exam1 - Exam 1 EE 323—04 j gh Name S E l WHO Vl closed...

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Unformatted text preview: Exam 1 EE 323—04 j gh 4/26/06 Name . S E) l WHO Vl closed book and notes; no calculator; (50 min.) 100 points I have neither given nor received unpermitted assistance during this test. (signature) 1. The following question addresses the characterization of a system; circle the correct term. (4)a) y(t+1) = 4x(t+2) — 2 linear/.-time varying causal. static- (4)b) Y’C‘ + y(t) =_ X’(t-1%. . . _ ' ,, . 4 @non—hnear th time varying .non-causal static. ‘ (10) 2. Given that the impulse response of the LTI system is h(t) = 2exp(-t)u(t) and the input to the system is x(t) = 3u(t), then determine the response y(t). Note that u(t) is the step function. do I '2 A t6 k366i £0 34 (-t Rhos) ax : awn no _ ‘3 ‘95 l _\ = joosulexme mm A/\ We” um t» A _ A t ”—E: 2 30436 c\/\ ' 6(“6 30 O "E r,\ ': (a (‘e—JC “h'e—O» mam: Mx-gfiwa 0r wag: jao 7mg \K\(":’>‘)0\/\ 2 5 EULNZQ, {AH—>3 5% "fi ’ _w l i “E L—t—M ‘ ~Ht t ) , ' A be ;\ — 6 g Q cl» $06 A 34731410 +4 3. A linear time invariant system is described by the ordinary linear differential equation with constant coefficients y'(t) + 4y(t) = 2x(t) with initial condition y(0) = 3 with input signal given by x(t) = u(t-1), where u(t) is a step function. Determine the following. You must show all steps and work in your solution to receive full-credit. @ (6)a) impulse response for the LTI system, h(t) 1 AM my. 4Lot+3= o «dd/K \no(m= 1. 4% -4Jc S+4=O => 3: ’l hokwl = K9. npf‘f’lto 4(0)) _ ‘ _ _4t ,. 14%) 1 Ke,4t (Mal law) : 1 2 Me‘ -- «:5 (Lil ., MOW)» a Lil-6) l ‘ o - t Mm : 2km 2 2 eJ’th/‘y cum: 5m = f Mom -03 (6)b) step response for the LTI system, s(t) @2 C . S- M : K fi—A‘t ° k ~ \ ~ sue a +wl SQLkiirA :3 4A=1 :> A": ‘ hum - \ :“ mi:- KW“: +21 ’ sow): Ke‘ «4/4 «=3 K— 4» so L+\ : (i 2117?“) uté’) Asa/i * 3m : 239R) : (ii \ iie-A‘t) um cued“ \‘ua : :9: (6)c) zero input response, y21(t) @ I cvéi‘rw+ 4 ‘13ng : 0 work 1360323 (is .. 415 => Laiq (H 2 K63 _ -4L0) 3 m: 3 mews” KG 3 _ _ 4t watt) 2 3e we (6)d) zero state response, yzs(t) (a n @7— ‘ d"? v 7:) \éng-’\ : SL‘E-\\ 4rHe~\\ — i ML~E~\ - AAZ ‘ 2, e‘ ) 3 (6)e) complete response, y(t) +3 «am : léagC—Q + was; ‘ElL'H: 3i; Adz-putt) + (42:- liesért’cflh) [ABE-l) (20)4. Determine the output y(t) of a LTI system with the impulse response h(t) and the input signal x(t) given in the figures below Show all steps in your solution, and provide a sketch of the your solution for y(t). 131.9: 5: um) 34’:- —\3ci,\ LQ pmr‘wu—Q Exam {0 \ 2} +§Lo l}: {012)3} ”KL-6 VAL-e) Lt} 4:<o or t>3 => 13(4320 it we: oétél \é-ELZ 24£L3 MOW lAC‘M lat» ‘ 1 7. . L w \ \W \ i ‘ 2 t ‘—1—-—*——i—'~’ ' by mac) , )\ l 2 II 1-, e 1 fits E t" J: x km XIV-xx M\\XL+«,\3 “E ’\ £4 \ t2 12:36 \t-n. - ”1 132 «am 424th Mcm- 2t Lava—[x—L- “ —t +t’h—1)\ " 3 = 3—71: @‘t Met): 2 but) -- Utt- 43- utt— A an} x kit—>603 A\ I J \ U: méx= 533M“ “(\‘fi—VW’WK‘QL‘ 3 (30¢ ‘ I: I“ \ a, +fHLX"\u(f—>—g a‘f)‘ Jaw “L‘EAVV “ll awe—MW SVLR-13Mt'>~\°*w w = ‘LjWDAX r F - _ 30 ~60 JO ( I 'L)M(t‘>\*\)él\ I A” .,_ 3‘ umfiui—é-Mchr UL) +4 4:. t-\ t t“ -a) ,6; FE \ (ix-2S A» fit» flow— {at aria, : 2 30 o 1 \ '2 Z ‘1)Ld :18 + L£-3> {AL-e- a x u: \wtt'x) + e—zsut+:?:>/;_g: ________ / : theta—3 — 2(£~\ Mi“) - r \J/ :a Me) = u WA + sit—wet) + cat-WW” 5. Shown below is the one-sided spectrum for the first few terms of a Fourier series of a periodic function xp(t), i.e., the Ck and Ok for up to k=2. Showing all steps in your solution for the terms shown, determine: C (4)21) fiindamental frequency f0 of xp(t) 9L 4 So = (web 320,40} = 20 2 (6)13) polar form of the F ouriggr series of xp(t) 1 Equal 2 Co 4- 2. C9,. ¢R(ZCY’OL+°"E Are-t) ‘0 20 3° 4'0 H95) I}.=\ . 60L rate = 2 + 4m3(2w(208i~3 i -' “ 4. 4. ca§( zthBQOIt-s- (—m) ID 10 30 ;(HA3 ~_ 2 + 4 605(40Wé) 4- cos{ eon-7541) 4: DL“ 0):,2 (6)c) exponential form of the Fourier series of xp(t) do we : 7 we cam/“L“ we we : EgZ/QIL w XMD = X911? “—4 :- ~ - _ c ”- “We . ~7i‘_/;e& )2: 0 1‘2 2:) fl 3 ‘ 3F _ Alfitaflloyt -30 _ )2\sba)f ~. . 349M : ‘59. Q 4— 2 Q 6 ”(flag)“ JO - sax-Vac)“: J. é’JWe J ‘. + Z + 2e 6 + 2 -AW 39ml: \ m - Wart— - 340% e 340w: ie 6, YpC'fi’Sttsz e 5-26 +Z‘l'2 .1 (6)d) two-sided spectrum, i.e., draw figures for the magnitude and phase of X[k] l “(0%,“ 7' 2 2 i J l a 5‘— (Vs ‘3 wlko—Ba-Zo-m m 7,0 30 40 4 5 (6)8) trigonometric form of the Fourier series of xp(t). «luv wk” 4am“ ut’vk tome-TA =~ rose? :3, Q0: 2 , (1.24 3 Q2: ‘\ w Md b); 2 o Viz yptex : CLO -\— 2(QflLwSN/acgbt—l-lofilsgwa9fi) 17.3»: y—etfi ~ 2 + 4m5<2n~cixoofld -\—‘ (M 603 (mmmw 2 4. 4 405 4on-b — Cos 950W (4)0 power in the spectrum, Pas ”I. p : c; + ‘3: 2% 2 r7 22+;Z.(42+\?) 2 4+5:- \Z.S y...
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