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Exam2F11Soln

# Exam2F11Soln - NAME 31 Oct 2011 ECE 538 Digital Signal...

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Unformatted text preview: NAME: 31 Oct. 2011 ECE 538 Digital Signal Processing I Exam 2 Fall 2011 Cover Sheet WRITE YOUR NAME ON THIS COVER SHEET Test Duration: 60 minutes. Open Book but Closed Notes. Calculators NOT allowed. All work should be done in the space provided. EE538 Digital Signal Processing I Exam 2 Fall 2011 Problem 1. Consider four different signals with the spectral shapes shown in Figure 1 on the next page, each of which is sampled at the Nyquist rate. Digital subbanding of the the DT signals is effected via the system in the block diagram in Figure 1. The respective impulse responses for the four ﬁlters fk[n] = ej(k%)”th[n], k = 0, 1, 2, 3 are deﬁned in terms of the ideal lowpass ﬁlter below sinﬁn) th[n] = 4: (1) 7TH (a) Plot the magnitude of the DTFT Y(w) of the ﬁnal sum output y[n] over —7r < w < 7r. You will lose credit if you just do the plot with no supporting work and/ or explanation. At the same time, though, be concise and don’t overdo it as you have to save time for the other parts of this exam. Note that (if = 377‘ is the same DT frequency as 27r 7r “4‘ = ‘5- x my a 1:1 F — B B xa 1“) Ideal AID Xf") ——’ F = 23 “’@ S xasz) A M F “B B x82“) Ideal AID X41) ' FS=ZB . x (F)‘ as F. - B B x33“) ldealA/D X3 F3: 23 Figure 1 Digital subbanding of four real-valued signals each sampled at Nyquist rate. kx‘gkpass p.156“ (b) The four polyphase components of th[n] are deﬁned as MM 2 th[4n + E], 6 = 0,1,273, such that in 35471 E sin 7TH g hg[n]=4%&=—f—(;+—EQ, £=0,1,2,3 (2) The frequency response of each ﬁlter is Hg<W> ( the DTFT of MM) 6 = 0, 1, 27 3. Plot the phase of each ﬁlter ZHAw) over —7T < w < 7r, 6 = 0, 1, 2, 3. You can either do them all on one plot or on separate plots. Clearly indicate the slope in each case. \(UQB z. *1 'TT 303 4 H} (be) 1 25' . U.) 4'. H1O“): 7* 4H‘Cw3 7‘ i R Pour 4 HOCw‘) “=0 :6 2:0 TV 7°“) «’TC (c) Digital subbanding is alternatively effected more efﬁciently with the polyphase ﬁlters via the block diagram in Figure 2 where the respective input signals are deﬁned as below. Determine the value of each of the amplitude coefﬁcients 0%, k: = 1,2,3, ﬂ = 1, 2, 3 so that the output y[n] is the SAME as the output in Figure 1. Show all work on the next page but input your ﬁnal answers in the table below. You will lose credit for answers that are not justiﬁed by work and/ or explanation. 50M 2 atom —— mﬂn] + :172 [n] + \$3M] 51M 2 330M —— auxﬂnl + algmﬂn] —- a13x3[n] 32M 2 \$0M + 0(215U1[ ] + 0622\$2[7”L] + d23x3[n] 53M 2 930M — a31x1[n] + a32\$2[n] —— 0433mm] 063k [€21 [6:2 kZTi 6:1 L5!" J;': ‘3”; 6:2 L52" j - “3.2-3 6: J3" J3‘1- \JJ-‘l U )M MUD ,__ Ozﬂ L‘Qm) 30m]: x O[n]+ x [n] +x2[n] +x3[n] “‘0 S1 [n] =XO[n]+ (x11 x1 [n] + interleaver y[n] \$2[n]= X0[n]+ a21x1[n]+ 0:22 x2[n]+06 23 x3 [n] h ——>———On] 33m] = xO[n]+ d31x1[n] + a32 x2[n] + 0:33 x3 [n] a Figure 2. 5 (((d) The system depicted in Figure 3 is employed to recover the four original signals back from the interleaved sum y[n] in Figure 2. Fill in the values of ,8“ in the table below so that zk[n] 2 might], k = 0, 17 2, 3. Show all work on the next page. 20W = 90in] -- 91in] + 92 [71] + gs[n 21M 2 90M —— 51191 [71] + 31min] “ 51393lnl 22M 2 90M + 52191 [n] + ﬁ2292lnl + 523%[71] 33W = g0[n] + B3191lnl + 632mm] —— 53393lnl 5kg 5 =1 2 =2 6 =3 k = 1 J A j k = 2 A \ ,_ g k = 3 -_i Al ‘3 ~ 3.3% {4% +9—> . J 41, ﬂ SOMQ f. 6 kLP E4h’r D anaiygﬁ as is? F0”? J 3;? 21 344“] —>-—90[n] 20W: 90 ”1+ 91inl+gztn1+ 93m] Y[4n'1l—>-—g1[n] , z1[n]= go [n1+B11 g1 [n1+ B12 92[n]+ B1393[n] y[4n-2]_- 92m] 22m]: go [n1+ [521 g 1[n1+ 52292[n]+ B2393[n] y[4n-3]_.-_93[n] 23[n]= go [n1+ B31 91 [n1+ 33292th Bssgsln] Figure 3. (e) Consider 8 different real—valued signals, ”klnl, k = 1, 2, ..., 8, that were sampled near the Nyquist rate. Let 23,471] denote the respective Hilbert Transform of each signal. We form the following four complex—valued signals: ' .4 w «(Rf 930W = (vilnl--jﬁ1[n])—-(v2[n] —[email protected][n]) m «T? {4 4 ,mqguxam/q min] = (vslnl ——j2>3[n1 -(v4[nl —j®4[nl) m .. _ (v6lnl —]?AJ6[TL]) )rvx ‘ﬁ <Uu— 2;: Cwnk §ﬁ<w< Ti‘ )_ 552W = (Uslnl --j135lnl) — \$3M = (U7[n]+j177[nl)+(vs[nl *ﬁJSlnl) {vx ABTVM < w ~1T/4 These four signals are the respective inputs to the system in Figure 1 to form the sum signal y[n] For each of the 8 signals, Uklnl, Is 2 1, 2, ..., 8, indicate the frequency band that it occupies in the sum signal y[n] within —7T < w < 7?. I ﬁlled in the ﬁrst one for you. You have to put in the correct answer for the remaining seven signals. Signal Located in Frequency Band: 111M 0 < w <3:— v2[n] JRH; <w<o 113M ’W/a < Ou< 37174 Mn] wl4<ou<TWz U5[n] ‘ Tr < w< “SW/4— v6[n] ”gs/4 < Luc "1T" vlel -—mz <w< mm vslﬂl ~3ﬁlf+< oo<—’h-/z W“ *k:.\f\\( " f as ercgnal {MleX ...
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Exam2F11Soln - NAME 31 Oct 2011 ECE 538 Digital Signal...

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