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Unformatted text preview: EE 350 EXAM I 25 September 2008 Last Name (Print): 3 olv E! OHS First Name (Print):
ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO EL
4 25
_ Total 100 __l_ Test Form B 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” 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. (7 points) Determine if the signal
f(t) = 2e"ll [u(t + 1) — u(t — 1)] is an energy signal, power signal, or neither. If the signal is an energy or power signal, calculate E f or Pf,
respectively. "F09 "\ +1 BQCowse Em lpCt)l’—90J 24 um” be— pin/£2. lit)0
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’0 Onotzﬁeigﬂae When£20 1C (1:) 2. (11 points) A system with input f(t) has the zero—state response W) = f(2t— 1). o (4 points) Is the system zerostate linear or nonlinear? To receive partial credit you must justify your ' ﬁm 613—» (W = neato
has) a? #266 = ﬁatﬂ) {3306) = l ' F (Z‘E’I)
,9] (—9 t)  OCﬁttbr {aﬁm #3 3 No‘ka— #3619: 19" o—az ‘ché‘ranJ ‘ﬁnle ‘FL It) Q
+oll0uq§ 1955 the. 552me IS CQrO'S‘éu—ée Menu“. 0 (4 points) Is the system time—invariant 0r time—varying? To receive partial credit you must justify your F,(+.) E—a g. It)
got) “Goa—1°) agape/e) No'ée 'tl'uu'é
(9,112) = 44313) = "F. (Zt'lT) 7‘
941(1) =£(1—1) any» $.69: ‘ﬂfZE D _—_9 $515573 g. 41' (1“: ,T) ,1) _._. G (24: .4 27) 82mm. 714;) . F(Lt'lT) 755M477) = «She—psz the £34217) ‘15 ’lee.~vou*imiP o (3 points) Is the system causal or noncausal? To receive partial credit you must justify your answer. Suepon we Calculaiaq aa at) oak; amt 4; =10 sec“ the” (m: 430°!) :évr‘
& J Ira N) a“
thug}; bl “(U/{aim % of; {um 15:10 sec 3“ f—( ) i 3. (7 points) Figure 1 shows the input f (t) and corresponding zero—state response y(t) of a linear timeinvariant
system. f0) 31(1) 2 0 2 0 2
Figure 1: The input f (t) and resulting zero—state response y(t) of a LTI system. A new input f1(t) is applied to same system, and ﬁgure 2 shows the resulting zero—state response y1(t). MU) Figure 2: The input f1 (t) to the LTI system results in the zerostate response yl Find an expression for f1(t) in terms of f(t), and then carefully sketch f1(t) in ﬁgure 3. 8&1 “peaL,”  2. 0:3 2 (72) Because, {the SdSiLEon 75 LTI, 05 J TS cussva [fl let/EM J £uy‘ a" L_ T]: 5d s'bem 4m —{:}~9 alts &035 not guaran‘léﬂe, ‘Szct/ ;) (*5/92) ~ Anti (:9) —;,:: ~ a ‘n (Gucci
mi My: Jae/z) 0 mg? k v m ' 6‘ Figure 3: Sketch of the input f1 Problem 2: (25 points) 1. (10 points) When solving for the zero—state response of a particular LTI system, it is determined that the
homogeneous solution has the form yh(t) : 6—“ {A cos(4t) + B sin(4t)} + Cest
for t 2 0, where the coefﬁcients A, B, and C are nonzero realvalued constants. o (5 points) Using ﬁgure 4, indicate the location of the roots of the characteristic equation in the Aplane. .. —\T"l:
 744:; ‘IJZrm egtiAcosmaibammYS: e {Awsw£)+ Kwanzaa] @02on Complem cirydﬁwlzo. routs: 7shL: '5 '3':ng 4:
' ten" Caz; Wueavk a, rec/t “E ()3 = +3 Figure 4: Location of the characteristic roots in the A—plane. o (5 points) State the characteristic equation in the standard form
An+an_1)\”_l +~~~+a1A+ao :0 by specifying the coefﬁcients ai. Qm= (ammm cw w = (7\+6‘7~/)(7~+€+/~1)(7“3)
= (21+ lo7\ r HOW3) 7‘3 + (Io37 79' + (—30 + V137 ' '23 QC» = 7‘3 + 77? +[l7l "[23 :0 2. (15 points) A LTI system with input f(t) and output y(t) has the ODE representation
:9 + ay+ (2a —4>y = (2a — 4m
where a is a real—valued constant. c (3 points) Specify the roots of the characteristic equation in terms of the parameter a. = ’32. \lmz __ —orivacz_¢+.5
“1,2 3 + _(2¢‘I) _ 7’ 8
   —cc°C41
: we: (cc~07— _ 2—1311 z Jeri—i; —————
2 " z. 2. 7.. 2='2J 322—m‘l‘1 o (2 points) For What values of a is the system unstable? 0 (2 points) For what values of a is the system asymptotically stable?
ﬁe sastem is dual mp'ba't1cg9 stable. when both 7“ CWQ 77— are’ Yuaw'tlw. For a; = d‘ + 2' < " o (2 points) For what value of a is the system marginally stable?
\ :. 'tMS
we siéte'n 5 may] 0Q”? «Ul‘m n A 17" 2' qnﬁ 7" o) (dﬂ‘lL Lon hows when 10L=+Zl o (2 points) For what value of a is the zero—state unitstep response critically damped? 7%9. rasponse Ts crdmcdul? gain/kale taken 7“ uni 37, are. real) ﬂejvb'elV‘Q—J am9— [ﬂent( This (Maﬁa4.1m kdﬂs “Fur” , when ﬂ‘ =. '/\7__ = “Z.
0 (2 points) For what values of a is the zerostate unitstep response overdamped? anQ Juana") this (means W‘ a— oC. > Z) excep'é at. = 4“! _
(0C = 4“! is exolchnD— because. ﬂier matf “VG— 19522.ch— ) o (2 points) For what values of a, if any, is the zero—state unitstep response underdamped? WM 9» “luck 1. ME fume, Comﬂleit Cavﬁaﬁg With an. M “ﬂue, reui pa/‘é. Problem 3: (25 points) 1. (12 points) The circuit in Figure 5 is driven by two independent voltage sources7 each having a constant
strength Vs. Prior to the switch closing at time t : 0, the current and voltages within the circuit have reached
steady—state values. The current y(t) through the inductance L represents the circuit output. Figure 5: The switch in the passive RL circuit is closed at time t 2 0. o (3 points) Find an expression for the current y(0+) in terms of the circuit parameters. Justify your answer. t=—o— mgr/OE.» apnea,5 «.5 a. Shur‘ﬁ 'a(0*)=a(o') bum Curranl: Cuom'l: change
ansbunbunebdla. in L. o (3 points) What is the steadystate output current, g/(oo)7 in terms of the circuit parameters? Justify your
answer. In .sLeawo "shit; ;_ bec°me$ ax. Confewn'é C(n‘Q. So the Iﬂca/c‘éuﬁ
“'FPQMJ 4L3 ov shark cwcvs‘t‘: Cl L7... 30”) = c. + CZ %_ F—CL—
c ,S _ I/ \ Lot
V5 "42: ’3' 5? ,Obms V Vs
igﬁﬁ :1 0i 0 (6 points) Find an ODE that describes the behavior of for t 2 0 and express your answer in the standard form dy
E + (10 31W ~ be by specifying the coefficients a0 and be in terms of R, L , and Vs. 2. (13 points) A LTI system with input f(t) and output y(t) is represented by the ODE The system is driven by the input and the initial value of the output is y(0) = 3. o (6 points) Determine the zeroinput response yzi(t) for t Z O. _ at 1+ 2.=—o =‘> ’7‘.='2 ‘1'“9‘ yhc'gr'c’e
7» grape tzo 1172$,
Char. 96 n. : pavhmuw “khan: arc1;): @. 4) éé+ zzr/g =0 32:33: ahit3+gfﬁﬂ = Ca'Zt + o t 20. To swingI 53C (0) :3.) [games 32:” t 20‘ o (6 points) Determine the zero—state response yzs(t) for t 2 0. ._.. "Z‘b
4.5 a—LDO‘Q'J #5053” C— e 133° graz): cc. +GL't + b’tz J £20
7.. 2. a t
a 2m F.,_——Jz’
a? +Zyp 2 2t.“ *9 Z745? + +22 72+ 2+(t)
if. yr: rd a :=> iz“*’”‘° ’M‘l’
[+3_¢, =0 =>d—=’7' V=l
\b =
Y(2.t‘) + 2.(7+/5 t + CONN*3 at :0 =0 arm: J5 I=+t7 tZo #aS(e)= ghﬂs)+y/(£>: E e _c.'“‘ 798: gisav): O 2.
.. 2.
#3566): “is 21— + — +1: 153" Problem 4: (25 points) 1. (8 points) The RLC network in ﬁgure 6 is driven by a constant current source with strength [0. The switch
has been opened long enough so that all the voltage and currents within the circuit reach steadystate values
before the switch is closed at time zero. Figure 6: The switch in the passive RLC circuit is closed at time t = O. o (2 points) Determine the voltage, y(0+), across the capacitor immediately after the switch is closed. Clrcglt out: i=0“: (Micah ail/tears K3 Shari: QK‘EJ C‘s/)cLUL" as 0/2,, 0.5140. 05ml) Louu
‘lk E —9 one 02
V09) RH'QI %~RL o (2 points) Determine the derivative of the voltage, y(0+), across the capacitor immediately after the switch
is closed. 36 (Wr'n47 J‘VI‘5I9n) van; the ohun when) (Lfo’ = ——K—‘ 19 ‘—‘ Ezra ‘ ‘L(o+)= (Lro‘) = Jf‘lo J as fur/211$ cannot de’ye’ 'nsbun'ecmm‘d' r" L UM? KCL ‘17 Name, 14 in 1:ng 6 aﬁ ﬁrm tzo+ vials“ —Io + (is? 4 {009*} + (Lid’) =0 Ia + 351:: + (C(O'r) + Sir—Ia $0 => Cb(o+)=o 0 (4 points) Find an ODE that describes the response of the capacitor voltage y(t) for t Z O, and express
your answer in the standard form y+aiy+aoy=bo by specifying the coefﬁcients a0, a1, and 170 in terms of the circuit parameters R, L, C, and IO. 10 A (LCE
I0 6 L
VCL 01E hogQ, :4 d‘dﬂs (Mb)
4 E
_ Io + 2) + c.3112, + LLfo+)+ {Saccxgt =67 D‘¥X’o’€nﬁw‘[ﬁe Ext“ Suges % b“. galVi: a «drawn mg rQ§pafE '57 {Sher 11 2. (9 points) A circuit with input voltage f (t) and output voltage y(t) is represented by the ODE dgy dy
— 6— 9 t=1 t.
dt + dt+ m 8m The circuit is driven by the input voltage
f(t) = (1 + 2e‘t) u(t), and the initial state of the system is speciﬁed by y(0) = 1 and = —1. Determine the response y(t) for
t Z 0. ' E m: ﬁnal67+? = (7+332'=o kHz“=01; + Czte‘ £20
5 % I ' Becqum O an  an rah rayEs 04: the. charmLerufio eava'éloﬂ,
awe): «+1959 1:20, 54. sable. was) [2.5 cm ODE £0 99 09,4: — —  4.
et *ééet + 7¢+7@2t, = l3+ 362,
a"? Jr W a. :2 12 3. (8 points) Figure 7 shows an mﬁle for generating the zerostate response, y(t), of a circuit driven by some
input. A = 1000; B = conv( [1,16,100], [1,5] );
D = [0: 0.0001 : 31’; C = 2*ones(size(D)); y = lsim(A, B, C, D); Figure 7: MATLAB m—ﬁle for calculating the zerostate response y(t). 0 (3 points) What numeric values would the MATLAB command roots(B) return? L112, 2. acnedies ihe pobnomwQ, Q(;\) (a, GM ace
Q(7~\: ('7‘z+l€>’7‘+loo3C'/W+S) : BC?) ‘ 7‘3 4 2 7cz‘ 44802 4 :00 \r/ the 005 ,
ﬁn. (ammungv rth (8) return: the, oWw‘éﬁﬂS'bt rifts! 7[ + w are.
m “A: W rans pl the, SchnQ—aruoqf ﬁerm) a + 67 l J r——‘ .. r.  i: 6
21,1: ‘8': 67‘109 '' '8: 56 8 ﬂ‘ 1 «Jule. the. '90‘35» (3.9;) ha) a. roy‘t ’33 = —$‘. ﬁnd, 50
WI” roéwA root (8) o (2 points) Over what range of time values is the zerosstate response calculated? T/me, 'U'Wr‘q, ,OMQ. genera. J the 15th 9654* VSQQ/g~ QS'M' I b .1: E0: o.ool ‘ 3’] Raﬁ. So the. 434,02. Pct/1&9, % 116$,mgloc'étvﬂ :3 oft £3 ﬂ. «Q.» to.
14/50 nafe that tho, (.011: ,01371. fun, the (7 (49/7/‘Om‘ 0'55) [9(b)—..hoc>J
thQ. the. 'I'Wrth ,Qmo, specnqe, ﬁg IﬂﬂJé
~51 Hil ; LvLIt), o (3 points) Assuming that the response y(t) reaches a steadystate value over the simulation time interval,
what is the expected steadystate value of the zerostate response y(t)? 7%., ODE represenLuém % 1k. Lop/c.4294 (as’3'me Q‘Dlgzpmypj or C O. a 4» ll? +soa:/ ffoaﬂ : 200.9
In J'Leagi5£a—£ea 524:0) ghee505006155 :25“) “W55 13 ...
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