EE20-2007-10-11-Midterm-02-SOL

EE20-2007-10-11-Midterm-02-SOL - EECS 20N Structure and...

Info icon This preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
Image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EECS 20N: Structure and Interpretation of Signals and Systems MIDTERM 2 Department of Electrical Engineering and Computer Sciences 11 October 2007 UNIVERSITY OF CALIFORNIA BERKELEY LAST Name Age—{L FIRST Name H I ”’1 0 Lab Time ~ [3 l o (10 Points) Print your name and lab time in legible, block lettering above AND on the last page where the grading table appears. 0 This exam should take up to 70 minutes to complete. You will be given at least 70 minutes, up to a maximum of 80 minutes, to work on the exam. 0 This exam is closed book. Collaboration is not permitted. You may not use or access, or cause to be used or accessed, any reference in print or electronic form at any time during the exam, except two double-sided 8.5” X 11” sheets of handwritten notes having no appendage. Computing, communication, and other electronic devices (except dedicated timekeepers) must be turned off. Noncompliance with these or other instructions from the teaching staff—— including, for example, commencing work prematurely or continuing beyond the announced stop time—is a serious violation of the Code of Student Conduct. Scratch paper will be provided to you; ask for more if you run out. You may not use your own scratch paper. 0 The exam printout consists of pages numbered 1 through 10. When you are prompted by the teaching staff to begin work, verify that your copy of the exam is free of printing anomalies and contains all of the ten numbered pages. If you find a defect in your copy, notify the staff immediately. 0 Please write neatly and legibly, because if we can’t read it, we can’t grade it. o For each problem, limit your work to the space provided specifically for that problem. No other work will be considered in grading your exam. No exceptions. 0 Unless explicitly waived by the specific wording of a problem, you must ex- plain your responses (and reasoning) succinctly, but clearly and convincingly. 0 We hope you do a fantastic job on this exam. MT2.1 (40 Points) The block diagram below shows an architecture for implement- ing an amplitude modulator using signal adders and square-law devices (SLDs). The real-valued signals x1 and x2 are the inputs to the amplitude modulator and y is its output. AMPLITUDE MODULATOR Each square-law device is characterized by the following parabolic input-output graph, where xsu) denotes the input to the SLD and ySLD the output. The parame- ter K is a positive constant. ySLD K 2 YSLD(t)=K XSLD (t) 0 XSLD Formulas and Facts of Potential Use or Interest: cos(a + B) 9 = 00804 cosfl — sina sinfi cos (1 cos B = %[cos(a + 3) + cos(a — 3)] (a) Show that if K = i, the amplitude modulator output y is characterized by W) = 5610000205), Vt e R. 3: Klb‘fixaf— @4311 :lKX‘X’Z 2 :5 d : X‘X‘Q‘ K: i- L\— (b) Select the strongest correct assertion from the following choices. Explain your reasoning succinctl , but clearl and convincinl . (I) . The am ulitude modulator above must be memoryless. (II) The amplitude modulator above can be memoryles. (III) The amplitude modulator above cannot be memoryless. Til-t SailEM \sz V\c> mamofi‘ér Qlflmflfik’. T3 SLD ”‘3 QO00 \{gg lDQCMSQ ITS (“pr IS ‘lfls mm mngowgl Medal l3 \l-S owvvk’x;l\olrlers 0w: memoralQSS, A\Sq “CAVQ fiCCY Xt§1] l8 we, («\Swf ofifi SEEM 3(f\:£(xa’l):x(+\)lllf\ (c) Select the strongest correct assertion om the following choices. Explain your reasoning succinctly, but clearly and convincingly. \ (I) The amplitude modulator above must be causal. II The amplitude modulator above can be causal. (III) The amplitude modulator above cannot be causal. Even memorale SS % 63—th \\ 3 waswl . (D0 noTQ,\\0on\/Vcr) Tem'f fig CWNU‘SQ is vxollfxzxt» (d) Select the strongest correct assertion from the following choices. Explain your reasoning succinctly, but clearly and convincingly. \ (I) The amplitude modulator above must be time invariant. (II) The amplitude modulator above can be time invariant. (III) The amplitude modulator above cannot be time invariant. EROfi "\QMota\C$S EAEM (ms out: AC§CAQ Tlptfl as) Q ox‘CSC) '\ '3 Moi \«woxcwwfl’. xlt) X Y9K CR“ «\39 «Pa-kt “2&5 wad 5.5L): .1 ->3€H=Nm ,zlfl LJ im = 5 m ljfll ) as 5m Amgm letflxatt—t.) : alt-t) \\ V, M is m (e) Select the strongest correct assertion from the following choices Explain your X Q“) reasoning succinctly, but clearly and convincingly X M‘- L: "U‘ ——> xx GIL ‘3 :zml 73~ «my XL") (II) The amplitude modulator above can be linear. X :(Jfl'jsafsmgN a M“ “1M5 (9) III The am litude modulator above cannot be linear. EZ:::::\ gum - .. x 1311a,. 31“ (f) Suppose the input signals x1 and 202 are sinusoids characterized by instanta- neous values (I) The amplitude modulator above must be linear. x1(t) = co‘s(w1t) and $205) 2 cos(w2t), Vt, 1 where O < wl < %' For simplicity, assume K = —. (i) Provide a well-labeled sketch of the spectrum of the output signal y. Be sure to explain your reasoning succinctly, but clearly and convincingly. All/B: Cos «Ad'— cns w;T :— %{C03Km1+wfl‘fi+ C03‘(W:~_w‘)f}i X; y‘t yq—t 5. ~45 » Mw w w J~ a2 - W141?) (ii) Prove that the output signal y is periodic if, and only if, the ratio of the frequencies cal and tag is a rational number, 1. e , ?Q(‘\\ 9 A Yd. ?€m9c\1‘ _ E Q W ' T‘Q SKM 9'? 1200 VKNMLQ Cl “Lb-«4&8 \S QV’WQ‘X V9 «Km-i Oflglgr) flaw 9C3£>Q€lflyfi ”gaklcche “tack Fitmo‘ks Mt ”Emmi 1;le “ES 0’8” {AAA 0% Y9; .... m M '74., l) _ g“ MAS—l. how“? if?“ I: )n EN > 0K. U’ \ I 4 {W\ X Wa—w“\ _ [VA-w» flW~QW:Mqun/\J f _ 2“ firm fl,—~B I n a \ ‘1 \ F“ ur-tJ\ 13?“: «x “Elfin A WE’VKN W n—M 6Q "—-'=—\ l““"‘\ Wat” KMMW 1': 73"”? “T“ I MT2.2 (35 Points) The following diagram shows a combination lock. 4 The lock can be opened only if an ordered sequence of three numbers—selected in strict adherence to the following steps—matches the lock’s unique ”combination”: Step I. Turn the dial clockwise two or more whole turns, and stop at the first num- ber of the combination. Step II. Turn the dial counter-clockwise one whole turn past the number in Step 1, and stop at the second number of the combination. Step 111. Turn the dial clockwise and stop at the third number of the combination. The combination for this lock (i.e., the only sequence of numbers that opens it) is (1, 3, 5). This means that to open the lock, a user must stop at 1 at the end of Step I; stop at 3 at the end of Step II;“and stop at 5 at the end of Step III. The thick vertical arrow at the top of the diagram is fixed, and it marks the number selected by the user at the end of every step. The combination lock can be thought of as a mechanical system, where the ”input signal” is the sequence of three numbers x 2 ($1,332,553) 6 {O,1,2,...,7}3 selected by the user who rotates the dial according to the steps and rules described above; needless to say, the sequence of numbers selected by the user may or may not match the lock’s combination (1, 3, 5). The ”output” signal y = (y1,y2,y3) shows the the sequence of states of the lock cor- responding to the input sequence (3:1, 332, $3). The state of the lock upon completion of each step described above is either Locked (L) or Unlocked (U). By way of example, if the input signal is (1, 2, 4), the corresponding output signal is (L, L, L), which means that the user has failed to open the lock. If, on the other hand, the user applies the input signal (1, 3, 5), then the output signal will be (L, L, U), which means that the user has succeeded in opening the lock. We can describe the combination lock by the function CombinationLock: {0,1,2,.. . ,7}3 —> S {(L,L,U) if 1:: (1,3,5) 3/ Z (L, L, L) otherwise. The set S is something you will determine below. In tackling this problem, consider only dial rotations that conform to the rules described above. (a) Determine the size of the input signal space; that is, determine how many valid input signals ($1, 3:2, 2:3) exist. 3 ‘ 3 . K - A \gé)’\)"‘)7§ l A: 8 E4CL\ XL“ \f\a\$ 8 (DOSS\\0(\/tj(ES . ‘ (24,255) Also, determine 5 so that the description of the system is an onto function. What is the size of 5? Tight are gfi\&_Two ()ossiL\Q 3Q \KivchS A; 3&3 1 LLJLIL) ““J‘ LL‘)L)\A>‘6 3—3,th law) (b) Select the strongest correct assertion from the following choices. Explain your reasoning succinctly, but clearly and convincingly. (I) The combination lock must be a memoryless system. (II) The combination lock can be a memoryless system. ‘ (III) The combination lock cannot be a memoryless system. ! [ntadlfixml 1 \SQ knew) fioj'fifi \ché va's‘l— keg? Jfi‘aclg all WC SQTWCWCQ cl" TRQ Co alts ew‘l/thol, 39 ll, mwsj how“: Mew/Leta. Vloft, CDAQPC—l’fl\ ) CéoSlOle Flafi {Bl/lowing l/NQ My“:w NRC/l 031st Si WQQRQQ ealFS‘. A \va’ /\ 1 Q” 3:5,) w? (L)\~)U\\13‘ K31X$ H‘s—#65 X i9 t ‘H, 32: 032)?) “—3? LL)L)\—):3 50 tfiqsgm Ck“ Q me mocfi e33 ~ (c) Select the strongest correct assertion from the following choices. Explain your reasoning succinctly, but clearly and convincingly. l (I) The combination lock must be a causal system. l (11) The combination lock can be a causal system. l\ J— (III) The combinatlon lock cannot be a causal system. i’ A ‘ ‘ ‘ . mamas w wt . mwew ..T.,..i...\.ig lfi‘ilr Y‘QSFchlVE’RFSTinQsX‘ «we X, ) fmt’thCQ Q‘Nfi‘fiw OWE lt‘lflm‘lTCA . . Mi . A W? Ar") °"\°\ "\dwollma} «bf ms‘necl—Iwc hrlidfli El «Ml 3‘ 1 '\ \ alO‘l‘flc‘m :31LL PK? *) y 3\ 3 . a EV“ 0 in . N g“ n .. 03‘ “A A first ox? \olechA m t «Allncl‘vllm, Q a El VGA V F x rlx X ) Ycoclulce 0&1?le amp ’\ “flair wesfie’d—(VQ SQLonoX Qr{\r(‘\‘€,§, * Xa mmA X92 6 «VAX/«Apt lkfiwfii (Lax ugh) «WA lAC\U\l>l:v\ )TRQ;(‘ VQSQQLWQ $6CQQ1€IXV$QS El “AA 1 3 L L W A .; Q. L l J all?» 5.1V 3 . . .. a 3, ~ m x “ No (MW Q'l’ Aisfiacl'1wi3?s Cam lea {olewfiéflx‘rx MAIEWQQ Qw—lflQS E;‘ "5:, 3:31 MT2.3 (30 Points) Consider an ideal interlocking pair of rotating mechanical gears shown in the figure below. By ideal we mean that you can ignore friction and slip- page between the gears. —R2 0 Y“) R2 The respective radii of the two gears are shown in the diagram, and are related according to 0 < R1 < R2. The smaller gear is the driver. The projection x can be thought of as the input signal to this mechanical system. The larger gear responds to the rotation of the smaller gear. Accordingly, we can think of the projection y as the output signal of this mechanical system. The smaller gear is shown rotating clockwise with angular velocity cal radians per second; the larger gear rotates in the opposite direction. At the contact point P where the two gears touch, they must have equal tangential velocities. Therefore, wlRl = (.ngg. An input-output depiction of the mechanical system is shown below. X Fay The instantaneous values 55(t) and y(t) of the input and output signals are shown in the figure and are described by $0?) = R1 cos(w1t) and y(t) = R2 cos(w2t), Vt. In answering the following questions, assume col = 7r / 3 and R1 = R2 / 2 (a) Select the strongest correct assertion from the list below. XU’) _. .——- c a 5 C21“) (i) F must be a time-invariant system. 3“,.) 1 R ,1 co 3 Gill“) (ii) F can be a time-invariant system. \ (iii) F cannot be a time-invariant system. 3 —-——\ xtfl- "(03(L(+'é§) :P:C 043:6: X(+> Lél' Qbfi 1 x (l1 (a) 1’9 5853?“ werefi,fi&fli\ a U2) “(1’43 R2 C95 (13%)): ~€l COSEV\ ‘\ Bud'siace 1‘ :5 \“ 363$“ W\S\\$\Q§§0m >9“:er gmws‘l'loe IJCvCchiJFé \hea‘t no . (b) Select the str gest correct assertion from the st below (i) F must be a memoryless system. (ii) F can be a memoryless system. A’ \s a new; 3 «6‘ (iii) F annot -e a memoryless system. ”WT—X ’==B __‘ M X(0\) : ley "z T; Gol’ TI) QQ’VMQMM less) at) = Ea + 4%; --— 300) (c) Select the strongest correct assertion from the list below. «A3 (i) F must be a causal system. , EV 6'“ "8‘ 0 “Q “L {’fie :6 'l’ 0 aflqw (‘C \Ook‘ \’\ G (ii) F can be a causal system. Wfilcwi/s 565th Q’QJ’ (iii) F cannot be a causal system W& 0 T— $6§\/QW\ ‘ 5 CMWMCKEYWLQA bfifi 9&«r aft CC TEC Mr S\&:: earmmwiiér ‘5‘ \S S QCJ \{JI TEE QNJYFQ 0 kai/ :3r 3 R0 \ch . \Té {hm “QQA nq‘l’ lcék Oslneocell vv 0 lee )“ “i7 l) a l 5? A o Golfer‘s/MAE Q 9W1? “fir G f A if Jfilbvw/DS lfia-j’ a You Come look C :17 l‘Rls a; A5 we.“ we owfi 7:10 inf ME (ital/(CA “~00 l/O 0:?wpfa::\?j:§ja ~lr/A/(e N”? cos 9J1“ AAA Cos («Kl/3- fit FMJWCE ”(an/IL ;% LAST Name AL FIRST Name M ’ ”I 0 Lab Time ?! 10 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern