EE20-02-Midterm-01-2010-09-21-SOL

EE20-02-Midterm-01-2010-09-21-SOL - EECS 20N Structure and...

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Unformatted text preview: EECS 20N: Structure and Interpretation of Signals and Systems MIDTERM 1 Department of Electrical Engineering and Computer Sciences 21 September 2010 UNIVERSITY OF CALIFORNIA BERKELEY LAST Name <46 L FIRST Name D Lab Time 7 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 one double—sided 8.5” x 11” sheet 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 8. 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 eight 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. MT1.1 (45 Points) The instantaneous position of a particle on the complex plane is described as follows: Vt E t, 2(t) = sin(t) e”. (a) (10 Points) Determine reasonably simple expressions for the instantaneous Cartesian (rectangular) coordinates of the particle; that is determine x05) and [1 K A 1 y(t), where 2(t) = x0?) + iy(t). X (fl 34+) 2 a '. ZCl’W: Sinkk‘ (cos T-H- Slnl\ :— SEAT COS T +1. Sin: 1:— :|s;w\(:tl +lil—<°;Q\+)) WW4”; 41“ [t “It Mom {who —\ ‘ —c l’ _\ -r ' ”44/7/91 L& T C _ <0")an it C 1 fr" 4:“ 1 L g L l . "fclY‘QMaue. Tfile L from fie JCVIOMM'JZ‘C' (if) A. - fl 1 ' - ' Wary—(m smbd’) -l—C°s u m H arm 2“,». —Lcos(9~fl-\-(-Ll'-5l'\. _ "TL / ‘ “(4) ‘ .2 2. KS «Lame. (b) (15 Points) Provide a well—labeled plot of the trajectory of the particle on the complex plane. That is, indicate the path and the direction of the particle’s motion. To receive credit, you must provide a succinct, but clear and con— vincing explanation of your work. Please note that we are not asking you to plotx(t) and (t). Lill— l ’j ['21— ' NC. kna") W 7&4” ZU’)—_€_’:_= if ’—>———+ - m i— “/1 _L_ I ' 11-h; — _ -+ ,1 Ll’ Wflg}: ed; / lJeml'c-Els fiefi $06k) l‘ns avx-J'ameqwg laggfiin oft/omr—flc. C in {e E7; Cam’oltx FLULQ I ll" NICOLAS final, #6 fq/‘fic e Fade/Wes «CI/"j- | _ _ ‘E‘ZK —¢ fall}; Z1 «TA/fie, ml? of'1“A'“‘%&L“flléf-clac¥m¢/Still/1a-lL/‘OWI 70k) :i’e : 7%: LH’): Wfiri;z S" 7761 7776732? firzm _&W. '-l Sl.M/’[Z’#C MC 'Far LFM Shlfl’cal ml? 41”“ Shh-1A ? E. MW) («161 G’I’C’ler £1th ('n r‘E ): [TX Set. fM'T (q (C) 5113:3111:1l:)itsshfi:1dt::f thte lparticlg’ 5 position exhibits periodic behavior, and en a perio p and fundamental frequency L00 WC. slxawtol l‘R-Jl/ Zfi’1:i_l_ejaf-$_, . fTs kamial be 6 \CGJ‘ TfiJl’Z l5 [De/Wall“; lOECA-W—sfi, gill} :5 A. rklkgar‘ “11,12 conn'rerclch-«JISE Inn. cIrc\e Tame. as ‘hus'l’c as fie. Flimstn" ed, 5 we, CXfec-l’ 773E 7£mvlmerjrl lam-Tmemca To lot, 1'4”“ 9“ s‘cfin'l/‘Wtsrnmlma To ferIa'l' 3L f‘aI’L "T ""45 ‘1:le S qu 1f“? MM‘C 'FWM‘I-ll HQ, ”My: 7C“"l " a 75ML‘1gJ’za—‘lf1:z(l) 1L2(‘HFZ/CL::V//L2F ‘ ‘RF’l—emr-k—fi “ml/C 5“ The” ' 1’ i-lflI/C L '5 l ’7 P‘Tr r7il/ fk % 17:1 0/”! {MAKES F05 l (d) (‘1t’0 Points) Provide well— labeled plots of |2(t) )land 42(15 Z6”: sl‘AlT1tl ”>3 \Z.(fl:\ ls';( (H1 AS +0f fie) Pkfise “’sz (+1} . k me Im zzfl 11% it ' "m” 1141“) m T3 >0 O Slnl'kW I1 “\Smlfllc l’f 5M( fl< lsmd’l lc ('(f'f'm S'U’IA NOEL TA. %6 \m‘l’eNJ 'H<+<T s7nt>0 far 0<t<rml 51n+<o {’19P —T<r<0 4:1 3' 1 “ lZL‘l’)]=\S1nL1’)\ 5° ¥zmz t’ 0<t<fi’ .:rr I0 ‘2‘ Ir 7- { iii—IT 'Tr< 7" < O W m Lfifi’fi/ 62:31:44”, ‘T o f ATF 7— An. crud 1-3 l’g 1101/1sz 9“ 71$ ZH’lWI Zlfl 1041's Alwyn” Canal,“ W1 ff— "WASH-1’02.“ MT1.2 (30 Points) Consider two complex numbers I and y. (a) (7 Points) Prove that (I + y)* = x" + y*; that is, the complex conjugate of the sum of two numbers is the sum of their individual complex conjugates. L0” “"U’xr ”‘4 WW3: ”:5 “#Umwiéxfiéa =5 * . _ _ (Mal :<XR*3K3‘ L (xrlétrl Zea“ "l’> "llrcfl 3:) => XY 3* we Cam Casi/J. CX7E’11 7%.; frgwhl’ 7; 0L I‘M“ 07C Mof‘C 44 750 Cam/){ex number‘s, Er Cum/3’6, RSI-v1) 7306 fame mcflga/ 45 ”£6006 quUC X 34' NC Cam Shad 773+ %Z1¢> : §zk (b) (7 Points) Prove that (xy)* = x*y* ; that is, the complex conjugate of the prod— uct of two numbers is the product of their individual complex conjugates. For TfI-S [Dot/‘7" we [”7" exfr‘ffl X amig‘l [fife/Ar firm/U45 A PAlC 01L newmlo} ma/tfklxig 7:) Cam/FAX nvw‘ c/‘f IS Mqrf €06]? ’lnnc UL ficfli'f‘fi DY: r‘csscol {A polar 73pm. Dfaflfcwlaf‘, ’67:... __ [0? X :Rxel'ex/ wLere szijivl 63¢ 22$.» Mule/£7" 0.1- '66 ’9- .Jher‘e Rail“ «Ml 6‘ 3 X3” [AC/1 XJf-ZKXC x Rafi 556x801 CL( Xl 63 ’LF 93‘ 0‘ «[9 "6‘ I fixer-413797 (xajx: fixfa C < T 3‘ 6(Rxe jOQJ'e d’) _ So at 0C SllOW'N lid/w V? T 67’ JL iorc 776m 7:}: nmLE/‘f We 5“" ail—6AA TAPS 7—0 °’\— film/"4 “W W mm; (C) (16 Points) Consider a cubic polynomial 3 Q(Z) =2:0Lk.2'k=0L0+0L12+0L222+0L3237 k=0 where every coefficient ah. is real. (i) (8 Points) Prove that if 2 is a complex root of the polynomial (i.e., Q(2) = 0, and 2 g: t), then so is 2*. zls K mt .{v am :3 Q(L\: iquzkzo vs 3 =0 @71\:(k:mkzij—_ O . LL37 hf: I‘CSvLH—off‘rT (“l/”Q '5 no“) =0 * k __ (,va I 1% C ”baf— k M a(zy=7;(qkz)x_os,/U W cm \— f” “’3‘- dm: 5W: :«kwlfiaéz’ro a ‘ [Ct—O T. We leNC slmwn TR Akrugk lo/c 4k€lK (ii) (8 Points) Prove that Q(2) has at east one real root. &(z\ lms Thee, Mail’s) so we Cam er—l’e 61m sag Lz—z.\ (z; a (HA, Mere am —.é2(zz\=&(zs)=oa 3+ zl g all/as a ”mm/4%“ g e 03K 55 who 4 ”5.7L: C~U Wm ran—l’ 1;)1’6‘4" "5 21,771.? £[email protected]\:0\3(z—Z\3(z —z‘*)(z~Z-5l- IDCP force, <3 64/: beC‘MSE ‘7: 236(C‘HC,7%c/1 Z3 MKS-Cf,“ 6“: - (04774 [6 IM- Foal/j Lgf 7f: affier 7:“, ’2197'} «pg film 7: 1001 z, 4412,) _/h/erc7%ne Z 618- “ . . generally jug. 9’l’l'°'l'lcrfa/i~’l°m'“/h”mj— “"V/ 0573;0ij I'MS 47L, Inns-l. MC le’I/Iv/WC Wadi MT1.3 (30 Points) For A > 0, let M A t) be defined by ift<—A if—Agtg—% - A A 1f—3<t<3 ifggtgA ifA<t. HODIHO A 0 (a) (7 Points) Provide a well—labeled plot of UA. 1 I A E “am -A 1/20 5/; A + 3“. 3y \ (b) (7 Points) Letsz be a function that is continuous at 0. Determine l - ;_ 00 ‘ lim/ f(tv) td)t Fr A smile/1a ajufflfi'rdm 1’1; we «VIOLWCWJT Q. "Al TO B, “pg“, S‘M‘A“ '4 L ‘FIaMET Lclw: ( For A smxli 9.“th "We [AU M’o’f Lack boxejlsb ‘— 65¢qu \0; "MC fifff‘oXln/Ml AAA 0 b‘mn—h “TTKC N A A $63 {(5:99} 1% (—3,) WM 3A V "éf‘ M32? T‘f‘ifrfim :ns-AffflfTJf imam rs TUAA w ”W M m fiq‘ 7!:5 (04+MM5 9:} 0 :53 Z-mggzygffli yyfimwfl” )=1]L[03 if?” A-ao =5 Z»! W QM: 75/0) AaO-da (C) (6 Points) From the preceding part, how should we represent $11111) UA(t) from the point of View of how it behw inside integrals? We've Seth firs/.1— k 7%”)V5151'W:£[f [mofi&]#— fpyé M ‘26”)— $60) A—ea ”M“ w M 1351;111:er fin» T16 QTVUJ Ii Slim/J1" 0'1— LE meef‘r-foovl 7‘; L9 [OM-1369., rn-MCI’ 7L5 «£1 my sense 1:11» [Ln +)] ,11; 1:1 {+381 {W 1/147’ 15,1)» «@111 “1/ 2(1) “We” Tfic 5.»: we? wmm fame ,4 ”$111111 1’es1’ 5-50 9(d) (10 Points) Let f and g be two functions, each of which? 1 continuous at t: 0 W44 [11 y‘ 'f (1 ft: 1. D t an a e ermine LCTM km ’H‘h 341:) 111946, +00 an /_ W) + 6(t— 1)]f( 19W, .‘ A 1’: m1. where 6 denotes thew Dirac delta function. 1131113 31111111111111 1112111111141111 - WWW)“ -°° [Ed-21’) 1W —1 1111 )0") J1“: 1:15131‘1'1 *1} (“411111151— M 111(1) 2:) 1 £1111+81+v1\11(+1&111 —. HEM 11’1‘111'1 7 LAST Name flack D 7 FIRST Name Lab Time \ Problem Points Your Score Name 10 [O 1 45 Ll’ S— 2 30 3 O 3 30 3 0 Total 115 1 l g’ You may or may not find the following information useful: cos(0z W B) = sin(a W B) = cosa + cosfl sin oz + sinfl 6(at) cosoz-Cosfl —sin0z-sinfl. sinoz-cosflJrcosoz-sinfl. 2cos< 2sin< oz+fl> -COS< 2 oz+fl> -COS< 2 04—5 2)- 04—5 2). ...
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