hmw6_sol_sp08

hmw6_sol_sp08 - i ECE220 Homework Assignment #6 Solutions...

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i ECE220 Homework Assignment #6 Solutions Problem 7.10 Consider the differential equation obtained from an RLC cir- cuit d 2 v ( t ) dt 2 + R L dv ( t ) dt + 1 LC v ( t ) = 1 LC v s ( t ) (0.1) where R = 10 ,L = 1 , C = 2 and the initial conditions are v (0) = 0 , ˙ v (0) = 1 . Let v s ( t ) = 10 u ( t ) . (a) Find v c ( t ) . (Note: this was part of homework 5; solution repeated here for completeness.) (b) Find v p ( t ) . (c) Find the total solution v ( t ) . (d) Plot all three solutions. (e) Is the system underdamped, overdamped or critically damped? Solution (a) Substituting R = 10 ,L = 1 ,C = 2 , the second order differential equation becomes d 2 v ( t ) dt 2 + 10 dv ( t ) dt + 1 2 v ( t ) = 5 u ( t ) To find the complementary solution, we set v s = 0 and follow the procedure in section 7.4.1.3, page 354. We have b = 10 ,c = 1 2 , and thus ζ = b 2 c = 5 2 > 1 This is the overdamped case, with two distinct real roots. To find them, we must solve the characteristic equation λ 2 + 10 λ + 1 2 = 0 The roots are λ 1 = - 10 + 100 - 2 2 = - 5 + 7 2 2 ≈ - 0 . 05 λ 2 = - 10 - 100 - 2 2 = - 5 - 7 2 2 ≈ - 9 . 95
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ii Therefore, the complementary solution is given by v c ( t ) = C 1 e ( - 5+ 7 2 2) t u ( t ) + C 2 e ( - 5 - 7 2 2) t u ( t ) C 1 e - . 05 t u ( t ) + C 2 e - 9 . 95 t u ( t ) (0.1) A plot of the complementary solution is given in Figure 0.1. -10 0 10 20 30 40 50 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 y c (t) Time (t) Figure 0.1. Problem 7.10, part (a). As we can see from the plots in Figure 0.2, the signal C 1 e - . 05 t u ( t ) really dominates in the complementary solution 0.2, since the signal C 2 e - 9 . 95 t u ( t ) decays really fast. (b) In order to find v p ( t ) , we set 1 LC v s ( t ) = 5 u ( t ) and follow the procedure in section 7.4.2.1, with A = 5 . We guess v p ( t ) = Bu ( t ) ,B 6 = 0 . For t > 0 , we have d 2 v p ( t ) dt 2 = 0 , dv p ( t ) dt = 0 and get 1 2 B = 5 Therefore, B = 10 and thus the particular solution is given by v p ( t ) = 10 u ( t )
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-10 0 10 20 30 40 50 0 0.2 0.4 0.6 0.8 1 λ 1 = -0.05 -10 0 10 20 30 40 50 0 0.2 0.4 0.6 0.8 1 λ 2 = -9.95 Time (t) Figure 0.2. Problem 7.10, part (a), the two parts of the complementary solution. (c) The total solution is given by v ( t ) = v c ( t ) + v p ( t ) = C 1 e - . 05 t u ( t ) + C 2 e - 9 . 95 t u ( t ) + 10 u ( t ) In order to determine the unknown parameters C 1 ,C 2 , we plug in the initial conditions v (0) = 0 , ˙ v (0) = 1 and get C 1 + C 2 + 10 = 0 ( - 5 + 7 2 2) C 1 + ( - 5 - 7 2 2) C 2 = 1 The solution of this linear system is C 1 ≈ - 9 . 95 C 2 ≈ - 0 . 05 (d) The matlab code titled problem7_10.m is in the course locker. Plots are shown in Figure 0.3. (e)
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This note was uploaded on 03/30/2009 for the course ECE 220 taught by Professor Nilson during the Spring '08 term at N.C. State.

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hmw6_sol_sp08 - i ECE220 Homework Assignment #6 Solutions...

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