it cannot be differentiated there. However, we raise the question – which function when integrated will yield unit step function? The answer is that it is the impulse function. Therefore, we can consider δ(t) to be the first derivative of u(t) in the 'anti-derivative' sense.

KVL equations are true for all t. Both sides of an equation which holds for all t can be differentiated with respect to time. From this observation, it is easy to see that the following is true.

 

The zero-state response in a linear circuit for differentiated input function is equal to the derivative of the zero-state response for the input function.

Therefore, we can get to unit step response and unit impulse response in any linear circuit by successive differentiation of its unit ramp response.

Example: 10.7-1

An inductor, resistor and a unit impulse current source are connected in parallel. Find the zero-state impulse response for resistor current and inductor current in this circuit.

Solution

The impulse current cannot flow through L and hence it flows through R producing an impulse voltage Rδ(t) across the combination. This impulse voltage has an area content of R V-s. R V-s dumped into an inductor of L Henry will produce a change in its current by R/L A. Since initial condition for zero-state response is zero the current of inductor at t = 0 + will be this change amount itself – i.e., R/L A. Thereafter, it is a source-free circuit since impulse current source is an open circuit after t = 0 +. Therefore, the current in both R and L will be (1) e–t/τwhere τ = L/R.

Example: 10.7-2

An inductor, resistor and a 1A DC current source are connected in parallel at t = 0. The initial condition for inductor is specified as Io. Find the time-domain expressions for current through L and R and the voltage across the combination. Also, find the total energy delivered by the current source, energy consumed by the resistor and the energy stored in the inductor for a special case where the initial condition is zero.

Solution

We use the result from previous example in solving this. The total response for inductor current is the sum of zero-input response and zero-state response. A series RL circuit with some non-zero initial condition gets formed when current source at input is set to zero. Therefore, the zero-input response of this circuit is same as the zero-input response of an RL circuit we have discussed so far. The zero-state response for switched DC current input can be found by integrating the impulse response with current input.

The total energy delivered by source is found by integrating the power delivered from 0 to ∞. The power delivered is given by voltage across the combination × source current value. Similarly, the energy dissipated in R and stored in L can be found by integrating vR iR and vL iL from 0 to ∞. The integral must return L/2 in the case of inductor because that is the stored energy in an inductor carrying 1A. Let us call these three energy values ES, ER and EL respectively. Then,

If the current source had a magnitude of I instead of 1, all the three energy terms are to be multiplied by I2. Hence, we see that (i) it takes LI2/2 joules of energy dissipation in the parallel resistance whenever we charge an inductor to I A and this dissipated energy is as much as the energy stored in the inductor and (ii) this energy dissipation is independent of R value. The value of R will decide the time taken to dissipate this amount of energy but not the amount of energy.

Example: 10.7-3

Find the inductor current response for t ≥ 0 + with the specified initial condition in the circuit in Fig. 10.7-4 (a). The part of the circuit that is needed to create this initial current in the inductor is not shown in the diagram.

Fig. 10.7-4 Circuit for Example: 10.7-3

Solution

If the unit impulse voltage source is the only source in the circuit, the inductor current at t = 0 can only be 0. The fact that a non-zero current is specified at t = 0 indicates that there must have been other sources and elements that were active for t < 0 in the circuit and that they were switched out before t = 0, leaving the inductor with a trapped current of 2A.

The time constant of the circuit relevant for t ≥ 0 + is found by deactivating sources and finding out the equivalent resistance connected across the inductor. The equivalent resistance is 15Ω and the time constant is 13.33ms. The Thevenin's equivalent of the circuit connected across inductor is shown in Fig. 10.7-4 (b).

The 0.5δ(t) voltage source appears across the inductor forcing its current to change by 0.5/0.2 = 2.5 A. The inductor current at t = 0 is specified as 2 A in the opposite direction. Hence, the inductor current at t = 0 + = –2 + 2.5 = 0.5 A. After t = 0 +, the circuit is a source-free series RL circuit and the inductor current will be 0.5 e– 75 t A for t ≥ 0 +. The inductor current for t < 0 cannot be determined since there were unknown elements active (otherwise, the current in the inductor current would not have been non-zero at t = 0) in the circuit during t < 0.

Example: 10.7-4

Find the inductor current and voltage across current source as a function of time in the circuit in Fig. 10.7-5 (a). The part of the circuit that is needed to create the initial current in the inductor is not shown in the diagram. Also, find what must be the initial current in the inductor so that the inductor current will be transient-free for t ≥ 0 +

Fig. 10.7-5 Circuits for Example: 10.7-4

Solution

Time constant of the circuit is 0.01s.The zero-input iL response is 1e100 t A.Zero-state response when the voltage source is acting alone is iL = (20/20) (1 – e–100t) A. The zero-state response when the current source is acting alone is iL = 1(1 – e–100 t) A.

∴ Total response for iL = 2 – e–100 t A for t ≥ 0 +

Zero-input vcs response = voltage across the first 10Ω resistance due to zero-input response in iL = – 10 e–100t A.

Zero-state vcs response when the voltage source is acting alone is 20 – 10 × zero-state iL for the same condition = 20 –10(1 – e–100t) = 10(1 + e–100t) V

Zero-state vcs response when the current source is acting alone is 10 × (2 – zero-state iL for the same condition) = 10 × (2– (1 – e–100t)) = 10(1 + e–100t) V

∴ Total response for vcs = 20 – 10 e–100t V for t ≥ 0 +

We could have obtained vcs as (10 × total response in iL + 0.2 × first derivative of total response in iL) too.

Inductor current will be transient free if the natural response terms in zero-input response cancel out the natural response terms in total zero-state response. The natural response terms in total zero-state response in this case is –2 e–100 t and hence initial current must be 2A for transient-free inductor current in this circuit.

Another point of view would be that the inductor current will be transient-free if the initial current and final current (i.e., steady-state current for DC steady state) are the same. In general, inductor current will be transient-free if the initial current at t = 0 + and the value of forced response component (this need not be a DC component) at t = 0 + are equal. The DC steady state may be obtained from circuit Fig. 10.7-5 (b). Applying superposition principle, we get the steady-state value of iL as (20V/20Ω) + 2 A × (10Ω/20Ω) = 2 A. Thus, the required initial current in the inductor for transient-free response will be 2 A again.

A single-inductor circuit can be described by a first-order differential equation on inductor current. It will be possible to work out the other circuit variables from inductor current alone. Therefore, it follows that if inductor current is transient-free, so will all the other circuit variables be.

10.8 SERIES RL CIRCUIT WITH EXPONENTIAL INPUTS

We take up the study of zero-state response of series RL circuit for exponential inputs and sinusoidal inputs in this section. We do not worry about the zero-input response anymore since we know that it is I0 e–t/τ where I0 is the initial current specified at t = 0. The total response is found by adding zero-input response and zero-state response together.

We permit exponential inputs of the form est u(t) in this section where s can be a complex number s = –σ + where σ and ω are two real numbers. The reason for generalising the exponential input in this manner will be clear soon. We put a -ve sign behind σ since we want the real part of s to be negative when σ is positive. This signal is a complex signal with a real part function and imaginary part function as shown in Eqn. 10.8-1.

We have used Euler's identity in expressing complex exponential function in the rectangular form involving trigonometric functions. No signal like this can be actually generated by a physical system since there can be nothing 'imaginary' about a physical signal. Therefore, this signal does not represent a physical signal and hence it cannot really be applied to any circuit. But it can be the forcing function in a differential equation – the mathematics of differential equation will not complain. Moreover, its real part and imaginary part are real functions only and can be made physically.

We know that zero-state response of a linear circuit obeys superposition principle. Let us apply a signal x(t) to the circuit and let the zero-state response be ix(t). Similarly let the zero-state response be iy(t) when a signal y(t) is applied to the same circuit. Now, what is the response when jy(t) is applied where j = √–1 ? By superposition principle it will be jiy(t). When we apply x(t) + jy(t) as input we must get ix(t) + jiy(t) as response by superposition principle again.

 

The real part of zero-state response for a complex signal input is the zero-state response for the real part of input and the imaginary part of zero-state response for a complex signal input is the zero-state response for the imaginary part of input. This is a direct consequence of linearity of the circuit.

Now we can see why we opted for this complex signal. We can obtain the zero-state response of circuits to sinusoidal inputs by taking the real or imaginary part of zero-state response for complex exponential inputs with σ = 0. It is easier to deal with complex exponential function than with sinusoids when it comes to solving differential equations. Let us solve for the zero-state response of series RL circuit with a complex exponential input.

This equation has to be true for all t and in particular for all t ≥ 0 +. This is possible only if the shape of functions on both sides of the equation is the same. This will imply that the trial solution for particular integral has to be Aest. Substituting this trial solution in the differential equation in Eqn. 10.8-2,

We have to form the total solution by adding this particular integral to complementary solution. Note that finding zero-state response involves finding a particular integral and a suitable complementary function also – because zero-state response involves both forced response and natural response terms.

Substituting initial current value at 0 +,

10.8.1 Zero-State Response for Real Exponential Input

We consider a special case of complex exponential with ω = 0. The input source function is then of the form e–σt u(t). With a positive value of σ, this input function has the same format as that of the impulse response of a series RL circuit (impulse response of any first-order circuit for that matter). Hence, the problem we are trying to solve may be thought of as a situation where the impulse response of one circuit is applied to a second circuit as input. Such situations arise when we cascade different circuits. The zero-state response is obtained by putting s = –σ in Eqn. 10.8-4 and is expressed as in Eqn. 10.8-5.

We note the ασ in the denominator and raise the question – what is the response when the real exponential input has the same index as that of impulse response of the circuit ? It is not infinite as Eqn. 10.8-5 would suggest. The correct answer is that this is one situation under which our assumed trial particular integral is incorrect. We have to find the particular integral afresh. There are well-established methods to arrive at particular integral under this kind of situation in mathematics. But we do not take up this case here. We will get at it later chapter and solve it without having to learn a new method for that. Therefore, we qualify the solution in Eqn. 10.8-5 by specifying that solution is valid only if α σ.

The input waveform is exponential and we can think of a time constant for this waveform too. Let us denote this time constant as τe with the subscript reminding us that this is the time constant of an excitation waveform. Obviously, τe = 1/σ. Further, we normalise the current in Eqn. 10.8-6 using a normalisation base of 1/R A and time by a normalisation base of τ s to put the solution in terms of the normalised variables as in Eqn. 10.8-7 where the subscript n indicates normalised variables.

The input functions and iL waveforms are shown for two cases in Fig. 10.8-1.

Fig. 10.8-1 Zero state response for exponential input (a) Input wave (b) Normalised inductor current

The time instants at which response reaches maximum and the maximum response are marked. General expressions for these may be derived by differentiating the function in Eqn. 10.8-7 and setting the derivative to zero. Note that sending τe to ∞ amounts to applying a unit step input and zero-state response indeed approaches the unit step zero-state response as expected.

10.8.2 Zero-State Response for Sinusoidal Input

The input function in this case is vS(t) = sinωt u(t) V. We can get sinusoidal zero-state response for sinusoidal input by two methods. In the first method, we use the result in Eqn. 10.8-4 with s = jωt and take the imaginary part of the result as our desired solution.

In the second method, we represent the input sine wave by combining two complex exponential functions by using Euler's identity as below.

Then, we get the zero-state response for the two exponential inputs separately and use superposition principle to arrive at the solution for sinusoidal input.

The angle ϕ in Eqn. 10.8-9 is defined the same way as in Eqn. 10.8-8. Both methods lead to same expression for the final response, as they should. The final expression may be recast in the following form where k = ωτ and the current and time are normalised with respect to 1/R and τ, respectively.

This waveform for a case with k = 4 is shown in Fig. 10.8-2. The number k can be interpreted as a comparison between the characteristic time, i.e., the period of the applied voltage and the characteristic time of the circuit, i.e., its time constant. k can be expressed as 2π(τ/T) where T is the period of input. Sinusoids undergo a full cycle of variation in one T and hence the value of T is indicative of the rate of change involved in the waveform, i.e., the speed of the waveform. Time constant is a measure of inertia in the system. Therefore, an input sinusoid is too fast for a circuit to follow if its T is smaller than the time constant τ of the circuit. Similarly, if input sinusoid has a T value much larger than time constant of the circuit, the circuit will perceive it as a very slow waveform and will respond almost the same way it does to DC input. These aspects are clearly brought out in Eqn. 10.8-10.

Fig. 10.8-2 Unit sinusoidal response of RL circuit with k = 4

The amplitude of forced response component, or equivalently, the amplitude of sinusoidal steady-state response, is a strong function of k. The amplitude decreases with increasing ω or increasing τ. In addition, the steady-state current lags the applied voltage by a phase angle that increases with ωτ.

Let us imagine that we conduct an experiment. We apply a sinusoidal voltage of 1V amplitude to a series RL circuit and wait for enough time for the transient response to die down. After steady state is satisfactorily established in the circuit, we measure the amplitude of current and its phase with respect to the input sine wave. We repeat this process for various values of frequency of input, keeping its amplitude at 1V always. We ensure that the circuit is in steady-state before we measure the output every time.

The data so obtained can be plotted to show the variation of ratio of output amplitude to input amplitude and phase of steady-state current against k (= ωτ). Such a pair of plots will constitute what is called the AC steady-state frequency response plots for this RL circuit. The ratio of output amplitude to input amplitude is called gain of the circuit. Its dimension will depend on the nature of input and output quantities. If we define a complex function of ω with the magnitude of function equal to the gain described here and angle of the function equal to the phase angle by which the steady-state output leads the sinusoidal input, the resulting function will be what we termed the frequency response function H() we described in Chapter 9.

Such an experiment can be performed on any circuit to get its frequency response data. However, if the differential equation of the circuit is known we need not do the experiment. The frequency response plots can be obtained analytically in that case. The frequency response of the series RL circuit is shown in Fig. 10.8-3 as an example.

Fig. 10.8-3 Frequency response plots for series RL circuit

We make the following observations on the sinusoidal steady-state response of series RL circuit from Eqn. 10.8-10 and Fig. 10.8-3.

  • The circuit current under sinusoidal steady-state response is a sinusoid at the same angular frequency ω rad/s as that of input sinusoid.
  • The circuit current initially is a mixture of an exponentially decaying unidirectional transient component along with the steady-state sinusoidal component. This unidirectional transient imparts an offset to the circuit current during the initial period.
  • The circuit current at its first peak can go close to twice its steady-state amplitude in the case of circuits with ωτ >> 1 due to this offset.
  • The amplitude of sinusoidal steady-state response is always less than corresponding amplitude when DC input of same amplitude is applied. This is due to the inductive inertia of the circuit. The amplitude depends on the product ωτ and decreases monotonically with the ωτ product for fixed input amplitude.
  • The response sinusoid lags behind the input sinusoid under steady-state conditions by a phase angle that increases monotonically with the product ωτ.
  • The frequency at which the circuit gain becomes 1/√2 times that of DC gain is termed as cut-off frequency and since this takes place as we go up in frequency it is called upper cut-off frequency. Upper cut-off frequency of series RL circuit is seen to be at ω = 1/τ rad/s. The phase delay at this frequency will be – 45°.
  • Circuit current amplitude becomes very small at high frequencies (ωτ >> 1) and the current lags the input voltage by ≈90° at such frequencies.

We assumed that the applied voltage is vS(t) = sinωt u(t) throughout this analysis. This means that the sinusoidal voltage happened to be crossing the time-axis exactly at the instant at which we closed the switch to apply it to the circuit. Though technically it is possible to do such switching (it is done in some applications that way), that is not the way it takes place in many practical applications. The sinusoid may be at any value between its maximum and minimum when we throw the switch. Therefore, we must analyse the response with vS(t) = sin (ωt +θ)u(t) for an arbitrary θ.

The function sin (ωt + θ) is the imaginary part of ej(ωt+ θ) = cos (ωt + θ) + j sin (ωt + θ). The zero-state response when est is applied to the circuit is given by Eqn. 10.8-4. So, shall we substitute s = j(ω + θ) in Eqn. 10.8-4 to get the required output? No, that will be wrong since it is only ω that gets multiplied by t, not θ. We should (i) interpret ej(ωt+ θ)as e ejωt, (ii) solve for the zero-state response for ejωt, (iii) multiply the response by e (we apply principle of homogeneity there) and (iv) take the imaginary part. We skip all that basic algebra and give the result below.

By substituting θ = 90°, we get the solution for vS(t) = cosω t u(t) as

Eqn. 10.8-11 indicates that it is possible to switch on an AC voltage to an initially relaxed series RL circuit in such a way that there is no transient response and circuit immediately goes to steady state – the switching instant must be such that θ = ϕ. This principle is used sometimes in switching of heavily inductive power equipment. In such cases, the angle ϕ is close to 90° and transient-free switching is possible if the voltage is switched on to the equipment at positive or negative peak.

Note that the sinusoidal steady-state response part of zero-state response in all the above cases could have been readily obtained by employing phasor method we studied in earlier chapters.

10.9 GENERAL ANALYSIS PROCEDURE FOR SINGLE TIME CONSTANT RL CIRCUITS

We have analysed the simple series RL circuit exhaustively. We would like to arrive at a generalised procedure for analysis of RL circuits that may contain more than one resistor. The circuit may also contain more than one inductor; but in that case we assume that the inductors will enter in series or parallel combinations and can be finally be replaced by a single inductor. If that is not possible, the circuit will have more than one time constant and we cannot handle it by the method we develop here.

 

A circuit with a single energy storage element can have only first-order dynamics and only one time constant. A circuit with only one time constant can have only one term in its natural response for any circuit variable. That term is of the form Ae–t/τ where A is to be adjusted for compliance with initial condition at t = 0 + for that particular circuit variable.

The single time constant that describes the natural response of the circuit is a property of circuit alone and does not depend on source function values. Therefore, it can be found from the dead circuit obtained by deactivating all independent sources. We get the dead circuit by replacing all independent voltage sources by short-circuits and all independent current sources by open circuits. Now, that circuit will contain only one inductor and possibly many resistors (and dependent sources, if they were present in the original circuit). We can find the equivalent resistance connected across the inductor by series/parallel combinations and by star-delta transformation if necessary. If the circuit contains linear dependent sources we may have employ unit current injection method or unit voltage application method detailed in Chapter 5 to find the equivalent resistance across inductor. Once we get this resistance, we can find time constant by τ = L/Req.

Next step is to check whether there are impulse sources in the circuit. If there are, the amount of V-s dumped on the inductor at t = 0 has to be evaluated and the change in inductor current at t = 0 has to be found out. This change in current added to the initial condition specified at t = 0 will give us the value of inductor current at t = 0 +. But we need the initial condition for the particular circuit variable we are solving for. Therefore, we have to carry out a DC circuit analysis at t = 0 + in which the inductor is replaced by a DC current source of value equal to its current at t = 0 + and all the independent sources are replaced by DC sources (if they are not DC sources already) of value equal to their values at t = 0 +. Solving the resulting circuit will give us the initial condition for the particular circuit variable of interest.

In the third step, we work out the forced response for all independent sources in the circuit. Since the circuit has a steady-state for DC and AC, we use steady-state analysis for this. In particular, if the sources are DC sources, we replace inductor by short-circuit and solve the resulting resistive circuit for the variable of interest. We may use superposition principle along with mesh or node analysis for this purpose. We employ phasor analysis to solve for steady-state if there are AC sources. The steady-state solution for all the independent sources are added up.

We add the total steady-state solution to the natural response, apply the initial condition for the relevant variable and evaluate the arbitrary constant in the natural response term in the last step. We can also use the zero-input response plus zero-state response method instead of transient response plus forced response method. These methods are illustrated through a set of examples that follow.

Example: 10.9-1

Find ix (t) in the circuit in Fig. 10.9-1 (a). The initial condition in the 0.2H inductor is 1A in the direction shown at t = 0. The circuit elements that create this initial current are not shown in the diagram.

Fig. 10.9-1 Circuit for Example: 10.9-1

Solution

The circuit contains two switched DC sources – a 2V-voltage source and a 0.5A current source. There is only one time constant.

Step 1 – Find the time constant

The independent voltage source is replaced by a short-circuit and the independent current source is replaced by an open circuit to get the circuit in Fig. 10.9-1 (b). The time constant is found by evaluating the equivalent resistance across the inductor.

Step 2 – Find the initial value for ix at t = 0 +

The value of voltage source at t = 0 + is 2V. The value of current source at t = 0 + is 0.5A. The inductor current at t = 0 + is same as at t = 0 since there is no impulse source in the circuit. Hence, inductor is replaced by a current source of 1A for t = 0 +. The resulting circuit appears in Fig. 10.9-2 (a).

Fig. 10.9-2 Circuits for Step 2 of Example: 10.9-1

This circuit contains three sources and may be solved by using superposition principle. The three circuits in which the sources act one by one are shown in Fig. 10.9-2 (b)-(d). Three contributions to the current ix may be found from these single-source circuits by simple circuit analysis.

Step 3 – Find the steady-state value of ix

Both sources in the circuit are DC sources and hence they can be handled together for steady-state calculation. The inductor behaves as a short-circuit for DC steady state. Replacing the inductor by short-circuit and using the known values of source functions we get the circuit in Fig. 10.9-3 (a).

This circuit involves two sources and can be solved by using superposition principle. The sub-circuits needed for that is shown in Fig. 10.9-3 (b) and (c).

Fig. 10.9-3 Circuits for Step 3 of Example: 10.9-1

Step 3 – Form the total solution and adjust initial condition

The transient response is of the form A e-10 t since time constant is 0.1s

The current waveform is shown in Fig. 10.9-4.

Fig. 10.9-4 The plot of current waveform for Example: 10.9-1

We have calculated the required current. We obtained the solution in the form of transient response + forced response (or steady-state response). However, this form of solution is not the appropriate form if we are required to answer supplementary questions on the problem. For example, can we work out the current if the initial condition of inductor is changed to 2A? Or if the voltage source value is changed to 5V? We will have to rework the problem almost entirely to get the answer for that. Note that it is not possible to decompose a solution into zero-input response and zero-state response components if the solution is available in the transient response + forced response format. However, transient response and forced response can be obtained from the solution in zero-input response + zero-state response format. This is where the superiority of zero-input response + zero-state response approach becomes evident. We work out the same example to see these two components in the solution in the next example.

Example: 10.9-2

Obtain the solution in zero-input response + zero-state response form for the problem stated in Example 10.9-1 and modify the solution for (i) initial condition changed to 2A, (ii) voltage source magnitude changed to 5V and (iii) voltage source magnitude changed to 4V and current source magnitude changed to 1A along with initial condition changed to 2A.

Solution

When we want the solution in zero-input response + zero-state response format, we have to split the given circuit into two sub circuits right at the outset – one containing all independent sources and with zero initial condition for inductor current and the second with all independent sources deactivated and initial condition for inductor current at the specified value.

These two circuits are shown in Fig. 10.9-5 (a) and (b). The solution of Fig. 10.9-5 (b) gives the zero-input response. The solution of Fig. 10.9-5 (a) gives zero-state response. Zero-state response obeys superposition principle. Hence, the zero-state response of circuit in Fig. 10.9-5 (a) with the two sources acting simultaneously can be obtained by summing the zero-state responses when they are acting alone. The circuits required for finding out these individual zero-state responses are given in Fig. 10.9-5 (c) and (d).

Step 1 – Find the time constant

This step is the same as in Example: 10.9-1 and the value of time constant τ = 0.1 s.

Step 2 – Find the zero-input response

The initial value of ix at t = 0 + is 0.5A from circuit in Fig. 10.9-5 (b). Since there is no source, the particular integral will be zero. Therefore, the zero-input response is 0.5 e–10 t.

Step 3 – Find the zero-state response in circuit in Fig. 10.9-5 (c)

The initial value of ix at t = 0 + is 0.5A from circuit in Fig. 10.9-5 (c). Final steady-state value is obtained by replacing the inductor by short-circuit and solving the resulting resistive circuit. 2/(2 + 2//1) is the current from voltage source. This gets divided between 2Ω and 1Ω. The value is 0.25A. Therefore, the solution is = 0.25(1 + e–10 t).

Fig. 10.9-5 Circuits for solving zero-input and zero-state responses in Example: 10.9-2

Step 4 – Find the zero-state response in circuit in Fig. 10.9-5 (d)

The initial value of ix at t = 0 + is 0.25A from circuit in Fig. 10.9-5 (d). Final steady-state value is obtained by replacing the inductor by short-circuit and solving the resulting resistive circuit. Applying current division principle, we get the value as 0.125A. Therefore, the solution is = 0.125(1 + e–10 t).

Step 5 – Find the total zero-state response in circuit Fig. 10.9-5 (a)

Total zero-state response is obtained by adding the two zero-state responses obtained in the last two steps. It is 0.25(1 + e–10 t) + 0.125(1 + e–10 t) = 0.375(1 + e–10 t).

Step 6 – Find the total response in the original circuit

This is obtained by adding the zero-input response obtained in Step-1 and total zero-state response obtained in Step 5. It is 0.5 e–10 t + 0.375(1 + e–10 t) = 0.375 + 0.875 e–10 t. It is the same expression we obtained in Example 10.9-1.

  1. If initial condition is changed to 2A

    This change will affect only zero-input response. Zero-input response obeys superposition principle. Therefore, the zero-input response becomes (0.5 e–10 t) x 2/1 = e–10 t.
    Therefore, ix (t) = e–10 t + 0.375(1 + e–10 t) = 0.375 + 1.375 e–10 t A for t ≥ 0 +.

  2. If voltage source value is changed to 5V

    This change will affect the zero-state response contribution from voltage source only. It was 0.25(1 + e–10t) when voltage source value was 2V. Therefore, applying superposition principle, it will be 2.5 times this function with voltage source value of 5V.

    Therefore,

    ix (t) = 0.5 e–10 t + 2.5x 0.25(1 + e–10 t) + 0.125(1 + e–10 t) = 0.75 + 1.25 e–10 t A for t ≥ 0 +.

  3. If voltage source value is changed to 4V, current source value changed to 1A and initial condition changed to 2A

The solution will get affected in all the three components. Zero-input response gets scaled by 2/1, zero-state response from voltage source gets multiplied by 4/2 and zero-state response from current source gets multiplied by 1/0.5.

Therefore,

ix (t) = 2x0.5 e–10 t + 2x 0.25(1 + e–10 t) + 2x0.125(1 + e–10 t) = 0.75 + 1.75 e–10 t A for t ≥ 0 +

Example: 10.9-3

Solve for vo(t) as a function of time in the circuit in Fig. 10.9-6 (a). Also, identify (i) transient response and steady-state response components, (ii) zero-input response and zero-state response components in the total response and (iii) contributions to zero-state response from the individual sources. The source functions are

 

Vs1(t) = 10sin(10t+ π / 4) u(t) V
    i
s(t) = 2cos(10t – π / 3) u(t) A

Fig. 10.9-6 Circuit for Example: 10.9-3

Solution

There are two switched sinusoidal sources in the circuit – one independent voltage source and one independent current source. They are of same angular frequency of 10 rad/s. Since we are required to identify the zero-input response and zero-state response components we have to solve two circuits – one with all sources and zero initial condition for inductor and another with all independent sources deactivated and specified initial condition for inductor current.

Step 1 – Find the time constant

The deactivated circuit for finding the time constant is shown in Fig. 10.9-6 (b). The time constant is 0.25/(10//10) = 0.05s.

Step 2 – Identify circuits for zero-input response and zero-state response and solve for zero-input response.

These circuits are shown in Fig. 10.9-7. Circuit in Fig. 10.9-7 (a) has to be used for determining zero-state response and circuit in Fig. 10.9-7 (b) has to be used for zero-input response.

The circuit does not apply any impulse to the inductor. Therefore, the inductor current at t = 0 + is same as at t = 0 and will be 1A in the marked direction. But what we need is the initial condition at t = 0 + for our response variable vo. The 1A initial current in inductor at t = 0 + gets divided equally between the two equal resistors and develops a potential of 5V across them at that instant. Hence, the required initial condition for vo = 5V.

Fig. 10.9-7 Circuits for determining various response components in Example: 10.9-3

This is a single time constant circuit and hence zero-input response in any circuit variable will be of the form Ae–t/τ where A has to be adjusted for compliance with initial condition on the chosen output variable. Therefore, zero-input response component in vo is 5e–20t V since initial condition is 5V and time constant is 0.05s.

Step 3 – Identify circuits for obtaining zero-state response components and solve them.

Three sub-circuits derived from circuit in Fig. 10.9-7 (a) for evaluating the zero-state response contributions from individual sources are shown in Fig. 10.9-8.

Fig. 10.9-8 Circuits for obtaining zero-state response components in Example: 10.9-3

We employ phasor method to solve for the sinusoidal steady-state response in each circuit, add a transient response term of Ae–t/τ format and evaluate A by substituting suitable initial condition.

Step 3a – Zero-state response in Fig. 10.9-8 (a)

The circuits for evaluation of initial condition and sinusoidal steady-state response for this circuit is shown in Fig. 10.9-9. Circuit in Fig. 10.9-9 (b) shows the circuit for initial condition evaluation. The voltage source is replaced by a DC source of value same as the value of vS(t) at t = 0 + i.e., 10 sin π/4 = 7.07 V. The inductor is replaced by a DC current source of value equal to its initial condition. But in a circuit for evaluating zero-state response initial condition for inductor current will be zero. Therefore, it is a current source of zero value, i.e., it is an open circuit. Therefore, vo at t = 0 + is 7.07/2 = 3.535 V.

The angular frequency of vS1(t) is 10rad/s. Therefore, inductor will have an inductive reactance of 10 x 0.25 = 2.5 Ω. The amplitude of vS1(t) is 10 V. In phasor equivalent circuit, we usually specify rms value of sources. Hence, 10/√2 at 45o is the voltage source value in the circuit used for sinusoidal steady-state response evaluation. The sinusoidal steady-state response is evaluated as below.

Fig. 10.9-9 Circuits for zero-state response due to voltage source in Example: 10.9-3

The zero-state response due to voltage source is obtained by adding this solution to transient response term and evaluating the arbitrary constant in the transient response term by using the initial condition for vo(t) we have already calculated.

Step 3b – Zero-state response in Fig. 10.9-8 (b)

The circuits for evaluation of initial condition and sinusoidal steady-state response for this circuit is shown in Fig. 10.9-10. Circuit in Fig. 10.9-10 (a) shows the circuit for initial condition evaluation. The current source is replaced by a DC current source of value same as the value of iS (t) at t = 0 + i.e., 2 cos (–π /3) = 1 A. The inductor is replaced by a DC current source of value zero. Therefore, vo at t = 0 + is 5 V.

The angular frequency of iS(t) is 10-rad/s. Therefore, inductor will have an inductive reactance of 10 × 0.25 = 2.5 Ω. The amplitude of iS (t) is 2A. Hence, 2/√2 at –60o is the current source value in the circuit used for sinusoidal steady-state response evaluation. The sinusoidal steady-state response is evaluated as below.

Fig. 10.9-10 Circuits for zero-state response due to current source in Example: 10.9-3

The zero-state response due to voltage source is obtained by adding this solution to transient response term and evaluating the arbitrary constant in the transient response term by using the initial condition for vo(t) we have already calculated.

Step 4 – Get total response and identify various components

The zero-input response = 5 e20t A

The zero-state response due to vS(t) = 1.41 e–20t + 2.24sin (10t + 108.44°) V

The zero-state response due to iS (t) = 0.54 e–20t + 4.47cos (10t+3.440) A

... The total response and its components are

Example: 10.9-4

Find expressions for inductor current and voltage in the circuit in Fig. 10.9-11 (a) for t ≥ 0 + and plot them in the range 0 + ≤ t ≤ 1s.

Solution

The voltage source in this circuit is specified as 15 u(0.5–t). u(x) is 0 for x ≤ 0, 1 for x ≥ 0 + and undefined at x = 0. Therefore, u(0.5-t) is a time function which is 0 for t ≥ 0.5+, 1 for t ≤ 0.5and is undefined at t = 0.5. This waveform is shown in Fig. 10.9-11 (b). Physically, it implies that 15V DC source was connected to the circuit at infinite past and it is removed and instead a short-circuit is put across the circuit at t = 0.5s.

Since the 15V source was connected long back the circuit had enough time to reach DC steady-state by the time t becomes zero. Therefore, the initial inductor current at t = 0 can be obtained by solving the resistive circuit with inductor replaced by short-circuit. This will be 0.5 A. The relevant circuit appears in Fig. 10.9-12 (a).

Fig. 10.9-11 Circuit for Example: 10.9-4

The switch closes and puts 15 Ω resistance into the circuit at t = 0. A new steady-state will be established in the circuit if there is no further change in the circuit. But the circuit cannot anticipate that a structural change or a change in source functions is going to take place in future and modify its response in the present based on such anticipation. Therefore, the evolution of circuit variables will be towards the expected steady-state commensurate with the current nature and values of source functions. In the present example, we know that the circuit will not be allowed to reach steady-state since the input is going to go down to zero at 0.5 s and the circuit will start on a new transient at that instant. But that does not prevent us from asking the question – what would have been the final value had the circuit been allowed to reach there? – Because the circuit is going to move only along that waveform up to 0.5s. This expected final value, had the circuit been allowed to proceed to steady-state, can be worked out by replacing inductor by short-circuit. This circuit along with solution is shown in Fig. 10.9-12 (b).

Fig. 10.9-12 Initial and final value evaluation for the circuit in Example: 10.9-4

The time constant relevant for t ≥ 0 + can be found by setting the voltage source value to zero with the 15 Ω already connected in place in Fig. 10.9-11 (a). The equivalent resistance across inductor will be 7.5 Ω and τ will be 0.1s.

Now, the expression for i in the time range 0 + ≤ t ≤ 0.5 can be found as below.

At t = 0.5s, the input voltage goes to zero. A new transient in a source-free circuit starts at t = 0.5 +. Since there is no impulse voltage involved in the circuit, the inductor current will remain continuous between 0.5 and 0.5 +. The value at 0.5 can be found by evaluating i(t) at that instant using the expression for i(t) in Eqn. 10.9-1. Further, in a source-free circuit there is no forced response term.

The plots of inductor current and voltage across inductor are given in Fig. 10.9-13.

Fig. 10.9-13 Inductor current and voltage in Example: 10.9-4

10.10 SUMMARY
  • Circuits containing energy storage elements have memory in time-domain. They are described by linear ordinary differential equations with constant coefficients. We need to know the forcing function from some time instant onwards along with initial conditions in the circuit specified at that instant to solve such differential equations.
  • We focussed on series RL circuit in this chapter. RL circuit is described by a first-order linear differential equation. The past history of inductor is contained in a single initial condition specification in an RL circuit.
  • The solution of the differential equation describing the inductor current in an RL circuit contains two terms – the complementary function and particular integral. Complementary function is the solution of differential equation with zero forcing function. Particular integral is the solution due to the input function and is defined in the domain of input function. The total solution is obtained by adding these two. The complementary function has arbitrary amplitude that should be fixed by ensuring that the total solution complies with the specified initial condition.
  • The circuit variables in the RL circuit will contain two response components – transient response (also called natural response) and forced response. Natural response is the way in which the inertia in the circuit reacts to forcing function's command to change. Complementary solution gives the natural response and particular integral gives the forced response in a circuit.
  • The nature of natural response of a linear time-invariant circuit is independent of the type or magnitude of forcing function and depends only on circuit parameters and nature of interconnections. Natural response in RL circuit is exponential of the form A e–t/τ where τ = L/R is defined as time constant of the circuit. A is to be fixed for compliance with initial condition.
  • The initial current in an RL circuit at t = 0 and t = 0 + are the same if the circuit does not contain impulse sources and it cannot support impulse voltages.
  • Step response of a circuit is its response when unit step input is applied. In the case of an RL circuit, step response is a rising exponential, approaching a steady-state value asymptotically as t → ∞. The step response never gets done. However, it may be considered to be over within 5 time constants for practical purposes.
  • Time constant can be understood as the additional time required from the current instant for the step response to reach the final value, assuming that the rate of rise of response is held constant at its current value from that instant onwards.
  • Free response of an RL circuit is its response when input is zero and there is some initial energy trapped in the inductor. It will contain only natural response terms. The inductor current in this case falls exponentially towards zero.
  • If all the transient response terms are of vanishing nature, the only remaining response in the long run will be the forced response component. Then, the forced response component is termed as the steady-state response provided there are constant features describing the forcing function. Three kinds of steady-state are usually studied in circuits – DC steady-state, AC steady-state and periodic steady-state. Inductors can be replaced by short-circuits for DC steady-state analysis. AC steady-state analysis can be carried out using phasor analysis.
  • Impulse response of a circuit is its response when a unit impulse input is applied to it. Impulse response of circuits will contain only natural response terms.
  • The response due to initial energy and the response due to application of impulse are indistinguishable in an RL circuit and hence they can be replaced for each other. An initial current of I0 in an inductor of value L can be replaced by zero initial condition with a voltage source LI0 δ(t) connected in series with the inductor.
  • Step response and ramp response in an RL circuit can be obtained by integrating its impulse response successively.
  • Zero-input response of a circuit is its response when there is no input but there is initial energy. It will contain only natural response terms. Zero-state response is the response when the circuit is initially at rest (zero initial conditions) and input is applied. It will contain both natural response terms and forced response terms. The total response is given by sum of zero-input response and zero-state response.
  • Forced response (and hence steady-state response) obeys superposition principle with respect to input source functions. But transient response and total response do not obey superposition principle – neither with respect to initial conditions nor with respect to input source functions. However, zero-input response obeys superposition principle with respect to initial conditions and zero-state response obeys superposition principle with respect to input source functions.
  • Total response in single-inductor, multi-resistor circuits can be found with the help of superposition principle and Thevenin's theorem by evaluating zero-input response for the entire circuit and zero-state response for each source separately.
10.11 PROBLEMS
  1. What is the differential equation describing iL for t ≥ 0 + in the circuit in Fig. 10.11-1?

    Fig. 10.11-1

  2. A series RL circuit with non-zero initial energy is driven by a unit step voltage source. The circuit current is found to reach 75% of its steady-state value in one time constant. Express the initial current in the inductor as a percentage of its steady-state value.
  3. A DC voltage source of 1V is switched on to a series RL circuit at t = 0. iL is found to be 50% at t = τ. Was there any initial current in the inductor? If so, what is its magnitude and direction?
  4. Find the ratio E/V such that iL in Fig. 10.11-2 becomes V/R at t = 0.5 + τ where τ = L/R.

    Fig. 10.11-2

  5. What is the time required for the energy stored in the inductor in a series RL circuit to reach 90% of its steady-state value in step response?
  6. iL in the circuit in Fig. 10.11-3 is –2A at t = 0 due to other sources that are not shown, but are switched out at t = 0. It is found that iL(t) = 0 for t ≥ 0 + . What is the value of L?

    Fig. 10.11-3

  7. iL in the circuit in Fig. 10.11-4 at t = 0 is –2A. It is found that iL(t) = 0 for t ≥ 0 +. What is the value of time constant of the circuit?

    Fig. 10.11-4

  8. A series RL circuit with non-zero initial energy is driven by a 1V DC switched source. The inductor current is found to be (2 – e– t) A. What are the values of inductance, resistance and initial current?
  9. A series RL circuit with non-zero initial energy is driven by a 1V DC switched source. The inductor current is found to be 3A for t ≥ 0 +. What is the value of initial current and what is its direction relative to the 3A current?
  10. An AC voltage source = V sin (ωt) is applied to a series RL circuit from t = 0.The circuit current is found to be = 0.7sin (ωtπ/3) A. Was there any initial current in the inductor? If yes, what is its magnitude and relative direction?
  11. A series RL circuit with non-zero initial energy is driven by a 1V DC switched source. The inductor current is found to be 2A. What is the new iL in the circuit if (i) the initial condition is doubled and (ii) if a 2V DC switched source is applied with no change in initial condition?
  12. The inductor current = 0 for t ≥ 0 + in the circuit in Fig. 10.11-5. Find the values of I0 and IS. I0 is created by other circuit elements that are switched out at t = 0. They are not shown in the diagram.

    Fig. 10.11-5

  13. A series RL circuit is driven by vS(t) = δ(t) – δ(t – 1). Its time constant is 1 s. (i) Starting from impulse response, find the voltage across the resistor in the circuit when driven by the input vS(t). (ii) Using the result, derive an expression for voltage across the resistor when the circuit is driven by a rectangular pulse of unit amplitude and 1 s duration.
  14. Derive expressions for maximum voltage across resistor in series RL circuit when it is driven by e–αt u(t) with α ∊ 1/τ where τ is the time constant of the circuit. Also, find an expression for the time instant at which this maximum voltage occurs.
  15. An exponential input of kδ(t) + 2e–2t/τ u(t) V is applied to a series RL circuit with time constant of τ s. The output for t ≥ 0 + is observed to contain only e2t/τ waveshape. What is the value of k?
  16. The steady-state voltage across resistor (vR ) in a series RL circuit has an amplitude of 7V when the circuit is driven by a sinusoidal voltage of amplitude 10V and angular frequency ω rad/s. (i) Find the phase angle of vR with respect to the input sinusoid. (ii) If another sinusoidal voltage of 15V amplitude and 3ω rad/s frequency is applied to the circuit, find the amplitude and phase of vR under steady-state condition.
  17. The desired current in the inductor in a series RL circuit with initial condition as shown in the Fig. 10.11-6 is given by

    Find the vS(t) to be applied to the circuit if the initial condition is 1A as marked in the Fig. 10.11-6. Sketch the required voltage.

    Fig. 10.11-6

  18. What is the rise time of inductor current in the step response of circuit in Fig. 10.11-7?

    Fig. 10.11-7

  19. The switch S1 in the circuit in Fig. 10.11-8 is closed at t = 0 and the switch S2 is closed at t = 0.12 s. Both switches are ideal. (i) Find the current in the inductance as a function of time and plot it. (ii) Find the voltage across inductance and plot it. (iii) Find the voltage across the first resistor and plot it. (iv) What is the time required to attain 90% of final steady-state value of current?

    Fig. 10.11-8

  20. Initial current at t = –∞ in the inductor in the circuit in Fig. 10.11-9 was zero. Find iL(t) and vL(t) for t ≥ 0 + and plot them.

    Fig. 10.11-9

  21. The switch S in Fig. 10.11-10 is a two-position switch and starts at position-1 at t = 0. It is kept in that position for 10ms and then thrown to position-2. It is kept at 2nd position till the current in the inductor goes to zero. At that instant, it is thrown back to position-1. Then, the whole cycle repeats. (i) Calculate and plot two cycles of iL(t), iS1(t) and iS2(t). (ii) What is the frequency of switching in the circuit? (iii) What are the average currents in the two sources and inductance? (iv) What are the average power delivered to the 24V source and the average power supplied by 12V source. (v) If the idea was to charge the 24 V battery from 12V battery, what is the efficiency of this charger? (vi) Suggest a method to control the average charging current in the 24V battery.

    Fig. 10.11-10

  22. Repeat the steps in Problem 3 with the circuit in Fig. 10.11-11.

    Fig. 10.11-11

  23. The switch S in Fig. 10.11-12 starts at position-1 at t = 0. It is kept in that position for 10ms and then thrown to position-2. It is kept at 2nd position till the current in the inductor goes to zero. At that instant, it is thrown back to position-1. Then, the whole cycle repeats.

    Fig. 10.11-12

    (i) Calculate and plot two cycles of iS1(t) and iS2(t). (ii) What is the frequency of switching in the circuit? (iii) What are the average currents in the two sources? (iv) What are the average power delivered to the 12V source and the average power supplied by 24V source? (v) If the idea was to charge the 12V battery from 24V battery, what is the efficiency of this charger? (vi) Suggest a method to control the average charging current in the 12V battery.

  24. The switch in the circuit in Fig. 10.11-13 was closed for a long time and is opened at t = 0. Find and plot the current in inductance and voltage across it as functions of time.

    Fig. 10.11-13

  25. Find the impulse response of the voltage variable v(t) in the circuit in Fig. 10.11-14.

    Fig. 10.11-14

  26. (i) Find the zero-input response, zero-state response and total response for iL(t) and v(t) in the circuit in Fig. 10.11-15. (ii) Obtain iL(t) and v(t) if the current source on the right side is made 2 u(t).

    Fig. 10.11-15

  27. What must be the value of k in the circuit in Fig. 10.11-16 if v(t) = 0 for t ≥ 0 +?

    Fig. 10.11-16

  28. The switch S in the circuit in Fig. 10.11-17 operates cyclically at 10kHz spending equal time at both positions in a cycle. Estimate the average power delivered by the current source and the average power dissipated in the resistor. Averages are over switching cycles.

    Fig. 10.11-17

  29. In Fig. 10.11-18 Vac = 10sin (314t) V. (i) Find the steady-state output voltage waveform for vo(t) and plot it. (ii) What is the percentage peak-peak ripple with respect to average value in the output voltage? (iii) Repeat (ii) if inductance is changed to 2H. (iv) Justify the following statement – “The steady-state output voltage across resistance in an R-L circuit will be more or less constant at the average value of driving voltage even if the driving voltage has a.c components if the product ωL >> R where ω is the angular frequency of AC components.”

    Fig. 10.11-18

  30. Let Vs (t) an arbitrary time varying periodic voltage source with a cycle average value of Vdc. This means that Vs(t) can be written as Vdc + Vac(t) where Vac(t) is a time varying periodic component with equal positive half-cycle and negative half-cycle areas. Let that area be A V-s. This waveform Vs(t) is applied to a series R-L circuit and output voltage is taken across R. Assume that L/R >> T where T is the period of Vs(t). Show that under periodic steady-state the (i) average value of output voltage is Vdc (ii) the peak-to-peak ripple in output voltage ≈ A/τ V where τ = L/R. (iii) calculate the quantities in (i) and (ii) for the three inputs given in Fig. 10.11-19 if τ is 25ms.

    Fig. 10.11-19

  31. Two series RL circuits are connected in cascade using a unity gain buffer amplifier as in Fig. 10.11-20. A buffer amplifier is an electronic amplifier that presents infinite resistance at its input and behaves like an ideal voltage at its output. With a unity gain, its output voltage is same as input voltage. Buffer amplifiers are used to interconnect circuits that will interact with each other otherwise. The initial current in the inductor of first stage circuit is 0.5 A and that of second stage is 2 A. Find vo(t) for t ≥ 0 +.

    Fig. 10.11-20