Massachusetts Institute of TechnologyDepartment of Electrical Engineering and Computer Science6.002 – Electronic CircuitsFall 2005Homework 9 SolutionsHelpful readings for this homework: Chapter 12Exercise 9-1: Using one 3-nF capacitor and two resistors, construct a network thathas the f oll owing zero-state response (ZSR) to a 1V step input. Provide a diagram ofthe network, and specify the values of the two resistors.Massachusetts Institute of TechnologyDepartment of Electrical Engineering and Computer Science6.002 - Electronic CircuitsFall 2004Homework #9Handout F04-046Issued 11/04/2004 - Due 11/12/2004Helpful Readings for this Homework: Chapter 12Exercise 9-1: Using one 3-nF capacitor and two resistors, construct a network that has the following zero-state response to a 1-V step input. Provide a diagram of the network, and specify the values of the two resistors.1Vt1Vt23--VNetwork23--V13--Vet 20µs⁄–+v2(t)+-v2(t)v1(t)v1(t)+-Exercise 9-2: Exercise 12.4, Chapter 12 (p. 985)Exercise 9-3: Consider a linear time-invariant system. Suppose its ZSR to a unit step applied at t 0= is A 1 et τ⁄––(). What would be its ZSR to the input SMt+ , applied at t 0= , where S and M are constants?Problem 9.1: Problem 12.6, Chapter 12 (p. 991)Problem 9.2: In the network shown below, the inductor and capacitor have zero states prior to t 0= . At t 0= , a step in voltage from 0 to V0 is applied by the voltage source as shown.a) Find vC, vL, i , and didt----- at t 0+= .b) Argue that i 0= at t ∞= so that it() has no constant component.c) Find a second-order differential equation which describes the behavior of it() for t 0≥ .From the graph, we get the following values.v2(t = 0) = 1[V ] (1)v2(t → ∞) =23[V ] (2)ReqC = 20[µs] (3)We are looking at the ZSR of this circuit for finite voltages and currents. Therefore, no voltageis initially across the c apacitor. From this and the initial condition of v2, the capacitor must be inseries with v2, and not connected across the terminals of v2.As t → ∞, the capacitor acts like an open circuit. For v2to be23[V ] when the capacitor actslike an open circuit, one resistor has to be across the capacitor and the other across the terminalsof v2. This is shown in the following figure.Now solve for R1and R2using the final condition for v2and the time constant.+-Network+-V1(t) V2(t)R1R2CThe final condition for v2is a simple voltage divider relationship.1VR2R1+ R2=23[V ] (4)For the time c onstant, kill the voltage source by shorting it, and solve for the equivalent resis-tance.Req=R1R2R1+ R2(5)R1R2R1+ R2∗ C = 20[µs] (6)For C = 3[nF ], R1= 10[kΩ] and R2= 20[kΩ].Exercise 9-2: Exercise 12.4, Chapter 12, page 695.+-V1(t)R2CR1LFor ZIR, short out the source.V1 = 0R2CR1LR1can now be ignored, and the circuit is a series LCR circuit.R2CLNote that the current in this circuit is ic= il= iR2and ic= Cdvcdt.Doing KVL around the loop givesLdicdt+ vc+ R2ic= 0 (7)LCdvcdt+ vc+ R2Cdvcdt= 0 (8)dvcdt+R2Ldvcdt+1LCvc= 0 (9)α =R22L(10)ωo=r1LC(11)Plugging in for known values of R2= 15Ω, L = 1µH, and C = 0.01µF , we see that α =7.5e6[s−1] and ωo= 1e7[s−1]. Since α < ωothe circuit is underdamped.Exercise 9-3: Consider a linear time-invariant system. Suppose its ZSR to a unit stepapplied at t = 0 is A(1 −e−t/τ). What would be its ZSR to the input S + Mt, applied att = 0, where S and M are constants?vIN(t) = S + Mt (12)= Su(t) + MZtou(t)dt (13)vOU T(t) = S[A(1 − e−t/τ)] + MZto[A(1 − e−t/τ)]dt (14)= SA(1 − e−t/τ) + MA[t + τ e−t/τ]to(15)= SA(1 − e−t/τ) + MA[t + τ e−t/τ− 0 − τ] (16)= SA(1 − e−t/τ) + MA[t + τ e−t/τ− τ] (17)Problem 9.1: Problem 12.2, Chapter 12, page 697.C2C1A+-+-V2V1+ -VAAt t = 0, switch closesV1(0) = VV2(0) = 0VA(infinity) = 0a) Compute the initial charge of the systemq(0) = C1v1(0) + C2v2(0) (18)= C1V (19)b) Find the voltage across both capacitors a long time after the switch has be en closed. Re-member that the total charge of the system must be conserved.A long time after the switch is closed, vA(t → ∞) = 0. Therefore, v1(t → ∞) = v2(t → ∞).Let Vfinalbe this final voltage on both capacitors.q(0) = q(t → ∞) (20)C1Vfinal+ C2Vfinal= C1V (21)Vfinal=C1VC1+ C2(22)c) Find the energy stored in the system after a long time.Efinal=12C1V2final+12C2V2final(23)=12(C1+ C2)(C1V )2(C1+ C2)2(24)=12C21V2C1+ C2(25)d) Find the ratio of the final stored energy to the initial stored energy. Where did the rest ofthe energy go?Efinal=12C21V2C1+ C2(26)Einitial=12C1V2(27)EfinalEinitial= (C21V2C1+ C2)(1C1V2) (28)=C1C1+ C2(29)The energy was either dissipated or stored in element A.e) Assume element A is a resistor R. Find its voltage or current, and from that, find out theenergy lost in it.C2C1+-+-V2V1+ -VANote that the natural res ponse to this circuit can be found by combining capacitors in in series.This gives a time constant of RCeq.Ceq= C1||C2(30)=C1C2C1+ C2(31)τ =C1C2C1+ C2R (32)vA(t = 0) = V (33)vA(t → ∞) = 0 (34)Combine the initial and final conditions with the time constant to solve for vAand iA.vA(t) = V e−tτ(35)iA(t) =VRe−tτ(36)Elost=Z∞0P dt (37)=Z∞0vA(t)iA(t)dt (38)=Z∞0V e−tτVRe−tτdt (39)=Z∞0V2Re−2tτdt (40)=V2R−τ2[e−2tτdt]∞0(41)=V2R−τ2[0 − 1] (42)=V2Rτ2(43)=V22RC1C2C1+ C2R (44)=12C1C2C1+ C2V2(45)f) Find the ratio of lost energy to initial energy. Is it what you expected? Does it depend onR?ElostEinitial= (12C1C2C1+ C2V2)(2C1V2) (46)=C2C1+ C2(47)Note that this expression does not depend on the resistance.In part (d), the ratio of the initial vs. final energy was calculated without knowing element A.If element A is a resistor,Efinal= Einitial− Elost(48)Therefore,EfinalEinitial= 1 −ElostEinitial(49)= 1 −C2C1+ C2(50)=C1C1+ C2(51)This expression is consistent with part (d) when element A was unknown.g) What would happen if an inductor was placed in series with R? Sketch the behavior of thecurrent.iA(t = 0) = 0 (52)iA(t → ∞) = 0 (53)vA(t = 0) = V (54)vA(t → ∞) = 0 (55)The voltage condition at t = 0 indicates that iAis initially positive.For an LRC circuit,α =R22L(56)ωo=r1LC(57)The values of L, R, and C determine whether the current is underdamped, critically damped,or overdamped. These cases are shown in the following figure.0 2 4 6 8 10 12 14 16 18 20−0.4−0.200.20.40.60.81Problem 9.1gtiA(t)UnderdampedCritically dampedOverdampedProblem 9.2: Problem 12.6, Chapter 12, page 698.The circuit before+-C+-LiLVc2VRFor t < 0, in steady
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