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

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Unformatted text preview: 2.003 Spring 2003 Quiz 2 - Sample problem Set 2 Solutions Problem A - RLC circuit analysis 1. Vo 1 = Vi LCs2 + RCs + 1 2. n L R L R = = = = 2 5000 = 31, 400 r/s = 1 = 0.001 H = 1 mH 2 n C 2n = 2 0.707 31, 400 44.4 1 LC 3. There are no zeros, poles at roots of s2 + 1000 1 s+ 0.001 1e - 6 1e - 3 s1 s2 xss = 1 = 0 -1e6 -1e3 dominant pole Step Response 1 0.9 0.8 0.7 System: sys Time (sec): 0.001 Amplitude: 0.633 Amplitude 0.6 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 4 5 x 10 6 -3 Time (sec) 1 0 Magnitude (db) -20 -40 -60 -80 -100 0 10 10 1 10 2 10 3 10 4 10 5 10 6 10 7 0 Phase (deg) -50 -100 -150 10 0 10 1 10 2 10 3 10 4 10 5 10 6 10 7 Frequency (r/s) 4. 5. s = -1e3 = s2 6. vo vi vo For circuit with R&C in series vi For circuit with R&C in Problem B 1. T (s) n 2n K 2. Vout V1 Vout V2 = = (s + 3)(6s + 1) (s + 3)(6s + 1) + (8s + 7) (6s + 1) (s + 3)(6s + 1) + (8s + 7) = = = = K s2 + 20s + K K 2 K = 20 100 = = R2 R2 + (R1 R2 C + L)s + R1 + R2 R2 Cs + 1 LCs2 + (R1 + R2 )Cs + 1 LCs2 3. Solve using superposition Vout (6s2 + 27s + 10) 6Vout + 27Vout + 10Vout = V1 (6s2 + 19s + 3) + V2 (6s + 1) = 6V1 + 19V1 + 3V1 + 6(V )2 + V2 2 Problem C The transfer function for this system is x(s) f (s) n 2n = = = 1 meq s2 + cs + k k meq c meq 1. From graph, we measure the following T d Mp A n meq c = = = = = = 2 1.0s d = = 6.28r/s T n 1 - 2 0.75 - 0.5 100 = 50 0.5 A = 0.215 2 + A2 100 ln = 0.693 Mp 6.43r/s 2 n = 4.8 5kg k 2n meq = 13.8N s/m 14N s/m Alternately, you could determine using the log decrement method. 2. meq I = = I r2 0.5 kg m2 m+ Problem D The transfer function for this system is (s) (s) Thus n 2n = = = k Js2 + cs + k k J c J 1. There are a couple of ways to solve this part of the problem. First, you can read r = 9 r/s and Mp = 5 dB from the bode plot and use the following 3 relationships Mp r = = 1 1 - 2 n 1 - 2 2 2 to find 0.3 and n 10 r/s. Or you can read n = 10 r/s directly from the phase plot ( = -90 ) 2. k = 1500 N m/r, c = 90N ms/r 3. = 1.1 r/s (t) sin (1.1t + 0) = 10 r/s (t) 1.58 sin (10t - /2) = 20 r/s (t) 0.3 sin (20t - 2.75) 4 ...
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This note was uploaded on 02/23/2012 for the course MECHANICAL 2.003 taught by Professor Davidtrumper during the Spring '05 term at MIT.

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