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Unformatted text preview: EE 350 EXAM II 1 March 2004 Last Name: gsglu‘ngni First Name: ID number (Last 4 digits): Section:
DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO
Problem Weight Score I
Instructions
1. You have two hours to complete this exam. asap This is a closedbook exam. You are allowed to use both sides of a 8.5” by 11” note sheet. . Calculators are not allowed. . 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 2.1) Continued.” No credit will be given to a solution that does not meet
this requirement. . Do not remove any pages from this exam. Loose papers will not be accepted and a grade of zero will be assigned. . 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. (12 points) A LTI system with input f(t) and response y(t) is represented by the ODE
1') + 4W) = f0) Find the impulse response representation of the system and, if your solution contains a term with a
delta functional, simplify that term. Funk the «Zero{tactic Vﬂl‘tC'bp mpg“ 66*?) 'Hﬁen huh :J&/£+:. — Char. E60. : )z'l' “1:0. Rodhs 7_ Char. £60", 2 = i: 2).
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a when #50 2. (8 points) A LTI system, different from the one considered in part 1, has the impulse response
h(t) : [1 — sin(2t)] u(t). Is this system BIBO stable ? Show work to justify your answer. b(t) 3. (5 points) The block diagram in Figure 1 contains three LTI systems whose impulse response functions
are h1(t), MO), and h3(t). The input to the overall system is f(t), and the output is y(t). Represent
the composite system using a. single impulse response function h(t) so that W) = f (t) * W), and express h(t) in terms of h1(t), h2(t), and h3(t). f(t) Figure 1: System comprised of several subsystems speciﬁed by their impulse response function. (0) got) : «~01:an  a.C£=7>rh3(t3 (O) 3.64:3 : F(b) at. in Hz) Cb) Wm = germHt) s was: ‘FHnuszBF“) subwuey can WV. (A) ("to (:2) 3M; 3C4.) : [:Ftt) 9: hot.) F(£)])the)  [Fztwt‘J—gﬂ» In It) at
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hlt) Problem 2: (25 points) 1. (14 points) Use the graphical convolution method to determine an expression for the zerostate response y(t) : f(t) * h.(t) for N)
h(t) —2 u(t) + 4 u(t — 1)
3 e—(t ’ 2) u(t — 2). In order to receive credit:
0 Specify regions of integration by writing the convolution integral with appropriate limits of
integration and specifying the range of t for which the integral is valid. 0 For each convolution integral, substitute for f and h using the expressions given above and
simplify the integrand by eliminating unnecessary unitstep functions. 0 For each region of integration, provide a sketch showing the position of f and h. 0 Do not evaluate the convolution integral for each region of integration. Firs‘l: sketch 46h) amp. 2115):" {‘ce 5“"
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° I 2. (5 points) Another system has the impulse response h(t) and input f (t) h(t) = "(t+2)—u(t—3)
m) = e3(t1)u(t)_e—3(t—1),,(t_4). What is the width (duration) of the zerostate response y(t) = f(t) * h(t) ? To receive credit, justify
your result or show how you applied an appropriate property. b (t) Lmﬂih (guru/é MA % h Lt) .4": 5. may; Cot/mam %
FHA 75 ‘4 (1.5m; the, W\£+; (gufw‘bmv PPOPQV—EJJ :4: a: m = ﬁt) a: hutA,
ﬂan .— uwﬂtkj a“) — (4].;ch % FC¢) 4, owjﬁ % 5/16) (#1th awry‘hm) ’ a (8) IS 7
I 3. (6 points) Another system has the impulse response h(t) and input f(t) Mt) : 3 cos(t) [u (i + g) “ “ (t _ 1)] 2
f(t) = 6(t+g)+6(t—g). Neatly sketch f(t), h(t), and y(t) = f(t) II: h(t). Use the. resulb ‘Flbvk 8'“? '7): {ItTl 8 (t +15) arm) + 80c— 1;)...Hﬂ gob): .Fct)a<\\lt)= : th+1§zD + th’g) Problem 3: (25 points) 1. (9 points) The circuit shown in Figure 2 has an input voltage f(t) and an output voltage y(t). Deter
mine the frequency response function, H (1w) 2 Y / F, and express your answer in standard form "<' bm(Jw)"‘ + l>m_1(1w)"“1 ‘ '  + In (w) + be HOw) = ~ : (Jw)" + (In1(Jw)"_1 + “1(Jw) + “0 "a Figure 2: Passive RC ﬁlter circuit. Skeﬂk 'tko. Circuit in ﬂ'Q— [Adar ﬁnal}: ml 053 ‘9 S'vfce J ‘EﬂmsCoIMUvizmq {=0 fin/L} thﬂ Ctrcmf'
R'ﬂL/(R+pz,> RIHKL = 1»: R1 ‘7. 4" (n+2; ﬂ 3 9’ 'gqu ’5 05:3 voHaaac, ‘jl’ét‘SlOn
V = “2* W £2— E
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/F R; CR.+R,_) + CR.+¢D/du/¢. +R.R.. a d—WCRJRL‘I" (2’; H RHQ
éthﬂsck.+¢p)+ﬂ.ﬂz3¢ + a 2— 10 2. (9 points) Consider a LTI system with input f(t) and output y(t) that has the frequency response function 0 001
J . w
H : —~——.
(3”) 1+ 10.001w
Using phasor analysis techniques, determine the sinusoidal steadystate response of the system for the
input f(t) = 7+ —% cos(1000t+ 15°). From your result, determine whether the system is a lowpass ﬁlter or a highpass ﬁlter. gCt) : 7 ”90) + T—Li: [H9 1000” C05 (locol: 4 15° dell“(710003) wkcrq,
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ferm % tkelbpﬂé 35 m4: ﬂaifél' {1,7}; #0. taulam. 11 3. (7 points) Write code for a MATLAB mﬁle that generates the Bode magnitude and phase plots for the frequency response function
30.001 w
H = —————.
(7”) 1+ 90.001“; over three decades starting at 10 rad /sec using a frequency vector containing ﬁve hundred logarithmi
cally equally spaced points. omega. = ﬂoaspcec i, LI, 500))
P = E 0.005 0‘];
Q : 0.00], I]; £0909. C an r] G?) omea a); 12 \ Problem 4: (25 points) 1. (12 points) Consider two realvalued orthonormal signals ¢1(t) and ¢2(t) that are deﬁned over the
interval [—T,T]. Form two new signals (1130.) and (3540,) deﬁned as ‘75305) a ‘75105) — 5 ‘75205)
‘75403) a ¢1(t) + b (7520), where a, and b are nonzero and realvalued. Determine all values of a and b for which ¢3(t) and ¢4(t)
are orthonormal. We need. ho choose. a uni b Judo tha" <¢3J¢a> 9b
<¢VJ¢~1>=lJ «ma <¢39 4’7) :0. 75‘56 Confirm4'5 (51/295;
the ‘Followly relw‘Etoaslth babe“ a “ﬁg 5" I = «33.4533 — <«¢.—L¢z, “5,. 591.): q lat/p,» z.b<¢/¢Q+b’<¢2/¢z>
I = q 7+b’ (0) I = <4“, a.» = <14 +b¢z a¢.+b¢z> = q "‘(abyiy qu <¢fg+k<¢z¢g 1‘ “1+5; (Jame as (0)) 2. (13 points) Consider the set of three mutually orthonormal signals $105), $203), $305) in Figure 3 that
are deﬁned over the interval [—1, 1]. These three orthnormal signals are combined to produce a signal 905) = C1 1P105) + 62 1P205) + Ca 1P3“), where c1, c2, and C3 are realvalued constants. Determine the value of c1, C2, C3 that minimizes the
energy of the signal g(t) — sin(7rt) over the interval [—1, 1]. w1( t) we w3(t) Figure 3: Three mutually orthonormal signals deﬁned over the interval [—1,1]. “€160 the Sid/sag 5.1% 7S ortk°v90“£J ‘Lo MIﬂIMIu
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 Fall '07
 SCHIANO,JEFFREYLDAS,ARNAB

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