EE4150_2004_hw2sol

EE4150_2004_hw2sol - 3.1 (a) X(z) = 2301):“- :

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Unformatted text preview: 3.1 (a) X(z) = 2301):“- : 325+6+z-1—I4z-2R0c:o<|zI<oo (b) i N V II II II A A H M| MI I-' A N v 3 v 3 v 3 N' N 3 3 II A MID-i N I J 3 + a II A \L OI ll i .L :6 O .Q E V MI 3.2 (a) X(z) = Z z(n)z'" II A g—n + EL N 3 II M N ; ,+ M a ; ROC: M > 1 w E. M. Na 3 | I z—l -—1 and §:nz-n z (1 _ -1)2 “Dc: '3' >1 -0 2 Th f - 1 _ 2-1 + z_1 ereore, z) — (1_z_1)2 (1_z-1)3 _ 4 - __ 2-1): (5) m 2 x(z) = 2(a"+a‘")z‘" ‘1 n=0 ‘ m m I E = Za"z'"+Za-nz-n I: I n=0 "=0 ‘ 0° 1 But 2:66.24. = mROCJz|>IaI oo 1 1 and a-..z—n __. ___.___Roc: z >— . (biz-1)“ ' ' '“l 1 1 Hence, X(z) — 1:77:71 '1'__ %z—l } 2—(a+f)z'l 1 1 —— ROCI I- (1_az-1)(1-&z-1) "'> mm” Ial) - l ‘ (C) °° 1 X(z) - :3; 2) z — 1 M > 1 _ 1+%z'1’ 2 (d) _ co X(z) = znansinwonzq —0 m . ‘ e-“”°" _ e-onn = na" _ z-" [ 2J- ] l aejWoz-l I ae_jw°z-l 21 .' [dz-'1 " (“z—1P] sinwo (e) m I X(Z) = Znancoawonz_n n=0 co . ‘ CJW‘m ‘J'Uon -= 2m" n=0 '52 l aej“"'z'1 ae'j‘”°z“ 5 + [112“l +'(az“)3] sinwo — 21122—2 (1 — 2acoswoz'1+ (122-2)2 , |z|>a (f) 00 X(z) = A Z r"cos(won + 4302"" n=0 0° . . . . eroneM + e-onne-M ~ 'I ‘n Ar [.____.__..__2 z A en 3-“ ‘2— l - reiwoz" + 1 — re'i'Wz'1 A [c0345 - rcos(wo -— ¢)z"] , M > r l — 2rcoswoz'l + rzz‘z ‘ (s) l m l 2 1' n—l —n 1 KM = 250‘ +n)(§) Z 1 n=1 °° 1 n_ _ l 32-1 -3 BMZME) lz l = (5322-1): z (1_z%z-1)2 n=1 f: n2(l)n-lz—n = z-l + $2-2 "=1 3 (1 — %z“)3 Therefore, X ( z) 1 —1 2—1 +124 ' zl 1 2 + l —1 3 2 (1‘52- ) ) z 1 l "'>‘ 3 (h) °° 1 °° 1 Km 29w“ — 2 gr?" n=0 n=10 l 10 —10 ._1__ _ (4).; l— éz‘l 1— %z’1 1_(;z-1 10 l = ’ ll The pole-zero patterns are as )ollows: 1 (a) Double pole at z = 1 anQero at z = 0. : (b)Polesatz=aa.ndz=i-. erosatz=0andz=-21-(a+%).. (c) Pole at z = —% and zero at z = 0. ‘ (d) Double poles at z = aej“’° and z = ae‘j'”o and zeros at z = 0, z = in. (e) Double poles at z = 021‘”0 and z = ae‘j‘”o and zeros are obtained by solving the quadratic L ,acoswoz2 — 2a22 + ascoswo = 0. l (f) Poles at z = rej‘” and z =®fjw and zeros at ,7 = 0, and z = rcos(wo - ¢)/cos¢. (3) Triple pole at z = § and zeros at z = 0 and z Hence there is a pole-zero cancellation so ‘ I (1': that in rea_.lity there is only a. double pole at z = § and a. zero at z = 0. (h) X(z) has a. pole of order 9 at z = 0. For nine zeros which we find from the roots of l_ (éz—r)lo = .0 . l or, equlvalently, (5)lo — z10 = 0 l '2" Hence, z" II I N '6 3 II ' u—- 3" Pl- Note the pole-zero cancellation at z = ' _ ‘ M 3(a) _= Z, X(k) [(3—40 3) Stu) —\gLV\—-\) 2 X”) __ -\ ) :7 : ‘pzfl '3.” (a) ‘ X(z) '= = 1+4z‘1+7z"+10z'3+... Therefore,z(n) = {%,4,7,10,...,3n+1,...} (b) X(z) = 22+5zz+8z3+... Therefore,z(n) = {...,-(3n+1),...,11,é,5,2,?} 3.\§ 52'1 X“) = _ -1 1 ' 1-2z'1+1—%z"“ Iflzl > 2,1:(n) = [2" - u(n) l “'5 < lzl < 2,1:(n) -(%)"U(n) ‘— 2"u(-n —- l) (%)"u(—n — 1) — 2"u(-n - 1) Iflzl < 315,201) (a) 1 X“) = 1 — 1.52-1 +0.52-2 2 1 = .1 - 2-! " 1 —o.5z-1 For |z| < 0.5,:(11) = [(0.5)" - 2]1_4(-—n - 1) For |z| > 1,z(n) = [2 — (0.5)"] u(n) For 0.5 < |z| < 1,:(n) =' —(0.5)"u(n) — 2u(—n - 1) (b) 1 X(‘) = (1 -— 0.524)2 0.52'1 5 = {m}2‘ For |z| > 0.5,:(11) = 2(n + 1)(0.5)"+1u(n + 1) (n + 1)(0.5)”u(n) -2(n + 1)(0.5)”+1u(—n -— 2) -(n + 1)(O.5)”u(—n — 1) mg 0 '1 E A .0 .9‘ a A a V II II II 3.4fi 1 _______———-———- ‘ 1 + all"1 + (122—2 It‘d? — 4a; < 0, there are two complex poles —a1 :1: j\/4az - a? :2 A N V II P1,: = 2 ‘ 2 ‘ /4 _ 5 IP1 2|2 = (El—)2 + a: '61 < 1 ' 2 2 => 62 < 1 If a? — Ma; 2 0, there are two real poles p _ -a1 i Val! - 462 1.2 -——-—"—‘——'2 ‘ ________.—-——- 1 d 2 < an —al — a1 — 4a: __.______._——— -1 2 > =>al—a2 < land Hal 12‘" ‘9’ ‘5 -1 Figure 3.2: ...
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This note was uploaded on 02/26/2012 for the course EE 4150 taught by Professor Wu during the Fall '10 term at LSU.

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EE4150_2004_hw2sol - 3.1 (a) X(z) = 2301):“- :

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