Exam3SolnSP2011 - S‘O\UL'l‘\rovx {Qfi/ EE301 Signals...

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Unformatted text preview: S‘O\UL'l‘\rovx {Qfi/ EE301 Signals and Systems III-Class Exam Exam 3 Thursday, Apr. 21, 2011 Cover Sheet Test Duration: 75 minutes. Coverage: Chaps. 5,7 Open Book but Closed Notes. One 8.5 in. x 11 in. crib sheet Calculators NOT allowed. All work should be done on the sheets provided. You must show all work for each problem to receive full credit. Plot your answers on the graphs provided. VIP Note Regarding DTFT Plots: The abscissa in each plot is the frequency axis. Ignore the numbers at the bottom. For each plot, the abscissa goes from —27r to +27? with tie marks every 77/ 8. There is a dashed vertical line at w = —7r and another dashed vertical line at w = +7r. You only have to plot over —7r < w < 7?. Multiple Choice Question 1. Circle EVERY answer that is correct: Primary differences between the CTFT and the DTFT include he time—domain variable for the DTFT is a discrete (integer—valued) variable for the DTFT whereas it is continuous—valued for the CTFT ® he DTFT is always periodic with period 271' whereas the CTFT is generally not periodic except under special conditions @The DTFT is a summation over time whereas the CTFT is an integration over time (d) The inverse DTFT is a summation over frequency whereas the inverse CTFT is an integration over frequency Tue ;h v in 59 D T FT {5 am i n ’79 t ra'i‘icv. M (e) The highest frequency for the DTFT is effectively 7r whereas the highest frequency for the CTFT is 00 Multiple Choice Question 2. Circle EVERY answer that is correct: Properties of the CT PT that are very different for the DTF T are T ime—Scaling Property (b) Convolution Property Differentiation—in—Time Property ‘ r ’r e’hxe/ answew 'q‘o/ fut“? One ((1) Frequency—Shift Property} U“ \\ a C WP ‘ (e) Time—Shift Property ( (f) )Duality Property Multiple Choice Question 3. Circle EVERY answer that is correct. When you pass an analog signal through an LT 1 filter, the Nyquist rate for the output signal is greater than the Nyquist rate for the input signal. hen you form the product of two analog signals, the Nyquist rate for the product signal is the sum of the respective Nyquist rates for the two individual signals. he Nyquist rate for the square of a signal is twice the Nyquist rate of the original signal. (d) The Nyquist rate for the derivative of a signal is greater than the Nyquist rate for the original signal. When you multiply a signal by a real-valued sinewave, the Nyquist rate of the resulting modulated signal is greater than the Nyquist rate for the original signal. Problem 2 (a). Consider an analog signal xa(t) with a bandwidth (maximum frequency) of W in rads/sec. The sampling rate is twice the Nyquist rate: ws 2 4W (where ws 2 27r/Ts.) Determine the range for acceptable values for the cut-off frequency, we, of the Ideal Lowpass Filter that allows ma(t) to be reconstructed perfectly according to the formula below. xa(t) : f: $a(nTs)h(t — nTS) Where: : TSSln(th) TL=*OO 7rt W< WC < (u;— \A/ w324W W<WC< 1W Problem 2 Consider a CT signal ma(t) with bandwidth (maximum frequency) W in rads/ sec. The sampling rate is chosen to be above the Nyquist rate at ws 2 3W, where ws 2 27T/T3. xa(t) is reconstructed perfectly according to the formula below. Let H (to) be the CTFT of h(t). Determine the respective values of cal and £02, both in terms of W, so that the CTFT H (w) is flat up to the bandwidth W and then rolls off to zero at ws — W. wad) = f: $a(nTs)h(t ~ nTs) where h(t) = TSlSiMwlt) 511190215) 601 7ft ’ITt and cos 2 3W n=—oo W‘Ay-UU12 CUS~W1 gW~W12W Problem 2 Consider the continuous—time signal Mt) below A discrete—time signal is 2 created by sampling 56(15) according to = 17(nTs) for T S = Plot the DTFT of mm, X (w), over —7r < w < 7r in the space provided below. W) = jZTs {51mm } sin(10t) . fig ' 1‘ T blah 3‘“ ((09’T€ DJ SxW (Wk) 5 TS % 5 fie ><<w '2 W” W’ “T k w+lu ><(xm:1m*(zo> co) 4 C) ‘ —\- “r. 4C3 80> w 1 may “'98: 20 NY‘CM‘ST- Ya 6 ho a\§a%\A8 .. 'L’B“ ’T\‘ (NM 7,0 Mapped ‘l'o UJMTS~ 10K .83.. .. : -12Tr ‘T O T 2T l I I I I I I I I I I I I I I I I I I l I I l I I I I I I I I I I I I I I I I I I I I I l I I I I l I l I I I I I I I I I l l I I I I I I I I I I l l I I I I l I I I I I I I I I I _ l | l l l J | | | I | I I I -6.§8IBB%MWW53M.W63317938585920.3QZEMMMIBBHZSEWQS4WZQQSMQBBZ Problem 2 Consider the continuous—time signal :c(t) below A discrete—time signal is 2 created by sampling x(t) according to = 33(nTs) for T8 : Plot the DTFT of mm], X (w), over —7r < w < 7r in the space provided below. 93“) = jQTs {8111(10t)} sin(10t) (UM320 w$=30 3o<4© => A\Cas;ngl_ ._ .— Th ave mq ed ‘l‘o‘ AM‘ASIAQ glrawfi a‘I’ U);"U~)M — '30” 20 ~ IO “59 Pl” Io. "m: 3. w and foes *0 2 29 —; I5 30 ‘ g 'L “L l. 10 \S- ET —_ TV“ Oink beVuW“ A“ W“ ' \ 3o ‘ZTT__ 4 “HM Tux 10 «g; ~ Eir- O \ 273* -4 ..... \ I I I I I I I I I l I I I I I I I l I I I I I 1 I I I I \ I I I I I I l l | I I | I | | | I | | l | | I I _ | | I I | | | | I | | | 45.55%mmma3mmmmawszemznsenmasvmmmmmsamgnmsmm32 Problem 2 (e). Consider a DT LTI system with impulse response hm : 8 {sin } 7TH 2 Determine the output y[n] for the input given by 27. Z’i‘r‘ 3m” FLQCQM’Y 0+ +(C0xngie I; 2 “‘7:- ‘i (0» 6% e> eta Jr H (-23 C?) +H<¥>Cos<¥A> + HUT} “5 @V‘B H ("it -1 H<T :\ Hi? :0 Uv‘k9w3 HMS—'3 (’5 " Tint/L517 yflrx): ; + Zecg <§n>+ (cg Workout Problem 3 Let H Lp(w) be the Discrete-Time Fourier Transform (DTFT) of the impulse response th defined below. ' 1 ‘ E71 s1n (8n) sm ( 8 n) ’ITTL 7F” thln] = 8 (a) Note: th[n] is real—valued and even—symmetric as a function of discrete-time. Thus, H L p(w) is both real—valued and symmetric as a function of frequency. Plot H Lp(w) in the space provided over —7r < w < 77 showing as much detail as possible. (b) hln] is defined in terms of thln] as: _ 7r h[n] = 2 thln] s1n (2) Note that h[n] is odd—symmetric as a function of time. Thus, H (w) is purely imaginary for all frequencies. Plot H (w) in the space provided. Note that the vertical axis values have the multiplicative scalar j = v-«l factored into them. (0) Consider the input signal below. = 16 cos Determine and plot the DTFT X (to) of the signal in the space provided. (d) Denote the output y[n] when the signal in part (0) directly above is the input to the LTl system with impulse response h[n] in part (b) defined by eqn. Determine and plot Y(w) in the space provided over —7r < to < 7r showing as much detail as possible. (e) Create a complex—valued signal as Zlnl = min] + jylnl Determine and plot the DTFT Z (w) in the space provided over —7r < w < 7T showing as much detail as possible. VIP Note Regarding DTFT Plots: The abscissa in each plot is the frequency axis. Ignore the numbers at the bottom. For each plot, the abscissa goes from —27r t0 +27r with tic marks every 7r/ 8. There is a dashed vertical line at w = —7r and another dashed vertical line at w = +7r. You only have to plot over —7r < to < 7r. (MI: 8x FIAI u Jig". El“ 6) 8 Plot your answer to Problem 3 (a) here. Show work above. ~’ZI“S‘ 2 _ I I | | -6.58E%M5‘fl I 24199253 | J | | I | mmmw 38818 I J 53920.39 I I I 23548511 I | I gazemsmmsz H (uh: j HLp<wn1§> I”) H“? (w+{£> Plot your answer to Problem 3 (b) here. Show work above. I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I l I I I I I I I I I I I I I I I I I l I I I I I _ I | I I | | | I | | | | | -6.%8E95m9flm53m.fl&5$$$7 WWBQZDBSZ’BSW 8317 MEEHZBEWQZMQBZGKBE32 J | | I | | | I I I I I \ \~\Q{%I\+:‘ 1% Plot your answer to Problem 3 (0) here. Show work above. I I I I I I I I I I I I I I I I I I I I I I 1 IILIII I IjIIJ IILIILIIII _ I L I | I | I —6.53memm5nm53mmmmwszeaeznsemm 851! MEHZBEWQEMQ'BZGKBEIZBSZ 10 Plot your answer to Problem 3 (d) here. Show work above. 3211||11| I I I | _ | | | l L L l | | l | | —6.§8mwmyam53mflz%mmwaze5392n39 masvmmmmmsemgmw’em32 11 OVQV O<UJ<TY> we have: I+3 (—Jvzl C)va "TT<Uk—J<'0> “’9 \ACWQ: \+;)(J>" 0 Plot your answer to Problem 3 (e) here. Show work above. | | I L l | | | l l | _ I I I I I I I I I I I I I I ‘6é3mmgflm534mflw GEESBQZDBGZIBMESVGEEHABEMQQMgbzmtsmmmz 12 NcJFE‘ Vet fQQQQEEEE _... WWWWW W. W h. p r a. c f C <— cu \ W::§Z§:::::f?§i TQafiQiQ$fiiE§§f?fléwwfmw . WNpTFT‘ v0 é——-—> \~ (mp lx’ca‘h'owg 6f 2 r , r ‘ ‘ I)" Mam ' . M M .w .- . ,w f m \95' ...
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This note was uploaded on 02/19/2012 for the course ECE 301 taught by Professor V."ragu"balakrishnan during the Spring '06 term at Purdue University-West Lafayette.

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Exam3SolnSP2011 - S‘O\UL'l‘\rovx {Qfi/ EE301 Signals...

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