BME211HW10 - Homework #10 Solutions P14.6-4 -vc + 12i L + 2...

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Unformatted text preview: Homework #10 Solutions P14.6-4 -vc + 12i L + 2 diL dt = -8 and i L = -C d vc dt Taking the Laplace transform yields 8 -Vc ( s ) + 12 I L ( s ) + 2 L ( s ) -iL ( 0 ) - sI = s I L ( s ) = - C c ( s ) -vc ( 0 ) sV vc (0) = 0, iL (0) = 0 Solving yields Vc ( s ) = 4C C s 2 + 6s + s 2 72 s ( s + 3) 2 (a) C = 1 F 18 Vc ( s ) = = c a b + + s s +3 ( s + 3) 2 -24 8 -8 + + s s + 3 ( s + 3) 2 0 a = 8, b = -8, and c = -24 Vc ( s ) = vc ( t ) = 8 - 8 e -3t - 24 t e -3t V, t (b) C = 1 F 10 Vc ( s ) = c 40 a b = + + s ( s +1) ( s +5 ) s s +1 s + 5 2 8 -10 + + s s +1 s + 5 0 a = 8, b = -10, and c = 2 VC ( s ) = vc ( t ) = 8 - 10 e - t + 2 e -5t V, t P14.6-5 vc (0- ) = 10 V, i L( 0- ) = 0 A d vc di and 400 i + 1 + vc = 0 dt dt i = ( 5 10-6 ) Taking Laplace transforms yields 1 - 400 -10 40 = I ( s) = 2 s + 400 s + 2 105 ( s + 200 ) 2 + 4002 400 I ( s ) + ( s I ( s ) - 0 ) + Vc ( s ) = 0 I ( s ) = ( 5 10-6 ) ( sVc ( s ) - 10 ) so i( t) = - P14.6-7 KCL: v1 5 + i = 7 e -6 t 1 -200t e sin ( 400t ) u ( t ) A 40 KVL: 4 di di + 3 i - v1 = 0 v1 = 4 + 3 i dt dt di 35 + 2 i = e -6 t dt 4 Then di 4 + 3i dt + i = 7 e -6 t 5 Taking the Laplace transform of the differential equation: s I ( s) - i (0) + 2 I ( s) = 35 1 35 1 I (s) = 4 s +6 4 ( s + 2)( s + 6) Where we have used i (0) = 0 . Next, we perform partial fraction expansion. 1 A B 1 = + where A = ( s + 2) ( s + 6) s + 2 s + 6 s +6 Then I (s) = 35 35 35 1 35 1 - i (t ) = e -2t - e -6t A, t 0 16 s + 2 16 s + 6 16 16 = s =-2 1 1 1 and B = =- 4 s + 2 s = -6 4 P14.7-1 6 - 0.010 1.2 - 0.002 s .003 .005 s I L (s) = = = - 5s + 2000 s( s + 400) s s + 400 iL (t ) = P14.7-3 -2 mA 3-5e -400t mA t <0 t >0 - 0.006 Vc ( s) + + s 2000 Vc ( s) - 106 .5s 8 s =0 - 6000 8 + 500 Vc ( s) + 0.5s c ( s ) - 0 V = s s 8 s + 12000 12 4 = - s ( s +1000) s s +1000 Vc ( s ) = Vc (t ) = 12 - 4e -1000t V, t > 0 P14.7-5 Node equations: Va ( s ) - VC ( s ) Va ( s ) 1 6 6 + = Va ( s ) = VC ( s ) + s 6 s s+6 s+6 6 6 6 VC ( s ) - VC ( s ) + VC ( s ) - 3 s s+6 1 s +6 s+3 + + + VC ( s ) - = 0 4 s s 4 2 After quite a bite of algebra: VC ( s ) = Partial fraction expansion: 44 1 6 s 2 + 56s + 132 9 V (s) = = 3 - + 3 c s +3) ( s + 2 ) ( s + 5 ) s + 2 s +3 s + 5 ( Inverse Laplace transform: 44 -2t 1 v (t ) = e - 9e -3t + e -5t V, t 0 c 3 3 6 s 2 + 56 s + 132 ( s + 2 ) ( s + 3) ( s + 5 ) P14.7-8 t<0 time domain Mesh equations in the frequency domain: 6 I1 ( s) + 6( I1 ( s) - I 2 ( s) ) + 6 I1 ( s) + 2 6 I 2 ( s) - - 6( I1 ( s) - I 2 ( s) ) s s Solving for I2(s): frequency domain 12 2 2 = 0 I1 ( s) = I 2 ( s) - s 3 3s 6 2 = 0 + 2 ( s ) - 6 I 1 ( s ) = 6 I s s 1 2 2 2 6 2 6 I = I 2 ( s) = + 2 ( s ) - 6 I 2 ( s ) - 1 3 3s s s s+ 2 Calculate for Vo(s): 1 6 1 6 1 -2 4 Vo ( s ) = I 2 ( s ) - = 2 = - - 1 1 s 2 s 2+ s s+ s 2 2 Take the Inverse Laplace transform: vo ( t ) = - ( 4 + 2 e -t / 2 ) V for t > 0 P14.7-17 Mesh Equations: 4 1 4 1 - - I C ( s ) - 6 I ( s ) + I C ( s ) = 0 - = + C ( s ) + 6 I ( s ) 6 I s 2s s 2s 10 6( I ( s) - I C ( s) ) + 3 I ( s) + 4 I C ( s) = 0 I ( s) = - I C ( s) 9 Solving for I C(s): 4 2 1 6 - = - + I C ( s) I C ( s) = 3 s 3 2s s- So Vo(s) is 4 24 Vo ( s ) = 4 I C ( s ) = 3 s- 4 Back in the time domain: v o ( t ) = 24 e0.75t u (t ) V for t 0 ( ) ...
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