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Unformatted text preview: EE202 Exam I February 16, 2009 Name: __________________________________ (Please print clearly) Student ID: _________________ CIRCLE YOUR DIVISION Section 01, 8:30 MWFSection 02, 12:30 MWFINSTRUCTIONS There are 12 multiple choice worth 5 points each and there is 1 workout problem 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. EE202, Ex 1 Sp 09 page 2 MULTIPLE CHOICE. 1. The transfer function of a particular circuit is H(s)=abs+a, a>0, b>0. The sdomain response of the circuit to δ(at) is: (1) abs+a(2) bs+a2(3) abs+2a(4) bs+2a(5) abs+a2(6) −abs+a2(7) bs+1(8) None of above Solution 1. Lδ(at)=1a⇒1aH s( )=bs+a. ANSWER (8) 2. The inverse Laplace transform of V(s)=8s+32(s+4)2+42is: (1) 4e−4tsin(4t)u(t) (2) 8e4tsin(4t)u(t) (3) 8e−4tsin(2t)u(t) (4) 8e−4tsin(4t)u(t) (5) 8e4tcos(4t)u(t) (6) 8e−4tcos(2t)u(t) (7) 8e−4tcos(4t)u(t) (8) None of these SOLUTION2. V(s)=8s+32(s+4)2+42=8s+4(s+4)2+42⇒v(t)=8e−4tcos(4t)u(t) V. ANSWER (7). EE202, Ex 1 Sp 09 page 3 3. If R=2 Ω, Z1(s)=s+2, and Vin(s)=4s+4, then the Thevenin equivalent voltage, Voc(s), seen at AB is: (1) 8s+4(2) 8(s+4)2(3) 8(s+2)(s+4)(4) 2(s+4)2(5) 4(s+2)(s+4)(6) 4(s+4)2(7) 2(s+2)(s+4)(8) None of above Solution 3. VAB(s)=Voc(s)=RZ1(s)+RVin(s)=8(s+4)2. 4. In the circuit below, vC(0−)=0 , C=2 F and L vin(t)=2s2+4. Then iC(t)=: (1) 2sin(2t)u(t) (2) 4sin(2t)u(t) (3) 2cos(2t)u(t) (4) 4sin(4t)u(t) (5) 4cos(2t)u(t) (6) 4cos(4t)u(t) (7) 0.5cos(2t)u(t) (8) None of these SOLUTION4....
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