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Unformatted text preview: ECE—20200 Test #1
September 17, 2009
Professor Talavage Professor Chen
8:30 AM MWF 1:30 PM MWF
Section 002 Section 001
t.
gPIease printclearly)
PUID:
INSTRUCTIONS This is a closed book, closed notes exam. N o scrap paper, calculators,
PDAs, or cell phones are permitted. Two transform tables are
attached with the test booklet. You are not allowed to place any items on the seat beside you. The test consists of 10 multiple choice (MC) problems'and 2 workout
(W0) problems. Please write legibly in the test booklet for the W0 problems.
Carefully mark your MC answers on the scantron form BEFORE THE
END OF THE EXAM PERIOD. Any work in the test booklet related
to the MC problems will a): be graded. When the exam ends, all writing is to stop. Any writing on the exam
booklet or the scantron form after the end of the exam is announced
will result in a grade of zero (0) on the exam. You will turn in both the scantron form and the test booklet. 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 test. 1 1. Multiple Choice Problems (6 points each) Compute the Laplace Transform of the following time—domain function: f(t) = e2e_2’u(z‘ —1) (E) None of the Above 2. Compute the inverse Laplace transform of the following function: 3S+2 F(S): s+1 (A) 3e’u(t) (B) 3 sin(t)u(t) ~ e"’u(t) (C) 3 cos(t)u(t) + 2e"’u(t) (D) Are—w) + 350) None of the IE 3. The inverse Laplace transform of (A) (B) (C) (E) x s2+4s+13 [2 + 263t sin(2t)] u(t)
[2 + 2e“3t cos(2t) ] u(t) 26(t) + 2e‘2t sin(3t) u(t) X. None of the Above 232+1Os+30 . IS 4. Listed below are the Laplace transforms and initial values of f1 (t) and 130) l 2 _
E(s)=~+—3, f1(0)=l
s s
2 _
F2 (s) = 7 and f2(0 ) = l.
s
The Laplace transform of dill?) + 2f2 (t) is (A) 1+—22—
S
(B) 9;
S
(C) 1+—47
S (D) (E) None of the Above 5. The input impedance, Zin(s), 0f the RLC circuit shown below is 29 3H Zm—> __ 1/39
3F } (A) 982 +15s + 7 3(s+1) (B) 3(s +1)
9s2 +153 + 7
(C) 982 +15s + 7
3(s2 +1)
(D) 3(s2 +1)
932 +155 + 7 (E) None of the Above 6. For the circuit below, derive the transfer function Vin (S) H(S) = = Z(S) I m (s) 2 R 1
s +—S+~—~
L LC
(A) S
S2
s2 +is+i
(3) RC LC
1 (E) None of the Above 7. /?¢i\
/ ,5 You are given, for the circuit in Problem #6, a transfer function that relates the input current, Iin to one of the three output currents (IR, IL or 1C): 2
S 2 1 1
s +—s+——
RC LC H(S)= To which current ratio does this transfer function correspond? I_R [in [in (B) [R
I (D) (E) None of the Above 8. In the circuit shown below, the initial capacitance voltage and inductance current are zero. Let V1(s), V2(s) and Vin(s) be the Laplace transform of v1(t), V2(t) and Vin(t). The nodal equations in the frequency domain are 1/29 V1“) 1/49 V2“) l
(A) (3 + ~S—)V1 + 2V2 = Vin
V1+(s+1)V2 = 0 IIIII __ (B) (3 +‘S‘)V1 _ 2V2 : Vin \) V1 — (s +1)V2 = 0
C (S + 3)V1(S) — 2V2 (5) = Vin (S)
( ) V1(s)—(s+1)V2(s)= 0
D (S + 3)V1(S) + 2V2 (S) = Vin (S)
( ) V1(s)+(s+1)V2(s) = 0 (E) None of above 9. In the Circuit below, the capacitor voltage at t = 0— is 2 V. Solve for Vout(t), for all t Z 0. 4n
+ 1sprout (t) 411“) 0.25 F (A) Mr) ~Ze"u(t) (B) 2W) (C) 4u(t) — 4611,“) (D) 6u<r>—4eIum ' (E) None of the Above 10 10. Consider the circuit below. The initial conditions are MO“) = 2 A and VC(O‘) = 0 V. 5 Q
l iL“) +
vino) 9 2 H 3 1: vcm The sdomain circuit, including the effects of the initial conditions, is (A) (B) e l/(3s) 9 1/(3s) (C)
5 o 9 l/(3s)
0 (E) None of above 11 Workout Problems (20 points each) I WOl. The Laplace Transform allows us to consider the solution of differential equations (such
as those that characterize the time—domain behavior of circuits) Via algebra. In this
problem, you will demonstrate your facility with this powerful tool. (A) (6 pts) Convert the differential equation below, including the initial conditions, into a frequency domain expression. 55(1) + 7x0) +10x(t) : 10w), x(0*) = 1, 540—) = ~1 (B) (4 pts) Solve (algebraically) for X(s). 12 (C) (6 pts) Obtain the partial fraction expansion (and corresponding coefﬁcients) for
X(s). (D) (4 pts) Obtain an expression for x(t) from you answer in part (C), using the Inverse Laplace Transform Table. 13 W02. In the following circuit, the initial capacitor voltage and inductor current are zero. Solve for Vout(t), for all t Z 0. 110‘) o 1 F 1 H 1 n Voutlit) (A) (8 pts) Construct the sdomain circuit. (B) (6 pts) Solve for the Laplace transform of Vout(t). 14 (C) (6 pts) Use the Inverse Laplace Transform to obtain Vout(t) 15 ...
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