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Unformatted text preview: EE 350 EXAM I 27 September 2007 Last Name (Print): 501014! 005
First Name (Print):
ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Problem Weight Score Test Form A _INSTRUCTIONS
1. You have 2 hours to complete this exam.
2. This is a closed book exam. You may use one 8.5” X 11’7 note sheet.
3. Calculators are not allowed. 4. Solve each part of the problem in the space following the question. If you need more space, continue your solution
on the reverse side labeling the page with the question number; for example, Problem 1.2 Continued. NO
credit will be given to solutions that do not meet this requirement. 5. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a
grade of ZERO will be assigned. 6. The quality of your analysis and evaluation is as important as your answers. Your reasoning must be precise and clear; your complete English sentences should convey what you are doing. To receive credit, you must
Show your work. Problem 1: (25 Points) 1. (10 points) Consider the signal a a o (3 points) Sketch the signal f (t) in Figure 1, and appropriately label the tick marks. I‘( l) Figure 1: Sketch of f(t). o points) Sketch the signal IX? g(t) : Z 3 f(t — 3n) 7L:OC in Figure 2, and appropriately label the tick marks. Figure 2: Sketch of g(t). o (4 points) Classify the signal g(t) as an energy signal, power signal, or neither. If g(t) is either an energy
or power signal, ﬁnd the corresponding metric, E9 or Pg, respectively. 3 T l
P; = ~;:S}Z(4=)145 = 13— g (34:3L1L + 13‘ (3)5“
a a Z
n l 3
—_ <7 £3 .1 4;! 2 .3. EL
il's‘lo + 3[ L 3 l 3 Pzw 2. (15 points) A system with input f has the zero—state response W) Z 3 f (W) + 2 o (8 points) Is the system zerostate linear or nonlinear? To receive partial credit you must justify your answer.
_—_ Hum +2
«om—+ —+ am 3
_, 30¢?Cm) +1 = 3 F 04:0 * Z ’
me = cc at» Fag“) r)" (a ' ijt berm.) lat) = 33 a—HH‘O
the QoMaﬂenlegﬁ Because. J‘ H53 4’ 0( a “73 back] ("gimp 4\ “st u the. sgstem is £33}: 22’0"
“\SJ at o (7 points) Is the system timeinvariant or timevarying? To receive partial credit you must justify your 7 {10:3 acﬁ= SFCHJ) 4;
F,(t5= {—ch ~ﬂ 3.01:3: 3¥‘(ltl)+l = 3‘C(H:)'T l" 1 Do€> Oq, (1:) saﬁf‘TD? an, JCLQMWG' /—‘§ “Wu/'1? "in 3'03 : 3 ¥Cltl’r) +2
yet,T) :: 3 (PE—TA 2 82.0,“,va ahbtl 26 3,0627%) t/uz, SdS‘l‘Ernfj 151mg Vagmﬂ_ Problem 2: (25 points) 1. (13 points) A LTI system with input f(t) and output y(t) is represented by the ODE d2 d ,
E15 + 8 d—i + 16 a2 y(t) = 16 f(t) where a is a positive real constant. 0 (3 points) State the characteristic equation for the system and provide an expression for the characteristic
roots in terms of the parameter a. 2 ..
databases $5“:me cs2th = 73‘”? +16% ° 7m and»er mac o (4 points) Expressthe dimensionless damping ratio C and the natural frequency (an in terms of a. ’AL+87\+164} s 77’* lav/‘7 +Wf?‘ Q (.70 3
ave/Ema. coe‘F'p?‘c\On'bS‘7 2°: a} use? ==> ﬂ) P=L=L “F20. 7\ ' 110W»: 5’ o (3 points) For what range of a is the unit—step zero—state response of the system underdamped? F” 0'0 unﬂar‘ganymﬂ' Peace/M. 0 4 ‘F 4. 5 this
Feb/{res than/l7 a. v I, o (3 points) Find the DC gain of the system in terms of the parameter a. SIﬂLQ “a? ﬂa'plnaﬂ a) be‘ a, (7051‘va ‘Hm. ,sds‘lzom LS Md. #7! \Lao‘E‘ Craig/V Staa‘ﬂli. C ‘Fof av = w) ‘70 :0
W m 6 VLQm 7.3 ﬁloT ma.“ 8 For a. ) o and}, gm'éej
a. Conitana‘b Inna“: Fe Mn“ QUOVK froﬂwq, a‘ Go")! "xwre ﬂ Zena.
_ Q ‘ Q00; Jean.
r; fonﬂ a; the OMTME risttc, m as + 8; _l_ gas355;} {:0 =9 DCgaln
i/ of 4 0 2. ( 12 points) Figure 3 show the roots of the characteristic equation for a certain LTI system. Figure 3: Location of the characteristic roots in the /\plane. 0 (6 points) State the form of the homogenous solution yh(t). Each characteristic mode in your expression
must be real—valued, that is, you cannot have terms of the form aexpmﬁ). ghUﬁ " C, + gal'(CLC‘OSL +C3smt) 1:20
W’
'9ch Fw‘b ‘Frdm complex Coqddgdke Yogi; wt dféaln o points) Suppose that you are calculating the zero—state unit—step response of the system. Specify the
form of the particular solution for t 2 0. in this cue the, {arena gun/{7mm TS a— wm'l'lw‘éi gecwhser zen: Is a— rooﬁ ﬁe, charmlswslyrc, QOJw‘bcon) the
‘ 6x2, \CWM parﬁlwlar bolu (or b g? at) 2 b 4:. o points) Specify whether the system is asymptotically stable, marginally stable, or unstable? To receive
credit, justify your answer in one or two short sentences. 7—»ﬂ1Q”$¢n’:€m Er marimwgz 54:4.le— bectwmr % the
Y‘ou'b 7‘136. U! Problem 3: (25 points) 1. (8 points) The circuit in Figure 4 is driven by an independent current source with a constant strength I 0. Prior
to the switch closing at time t : 0, the current and voltages within the circuit have reached steadystate values.
The current through resistor R3 represents the circuit output. R, W) n Figure 4: The switch in the passive RC circuit is closed at time t : 0. o (4 points) Find an expression for the current y(0+) in terms of the circuit parameters Justify your answer. GAYWit 6i t: O— 10 0  'UQCO‘D o (4 points) What is the steady—state output current, y(oo)‘? Justify your answer. GNU/[é 1h 45éedﬂf’5ﬁw‘k, thQ capau‘lsor‘ appears 66 W O
thﬁ— Cur r o :5 oomjgﬂ, Laﬁmon “bké banﬂ amSL 6'»
Con'butmy R7, anxﬂ/ Us”,a cgfr‘an‘la— can/lﬁonj
M/
(12. ~—
(003 '3 __,————— .Lo
9’ RL+9~3 2. (9 points) A LTI system with input f(t) and output y(t) is represented by the ODE d’l
—”’ + w fm‘ 2 W) The system is driven by the input
ﬂn=e‘”mw and the initial value of the output is 31(0) 2: . o (4 points) Determine the zero~input response yli(t). 9“): — Ga {:2
Q0): 1+2. ==O 7.) 7),=—2_ :7.) ybtt), : “E70 0
> :0 — .
" BQCOLJSQ, 40b) 110 £0.» 13—7/0) Ct) @
?Nﬁ’ —at; 7
' #3; Hr):— ths); 0.6 ;C.e. t,o
/ I
choose. a. go thatégawkh
‘21: t 70
ECC'Q; e ._
o (5 points) Determine the zero—state response gm“)  Z t; f 20
’ O 6
I35 wkaua, y, Lt\ ,. l 'L
6’ , o 2 3;, 24¢;
4r
. 523E 56) then % p
7
09W“ ‘3 t ’0
2t_+~ J; #20
a (53+ H53 e C’\e
' gag HQ’ g): cyan c4 50 thdb ya; Co);O.
7:; C s “"L‘ .
#25“)? C‘fJi so 3 I 2,.
—lt' J__ 7
; ~ lie + Z to 3. points) A ﬁrst~order system with input ﬁt) and output y(t) is represented by the ODE dy , \
(17 + a g(t) w ﬁt), where a is a real—valued parameter. For a certain input f (t) and some initial condition 31(0) it is known that
the zero—input response is yzi = 2 while the total response is 0 (4 points) What is the numeric value of the parameter a? an): 7i+w=0 “> “To” The. LOT/n by» the. Zero—mini re/Apoosells t
ilt ﬁov ‘570
~ 0 e ,
glam/3; gnaw) > QC / i
' kJ‘ are,. («Jen t QQOWUJQ/ we, a 3.£— f2 0}
a; (5} ; 26 If Follows £31, 2: +3.
o (4 points) What is the input ﬁt)?
The, 4:94:ch ru.pan& RM the, ﬁorm 31;
e +g€ [213%
any, UIQ K’now thd)‘: ——
.— yam ; WM 4*pr“
at __h. ~21?
%u>—3:: a; +6  t 4;
2.1: h a {310.1qu 401v tun
This means f%.) a Q, . J ‘3 [7 ’ must 5.4115? the, ODEJ diva So ' 3 = F115).
9’? + 07"
Selma? 'F'of‘ ‘Fl‘bl, Zt — 2—6 — e
“433: (2*3577‘ 2 Problem :1: ('35 points} l. (8 points] Figure 5 shows a remut photograph of David Salvia riding the Hkyt‘oastcr at Kennywnud amuseman
park. betllfEI'Pfl in a nylon harness and released from a height of 180 feel. David swings bark and forth until nmt‘lmnir‘ul damping arrests his motion. The ODE W26
0 '2
(1'0 % (18 4 5 0 0
_._. a. a — ~ — :
(th ‘ alt 100
approxin'iares the? (lyInimic behavior of the angle 6' deﬁned in Figure 6. For cert(Lin initial conditimis
r—r WJ‘
6(2‘.) : Ag rid/T} cos it ~— gm u.(t)
' 10D where Ag and g) are (.‘Ul'ltitaIltﬁ. Deternnne the numeric value. of T. Your answer should not (:(.)11t.ai11 square mm
synilmls ('JI' nunilmm raised by exponents: expregs your HIISWCI' as the ratio of two whole numbers. Figlll‘ll‘ F): David Salvia in motion. Figure (5: Skewh of the skycoﬂtvr. tin—E “the resf’ome‘ T5 ungergamﬂeoq arma 5" 75“ “’ng . gins, Cormko‘i: ﬁ+jséjJICufgdmﬁm hmﬁ' m 40w“ 3,11: — (PM t?¢.u;0_ It
ﬁe. 0 Grant. l ’9 mallow; {’th 'C 2 ' l/p‘eE'A‘JZB : if: .
mi”: (Sr/{0031 55:? “"7 = 57"“? l i  ﬂ, : Lf/loo “rm the sat«'ﬁwn '9)! 60(2) J ou o ~ I_ FL
Saba. gar (an Us“? the rein, «3le (#97 , U—m d 2. a 7' 1': MIL/DU}: W“(la: WL
mi: tun Chi”)? “"70 n FT “ i“ 705' ‘* loo 'From tho— ODE) 2. (9 points) The RLC network in Figure 7 with output voltage y(t) is driven by the current source f(t)i Determine
the ()DE representation of the circuit and express your result in the standard forin d'n, dn— 17 dm dm, 1
dt’E’J + “"4 drniij + ‘ ‘ ' + any = bm f + b f +w+bof. dtm ""1 dtmil f0) KCL. “it nodL. F} (,1) KVL. (“MR Q00? B" (Lina z. [Lita + L Erb— CZ) abuwiﬁo” 9‘3 3 glimmth 2.61;) in ebuufb‘m C2) “‘5’? . 2‘ From abUcv'E/‘Jn C”) C Lt) : 4 .. C
‘ £0 £2):
~°“U SdbS’tt’tute ﬁe, 5,77 re‘j‘slm LU)” K; (It?) In #61:): R9 —gcat1 + L5? LC—g C3) 10 3. points) An active circuit with input voltage f (t) and output voltage is represented by the ODE (13, (1231 (ll d .
—:U+5—1_.i+s—J+4y(t) : d—J;+16j(t). The circuit is driven by the input voltage I
f(t) = t e‘t sin(2t) Mt).
Write an m—ﬁle that o Determines and displays the roots of the characteristic equation. 0 Calculates the zero—state response of the system to the input f (t) over the time interval 0 S t g 15 using
1500 points. 0 Plots the zero—state response and input in a single plot, and distinguishes the two curves using a legend
as well as different line styles. The m—ﬁle should also add appropriate labels to the axes, and provide the
title Active Filter Response. “70 Repment 60:; m her/r15 al P well 6?
= E51613 Q; EMS—5 8» H33 7;; bet? rmme elm. rqwfsc ndﬁ‘o roots root5 ((9) ‘70 PmoQ, Zam's‘kwﬁe Flo/Sports?— t— :7. ’mepmﬂ, (0)15J Koo); .— {1 z t .H6 exPCt) .‘7F 310(236633 Oi : Qsim (gagging % mot robutts / /
Plot (’tJ‘FJ l’rl’) ,Qeaemﬂ' (‘ m‘nut ’J ‘ rezponsz’)
intle (’ A’o‘tlL/Q, PHIBe,— Raeslaonszl) XleQ/l L‘ “f:th 1:81!)
OnllcerQl C’ “ﬁlth1:91.!) 11 ...
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 Fall '07
 SCHIANO,JEFFREYLDAS,ARNAB

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