examIII_s00 - EE 350 EXAM III 10 April 2000 Name Soiu'honS...

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Unformatted text preview: EE 350 EXAM III 10 April 2000 Name: Soiu'honS ID#: Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO - - - - - Test Form A Important Guidelines 0 Answer each problem on the exam itself. 0 The quality of your written solutions is as important as your answers. Your reasoning must be precise and clear. Your grade will reflect the clarity of your presentation. Problem 1: (25 Points) For a. given signal at {¢1(t), ¢2(t), ¢3(t)} defined over the interval [0, T] <¢h¢j>=0 5%}. <¢21¢2>22 < ¢lr¢1 >21 < ¢3fifi3 >: 3. A certain signal f(t) can be exactly represented in terms of the given signal set as f“) = 3915105) + 4452“) + 3953(1) over the interval [0,T]. For t < 0 and t > T, f(t) is zero. 1. (5 points) Is the set signal set orthogonal ? Is the signal set orthonormal ? Justify your answer in one or two sentences. . TjQ Slgnag. set is orthogonag bacmse. =0 'Er L¢0L «All Siam-Q- 92:1; :5 not magma mom «was; 7L; «ma 953,43) i=1. 2. (10 points) Determine the value of for z' = 1,2, and 3. 3. (10 points) Find the energy of the signal f(t) by either directly evaluating < f, f > or using Parseval’s theorem. b. Because. the. enema Ea 3.3 the. error- mam/Q- u Ezra) 3 <49, {0 = c7; Mm.» ll <T<¢.,¢.> + C721<¢,,¢;> + <34¢:,¢3> = (if-l + (W‘- 2. + (3)23 3 ' «39) = 63 Problem 2: (25 points) A function g(t) is described as 9(1) = cos (3m + + cos (47d) cos (2m + . 1. (5 points) Determine the period To and fundamental frequency we of the signal g(t). ZII' 50k) = COb(31T't-r ~§-) + é" COS(ZIT+ '3‘?) + .212— (93(611-£ 1- _ .L a“: i Q = Co5(3rr-i:+ 31rTo+ 7—35) + Jacqui + 2W» ’-.,"-')+ zcdsén-b-P 611% + 15.) For act)=g{-h+ Tax) we, nepl 31TT. : ZFriJ 2W7}: 211,1) 5?; Tc: 27.76 (“here "a m.) Ong- P 6W1. minder? to FWU“ S’WV'wflebfili. eta er) => cur = :2- :5_ The 36%. pg, loaders Sw-Elsiyud the 0‘64: 01540"? "L n 32”) JP 11 M q i :1. This ewes To = Z anL We 2. (10 points) Determine the exponential Fourier series coeficients D” of the signal g(t). 3G5 :L J: cos (ZV‘t ‘35-) + w(37rt+ + i- C95(67F'é '4' = C; Co5(2uo.t +62) 4- C5Ce5(3ua.t'-|- 99 + CfCo5(6ub-t +96) arc- Thq- C 'E hiaonomefiric. chhQr SQr-HZS Coegicm'hr Dmpsur Co ‘-"- 0 Cal-t) has zero war-ave, Janine) = =3:JT- (=J‘)96$"‘ CL: 1;)913'L;) C3 .383 '31 c’ L v a 2- Co =- O Th9. erponoa'hfi FaurlQr Sena-S cafl‘c‘oafi '2’ alga" %' bo no: 2%. am MO. 13." = 93— e15" . filer/3 _ Lefi-T/‘f 0°: 0 D3 rite / I)“: 'J‘r/‘i .- ' 2.11 5 e- bz ‘ #e#W/li D_3 :2 d 3 b-é y. b 3. (10 points) A function f(t) has the trigonometric Fourier series representation f(t) : 2 + :2 3 sin sin (4m). 11211-2 2 The signal f (t) is applied to the system shown in Figure 1, where the frequency response of the system is specified by w H(w) : «28—Jfi for 11 5 |w| S 17 , 0 otherwise Determine the steady-state response of y(t). f(t) y(t) Figure 1: A system block diagram with input f(t) and output y(t). The. "Jpn-E S'lgnafl. 1M Jeff” 2 3—- .S:'0(L5r1')(o$(‘l of: ‘1.) n: S 2 ‘4 l nth-L .. DC. con-(amend: Eunflwmtfl £m6.eag_ We -’ q "24: The {Har- w.” pug pa.” compgnenfié # fit) 3 we :. [Zeal '1‘”, =- lb. Us"? JuPQrf95IIISIof'I) -55 ~ .1- Sm(£§ Q L 3 '5'" _:n_" a F :' 2.31 snug: fit sm(——-",_)Cos(qn-b L) n n £32.- ,1; 'g" Mob“. 'j l2. Si. 3%: 7_ Y = New: 9 a = 91/2... ism“ A Hfafl'No) E I. an”) :. .23: sr‘nP-E') “shun-h - 11%,‘3—15) I ’3 only. 3T°l “Mg Laymonl'C; '4 F 15) p155 throfi/A the. -F.H£r 0 stem (to —n s) + ems) 30:) = 5135*) "' as“) am = — .15- Comzt -1- n5) Problem 3: (25 points) 1. (9 points) A LTI system is described by the impulse response function (I./ h(t) = 2 er” u(t h 4). Find the frequency response function of the LTI system by direct integration of the Fourier transform integral. No credit will be given for other methods. I “M = i w“ (*3 e”“"£+ = arc-3+ mm H m «f w+3){7 t... -gw-flbt :: 2] e d = - i Q q d‘ ‘1 O Phase in Degrees 50 -100 —1 The magnitude of H(m). The phase 01H(co)r —10 —5 0 5 ID 15 min radisec Figure 2: The frequency response function, H (w), of a LTI system. 2. (4 points) Char acterize the system described by the frequency response function shown in Figure 2 as being either a lowpass, highpass, bandpass or bandstop filter. b Bee-wan. any Fur-cage"? wmr9n0n£ cure, paJSfig the add tens Ts N jakpMS 79 Hush 3. (4 points) A signal f (t) is applied to the system shown in Figure 3, where H(w) is specified in Figure 2. Will the characteristics of the filter allow an arbitrary signal f(t) to pass through the system with out distortion ? Explain your answer. Figure 3: System block diagram with input f(t) and output y(t). ' 81wa £142. pinup. anfl. m w.” 4.5:;ch we), 75$ 15 re‘sfonse- ARC”) ‘3 m4: £11.54»? 7» ‘4’) weapon” was); Is rad-é Constan'éJ the. {Hei- admin/Q. ? in) g: on“? (-e ~70) J 4. (8 Points) If f(t) is a real-valued function, show that F(w) = F‘(—w). Hint: It may be helpful to sisal-t with the expression for f(t) given by the inverse Fourier transform. .. .1. ‘° J-"t C FUD _ M {wage 09M; I) Became, Ts an. Lunc‘EIOr-I) «FC-t):-F*'f-b) when “° 4: °° wt 1“ —w i W =+"'(a = (ll—T, Lam? fit) = gag-:fiwm i Problem 4: (25 points) The demodulation system shown in Figure 4 uses a lowpass filter with the frequency response function b 10] OSwao H(w)= —103 —w,, 5 w < 0 0 [ml >wo cos(wst) 6 Km 11(0)) Y0) sin((no t) Figure 4: Demodulation system. 1. (6 points) Find an expression for the Fourier transform X (m) of 3(t), and express your result as the sum of impulse functions (your expression should not contain the convolution operator). X09 = costS-E) San (wot) ,._/ Sri‘QSCwsfi Si‘c‘votfl = '5'”; UL {G’swfij a»: sm 004:} X ('5 Ex H II ‘ ffgcw-ngq- SCI» w5)]*d21r [—Stw+wo)—-8(w-~Uofl 3’ 7L1- Scw— «15):: Siwr-wa + S’CW‘PWQ‘: A'LWWaD L I: .- 3 Cw -ws\\l: ['(W'Wo) " 5(WW5\ té'tuI-wa1 ":5 [8(w4wa_wg+ soul-we fws) - 8(u—kb-w3) —§-(lu—w¢ «us-)1 b 10 2. (4 points) Sketch X (m) using the graph provided in Figure 5. Clearly label the location and weights (areas) of each impulse function. X ((0) Figure 5: A blank graph for sketching X 3. (9 points) Sketch the Fourier transform Y0») of y(t) in Figure 6. Clearly label the location and Weights (areas) of each impulse function. Y(w) Figure 6: A blank graph for sketching Y0»). 11 b b 4. (6 points) Find y(t) by direct integration of the inverse Fourier transform integral (no credit will be given if you use a. known transform pair) From 95 mm. C You}: 511 S—(w+w°-w_s) + $11 9(W‘Wa1'u’5) a) ad.- 0109: mm? w 9° m on .U-l: W15 ‘= S—Cw +wb'ug)ev‘ Jay + % g &'(“V’Wo "Wfi Q'OL 0L '4? 0O S. acwo *WJ)‘L" "7: Q + 52:- - .- ‘- (“0 ‘WJ) ‘5 ei yC-E) = .5' cos (mo-ow )1: l2 ...
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