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Unformatted text preview: ECE 210/211 Analog Signal Processing Spring 2008 University of Illinois ' Basar, HasegawaJohnson, Trick Exam 1 Thursday, February 14, 2008 — 7:008:15 PM Section:
circle one Class: ECE 210 ECE 211
circle one Please clearly PRINT your name IN CAPITALLETTERS and circle your section in the boxes above.
‘ This is a closed book and closed notes exam. Calculators are not allowed. Please show all your work.
Backs of pages may be used for scratch work if necessaryL‘All answers should include units wherever
appropriate. ' ' ' ' ‘ 3 ‘ ‘ '
Problem 1 (25 points)
Problem 2 (25 points)  Problem 3 (25 points) Problem 4 (25 points) Total (100 points) Problem 1 (25 pomts) 6‘1 3 5 Nott¢1~l+gizr€/M a) [4 pts.] Find
0 o ' 7/2.
LL +j = /35 [(‘1+j)7= 79/5 —1+j= AD: (_1+j)7 = 92.
0/2 225° .
b) [8 ptS ] For the following circuit: — 1550
i) Find 10.0%?me
— W415: Io *4 V szI—Io —:S’V 0 11) Flnd the 1nJected or absorbed power by 2A independent current source
Q
BA 2 2A ' ”all; 1
2'4 ‘1}; : ft/V+ZSLI0 {V :> ’ ll,
w ¢ 4v 2:11 ’7'
5 ” b
c) [7 ptS.] For the following circuit, determine RT to the left of terminals a and b.
‘ u 12 +5
“ a a 3
‘ 4.
_ 2 =2”;
VT“ 5” T 5 E
_’> RT = 3 52'
Mie® muggy; fizw‘
d) [6 pts ] A circult is described by 03 0‘1 5": ~ 740" 2
dy
— + 2y(t) = 2 cos 2t y(0) = 3 b __
dt 5 3(0):}; k3 M
i) What is the zero input solution? S‘f‘ 3 é 2f
__ t =
”Li + 713169310; I 6—335 K6 40 yzx() —_
Wg+2)K€ t20:? 5"2 ,
ii) What IS the steady state solution? W ‘ 41%: .5: C Q (2.5‘é/6’D)
25 yss(t) = 4*
(ng 4— ll: 2 €42 wipe». 3a.): _‘/ 6“, 49
ﬂ é 2" ..L ’4‘9‘ 0 5 2%
a 52 : J: ’ J/__ 'WSO
@4‘4233’221 f‘lv‘fjie  ‘L 219450) Problem 2 (25 points)
3 Q ”909 a) Using nodal analysis, ﬁnd but do not solve two equations that could be solved for vA and V3.
The only variables in your equations should be vA and VB. Mam»: 0,1qu .1 2A 1)} __ 45a 3' 2/9
“’33 3 3n 337'
. 041/ J—w 1&1"; v are. 5” a)EQ.#l.: _ W ‘7):3 .. é _a)EQ.#2: "L54 +51% 55’ b) Using loop analysis, ﬁnd but do not solve two equations that could be solved for i1 and i2.
The only variables in your equations should be i1 and i2. lp‘oqrii *ZV+&3151+35>—Uz+m“52) "FM/=0} (26»351337 r c ~ :7 \ t ‘—
700‘VZ“ *31/4—352. (£22AL’> F0 ' “BL: +3l2 ' ? Problem 2 (continued) SQ W609 c) Choose RL in order to maximize the power burned in RL. Specify RL, and the power PL burned in RL. Your answer should be a ﬁmction of some, or all, of the variables vA, VB, 2'], or 2'2 (as they are
speciﬁed by the circuits in parts a)>and b)).
1 15¢ ZL (?4ri' 19}
00¢ U
L,MAx ‘5 4:21 3 54/05/51 : ’07?) 42/?/
0 R 3/2 52. pm; Elf/3T: ‘er : Ugtrufal C)RL= 4f
8/22 Problem 3 (25 points) a) Complete the circuit below to achieve the given output. You have two 1 k9 resistors, one
10 k9 resistor, and an ideal op amp available. v0 = ~10 (v31 + V32) t ﬁaHL (Bani? awe
_ [Miler/$81 ( Fm 0525 f 3145 {— ”72¢
_ a/O
00%) ’5 ‘/Oé j; QOVMM’” Fm ms :1? 542*“ e tea): ~ 30v 'Mél WW 3#5
= '30V+ {M59 (t '3”)
If R = 1 k9, L = 1 mH, and vs is as shown below, sketch Vo(t). Clearly label values on the axes. Problem 4 (25 points) Consider the circuit below. The switch IS open for a long time and the circuit is at the steady state. At t— — 0, the switch 18
closed. a) Find i(0—') and 1(00). :5: 0’ ms; {ng 1(0~)= % A
M t _ :> at»: W
b) What 1s the time constantt fort> 0 ofthis circuit? ‘ (WT/Lava?” r? C
T: L/R : $751,: 7/59
iatrhema 57; CH Ado». E: ”C: 7'15 5 \ L L00
c) What is i(t),t>0? £115). '2 K 6 it > ...
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