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Unformatted text preview: EE 350 EXAM II 11 October 2001
Last Name (Print): SQLL'EKD as First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Weight Score
1 I
2 25
I
4 25 Test Form B INSTRUCTIONS 1. You have 2 hours to complete this exam.
2. Calculators are not allowed.
3. This is a closed book exam. You may use one 8.5” X 11” note sheet. 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. If you introduce a voltage or current in the analysis of a circuit, you must clearly label the
voltage (current) in the circuit diagram and indicate the reference polarity (direction). If you fail to clearly deﬁne the voltages and currents used in your analysis, you will receive ZERO
credit. 7. 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) 1.1 (8 points) Figure 1 shows the block diagram of a system with input f(t) and output y(t) that is
comprised of three interconnected LTI systems with impulse response functions h1(t), h3(t), and
h3(t). Determine an overall impulse response function, h(t), such that y(t) = f(t) * h(t). + f(t) web)
X60 + (9—) W) Figure 1: Linear timeinvariant system with input f(t) and output y(t). xca = new: vsch 0)
w (a = “a an 11, (we  {ith) cz)
gun = h3C'9*<wca 44cm) (3) SubStIEt/‘Elna CI) ale. C2) in (3) aselﬂj (>109 = “git9*}: ﬁt) at Mat) 4%) + H183 unto] 7‘ Y (1:) an mama;  Hﬂadum +— ﬁt) a: Inca “13%) [hma‘hsca ~h9bb) + Lam Hum] #01:) = Haas“ In (t) 1.2 (12 points) Determine the impulse response function of a system with input f(t) and output y(t) that
is described by the ordinary differential equation 12(t)+ 500 110:) = 3000 f(t). h (*3 = it
0“ we mm. Sch +ct) =. “can, ' 62(76 = 'A+S“Oo=o =3 gut): [at /%+ $00 P5: 3000 ‘17—) 5" 6 f
3‘, Lt) t\ : é. CconsE) as {Ct} =— i 0
3' 6 ‘1: 26, 1.3 (5 points) Write the necessary MATLAB command(s) to generate a plot of the zerostate step response
of a. system described by ya) + 172 y(t) + 1920 y(t) = 3840 f(t) + 19200 f(t). mm? ( [38%) 192001) E1, I72, ncxzoj) Problem 2: (25 points) 2.1 (10 points) Determine the frequency response function of the passive circuit shown in Figure 2, and
express your answer in the form Y : bm (1w)m + bm1(3w)m‘1 +  ‘ b1(]w)+ be
P (Jw)" +an—1 (1“!)7“1 + ' ' '01 (W) + 0:0 . 26 “Til/3'0 H(Jw) = gmx ¢LW7L W 90000 Figure 2: Passive RLC circuit with input f(t) and output y(t). Usu’U‘ Jorbaaw" ‘9‘;"St'¢::K 7L
1 (U
V: j{,_:21:2;_,s E
__L__, + R tut/7L. WZQ Wﬁﬁfi...’ Voltqyz, “FOSS K ( wzg3c w'7L)
R+ayu7L +(R/W7L‘)(/WZCD 32.
“9”” E 2.2 (10 points) Figure 3 shows the Bode plot of a LTI system whose input is is
f(t) = 7 cos(1000 t — 45°) — 12 cos(50, 000 t + 25°). Determine the sinusoidal steady—state response y(t) of the system. The Bode magnitude and phase plot of H000). 20 ﬁ.
. . ( I :20413
10 .... " .. ’Hé‘ Wade
.2 o _
« HM mm —— ‘0
€40 ..................... .. d
5' Phase in degrees 102 103 10
Frequency in rad/sec. Figure 3: Magnitude and phase of the frequency response function. U82, Super!) 9 Sl‘tlon : 4‘ It) : ’7 (as (locob "15‘°) oo '0
gltt) : Witty/1%” C05(looo‘l: “VS°+ 5%‘3‘W’33 :1 '70 C05 (locot " 95°) 59:"
~FzCtb : —}2.Ccs($0)ooo't +ZS°3 I M 529000))
glue) 2 ~Illﬂy%oo)‘(os(foaoot +7.3” + 2t ~ l’2.60.3 (50,0004: ‘ 60°) COS(IOOO‘E’ ~9§°D ‘lLCoSCSbJoooE ~60°> 2.3 (5 points) A certain LTI system has the frequency response function 9130 ]w + 30600 H = .
(3“) (M5 +188 1w + 3060 Write the MATLAB command(s) to generate a Bode plot of this frequency response function. bafﬂe, ( [7190, 3060on F1, 188) 3060]) Problem 3: (25 points) 3.1 (5 points) Figure 4 shows the input f(t) and impulse response function h(t) of a LTI system. Sketch
the zerostate response of the system. f(t) MD (1) (2) —1 2 —2 4 Figure 4: Input f(t) and impulse response function h(t). 42m =  SCtH) + Zane2) 5mm = «Hanks = (a(t+o+25—cf.a\wa> [Enb) '2 th+1) + zMb—z31 3.2 (8 points) Another LTI system has the impulse response function
h(t) :u(t+4—a) — u(t—4—a).
o (4 points) Suppose a : 0 so that
h(t) : u(t + 4) — u(t — 4)
and the system is driven by the input
f(t) : u(t) — u(t — 2). What is the width (duration) of the zero—state response y(t) : at h(t) ‘? F (t) 1) Hz)
I 
t 4:
Z  H q
a (a = we) * hit)
( 3
JVFG. WA % act) = Jurw'écm % + ﬂur¢£on é I (4 points) For what range of values a is the system causal ? Hit) ‘The. system is Gala/ya \‘F hob) 75 a, cmﬂ SaﬂOQ. Choobe OC > H A 3.3 (12 points) A system with impulse response function
ha):e‘h4w is driven by the input
f(t) : u(t — 2) — u(t — 3).
Using the graphical convolution method, determine the zerostate response y(t) = f(t) 4: h(t). In order to receive credit, clearly specify the regions of integration and, for each region,
provide a sketch of f and h. 03
§Cﬁ  'Flbﬁkhlt) : fFCtﬂvfJ: 'C)oQ,t; F(z;) f hoec) A d)
d! D 096t>= Hakemac =0 téz
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t ~ao the '0 _£ ‘5 22m guan Ram 11: z< I= <3 iiiit) 4: t ‘ e;
‘ an.) =X ﬁzﬂg/tﬁczc . é jetlt
7 2. t
'C 1: 
22553 : e I:an = €110}ng
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—(l:2
owns): {~Q ) LZbél’s 10 Svmmmr12\n 11 Problem 4: (25 points) 4.1 (9 points) Consider two signals = (1
(1’20) = ,Bt deﬁned over the interval [—2, 2] where a and [3 are nonzero positive realvalued constants. Show that ¢1(t) and ¢2(t) are orthogonal and determine the value of a and ,8 so that the signals ¢1(t) and
¢z(t) are orthonormal. :2 ¢é[£ 2. a m “£2 ct; be) 12 4.2 (6 points) An orthonormal signal set {¢1(t), ¢2(t), ¢3(t)} is deﬁned over the interval [0, 2] and exactly represents a certain function over this interval as M) = 3 ¢1(t) + «/§ ¢2(t) _ 2 433a)
Using Parseval’s theorem, determine the energy of the signal f(t) over the interval [0,2].
=) haemvsq, the, tin/é 7) 3
<4, ¥> = 2» Q: <¢5i 0L) OrthonormwQ
L"! ‘ (337— + (53‘ + (25" =7+3+l4 13 4.3 (10 points) Suppose that ¢1(t) and 433 (t) are nonzero only in the time interval 0 g t S T and that they
are orthonormal over this time interval. Consider a LTI system with the impulse response function h(t) = ¢2(T — t).
In order to understand Why this system is called a matched ﬁlter for the signal 452(13), show that o (5 points) If ¢1(t) is applied to this system, then the output at time T is 0. 09
act) :2 qb,(h)~ubHsB'—= §¢.Cc)b(i:~t)ch : loo ‘D'm ¢zCT~ HQ oQ‘c
O‘LCT’) :2 5.00 (p; Lt) dJZC‘C)oQ‘c z 5T¢dm¢zctqgot :0 73912, n‘l:€(7V~VQ, Lg agro becausg’ the
“ﬁnds Am‘ orthonormng an hean er’ﬂdoaonaaz T
<ch «21> 2L <b,tt)¢zt+.\a0,t =0 0 (5 points) If ¢3(t) is applied to this system, then the output at time T is 1. a (t): (pic'mwhHrD SJ ¢2_C'c) h(tt‘)cQT
=j gbzca c1621 7—: 7': +t3Qt GMT? ._ f; due dam Q: = gemWWQC ._. ( 73kg ﬂayl: m ﬂrﬂ is oAQ, lchauJA. The Slynuk
are eY‘thonom/mJ.) *thw‘é 13 < @520 $2,) : jTgﬁZ L—e) $1 Hr)ch .2 L 14 ...
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This note was uploaded on 03/17/2008 for the course EE 350 taught by Professor Schiano,jeffreyldas,arnab during the Fall '07 term at Pennsylvania State University, University Park.
 Fall '07
 SCHIANO,JEFFREYLDAS,ARNAB

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