MidtermReview_Soln

MidtermReview_Soln - EE411 Spring 2008 Midterm Review...

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EE411 Spring 2008 Midterm Review Solution (Q1) 5.62 Let v 1 = output of the first op amp v 2 = output of the second op amp The first stage is a summer i 1 2 1 v R R v = o f 2 v R R ( 1 ) The second stage is a follower. By voltage division 1 4 3 4 2 o v R R R v v + = = o 4 4 3 1 v R R R v + = (2) From (1) and (2), i 1 2 o 4 3 v R R v R R 1 = + o f 2 v R R i 1 2 o f 2 4 3 v R R v R R R R 1 = + + f 2 4 3 1 2 i o R R R R 1 1 R R v v + + = () f 4 f 3 4 2 1 f 4 2 R R R R R R R R R R + + = (Q2) 6.46 Under dc conditions, the circuit is as shown below: 2 Ω i L 3 A + v C
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By current division, = + = ) 3 ( 2 4 4 i L 2A , v c = 0V L 2 1 w L = = = 2 2 L ) 2 ( 2 1 2 1 i 1J C 2 1 w c = = = ) v )( 2 ( 2 1 v 2 c 0J (Q3) 6.53 [] ) 4 8 ( 6 ) 12 8 ( 5 8 10 6 L eq + + + + + = 4 16 ) 4 4 ( 8 16 + = + + = L eq = 20 mH (Q4) 6.57 Let dt di L v eq = ( 1 ) 2 2 1 v dt di 4 v v v + = + = ( 2 ) i = i 1 + i 2 i 2 = i – i 1 (3) 3 v dt di or dt di 3 v 2 1 1 2 = = ( 4 ) and 0 dt di 5 dt di 2 v 2 2 = + + dt di 5 dt di 2 v 2 2 + = ( 5 ) Incorporating (3) and (4) into (5),
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3 v 5 dt di 7 dt di 5 dt di 5 dt di 2
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This note was uploaded on 10/26/2009 for the course EE 411 taught by Professor Lee during the Spring '07 term at University of Texas.

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MidtermReview_Soln - EE411 Spring 2008 Midterm Review...

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