Final sp10 - ECE-202 FINAL May 7 2010 Name: _SOLUTION_...

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ECE-202 FINAL May 7 2010 Name: __________ SOLUTION _______________ (Please print clearly) Student ID: _________________ CIRCLE YOUR DIVISION Section 1-202, 3:30 MWF Section 2-202, 7:30 MWF INSTRUCTIONS There are 34 multiple choice worth 6 points each. This is a closed book, closed notes exam. No scrap paper or calculators are permitted. A transform table will be handed out separately. Carefully mark your multiple choice answers on the scantron form and also on the text booklet as you will need to put the scantron inside the test booklet and turn booth in a the end of the exam. Nothing is to be on the seat beside you. When the exam ends, all writing is to stop. No writing while turning in the exam/scantron or risk an F in the exam. Again, turn in the exam booklet with the scantron inside the exam. There are two exam forms and this is necessary for the proper grading of your scantron. All students are expected to abide by the customary ethical standards of the university, i.e., your answers must reflect only your own knowledge and reasoning ability. As a reminder, at the very minimum, cheating will result in a zero on the exam and possibly an F in the course. Communicating with any of your classmates, in any language, by any means, for any reason, at any time between the official start of the exam and the official end of the exam is grounds for immediate ejection from the exam site and loss of all credit for this exercise. The professor reserves the right to move students around during the exam. Do not open, begin, or peek inside this exam until you are instructed to do so.
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EE-202, Final Sp 10 page 2 MULTIPLE CHOICE. 1. In the circuit below, L = 1 H, R = 1 , and V in ( s ) = s + 1 s + 2 . Then v out ( t ) = (in V): (1) - 2 e - 2 t u ( t ) (2) 2 e - 2 t u ( t ) (3) δ ( t ) - e - 2 t u ( t ) (4) ( t ) + e - 2 t u ( t ) (5) ( t ) - 2 e - 2 t u ( t ) (6) ( t ) + 2 e - 2 t u ( t ) (7) e - 2 t u ( t ) (8) None of above Solution 1. V out ( s ) = s s + 1 V in ( s ) = s s + 2 = 1 - 2 s + 2 which implies v out ( t ) = ( t ) - 2 e - 2 t u ( t ) . Comma ANSWER: (5) NC ANSWER: (6) 2. In the circuit below, R = 1, C 1 = C 2 = 0.5 F, and i in ( t ) = 4 u ( t ) A. Then i C 2 ( t ) = (in A): (1) 4 u ( t ) (2) 0.5 ( t ) - e - t u ( t ) (3) 0.5 ( t ) + e - t u ( t ) (4) 0.5 e - 2 t u ( t ) (5) e - t u ( t ) (6) 2 e - t u ( t ) (7) 2 e - 2 t u ( t ) (8) None of these Solution 2. I C 2 ( s ) = C 2 s ( C 1 + C 2 ) s + G I in ( s ) = 0.5 s s + 1 × 4 s = 2 s + 1 . Thus i C 2 ( t ) = 2 e - t u ( t ). Comma ANSWER: (6) NC ANSWER: (3) 3. The transfer function of an unstable nuclear meltdown transfer function (normalized to nice numbers) is H ( s ) = s 2 - 3 s - 2 s ( s + 1)( s - 1) The step response has a term of the form Ke t u ( t ) where K = : (1) 1 (2) 2 (3) –1 (4) –2 (5) 0 (6) none of the above
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EE-202, Final Sp 10 page 3 Solution 3. s
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Final sp10 - ECE-202 FINAL May 7 2010 Name: _SOLUTION_...

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