HW3%20solutions_EG

HW3%20solutions_EG - H O M E WO R K 3 S O U L U T I O N S 1...

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HOMEWORK 3 SOULUTIONS 1. First consider the time period s 5 s 0 < t . KCL at node A gives 3 2 1 i i i = + which then can be written 3 F 4 3 1 i dt dV V C C = + Ω In order to use the inductor I-V relationship, differentiate the previous equation to get dt di dt V d dt dV C C 3 2 2 F 4 3 1 = + Ω and substitute to get A C C V dt V d dt dV H 5 1 F 4 3 1 2 2 = + Ω Recognizing that C A V V = V 6 , this equation can be written in standard form as 3 . 0 05 . 0 0833 . 0 2 2 = + + C C C V dt dV dt V d From this form, we can see that 2236 . 0 = n ω and 1863 . 0 = ζ , meaning that this system is underdamped . The damped natural frequency is thus 2197 . 0 = d .

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The particular solution assumed to be a constant and is given by () 6 05 . 0 3 . 0 , = = t V p C This gives us the general solution: () ( ) ( ) 6 2197 . 0 sin 2197 . 0 cos 0417 . 0 0417 . 0 + + = t Be t Ae t V t t C Since the capacitor and inductor initially store no energy, ( ) 0 0 = C V and 0 0 3 = i . The second equation above then leads to the condition 0 0 = = t C dt dV Thus, we get 6 0 0 + = = A V C , which implies 6 = A Substituting in for A and differentiating gives t Be t Be t e t e dt dV t t t t C 2197 . 0 cos 2197 . 0 2197 . 0 sin 0417 . 0 2197 . 0 sin 3182 . 1 2197 . 0 cos 2502 . 0 0417 . 0 0417 . 0 0417 . 0 0417 . 0 + + = Substituting in the initial condition gives B dt dV t C 2197 . 0 2502 . 0 0 0 + = = = , which implies 139 . 1 = B Thus, the solution for V C in the interval s 5 s 0 < t is 6 2197 . 0 sin 139 . 1 2197 . 0 cos 6 0417 . 0 0417 . 0 + = t e t e t V t t C The final voltage during this time interval is V 9607 . 2 5 = C V
During the time interval s 5 t , the circuit looks like Thus, the two loops can be treated independently. KVL for the upper loop gives 0 3 F 4 = + Ω C C V dt dV In standard form, this is 0 0833 . 0 = + C C V dt dV Since the right-hand side is 0, there is no particular solution, and so the general solution is () t C Ae t V 0833 . 0 = Using the final value from the previous time interval, we get 5 0833 . 0 9607 . 2 5 = = Ae V C , which gives ( ) 5 0833 . 0 9607 . 2 = e A Therefore, the solution for V C in the interval s 5 t is 5 0833 . 0 9607 . 2 = t C e t V

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2. Using the method of Logarithmic Decrement, we get 832 . 1 V 50 V 2 . 53 V 50 V 70 ln = = δ Therefore, 28 . 0 832 . 1 4 832 . 1 2 2 = + = π ζ Using the measured period between two peaks, we find () rad/s 10 6 s 3 / 5 5 2 5 × = = μ ω d Then, we can compute rad/s 10 25 . 6 28 . 0 1 rad/s 10 6 5 2 5 × = × = n From KVL around the loop, the second-order equation governing the dynamics of the circuit is V LC V LC dt dV L R dt V d C C C 1 1 2 2 = + + Matching the third term gives 1.6nF 1 rad/s 10 25 . 6 5 = × L , which yields mH 6 . 1 = L Matching the second term gives mH 6 . 1 rad/s 10 25 . 6 28 . 0 2 5 R = × , which yields Ω = 560 R
3. First, consider the circuit before the switch moves:

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This note was uploaded on 10/02/2010 for the course MAE 3780 taught by Professor Wickenhieser during the Fall '08 term at Cornell.

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HW3%20solutions_EG - H O M E WO R K 3 S O U L U T I O N S 1...

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