Chapter 10 - 10.1 10.2 for; CHAPTER 30 36: 20:09: i— =—...

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Unformatted text preview: 10.1 10.2 for; CHAPTER 30 36: 20:09: i— =— a =—- 0F" = '- and 6: =bI.05783f51) 1.5 and 2019:;063 15 to r. 5; 0.1386 Equafion EqUadon is (10.3.15) gives N: {.5535. We choose N = 3" If we use (10.3.13), we. get: mg = 1412.5 rzdls. Tn: unnormn.!ized filter (1412.5)2 :1 +wf§(14115)3+(1412- 3): with fp =3kHzandN=2.wcsoIveforaJcas I HO) 2 me = “:11 HQ):- f F’ = 2662‘lmélse: «£25. — 5,3 and 1 - IN I} = fi[g§ - if = 62631mdfsec The filter transfer funcu'on is (26622)2 H(:)=——-——— - :1 + fi(26622)5 + (26622); We sti- égsffln Q nomcfgzeé. lam Pass. Plkr'c‘ilg, Z°“¢aJ‘-£>= -' , mbsmgz = 4‘; e c _ _ . ' 0.4; 5a {acne g(___o_mg47cg OP; 5;: canals. s a: -1; (to :3. IJ)0 e H'— rot). Q—l— “:1. {.9 ad" I HCd) ; 5‘4. («LN-2d -:-|. _ q. ‘ J r‘lit'r - Y'k-ct‘nQCJ ' '24 “Cpncamg ' ‘- “33mm! band pass :- ts Cb _ _ =1 — L 5‘1 —-——-"’° (4’ ,w-__>=./§ (d :. cmo ) B Ui- w‘ '6‘ fame"? 1" I Hid ) : L :— A 1‘; L a" 4- G)“;- ‘_ 1-4.1:,"(6 :x10 )4»! mod” mod" ' 231 ___.._—--——“_—‘—""- C P 4, 3' - ‘5' *- - i'aémc 6%.... 1:44-26 e-lzcl MO d ..v. 3. 433 med. 'tnj mplaimfi d1: - we Nermaific bi Baa-ma (JP : I so 39am} L05 rise 20 ice?” _.__‘-_.__..__. __. --.'.-‘.. (Sums & : .7g4.g ‘ J1+e=- '10 AISO 20:. 10leNC-J64-Q)-r-CCN-—l)+ ZON103I°( 3) 9ch5 M = I-7ZJ‘S . ' an'i...___i .sq—rs‘" Chaos: N: ;-_ . Atsa F5: :3 1 -764-8. -‘. Pale: O_F Ckcbjfikv’ {Sugar on: Ca;- x LES—02. = ~éoz-%t :J {me-Q7 HCA) : I L 2.. . (.4 +- 601-355.) 4- (.121‘3'q7) 10-b— me normahchJ Bukfi'r'novH‘n Elk-r- c4: etch-.1- 3 '5 H3 = ‘ 2 l 2- . ‘ (AL‘I’A-rl) (4-H) [Q+i)1+[€_) JQ+) TO Flficl ‘H‘u. normail‘uc.’ Chbdshv Fin-er, 52‘- iHml: I" =4.“ 4o +LaJ—-c—.=\ ‘ J2— \H-i-ez- So 4%.} f3: é. 9;. L4 (1) = -2332. CWP’m rocks a; 4hr: 0.31:5;sz Crlkv a“: 3%“ I: .‘3 A : fill S'hkC-2959)ij§CocL (-zciasr) = —'f4~‘i :Jfi‘ios; 232 10-; [$7 ‘5 ,o = -s;nL. (.1933) = 1% ‘ HC CA = . a 7.1- 3 (x) +1323]: (3—:— -|4.¢-])2' + c 4030"] r m'— l — 3::1 ord'a: utter4orth f 2 - 4th order Butterwcrth 3 - 3rd order Cebyshev 1-0 I _¥—i : O -E“ {AF-tn- F-“U'fl it“: ago”: kéjm-ML, “(7.3. c 2: OM-i‘“ J . I ' From the specifications. 20mg“, _ = — 1. 3. so that .5 = 0. 4825- XII—>53 Equation (10.3.22) gives ZOIogmL 4125) + 6(N - 1) +20N10gi0(5) = 15 from which N = 1.0703. We set N = 2, so that 5 I , _ i F,""~E“"N§ where ml, 2 2:: 1000 = 6283.2 rad/sec. 1 r I From Equation (10.3.26) 5 = ; sinh" 4175 ppies of H(s) are at - ' - -- 6283. 2 5 = — «5 = — 1.077x105 i:jI.1651x105 We therefore have =1.6187, so that the , v '. 6283.2 smh(1.6187):} ‘5 coshfl. 6 I 87) 1 H = _ _ . (S) _(5-i-1.077x103F+-(1.1631x103)3 233 10.3 (1412.5)2 :1 +~E(l412.5)5+(1412. 3‘12 Hfij= From me 10.5, ze'flsinm T) f“:)=-?~—j‘?"_£_":7? z -2.-.e'“ cos(on)+e '3 Where are = a = = 993. 73. mm 2- : L, 6mm Hm # .403: ‘ ‘ zz—l.67z+-7168 "uh I" =_—-L—, .0902: H(Z)=-—-—-—'——'—— :3- L801:+.8189 ‘0‘ ‘3 (a) 1-0 15(m)[ 234 a») R) 13(9) I --..—.r.. (b) Digital Filter with f5 = 6000 Hz Pmblan 10.9 (c) Digital filter with fs Problem 10.9 235 10 kHz 10-10 UJP :. \ _..._.. item 3000 g __ “MCI-3:. éaoa L / c t and 2 2 Lan(3° ° ‘30” )._., 32.77 ‘3! ‘/3000 2 Ram onbicm [D I I 5', :— -Zob‘57 I 3:. = 5'3i‘13, N"=l-%34- Choose. #22 and use Ei-(zo-a-Is) +0 Sci.- 005 = £15443 Hm =__I_«-________—— e1... J; (1151:, .aqjg (“.34-”); __. - 14 ___—-—--~——--—--—..-__________ 6’}. “232-27 3’ +- (“34...1q); Sal: f: 2 _ I—g’" had: (‘/30co) 1 +5" “(5): c.0233c(5+l)1' 31-1-4-‘H5 4.. .5244 Io-II Rom Problem “7.3 I +51 nmmaliuJ low- pass .- is For {'5 = 30° H15 I +4.: PrecilSI'O‘rkC' fchvencfcs arc. 0‘ __. ._ 10¢ - z 7.75 LOCI 600 Ear: (-600 D and — Goa Lam (500 3 - 32.7 73 C2. coo d _ a 4 50 44m!— 03, - Lac -uJL I — [3 w -..- uJCJ - 035:; —_~ [20-03 236 10.12 (a) H(Q) z “me—J"- n=—co -| 7 no , = {1(n)c"c”+h(0)-I— Z hfnk'lm : + + h(_n)ejfll:1 n=l If hm = AM), I Hm) = M0) + E 2h(n)cos(§2n) "=1 Iand H (Q) is purely real. I If Mn) = — h(—n), h(0) = 0, so shat mm = j E(—2)h<n)sin(£2n) n=| and is purely imaginary. N — l (b) Let N be odd and le: 30:) = h(n + ). Then 7 .‘V-l‘fl 6(9) = J"? H(Q) and m N—l 7 If Mn) 2 MN - l — n), 3(a) f: g(—n) and ArgG(Q) = 0, so that ArgG(Q) = ( )Q + ArgH(Q) ArgH(Q) 2 ~ (iv-51)Q . , ‘ if - It 1102) r. — hm —1—n). 302) = - g(-—n) and A:gG(.Q; = - 3-, so that V—l L I )9— x1 ArgH(Q) z —( so] 237 10.13 (a) 0 n n=0 ‘ :2— otherwise 1'! (b) “51.11 an 11—point response and a delay of 6 samples, we have 1 2-1 1 -l 1 -1 1 —1 h‘"’=§‘4_37’°‘1§‘3’2 51 5011121: 3 1 - _, H(:)=-:(1—z “5—55 w:‘°)+l( ' —:")-;<:3-- )+(:"-:"’) The 1 1-point Banning window is Wn(n)= [.067 .3 .5 .75 .933 1 se response of the differentiator is .1667 —-.375 .933 0 -.933 .375 —.1667 .0625 -.0134] .933 .75 .5 .75 .067] The corresponding impel h.(r.)=[.0134 —.06?5 'with transfer function ma) = 0. 01340 - z"°)-. 0625( :3) z" — 3-9) + o. 1607(2‘2 - — o. 3150:“3 - 2'7) + 0. 9330:" - :4) 10-14 (a) The ideal fiiter has impulse response h(n) = sin(zrnf6)lrcn.4 The lZ-tap filter response with a delay of 6 is given by h(n)=[0 .0318 .0689 .1061 .1373 .1592 1 .1592 .1378 .1061 .0639 .0313 01 The impulse response of the filter With the Banning window 15 hh(")={0 .006 .0268 .0649 .0119 Plots of 1mm! and 1mm)! are as s lZ-mp filter does not provide a goo desired filter response. .1513 1 .1513 .0119 .0649 .0268 .006 O} hown below. It is clear that a d approximation to the ‘ollar.l:..ll.onli. inu.llu!..!..lll..l 1:11.]...lalv 7 [fir 0.6 0.5 .i; ... .J 239 ...
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Chapter 10 - 10.1 10.2 for; CHAPTER 30 36: 20:09: i— =—...

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