HW1_solution

# HW1_solution - MAE107 Homework #1 Solution Prof....

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MAE107 Homework #1 Solution Prof. M’Closkey Problem 1 The circuit to analyze is R 1 C 1 C 2 R 2 + V in + V out V i 1 i 2 Let i 1 be the current through capacitor C 1 and let i 2 be the current through C 2 and R 2 (since no current ﬂows into the V out measurement system). The relations for each component are: V in V = R 1 ( i 1 + i 2 )( 1 ) C 1 ˙ V = i 1 (2) C 2 ( ˙ V ˙ V out )= i 2 (3) V out = R 2 i 2 (4) Manipulate these relations to eliminate V , i 1 and i 2 . Substitute (2) and (4) into (3) and (1): V in V = R 1 ( C 1 ˙ V + 1 R 2 V out 5 ) C 2 ˙ V C 2 ˙ V out = 1 R 2 V out . (6) DiFerentiate (5), ˙ V in ˙ V = R 1 C 1 ¨ V + R 1 R 2 ˙ V out . (7) DiFerentiate (6), ¨ V = ¨ V out + 1 R 2 C 2 ˙ V out (8) Substitute (6) and (8) into (7) to eliminate V and its derivatives, ˙ V in ± ˙ V out + 1 R 2 C 2 V out ² = R 1 C 1 ± ¨ V out + 1 R 2 C 2 ˙ V out ² + R 1 R 2 ˙ V out . Rearraging into “standard” form, R 1 C 1 ¨ V out + ± 1+ R 1 C 1 R 2 C 2 + R 1 C 1 ² ˙ V out + 1 R 2 C 2 V out = ˙ V in , (9) so, a = R 1 C 1 ,b =1+ R 1 C 1 R 2 C 2 + R 1 C 1 ,c = 1 R 2 C 2 ,d =1 . 1

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Problem 2 Simplify these equations into a relation between V out, 2 and V in, 1 : ˙ V out, 1 + 1 R 1 C 1 V out, 1 = 1 R 1 C 1 V in, 1 (10) R 2 C 2 ˙ V out, 2 + V out, 2 = R 2 C 2 ˙ V in, 2 = R 2 C 2 ˙ V out, 1 (11) DiFerentiate (10), ¨ V out, 1 + 1 R 1 C 1 ˙ V out, 1 = 1 R 1 C 1 ˙ V in, 1 . (12) ±rom (11), ˙ V out, 1 = ˙ V out, 2 + 1 R 2 C 2 V out, 2 , (13) which is diFerentiated, ¨ V out, 1 = ¨ V out, 2 + 1 R 2 C 2 ˙ V out, 2 . (14) Substitute (13) and (14) into (12), R 1 C 1 ± ¨ V out, 2 + 1 R 2 C 2 ˙ V out, 2 ² + ˙ V out, 2 + 1 R 2 C 2 V out, 2 = ˙ V in, 1 which is rearranged to, R 1 C 1 ¨ V out, 2 + ± 1+ R 1 C 1 R 2 C 2 ² ˙ V out, 2 + 1 R 2 C 2 V
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## This note was uploaded on 10/09/2009 for the course MAE 107 taught by Professor Tsao during the Winter '06 term at UCLA.

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HW1_solution - MAE107 Homework #1 Solution Prof....

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