EE4150_2003_samplefinal

EE4150_2003_samplefinal - EE4150 Di ital Si nalProcessin e...

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Unformatted text preview: EE4150 Di ital Si nalProcessin e " Dr. Hsiao-Chun Wu Final Examination, Spring of 2002 Time: 3~5 p.m., Tuesday, May 14, 2002 Please ALWAYS work on the FRONT SIDE of each page. No answer on the BACK SIDE of any page will be GRADED!_ Ask for additional blank papers if necessary! " You may check any textbook, classnote or other references alone during the test. HoWever, NO CHEATIN G or COLLABORATION is allowed and such violation of the _ university regulations will be reported. Please write down your name and social security number here: Full Namer /“ iii/.- 0 *2, Social Security Number: Partition ' I |_ ' Question 3 Question 4 Total _ l Question 1 (25%) A . . . . 0° .r—kT periodic triangular trains“): 2 m( 2) needs to be sampled and then k=-oo . m—H quantized, where {HI—L )={ Tm ’ Ill 5 Tm as below. T! 2 0, elsewhere A/D Quantizer 2.2g as" We sample .90) at t: 0, 1:, i , "MT, ..... .., to form x_(n)=s(nT/4) where n is 4 4 ’ 4 any integer. (a) What is the fundamental period Np of x01)? (5%) (b) What is the sequence x(n) within the fundamental interval 0.5 n S N p —1 ? (5%) (c) What is the mean squared quantization error between x0e) and the quantized signal y(n), if four quantization intervals are applied? {Hint A zgcmaX—ZIxm—ii and . A _ yo” = {xmin +kA+—2-, If xmjn +kA S 3601) < xmin +(kf1)A, k = 051,25 xmax a 35(3) = xmax } (15%) (a) yap-{f WEI-W0) .4 Answer to Question 1: .II 1‘ N‘ :- L :'4‘ .P .___ 4. ! S %(n) : [Jail—~30]:- ) 3 I __ l Xhax=lj XmIn: a 2 A ‘- K X01)! 9C“) M l l . .5: "3 3L 0 fig 1 ._$'_ T 3 -l_ X01}- (:1) 'MJ‘E: Tlgo =‘ ‘5: “New +69 +6$¢ : 3 . ' 23—6 Question 2 (25%) _ A linear-time-invariant system transfer function can be described as a rational Z- (2 — 0.5) (2:2 +1.12 +0.3) y(*1) ¢ 0, y(-2) at 0, y(-3) = yH) = y(-5) = = 0- (a) What is the correSponding difference equation for this LTI system? (5%) transform H(z) = . Initial conditions of the output response are given by (b) What is the zero-state response for a unit-step input sequence? (15%) (c) What is the output response for a unit-step input sequence? (5%) Answer to Question 2: (0“) W29 we —. 0.!) H(%)= __,.....-—- :: 2 _ Xe) (2411-24-03) -1 —2 H i N .. .— ‘ i ) (b) fld‘J‘M/mi' (4)01): .fl'Wm-U 4- WM I) 01-2/0? 3. gm: bgqn-l) + WM) gmh Lain-'1'): agm-1)— aégthuz) +g/(w) ~04” 701-1) _ aé?(¢—;) +X(h-!) [301): ((2H,) 20:") "0‘! ch'l) i :r) ar-O‘S' 'bT—0.6 J ...w(-'1)§ 3H) 492(4) w (~25 = 3‘") )1 h . ij WW“);- : agpmn-i’e-t) - Z 0-5 é‘g’r’n—fi—z) fi=o fizo . ' j _ n it h 1% '- 3 r a utmfi—I) - Z ogra ucnqg-a-J "I 1E=o I ' fito n-l % n-z {a : 2: a MUM) a. an? :01- ULch-z). 1%?0 ‘&:o 8 n—! n n— “ [-63 (low—I) - 0w? I Q UM‘U H I I‘m my! 3. W21“): a WM) um) 21-! n-H Wm): fl" bum-U— of “a “01-2) +66 WW) if“ l—a / '” 7% gym): 2: b (xi/(fl‘ié) fizz) fl n bgahh& ‘-' )a{n~&-i)—— 2: __ “(’1‘g4) E ((—Q fizo I (2 _ fi’ ' h Lfl/an'fi"! I In ('9 _ ~— a(n*rQ—L) Mtnfivzh 0.: Z: _ _0~.§- {_a g; l 6: 13, _ + ibanwww) WWW :0 n - . an n- - c;ejle,Ia--.)'U(hr|)— _____ _ J2. (uh—l) (f*l£)CI-b) 1—63 (“a n—I "-1 I (fith‘J _0,;.;;<l—b£{,) ugh-z) + 0.5 a b“ and; (I—a)(:—"b") "‘2 l“; bYIHhI “+1 I i -[J*‘E)v J a r- m I) r rfi ” ‘1'“ b UH”) / fiaW/h. 931/9 '“E (4)01) ’ h _ .4 [1— HM») J WM) 50.5)” [1— (1-2)”) a 7" any—I) at; . 0~3 _ h~l )3- n... "H f!‘(‘0.") J (dour) If /-(/‘2) IJ 4 5’, “(4‘l) -— _L( (it‘l) _ " J.6 [/ (12)“; fr}! *- 0 - (~0 5‘) [2(4) + 9.4 yf-wj (an? \2. _ f h gzdfim): 22J(J1)/ __ [/v (*0-6)j_L((A-I) 3190:9642}: 0 2 44 n n aw!“ ._ __ ‘ a - ‘6 g + (0.5") [!(/2)Ja(n_,_) _ [/(0)_, 0‘3 4‘5, Dim-2} n-j ’1" a; (b2) J MOP-2') 0.6 _ _ -n,+f (C) g2; 01): anybU Wm l.- (~o.6) gM) UM) _ I I I 3.01): yzflnfl” 321‘”) c. I [1*- (~o‘éjnj “(hi-l!) (_0‘f.)”__ (A3)”; mm + W “(h-J} 2a; _ v a“; ' - _ n—I - nth Ira (—0.651 fJ (—M) [Ff/~22 m - ~————-—---- (101‘2) ._ airwa 4&8 0.5 n+1 (“(#2) J nH Question 3 125%! A linear-time-invariant BIBO stable system can be described as a rational Z-transform H(z) = ——_~IL—~—_—1~— , where p, 91 can be any positive integers. The impulse (l—az )p(1—bz )q I response Mn) can be derived through the inverse Z-transform. (a) If 1>a>b>0, what is the requirement for r where r is the radius of contour C: |z| = r we need to do the contour integral? (5%) (b) If a>1>h>0, what is the requirement for r where r is the radius of contour C: [2! = r we need to do the contour integral? (5%) (c) What is the impulse response Mn) given that a>1>b>0 and p=q=2? (15%) 11 Answer to Question 3: 9+3 (5” HQ)? 2 (2~ 6UP (a w? 12 (C) - -13 H ~36er 72 a . ' %- Hm: a 3 + w - a- . (69-19) (2—6:) . (61-19,? (at-at) +. b3-362L1 .2 + [9; 2-K (Ira): (ELL) 'awa)‘ (2—— L)‘ _l . h =2" an I 2 )m ah? C (2~fl)m " "J C (if) , - - ma" 0')“ 2 lTI" c (376) J n~m+l : 1 hi fl {gm—MI): W (hm)! (n—m+i)i [3 N Write C 0. ) W fl 1*: Mrfdf C {j n <0 J —: “i— ----‘-"% in 01% ~44, >9. NJ fig: 0th 2 Mia ‘ l .. ..l.-.. - _ d% NJEE axe—p)“ ‘ m ’49- fl Am. (A ' a 1, ANW = Z *1? -+ :(s—p ‘ 3‘ "(g-fl) £51.- ? 4&251 ' 14 Ara: I. 'olwf’) I“ /' “(4?— kl) Hg ‘ d gap-m ,;_ (3%,? __ _-' ; ' - —n1~l ~ - -, v. . _ I ~m dial-Q: ‘ (—447 -fl+&.)'(—M‘§;Q?F@‘tfji _. _ C94,): .' I, (mfinfi '0‘, ova—EL) 2 . ‘ i . 2 ~73 I ._.(~_-£) (712-!) (—2-2) fl fl-fl “mi-19a. \ : (M'Wgz)! \ - I I ‘ ' (~9-W—99; -(~.Q~M+Q "'0'" 1 I _) z A _—.)n—,0+4. - . (F73. (-w-m-m rim—9+2) WU . I A “5—0- +£v2y ‘W’UQH I-m-PH' l I -1) Cm ‘ (—4) fl .. (Q_I)J‘ m (WI-4)]: firm-P"- E? “139+ ‘- /-\r'. v" C’QE’» ( ) fl. ~£~m+f (h-l) A2] : I (-4) MHZ-2W1 (M-U’ (fl—t)’ — .. | b. 0 2 m m ’H x I 3% 2 85—! _._OH, "r Mfflc W“ aft‘: r r a 2.11:) C (875)»: ,4“) 3f Hg J (JV-91:029L { (:) -.()3:?.0 ) —_ 5 8 J.“ .3 h-" - w-s-afww, 4» (b-a) (la-6?). 15 "b 16 3 . ~ __ __ L) ’1 ' (0. 3G a M («144) h... h a. (1841'). (av-bf (“40‘ ' ' 3 3 a, n-l. : (£3,623 gum) + l” né WW) (tray; - (E-UxJJII 3 z 3’ -3 L) “a! h (a 623 gawk) * 6? ha gel—m) (05b) (R‘EJQ ' .3 1 n 10 w w)éum)+ ' “(a W”) draws (la-a) . (a? 30619) ' ' “3-. H H) ._ n " ‘ __ diam—n) ._ aha “H 3 - (OP-b) Question 4 (25%) A digital signal x01) is downsampled to fonn a new sequence y(n)=x(2n). On the other hand, fin) is upsampled to form a new sequence 201) such that 2(Nn+k)=x(n), k=0,1,2. . ., (N—Il), where N is an arbitraxy positive integer. The discrete-time Fourier transforms for x(n), y(n), 202) aije X(aJ), 1(a)) and Z(co) respectively. (a) What is the relationship between x(2n) and x(2n+l) if X(a)) = Y(2a))[1+e_jw]? (15%) (b) What is 2m) in terms othw)? (10%) 19 filo (n+0 0&7L ..-9_(;(h): 7(2h+}) Mal 7(a): 9" ' V7202» v.4) [(u):': “101) 6 kw . 1 . a” ' “J‘{J‘J)n + Z 30') e 11:-” . — . “ l4) X;(2.gd)e" 4— Fund) be _.UH W ICbchz «(Lyme J mm»: “M .1} 5'! a”: VH4) - Z; an) e H=—be 20 Q’rzn/ ,XImJ): You) .fi Em): 17m») y. a. mp1): yr») (b) - - «2(0):. h:—m Z in")? r) 9‘“ij .: you; .5193; . 21 ...
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EE4150_2003_samplefinal - EE4150 Di ital Si nalProcessin e...

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