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Unformatted text preview: EE 350 EXAM II 18 October 2007 Last Name (Print): AS 0 [Vi.1 on First Name (Print): ID number (Last 4 digits): Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO ‘Problem Weight Score
1 I 25
2 i 25
l 3 . 25_L
4 25 j
Total L 100 j
Test FormA
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 example7 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) A LTI system with input f(t) and output y(t) has the ODE representation (Pg 513_ .dt—2 + dt — f(t). Represent the system as a signal by determining the impulse response function h(t). If your expression contains
a term with a Dirac delta functional7 simplify that term. Find. the. armada; unI'Eaﬁep mpomJ the” Alex =Jy/JH2. Q0): 7% + ’A = ’AC'AHB :0 =9 7"”; 7‘"' —t
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jco\:0* 0‘2“" (t3 : (.1 +0. 2. (9 points) A LTI system, different from the one in part 1, has the impulse response representation where
[11(1) : 6(t — 3)
mm : e—tua + 2)
o (4 points) Is the system causal?
o (5 points) Is the system BIBO stable?
Justify your answers.  (+7 33 03 on W 3 at 3
 $35 3 ~ _ —e = e. <00
[unlit = J9} 1+. = e. Xe 1+; " e. I
0, a '
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Because. I HH‘Hl‘US A in L
00 L {um£2. con) " [3 H30 5 'L’db‘e. 3. (6 points) Figure 1 shows the block diagram representation of a system with input f(t) and output y(t). The
block diagram contains three LTI subsystems represented by the impulse response functions hula), mm, and
mm. Derive an expression for an overall impulse response function h(t), in terms of h1(t), mm, and h3(t), so that W) I f”) * Mt) Figure 1: Block diagram representation of a system. w H.)
M &: 4%l‘2‘ + (F’kk."‘F):"l‘3 g= Frlhz.'ll.3+h‘*h33
W
th) Problem 2: (25 points) 1. (12 points) The signal
f(t) = etu(3 — t) is applied to a system whose impulse response function is h(t) = e3t [u(t — 1) — u(t — 2)]. Determine the zerostate response Mt) : ﬁt) * Mt) : k (t) «4366) = 3.25.3);Ct "tMQ'C using the graphical convolution approach. Do got sketch y(t). In order to receive credit, clearly specify the
regions of integration and, for each region, provide a sketch of f and h. $553ng l<3+4= <2. or 74'5‘5 7.
7.. 4'. 7.1
‘C a e no zeta t t‘c e 0Q?
H warm? JCtl 3 S 93 Q ~3H:
6 7.4.:
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2. ~3+L
R€n3 3+‘b>2 OY‘ £75“ amtangsa
ﬁg
~3—° Mam—c) '0 ’ »
when t )5 t 7 3
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1 oo
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Sum «Him a —e
m / '5: [a 1
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6mg» ‘‘ 2
12 7 5'
O 2. (13 points) The signal
ﬁt) : u(t _ 1)  W — 3) is applied to a system Whose impulse response function is
Mt) : 4 ['u(t) — u(t — 1)].
Specify the interval [t1 7 #2] over which the zero—state response y(t) : f(t) ac h(t) achieves its maximum value, along with the value of the response y 011 that interval. 409 h H.) We
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4: h
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The mammwn wwluo. a (4:) : he’ll“ ‘cL‘CU (M52; '0“: —'C) «ampleLe? overluf’J l+t + J°°+LQM£ C)£C oCcura when
tAereLf Maxi/ma”; the. drew: LCt‘r) MaMMUm auerlwr) 9“,er 1%,.
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’ —t = ‘l 5 Vft
W‘ l. JM‘Z “0‘ hi: Hui: .— = ‘114: C—I+t)] =7 m alimgm vdvlJQ. Problem 3: (25 points) 1. (10 points) Consider the network in Figure 2 that has input voltage f(t) and output voltage y(t). Figure 2: Passive RC bridge circuit. (a) (6 points) Find an expression for the frequeHCy response function H (3w) and place your answer in the standard form
bm(]w)m + bm_1(]w)m‘1 + ‘ ' ' + b1(]a}) + b0 HOW) : (9a))" + (In—data)“1 + ' ‘ ‘ + “1“”) + 00 U.Smai 0 olive. SJV‘ISIOD I _. R "’
link. AF; R E W F l\ (b) (4 points) Find and simplify expressions for the magnitude, H(]w), and phase angle, 1H (3a)), of the
frequency response function. lac “l“ &—————l’ “WI/m = W :3”:
é‘” ‘ Irv/Act me u 4 ('W “ T“”"( 7/“) I [P9 2. (8 points) Consider a system with the frequency response function Y(Jw) mow) H(]w) : F(]w) : 0.1(Jw) + 1’ where ()1 is a positive, real—valued parameter. In response to the input @500» i: + e  ‘(O°)
f(t) : cos(10t + 25°) + SUM, the sinusoidalsteady state response is ‘ 0 005 ((000 4: .. *1 go) y(t) = Acos(10t + (15) + 10 SinM 45°). Determine the numeric value of the parameters ()1, A, 9 and 45, and state any numeric approximations that
were made. Specify the angles in units of degrees. we) __. ”LOU”! ($05004: + 15%, 3 Ham) + luca_.°.o)\cos(mt + 9—734 4H9coooi) 7‘ n = 5‘ == 10
SJ“. ‘Cor 5. «Min I HCdJOooUl 1' IO, ‘ Ibi+10aol ‘ loco L, 2 Iooobi : l0
)ucamal  : ""00 lo.la_looo+l ’looyHl 6N8!) bl) 'FlnQ. 9" ‘0 ‘ L'EE—
/ ' r2? r7: HzllKouo)’: W = /}+// Fmg. d; %_ ﬁrst «Qe'lrewmmral tagHam} & i‘o’ : low), 42115 =
our __ 0
¢._ 25°+ AH’é/“ﬂ 1'") ¢‘7° LH’CdKQ = I000)
boa—lemme e L? Lat “LIA; LH—Ld moo :VX. l0 =1‘7OO
ziwdmoa 1 #zwj] 5 54;!er /
OJ [000 70°
‘1 so = e ‘_ 700 + “#106003 ﬂ) 3. (7 points) For a certain system it is known that the frequency response function has the form 520w)? + 510w) + 50 HOW) : a2(jw)2 + a1(jw) + do, but the six parameters b0, b1, b2, (10, a1, and (12 are unknown. In order to determine the value of the unknown
parameters, the magnitude H(jw) and phase angle AHQw) of the frequency response function is experimentally
measured for N different frequencies (U1 through am that span the frequency range over which the system
operates. For each frequency w, the experimenter applies the input W) = E co.s<wit + 02)
and observes the sinusoidal steady—state response
y(t) = Y, cos(wit + $2). This data is entered into MATLAB as the vectors w = [wthy‘ LUNl F = [FlaF2a' ,FNl Y = [Y1,Y2,'“,YNl
theta_F = [017 92, ' 101V]
theta_Y = [(1)17 $2, ‘ ’ ' 7 ¢Nl 7 where the angles are in units of degrees. Write a MATLAB m—ﬁle that: o (4 points) Estimates the values of the unknown parameters b0, b1, b2, (10, a1, and a2. 0 (3 points) Using the estimated parameters, calculates the magnitude and phase response of the system
from 0.1 Hz to 100 kHz using one thousand frequencies equally spaced on a logarithmic scale. Store the estimated magnitude, in dB, in the vector est_mag, and store the phase angle, in degrees, in the vector
estphase. ; H: (Yo/F) . :3 ex? (A ‘7' (thatu.7~ the'ku..F’> §‘F~/Ls{0>3
n: 2.3 {73:23 [13ng : mu‘preﬁs (H, w, mJ n); 'l‘ 3 2:355ch0. ("I3 S“J loco); Cat“‘27) gamma]: Leora (go, asp; *4);
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P 70 show a, Q 5.,eACC lent 10 Problem 4: (25 points) Consider approximating the signal f(t) as N) : 01¢1(t)+ 02952”) + 6U), where e(t) is the approximation error. The signals f (t), C(t), Q51“): and @(t) are real~valued and deﬁned on the interval
[t1,t2]. Table 1 speciﬁes several relevant inner products. For example, (451(t), ¢)1(t)) : 2, while (ﬁt), gbg(t)) : 8. Table 1: Table of inner products. 1. (5 points) Are the signals (1)] (t) and ¢2(t) mutually orthogonal, mutually orthonormal, or neither? Justify your
answer in a short sentence. < 4):, Sb; ) :o the. signals gbl (.5) “"90 90L”) Be— C “0.80.
Ora. Met/q“? 0/ th 06 of] “Q. Bﬁttxvﬂ <¢g¢n> #l aan— <¢2ﬂ ¢L7#/J o «I.
the, 5 la nmlS (1), H3) «mQ @196) are, rigE MV‘LUqu (,th oarm . 2. (8 points) Determine the numeric value of the coefﬁcients (:1 and (:2 so that the energy of the approximation
error 6(t) is minimized over the interval [1‘ 1, t2]. Because. the Signal) 515,96) dnaQ (ﬁblt) are, mu‘L‘de; OVt'l‘Odono/Q, the, caeLﬁtcmob C, any, C2, ﬁlm/E,
eLe) ave, the, energy, 0‘F the. array mnm'mﬁée. 11 3. (9 points) Suppose that <¢1(t),¢2(t)) : 2 instead of zero, as speciﬁed in Table 1. Using a standard calculus
technique, determine the numeric value of the coefﬁcients c] an Cg so that the energy ES : /t 2 [m —{C1¢i(t)+02®2(t))} F dt of the approximation error 60) is minimized. 44) (1;? ___....———v . Begausq, <¢U ¢L> #:O‘) W can 00+ 'F'anvL thQ. CL US)? CC : <¢LJ¢L7
Cheese. 0‘ omJL CL 30 {ska}: dEe/ﬂcq: Allie/ﬂlLsog J. J(7—
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j: 4160 qbdilaJe  Q: L4 (Wt) dyi—thQl: C7. 4:.ch It) (19%»? (419(1)) ‘ C, <¢IJ¢')‘ C2,. (dz) ¢\) :0 (1..) a U” 7. ppﬂlél Ma e169 w‘lzwn m flog, twc unlit—ﬂown) 7/ Z ZC — l0 (2.)
<¥/¢’§c.<d> ¢,)—cz_<d>qu7: =2) 201* 7, )
/J 3 7” ‘1 +2_<1,I‘1cz = 9 L3 4. (3 points) Comparing the solution methods in parts 2 and 3, what is the advantage, if any, of using mutually
orthogonal basis signals 9571“)? Explain your reasoning in one or two short sentences. when the Slanuls (Phib) are Iris/bog“? orthoaoﬂurq we, can Jolve, ‘Fur the, CL “Agapenooant? MSW? C; "' <‘FJ (b; >/4¢LJ do). J3 whon the 5‘57?qu ¢nltl are, no'& rug/bulb, orthyon
)‘LLIVIGDUJVIYJ we must b Olﬁ 6" .5 Cit, 7 n 650 aft/on) in CC 5 [mu '7‘.) Wkan the (ﬁn RV? mv't’aﬂa arthoﬂonuﬂj [5 LS chDef to Solve W the CL
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 Fall '07
 SCHIANO,JEFFREYLDAS,ARNAB

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