Massachusetts Institute Technology Massachusetts Institute ofofTechnology Department of Electrical Engineering andComputer Computer Science Department of Electrical Engineering and Science Electronic Circuits 6 0026 002 Electronic Circuits 2004 FallFall 2005 Homework 9 Homework 9 Solutions Handout F04 046 Helpful readings for this homework Chapter 12 Issued 11 04 2004 Due 11 12 2004 Helpful Readings for this Homework Chapter 12 Exercise 9 1 Using one 3 nF capacitor and two resistors construct a network that hasExercise the following zero state response to a construct 1V step input that Provide a diagram of 9 1 Using one 3 nF capacitor and ZSR two resistors a network has the following zero state to a 1 V input Provide a diagram the two network and specify the values of the two resistors theresponse network andstep specify the values of of the resistors v1 t v2 t 1V v1 t Network t Exercise 9 2 v2 t 1V 2 V 3 2 1 t 20 s V V e 3 3 t Exercise 12 4 Chapter 12 p 985 From the graph we get the following values Exercise 9 3 Consider a linear time invariant system Suppose its ZSR to a unit step applied at t 0 is A 1 e t What would be its ZSR to the input S Mt applied at t 0 where S and M are constants Problem 9 1 Problem 12 6 Chapter 12v2 p t 991 0 1 V 1 2 v2 t and Vcapacitor Problem 9 2 In the network shown below the inductor have zero states prior to t 0 At 2 t 0 3 a step in voltage from 0 to V 0 is applied by theRvoltage source as shown 3 eq C 20 s di Find at v C the v L ZSR i andof this at t circuit 0 for finite voltages and currents Therefore no voltage We are a looking dt is initially across the capacitor From this and the initial condition of v2 the capacitor must be in series with b v2 Argue and not thei terminals of v2 component thatconnected i 0 at t across so that t has no constant As t the capacitor acts like an open circuit For v2 to be 23 V when the capacitor acts c Find a second order differential equation which describes the behavior of i t for t 0 like an open circuit one resistor has to be across the capacitor and the other across the terminals of v2 This is shown in the following figure Now solve for R1 and R2 using the final condition for v2 and the time constant Network R1 V1 t C V2 t R2 The final condition for v2 is a simple voltage divider relationship 1V R2 2 V R1 R2 3 4 For the time constant kill the voltage source by shorting it and solve for the equivalent resistance Req R1 R2 R1 R2 R1 R2 C 20 s R1 R2 For C 3 nF R1 10 k and R2 20 k 5 6 Exercise 9 2 Exercise 12 4 Chapter 12 page 695 V1 t L C R2 R1 For ZIR short out the source L V1 0 C R2 R1 R1 can now be ignored and the circuit is a series LCR circuit L C R2 c Note that the current in this circuit is ic il iR2 and ic C dv dt Doing KVL around the loop gives dic vc R2 ic 0 dt dvc dvc LC vc R2 C 0 dt dt dvc R2 dvc 1 vc 0 dt L dt LC L o R2 2L r 1 LC 7 8 9 10 11 Plugging in for known values of R2 15 L 1 H and C 0 01 F we see that 7 5e6 s 1 and o 1e7 s 1 Since o the circuit is underdamped Exercise 9 3 Consider a linear time invariant system Suppose its ZSR to a unit step applied at t 0 is A 1 e t What would be its ZSR to the input S M t applied at t 0 where S and M are constants vIN t S M t 12 Z t Su t M u t dt 13 o t vOU T t S A 1 e Z t M A 1 e t dt 14 o SA 1 e t M A t e t to t SA 1 e t SA 1 e 15 t 0 16 t 17 M A t e M A t e Problem 9 1 Problem 12 2 Chapter 12 page 697 VA A V1 C1 C2 V1 0 V V2 0 0 VA infinity 0 V2 At t 0 switch closes a Compute the initial charge of the system q 0 C1 v1 0 C2 v2 0 C1 V 18 19 b Find the voltage across both capacitors a long time after the switch has been closed Remember 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 Vf inal be this final voltage on both capacitors q 0 q t C1 Vf inal C2 Vf inal C1 V C1 V Vf inal C1 C2 20 21 22 c Find the energy stored in the system after a long time Ef inal 1 1 C1 Vf2inal C2 Vf2inal 2 2 1 C1 V 2 C1 C2 2 C1 C2 2 1 C12 V 2 2 C1 C2 23 24 25 d Find the ratio of the final stored energy to the initial stored energy Where did the rest of the energy go Ef inal Einitial Ef inal Einitial 1 C12 V 2 2 C1 C2 1 C1 V 2 2 26 27 1 C12 V 2 C1 C2 C1 V 2 C1 C1 C2 28 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 the energy lost in it VA V1 C1 C2 V2 Note that the natural response to this circuit can be found by combining capacitors in in series This gives a time constant of RCeq Ceq C1 C2 C1 C2 C1 C2 30 31 C1 C2 R C1 C2 32 vA t 0 V 33 vA t 0 34 Combine the initial and final conditions with the time constant to solve for vA and iA t vA t V e V t iA t e R 35 36 Z Elost P dt 37 vA t iA t dt 38 Z0 0 Z Ve t 0 Z 2 V V t e dt R 2t dt R V 2 2t e dt 0 R 2 V 2 0 1 R 2 V2 R 2 V 2 C1 C2 R 2R C1 C2 1 C1 C2 V2 2 C1 C2 e 39 40 0 41 42 43 44 45 f Find the ratio of lost energy to initial energy Is it what you expected Does it depend on R Elost Einitial 1 C1 C2 2 V 2 2 C1 C2 C1 V 2 C2 …
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