ece366-06ex2sol - Name: Solutions Student ID: ECE 366 EXAM...

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Unformatted text preview: Name: Solutions Student ID: ECE 366 EXAM 2 November 17, 2006 N0 textbooks, notes or HW solutions. One page of hand—written notes. Calculators are allowed. Exam is 50 minutes. To maximize your score on this exam, read the questions carefully and write legibly. For those problems that allow partial credit, show your work clearly. Good luck. 1. [20] State whether the following statements are true or false. No partial credit will be given so you do not have to provide any explanation. a) b) d) e) D An audio signal with 101-112 bandwidth is transmitted over a channel using AM modulation. The minimum bandwidth of the channel should be 20Hz. .1. The Nyquist frequency of sampling for x(t) == sin C(SI) is 20 rad/sec. Xiufl 1: JIFQCT i (NS: \OY'ad/s-Fc ‘11:. 5' The signal x[n] = cos(3n) + cos(§7m) is periodic. 4' . "i: not pen and C ' For a real and even periodic signal, the Fourier series coefficients, Ck , are always real. T The power of u[n] is 1. W2) F The spectrum of a periodic signal is always periodic. ii’r’s discrete, ~— hot’ponociic 1“ g) The ideal lowpass filter is not used in practice because it is not causal. T h) If a periodic signal x(t) has Foufier series coefficients C k , the Fourier series coefficients for the signal x(—r) will also be equal to Ck . "F: i) The inverse Fourier Transform of e‘za’uMJ) is 1 . . .d 2 '4‘ Jr Duality e, um é 3+5 u.) l K’ 1;“: . 1mm Mireflwulw) j) If X(a))=tri(a)/2),then jx(¢)d:=1. 03—” ’t SxHScH: Xl0\é i§> w 2. [40] Consider the ideal sampling system given below for the continuous time signal with the spectrum given in Figure l (a). Let Q = 8x,a)2 = 1071'. a) [5] What is the Nyquist frequency of sampling for this signal? b) [6] Assume that this signal is sampled using the ideal sampler shown below with T = 0.5 . Derive an expression for the spectrum of the sampled signal, X p ((0) , in terms of X ((0) . c) [10] Sketch the spectrum of the sampled signal, X p ((0) , you found in part (b). Draw at least the first 5 terms in the summation, i.e. k = 0,:t1,i2,i3,i4. Make sure to label the frequency and the amplitude axes. cl) [6] Show that you can reconstruct the original signal from the sampled one. What are the values of A, a)“ , cab for the bandpass filter shown in the Figure? e) [7] If the sampling rate T is now increased to 1 sec, can you still reconstruct the original signal? Sketch the spectrum for the sampled signal, and discuss Whether you can recover the original signal. l) [6] Show that the sampling frequency used in part (b) is less than the Nyquist sampling frequency. Explain Why we can recover the original signal even though we sampled at a rate that is less than the Nyquist rate. XI 11.!) p(t) = goat—M) n=-w (t) plt) 1 T t H( m) A —wb —wa (9a “3b ‘0 (b) ! ------—-mvwmmmmflwngw‘ymmm"mammwlawman-pmw—---uwumummymwzrmmwmmgrmmth—m—v—mmmmwmmnwwm minwulWWm WWWmmWMH.WWH m Extra Page for Question 2: a) U.) 3 —: ZUOTT) : 201T rad/sec. I fiXkawQ {JO NVOWVC‘SJY Samph‘hS we can S‘hh rcconsWt‘r W b d QOTY ? WW g0 nag» becoLM/J e Ifi ‘5 a flarf’ow an W91 ‘ \J ‘ a} , h0+ bom d p083, gxgw" . 01$ {GIT . mmmmmmmmmmmmm -v—uWW.WWWW—_ 3. [40] Consider the system shown in the Figure below. Assume that sin(1 0721:) x(t) = sin C(ZOmf), g(t) = cos(87rt) + cos(16m‘) and Mt) = 72': a) [6] Compute and. sketch X ((0) . b) [6] Compute and sketch G(a)). c) [6] Compute and sketch H (w) . d) [6] Compute and sketch Z (w) . e) [6] Compute and sketch W ((0) . f) [10] Compute and sketch Y(a)). sin(x) Hint: sin 00:) = x 0O XKLUB: V96] V by Stub : WY 8‘0???“ +g{w+&fi\] 4*“ [Slow-SW} Skeet/61:3] (>th ’Y . U.) *iéh’ “SH 831 HWY mWWWWWWWWMWWMWWMWMWWWWW W ~2 .. . WNW WWW-VWAWquW ‘ (:\ hhfl, gm {tout} : 4OS\h(‘_UOTTJC\ *3’ refi T “W » x w fixon' 105i] 1-; i6 rec [jg->40“ MT cu - ‘ .86) +Stw+$sfi31 in g Q} : umtw K m C31 Km)H\wB —~ gr: en c " a 4 DC] *fiWkLU) mumphfcdoohmh‘m m My, my fit to 3 his mmmqu) ' Wt») 3}? :Z'Kto— 81‘) +1” %\Lo+&Tt)] sol.“ zfi Extra Page for Question 3: BL redrka Jarec‘r(wv‘<§‘w\ +~_\_ Y~€C+KW+<§T :1 3’0“ 2 BOW 20 16W T \ FZCJT (Hg; 3 [(61% \wwg-rr\ + rflfi w+§fi~\ UO 30W 20W 201‘7 Who) 3? AWWW“MWMWMWWMWMWWWWWMWUWWM ‘ w W annmww—MWWWWHW ...
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ece366-06ex2sol - Name: Solutions Student ID: ECE 366 EXAM...

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