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Unformatted text preview: EXAMINATION I EE 350 Continuous Time Linear Systems
February 5, 2001 Last Name (Print): @3— First Name (Print):
11). number (Last 4 digits) : Ofﬁcial Section number : Do not turn this page until you are told to do so Problem Weight _
u _
u 
I. 5 _
100 Test form B 25
25
25
2 Instructions
1. You have 2 hours to complete this exam. 2. This is a closed book exam. You are allowed one sheet of 8.5” x 11” of paper for notes as mentioned
in the syllabus. 3. Solve each part of the problem in the space following the question. Place your ﬁnal answer in the box
when provided. If you need more space, continue your solution on the reverse side labeling the page
with the question number; for example, “Question 4.2) Continued”. N0 credit will be given to solution
that does not meet this requirement. 4. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a
grade of ZERO will be assigned. 5. The quality of your analysis and evaluation is as important as your answers. Your reasoning must be
precise and clear; your complete English sentences shall convey what you are doing. To receive credit,
you must show work. Problem 1: (25 Points) 1.1 l ' i ' ' ‘
( 6 pomta) Cons der 3. continuous—awe mgnal {(t), not: ﬂute I; Hz) = O m) =(3—t)u(t—2). 3 gr cm? 4: < z
o (3 points) Is 3‘ (t) causal or noncausal? Why? 4%) is causal. beeswax. kHz0’ 1; <0. ( Lg. 43M) = a ) o (6 points) Determine the expression for the odd component, fo(t), and the even component, fe (t), of f (t) ﬂt’c) z: 1cth+ H4?) 2 == (3't)u(£'1)+ (“hut—17¢) W
2 «Com .— We) *‘R‘H 2 = (34) M({1'2) (3+{:)ul~t2)
2 o (7 points) Sketch f (3 «— t) in the ﬁgure below. ¥[3"t} .: t
‘33» ll (3 ~ (3~’c))u( (3—h2) t M143) ﬂ 1.2 (9 Points) For the halfwave rectiﬁer circuit in Figure 1 determine the power Pu and the root mean
square value yum. of the output y(t) due to the input f (t) = 4sin(t). Assume that the diode is ideal. You
may need the trigonometric identities: c032(;1:) = $0 + 608(231» and sin2(;t) = %(l  cos(2sc)) Figure l: Halfwave rectiﬁer circuit for Problem 1.2 wt) = { We) , Rt) 20 0 ¥Lt)<0 9 ~41 (ti t
o , T 91"
To a 11"
p1? = #0 OS Wthl alt = E14} 5 iesim’t 0H:
T “If
g g 1cos(2t)dt = 4 _. 3 it
= 4 as
.g
3,
H
W
a;
ll
:1
It
to
< Problem 2: (25 Points)
2.1 (10 points) Consider a ﬁrstorder system whose input f (t) relates to its output y(t) by 2% + y(t) = 10 f(t). If f(t) = eztu(t) and the initial condition y(0) : 1,
o (4 points) ﬁnd the natural response yn(t),t Z 0,
o (4 points) ﬁnd the forced response y¢(t), t Z 0, and o (2 points) determine if the system is asymptotically stable. 1) an): 23+1=o =3, ﬁzz;
, —t
M maximal mpome is W 10" mt Igm 3mm: c1: /2 2) Simex. ht) = git , i324), ﬁJ{MadeW All I'rnffu. 2t
¥o4xm «NEH: PX. mot 200499 + «Mm = 1010(t)
HT:— 4szt+P£2t=1012t => )3‘2
11W = 13n(t)+1j¢(t)_ (bertht2th
We): 1: c+2 =9 c=~l W) = “’0 31¢th We) 3,. 113a Agate/mid M foHc ﬂab/(i firmer, may <0. 2.2 (15 points) The switch in the circuit in Figure 2 has been opened for along time and is closed at t = 0. 120 3 mH Figure 2: RL Circuit for Problem 2.2 o (7 points) Derive an ordinary differential equation in terms of i(t) for t 2 0. PM i: >0 , H30. chew} bxcorrmo A vii) = LEE
L in") R “M 12 (1:): wt) 5 god
T R Eli?
KCL at A: 21ch gm = 0
id) + = 9' Lou
ea} 29;, is {m = QR. E} + 2000 NH :2 19000
Git o (5 points) Calcuiate the initial conditions i(0+) and v(0+) At ho‘ , Wu. u'nwfl' Meow, 129. tic") 125? 51?.
_ é: ito') 24V 657. 0 $24“ 54V
ice”) = 54+24 _
n+6
At t=0 , o (3 points) From the ODE in Part 1, compute the time constant 1' and the rise time tr. Sim“, am) .1: h+2ooo =_ o Problem 3: (25 Points)
3.1 (8 points) An RLC circuit in Figure 3, is described by the second—order ODE f(t) d, Figure 3: RL Circuit for Problem 3.1
J If L = 50 mH, Determine the value of the capacitor C and the resistor R such that the overall system is
underdamped with damping ratio C = 0.4 and natural frequency w" = 20 rad/sec. 2.
. “n a Llc ———> c _‘_._ * glues F. I}
L'W‘ 50x16 2o‘"‘ 3 ll
ll ._ _R_ ' a  . 
. zzwm‘ L e R 2391,01. — 2 o_4—zo50xlo 70
ll 08 :2 3.2 (17 Points) For the circuit in Figure 3, if R = 2.5 Q, L = 0.5 H, C =: 0.5 F, y(0+) = 6 V,
iL(O“') = "1.5 A and f(t) = 3u(t), 0 (3 points) determine the initial condition %§t:o+ it! = (ii(0+) = “11:5 : "3 V/S
dt t=o+ C 0.5 ' I (6 points) compute the zeroinput response yzg(t),t 2 0 Subsh'i'ufx. ma val/us a" R, 1.J a, 4&1. ODE for; Rm imam, 0123
OH:1 + 503.4 +43%): Hit)
0“: To carmFM’fx. aim(t) , 31.15 19d") =0 . =9 ypm=o aw) = 7"“+5A+4 = (2+1)(2\+4) = 0 => A,=I, fig4 O
t 4.t
mod 30, gm = when rm g C”, + c2;
. 4; ~41:
"j [’c) = C,2. —4c2»e
USL IFm'rLI'aL code'Hom . «9(0) = 6 = c1+c “410) ="3 1 x. o (8 points) compute the zerostate response yz,(t), t 2 D . Simu hf): 3, ’0 >0, Wm: p,
o o
d’f atg _ .
tL + 5‘ t + 4F  4 3
=9 [/3 = 3! _t 4t
gt“: 3+C'L +64 To condor)qu 0323M 31b , 11(0) == 0 , afﬂr 0. 10 Problem 4: (25 Points) 4.1 (7 points) Consider a linear timeinvariant system whose zerostate response y(t) due to the input f (t)
is given in Figure 4. ﬁt) y“) 3
2 L'II
System 1 0 l t l 0 1 t
gm
3
2 0 2 t Figure 4: A LTI system in Problem 4.1 o (3 points) Is the system instantaneous or dynamic? Explain in one sentence.
m Wi'm M wow/ma smu «3m , t z I, IS nmjsmo
bul' Wald 5? mo II’YIPMI' ﬁx. pd) =0, 75 >1 0 (4 points) Find an expression for the zerostate response yl (t) due to the input f1(t) in terms of y(t).
Obsunm. Hial' £6) = _3_(1C(t+1)+ $rt1))
2 met sima +131. me Ala LTI, «041(k) z .3. («3min yttUJ 11 4.2 (6 points) Consider a certain system Whose zerostate response y(t) due to the input f(t) can be
expressed as W) = f2(t 2) o (2 points) Is the system zerostate linear? Explain. 717a Wit/m «54 momb'rmow. Mam homey/midi? Mop/i}? {0.1923, 1%) ~——a> get): 'FZCJc—Z) 01 Ht) ——> oz‘J“(Jc—z) 4 agti) o (2 points) Is the system time invariant or time varying? Explain. H is hbou—I'oqku'omi' SfrnCX. HimT) —~———> —Fz({+T~.2) = W’s—FF)  (2 points) Is the system causal or noncausal? Explain. 1+ is caudal m'rna Val) (may ottme cm 1%.
Pauli WM of 1%. I'm/mi 73%» “Phi2) 12 4.3 (12 points) 0 (8 points) We desire to plot y = coe2 (t) over the interval 0 < t < (it using 300 points with the Matlab
mﬁle shown below. Carefully examine each line if corrections are needed, reunite the corrected line. Original m—ﬁle Your corrections t=iinspace(0,6pi,3); t = MSPaCL (0 , 6 * 9 300) y=cos(t)*cos(t); ﬂ? = GOSH?)  '1“ 00.9 (i?) J' l , ; . DOWN) ( t, q? ) , a (4 points) Write a complete Matlab command for ﬁnding the characteristic roots of the ODE 613:; dy _ dzf
2E + dt + 3w) .. 4m + 5 dﬁ. >> noots([2,0,1,3]) >> A00t3([2 o 13]) 13 ...
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