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Unformatted text preview: EE202 Exam II March 19, 2009 Name: __________________________________ (Please print clearly) Student ID: _________________ CIRCLE YOUR DIVISION MORNING 8:30 MWFAFTERNOON 12:30 MWFINSTRUCTIONS There are 12 multiple choice worth 5 points each and there are 2 workout problems worth 40 points. 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. Work on multiple choice problems and marked answers in the test booklet will not be graded. Nothing is to be on the seat beside you. When the exam ends, all writing is to stop. This is not negotiable. No writing while turning in the exam/scantron or risk an F in the exam. 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. Exam 2, ECE 202, Sp 09 1.For the circuit shown below vout(0!)=. The switch moves from position A to position B at t = 1 sec, back to position A at t = 2 sec, and finally back to position B at t = 3 sec, where it remains forever. Then vout(3.5)=(in V): (1) 0 (2) 2.5 (3) 20 (4) 15 (5) 5 (6) 10 (7) 7.5 (8) none of above Solution1:At t = 1+, 20vout(1+)=100!vout(1+)=5V. At t = 3.5, 20vout(3.5+)=5!10+10!10=150"vout(3.5+)=7.5V. ANSWER: (7) 2. For the polezero diagram of H(s)=Vout(s)Vin(s)shown below, H(2)=10. The response to vin(t)=2cos(4t)u(t)V is vout(t)=(in V): (1) 2 1!e!2t( )u(t)(2) 2e!2tu(t)(3) 8e!2tu(t)(4) 4e!2tu(t)(5) 4 1!e!2t( )u(t)(6) 4e!2tcos(4t)u(t)(7) 4e!2tsin(4t)u(t)(8) none of above Exam 2, ECE 202, Sp 09 Solution 2.(4) H(s)=Vout(s)Vin(s)=Ks2+16s(s+2). H(2)=10=K208!K=4. Vout(s)=4s2+16s(s+2)!2ss2+16=8s+2. vout(t)=8e!2tu(t). ANSWER: (3) 3. Given that H(s)=VoutVinin the circuit below where C=2F and gm=0.5S, the COMPLETE range of Rfor which the circuit is stable is (in Ω): (1) R>0.5(2) R<0.5(3) R<2(4) R>2(5) R<4(6) R>4(7) R<0.25(8) none of the above Solution 3: (5) GVin=(G+Cs!gm)Vout"H(s)=GCs+G!gmHence, BIBO stability requires that G!gm>or 1R>gm=0.5or R<2. ANSWER: (3) 4.A circuit with transfer function H(s)=(5+s)(5!s)s2+2s+(5)2is excited by the input vin(t)=5cos(5t+45o)V. The magnitude and phase (in degrees) of the output in sinusoidal steady state is: (1) 500, !45o(2) 25, !45o(3) 50, 45o(4) 50, !45o(5) 25, !90o(6) 50, !90o(7) 500, !90o(8) none of above Solution 4: (7) 5!H(j5)=5!...
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This document was uploaded on 07/25/2009.
 Summer '06

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