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# sample2 - 2.003 Spring 2003 Quiz 2 Sample problem Set 2...

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2.003 Spring 2003 Quiz 2 - Sample problem Set 2 Solutions Problem A - RLC circuit analysis 1. V o 1 = V i LCs 2 + RCs + 1 2. 1 n = 2 5000 = 31 , 400 r/s = LC 1 L = = 0 . 001 H = 1 mH 2 C n R = 2 �� n = 2 0 . 707 31 , 400 L R = 44 . 4 3. There are no zeros, poles at roots of 1000 1 2 s + s + = 0 0 . 001 1 e 6 1 e 3 s 1 1 e 6 s 2 1 e 3 dominant pole x ss = 1 Step Response Amplitude 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 System: sys Time (sec): 0.001 Amplitude: 0.633 0 1 2 3 4 5 6 Time (sec) x 10 −3 1

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Phase (deg) Magnitude (db) 0 −20 −40 −60 −80 −100 0 1 2 3 4 5 6 7 10 10 10 10 10 10 10 10 0 −50 −100 −150 0 1 2 3 4 5 6 7 10 10 10 10 10 10 10 10 Frequency (r/s) 4. 5. s = 1 e 3 = s 2 6. v o For circuit with R&C in v i v o For circuit with R&C in series v i Problem B 1. T ( s ) = n = 2 �� n = K = R 2 = R 2 LCs 2 + ( R 1 R 2 C + L ) s + R 1 + R 2 R 2 Cs + 1 = LCs 2 + ( R 1 + R 2 ) Cs + 1 K s 2 + 20 s + K K 2 K = 20 100 2. V out = ( s + 3)(6 s + 1) V 1 ( s + 3)(6 s + 1) + (8 s + 7) V out = (6 s + 1) V 2 ( s + 3)(6 s + 1) + (8 s + 7) 3. Solve using superposition V out (6 s 2 + 27 s + 10) = V 1 (6 s 2 + 19 s + 3) + V 2 (6 s + 1) ¨ ¨ 6 V out + 27 V ˙ out + 10 V out = 6 V
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sample2 - 2.003 Spring 2003 Quiz 2 Sample problem Set 2...

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