Chapter 5: Analysis of Time-Domain CircuitsThis chapter begins the analysis of circuits containing elements with the ability to store energy: capacitors and inductors. Wehave already defined each of these reactive elements as an energy storage device with a differential IV relationship – we beginby using no more than these IV relationships and Kirchoff’s laws to describe the time-domain solutions of DC switching circuitswith capacitors and inductors.Time-Domain Circuits using Differential EquationsThe analysis of any DC circuit containing reactive elements will depend on solutions to differential equations, which in turnbreak down to the solution of a system’s homogeneous or natural response without any input and the heterogeneous or forcedsolution particular to a specific input. The form of the differential equation depends on the number of reactive components;the order of the differential equation (based in either a voltage or current) is equal to the number of capacitors and inductors.Typically, we are interested in some form of switching phenomenon; that is, at some fixed time (for simplicity translated to0) a switch is either opened or closed changing the charging/discharging properties of the circuit elements. The result can benumerically calculated by solving the circuit’s associated differential equation. To set up these differential equations, we need todetermine the set of equations (usually KVL loops or KCL sums) along with the appropriate initial and final conditions. Initialconditions are the values that a particular component(s) takes after a long time of being settled in the given state and just beforeswitching at time zero. If this were the voltage in the capacitor before being hooked up to another circuit, we may represent itmathematically as VC(t = 0−). The final condition, often also called the steady state value, is the component’s value a long timeafter the switching phenomenon has taken place; the notation for such a capacitor voltage value would be VC(t = ∞).Capacitor and Inductor Time PropertiesThe most basic time-domain circuits with reactive elements consist of a current source charging a capacitor and a voltagesource charging an inductor. The capacitor in Figure 5.1 is assumed to have an initial voltage VC(t = 0) before the switch closesat time zero.Figure 1After the switch closes, positive charge flows from the current source to the capacitor effectively charging it. Using thetime-domain IV relationship for the capacitor, we write the differential equation to obtain the capacitor voltage.IDC= IC(t) = C ·ddtVC(t)Resulting in:VC(t) =IDCC· t + VC(t = 0)The second circuit shown in figure 5.2 is a voltage source in series with an inductor.12Figure 2After the switch closes at time zero, the voltage source induces a constant voltage of VDCvolts across the inductor, therebycausing a constantly increasing current.VDC= VL(t) = L ·ddtIL(t)IL=VDCL· tNotice that neither one of these circuits is stable or fully realistic. As t → ∞ both will store infinite amounts of energy; further,reopening the switch at any finite time leaves no path for current to flow and the reactive elements to discharge.3Example: Consider the charging/discharging of an initially uncharged 1 µF capacitor.(a) Determine and plot the current for the capacitor if the voltage is 0 V for t ≤ 0, 3t2Volts for0 < t < 2, and 12 Volts for t ≥ 2.(b) Plot the instantaneous power of the capacitor.(c) Plot the energy stored in the capacitor as a function of time.(d) Verify that the instantaneous power is consistent with the changes in energy storage.(a) Using the differential IV relationship to solve for the current IC(t).t ≤ 0 : IC(t) = (1µF ) ·ddt0 V = 0 At ∈ (0, 2) : IC(t) = (1µF ) ·ddt3t2V = 6 t µAt ≥ 2 : IC(t) = (1µF ) ·ddt12V = 0AFigure 3(b,c) The instantaneous power is simply the product of voltage and current, while the energy storedin the capacitor can be most easily calculated as Ecap=12CV2C. Both waveforms are shown inFigure 5.4.Figure 4(d) The final step is to make a sanity check on the various solutions; differentiating the waveformfor the stored energy, we do again obtain the power waveform. 2Exercise: Consider the charging/discharging of an inititally unenergized (IL(t = 0−) = 0A) 1 mH inductor.4(a) Determine and plot the voltage across the inductor if the current is 0 A for t ≤ 0, 2t Amps for 0 < t < 3, and 6 Amps fort ≥ 3.(b) Plot the instantaneous power of the inductor.(c) Plot the energy stored in the inductor as a function of time.(d) Verify that the instantaneous power is consistent with the changes in energy storage.First-Order CircuitsTo transition from the idealized energy storage properties shown above to more practical circuits, we must introduce series andparallel resistances (real voltage sources have some finite resistance, and the shunt resistance of any real component is large butnot quite infinite). We will also see that these resistances affect the time-domain transient (initial or rapidly changing) response,but not the steady state values. We will start out with first-order circuits consisting of only one reactive element.First-Order RC CircuitsThe simplest first-order RC circuit is a voltage source in series with a single resistor and uncharged capacitor, as shown inFigure 5.5.Figure 5To analyze the first-order RC circuit, we need only write a single KVL loop equation. Since the relationship between the voltageand current in the capacitor is differential, this loop equation governing the system will inherently be a first-order differentialequation, hence the title first-order circuit.−Vx+ VR+ VC= 0To solve a differential equation, we must have the entire equation in a single variable and its derivatives. We may substitutefor VRusing Ohm’s law and the series current IC(t) to obtain a term dependent on VC.VR(t) = R · IC(t) = R · C ·ddtVC(t)Replacing the original VRby the term above yields the governing differential equation in one variable.RCddtVC(t) + VC(t) = VxThe final step before solving is to determine any initial or final conditions. Since the capacitor is uncharged, the initial voltageis zero; the final, or steady-state, voltage occurs when all time-derivatives are set to zero, giving VC(t → ∞) = Vx. Another wayto obtain this condition is to just look at the circuit: as long as the capacitor voltage is less than the source, positive currentwill flow clockwise through the resistor and the capacitor effectively