Homework 8 Solution

Homework 8 1) What is vc(t) if the switch closes at t=0? TCLOSE = 0 2 1 4 2 + 1 U1 2H Vc 0.5 3 - Ans: The initial conditions are found by assuming the capacitor is discharged, so: vc (0)  0 The second order circuit will require a second initial condition and we are looking for vc, we need the first derivative of vc. This is determined by the current at t=0. However, since there is an inductor in series with the capacitor and the current before t=0 is 0, the current through the inductor at t=0 will also be 0. Therefore: iC dvc 0 dt dvc 0 dt Since we are looking for a voltage, it would be best to use nodal analysis, since it gives us equations in terms of voltages. With the switch closed, the equation is: 3  vc v  c 1 4  2s 0.5s 2 3  s vc  2svc  vc Note: you could also do mesh here since it is a simple series circuit: 3  4i  2 di  vc  0 dt dvc dt 2 d v dv 3  2c  2 c  vc dt dt i  0.5 Which gives the same characteristic equation. The differential equation is non-homogeneous. The characteristic equation has one root, s=-1, therefore this is a critically damped system with a homogeneous solution of: vh  Aet  Btet The particular solution is a constant and substituting that back into our original equation gives: vp  C 3  0  2*0  C 3C Our total solution is then: vc  Aet  Btet  3 Applying the initial conditions: vc (0)  0  Ae 0  B *0e 0  3 0  A3 A  3 dvc (0)  0  3e 0  B *0e 0  Be  t dt 0  3 B B  3 The final solution is: vc(t)= -3[e-t + te-t] +3 2) What is vc(t) if the switch, U1 closes at t=0 and U2 opens at the same time? TOPEN = 0 2 1 4 U1 2 2 + 1 U2 2H 1 Vc 0.5 TCLOSE = 0 3 Ans: This circuit is the homogeneous version of Problem 1, but has different initial conditions. If U2 has been closed for a long time, the capacitor will have completely charged to 3V. Therefore, vc(0)=3 Since the capacitor has completely charged, the current through it will be 0. This current is also the current through the inductor. The current will not change rapidly, so i(0)=0. Therefore: i0C dvc dt dvc 0 dt With U2 open and U1 closed, the circuit is the same as above but without the voltage source. We can write the differential equation as: vc v  c 1 4  2s 0.5s 2 0  s vc  2svc  vc Which gives the same characteristic equation. The differential equation is homogeneous. The characteristic equation has one root, s=-1, therefore this is a critically damped system with a solution of: vc  Aet  Btet Applying the initial conditions: vc (0)  3  Ae 0  B *0e 0 3 A dvc (0)  0  3e 0  B *0e 0  Be  t dt 0  3  B B3 The final solution is: vc(t)= 3e-t [1+t] 3) What is iL(t) if the switch closes at t=0? U3 2 1 TCLOSE = 0 2 0.2 2 1 2A 1 IL Ans: There are a number of possible approaches to this problem. You could Thevenize the current source and 2 ohm resistor into a 4V voltage source in series with a 2 ohm resistor and solve the problem in exactly the same manner as Problem 1. Or, since the solution is a current, you can use mesh to analyze the problem as is. Either way the initial conditions are that, with the switch open for a long time, iL(0)=0. For the first derivative of iL, we need the voltage across the inductor. This involves some thought. The capacitor is completely discharged so has no voltage drop across it. That means that, when the switch closes, any voltage across the 2 ohm resistor will appear across the inductor. To determine that voltage, we look at the current sharing between the 2 ohm resistor and the leg containing the capacitor and inductor. With the inductor in the circuit, the current through this leg must be 0 at t=0, since it is 0 just before the switch closes. Therefore all the current from the current source flows through the resistor. 4V appears across the resistor and, consequently, across the inductor (remember: there is no voltage across the capacitor). This gives the second initial condition: di di 1 dt dt di 4 dt 4L Writing the mesh equation for after the switch is closed, using generalized impedances, gives: i  si  0 0.2s 0.4siL  0.2s (2)  i  0.2s 2i  0 2(iL  2)  0.2s 2iL  0.4siL  iL  0 Notice that the application of the s operator to the constant in the equation gives 0. This is a homogeneous equation with two complex roots: s  1  j 2 It has a solution of: iL  et  A cos 2t  B sin 2t  Applying the initial conditions gives A=0 and B=2 and the total solution is: iL(t)=2e-tsin(2t) 4) What is vL(t) in Problem 3 Ans: Using the relation for current and voltage in an inductor: vL di dt We differentiate the solution to Problem 3 and multiply by 1 to give: vL(t)=e-t[4cos(2t) – 2sin(2t)] 5) What is vc(t) if the source is 3e-t? 2 0.2 1 Vc 3e(-t) Ans: Using nodal analysis and generalized impedances, we can write the equation for vc as: 3e  t  vc vc  1 2 s t 3e  2svc  vc This is a first order, non-homogeneous equation with a root of s=-0.5. The homogeneous solution is: vh  Ae0.5t The particular solution has the form: v p  Bet Applying this to the original differential equation gives: 3et  2 Bet  Be t 3  B 3  B Writing the total solution and applying the initial condition, which assumes the voltage source was at 3V before t=0, gives: vc  Ae 0.5t  3e t vc (0)  3  Ae0  3e0 6 A And the solution is vc(t)=6e-0.5t -3e-t 6) Find vc(t) for the source V=2sin 2t, starting at t=0. Use Excel to graph out the response up to t=5s. Notice how soon the response settles to a steady state response. 1 3.2 2 sin(2t) 2 0.8 0.2 Vc Ans: This problem is a perfect example of why phasors were invented. The initial conditions are the same as problem 1 for the same reasons. vc (0)  0 dvc 0 dt Using nodal analysis and generalized impedances gives us the differential equation for vc: 2sin 2t  vc v  c 1 3.2  0.8s 0.2s 2sin 2t  0.16s 2vc  0.64svc  vc The roots of the characteristic equation are: s  2  j1.5 The homogeneous solution is: vh  e2t  A cos1.5t  B sin1.5t  The particular solution has the form: v p  C cos 2t  D sin 2t Inserting this into the original differential equation gives: 2sin 2t  C cos 2t  D sin 2t  1.28C sin 2t  1.28D sin 2t  0.64C cos 2t  0.64D sin 2t Summing the cosine terms and setting them equal to 0 and summing the sine terms and setting them equal to 2 gives two equations and two unknowns: 2  0.36 D  1.28C 0  4.511D  1.28C 0.407  D 1.448  C We can write the total solution as: vc  e2t  A cos1.5t  B sin1.5t   1.448cos 2t  0.407sin 2t Applying the initial conditions: vc (0)  0  e0  A cos 0  B sin 0  1.448cos 0  0.407 sin 0 0  A  1.448 A  1.448 And finally: dvc (0)  0  2e0 1.448cos 0  B sin 0  e0  2.172sin 0  1.5 B cos 0  2.896sin 0  0.814 cos 0 dt 0  2.896  1.5B  0.814 B  1.388 Giving the final solution as: vc(t)= e-2t [1.448 cos1.5t + 1.388 sin 1.5t] – 1.448 cos 2t + 0.407 sin 2t Notice that the time constant for the first exponential term is 0.5seconds. That means after 2.5 seconds, the transient response will have completely died out and we would be left with the steady state solution.