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Fall08Exam3 - ECE 210/211 Analog Signal Processing Fall...

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Unformatted text preview: ECE 210/211 Analog Signal Processing Fall 2008 University of Illinois Trick, Basar, Franke Exam 3 Thursday, November 20, 2008 — 7:00-8:15 PM Section: circle one Class: ECE 210 ECE 21 l circle one Please clearly PRINT your name l'N CAPITAL LETTERS and circle your section in the boxes above. This is a closed book and closed notes exam. Calculators are not allowed. Please show all y0ur work. Backs of pages may be used for scratch work if necessary. All answers should include units wherever appropriate. Note that the problems are not weighted equally and 50, budget your time accordingly. Problem I (25 points) Problem 2 (25 points) Problem 3 (25 points) Problem 4 (25 points) Total (100 points) Problem 1 (25 points) a) The signal y(t) = fit) - g(t). If {(t) has a bandwidth W. rad/s and g(t) has a bandwidth W2 rad/s, what is the bandwidth of y(t)? Fa») Me» = Y0” L’Zfllf AW .44 V: w 4‘11. Id; -—@,+uf,) Wfi-WL BW= W, {-11}; b) Find the impulse response of a LTI system whose step response g(t) = (l — 4t e'z') u(t). Simplify your result. Mil : 6%: : (—z/ 13‘”th $664+) ufle) + (I—stée’z‘é) 86H . . ' “7,; 7,“,1‘,» :5: rhmm We 4% 41‘ h(t)= — e +546 )4-569 c) A low pass filter is given with its impulse response function h(t) = 6“ u(t). What is the —3 dB bandwidth % = % ofthis low pass filter? (cm-mp6; 741 H80) = 33;?» 14“ w l I ___I_,.. = .J. ’- film) a (21+ “my = “vii '3) 21W}2 1‘5) B3 dB = L V‘dg’a OIL-it; #2? d) For the system descriptions below, oircle‘ivhether the systems are linear or nonlinear, time invariant or time varying and causal, or noncausal. ' y(t)? 3f[(t-—l)2] @0: NL TI or® C [email protected] ”men 1 k! yl(t)filatfi+) spells-01 +4; ak HY} jafiQ 7W; {GP “9:3 figs—:31} ‘ 4; e = ate—ta) «v M3: 3 s [cell—ta] Mike) @ .M“ sec): sud-WM” Problem 2 (25 points) a) The output of at UT] system 1 H(o>) W) = 110) * im- l l I co The frequency response of the system HQ») _106 0 106 and the frequency spectrum of the input signal F (m) are shown. Below sketch Y(u)), F a) “my” D: Clearly label the values on the axes. a) 407 0 lo7 1 Y“°’ </—' flu») = He») qua) fl - 7 (e co RN): Rafi/#029) pamwwova/s — (a )0 M WTo/MLZZ'S 7 . 7%— ¥ft): _/_<_> wave/0 f0 77 7 t "/7? W ia') = 7% 7 [— 7r ' .‘ 0f=7T 3> =._ 1 lug/MW 1' ,07 :1: T!" “ff-7"?!— 107 )0? YGO) 3 {‘eci'C/O w) f“ w; mmys Z FmTaMl—TLS’ 6 6 3/0) c 0 'rr _¥@—); % W / {— e 3, /D m 1: t M N) /r /D“’ 10 Wt 3%ng _ 77' 4 ww£4fiQWWAWL3aWGj /o Wweamw Problem 2 (continued) b) An amplitude modulated sinusoid s(t) = (f(t) + a) cos lOl is the input to the following receiver. multiplier ( If f(t) = cos 2t) what is the output y(t)? (a is a constant.) 1 Hint: cosa - cosb= 5 cos (a+b)+ é- cos (a—b). SQ) : (£61175 +05 (£96,101: : "Z(co¢ zlzLJrCo/LH) 1— qua/oz 3®= [flux/2% mm) + «mlflilmu : 7:,» (m 13% 1%me + tea/4% + £01219) +37: (col/éfi- (314/75) LMWWMmJg Wlwlf {MAJ/u)— ya): jmxfl Emu/i Problem 3 (25 points) multiplier —> Y0) Lct f(t)= sinCZG )H FE”): 3722.03 (w) 9e [1r 8(a) 3) )+7r3‘(w+3)] g(t)= 2 sinct- cos 3t G‘Q")- « 177' Ma‘é lml<2 f; 71’ .é -3 —é( .3) = re: 69....) Mc 69;, “1“”) {0, |co|>2 [ ( 7— + z 1 a) Sketch the Fourier transform of the signal at point (a). [email protected])+ 60») 117 .1,— 77’ m -5—4-3—2~I°l25t{5' b) Sketch the Fourier transform of the signal at point (b). a. \ 0 l c) For T = —,determine and sketch the Fourier transform ofy(t) for —5 < to < 5. 5M Li¥=i‘ 3 #) yea): i Féo)— was) azm M 505 ~27T£3 «Nd/5 YE") 17f ~5‘4—3-z“°' 2 3 75’ l<t<3 2, ~2<t<—l , _ are glven m the figure below. 3 — I, 1 Suppose f (t) = and h(t) = e 0, else 0, f(t) h(t) 2 t 1 3 —2 -1 I And y(t) = f(t) * h(t). Find y(O), y(l) and y(2). Mt) Mr) LL]; LL -7. —-l o 1' ‘7- “ O Q ? ‘ :e-tmw g‘ijwfl -3 -z —t o ’t -z -‘ O T q mam-‘3 P, WWW-’6) Si, iii; -2 ‘4 o -2 -n o I: 3635: XXHEthxz :3 gms-‘iKH‘Z' I Y0) = 0 ...
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