EE4150_2004_hw3sol

EE4150_2004_hw3sol - (a) _ 24(1) = Ae‘“‘u(t), a>0...

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Unformatted text preview: (a) _ 24(1) = Ae‘“‘u(t), a>0 oo / Ae-ate—jthtdt 0 E 3.3 u A e—(¢+j21F)t] 0°_ ' -a - j2wF o A a + j2wF A Ma” + (21rF)2 IX¢(F)I = 21rF -1 '1— an (a) 4X.(F) Refer to fig 4.2 (b) A=2~,a=4 phase of ’Xa(F) O —5 o ‘5 1o -10 45, o' 5 10 Figure 4.2: a; w X¢(F) = Ae“e"j2'ndt +/ Ae“‘e"j2-'F‘dt /.,@9 0 up A A ‘ - a-j2«F+a+j21rF 20A 412 + 21F2 20A 412 + (21F)2 0 |X¢(F)| 4X.(F) Refer to fig 4.3 4.3 (—— (3) Refer to fig 4.4. _ t 0, otherwise 0 T X¢(F) = (1+ 1)e—j2rndt+/ (1_ sequin“ _, 1- o 1. Alternatively, we may find the fourier transform of 1 _ I _ -, n- Then, ? - Y(F) = / y(t)e"2'ndt —1 = /0 le-jZ‘lFtdt+ 7(—Tl_)e—j21rf'ldt -‘rT 0 _ 28in21rF1' — ——m‘r_ l andX(F) = j21rFY(F)~ _ .«n'mrF‘r)2 — «Fr ' 2 .simrF‘r |X(F)l = ,F, ) = 0 (b) c‘, = i/Td2 z(t)e‘j2""/T'di TP_ ‘Tplz _l_ o (1 + 1)e-j2!k!/Tpdt + /7(1 _. 3.).e-j21rk1/LTpdt] T, _, T o 7' 1'. simrl'.‘r/T,,]z T, «hr/T, I (c) From (a) and (b), we have Ck = fixed?) Anal-6 0.7 ' 0.5 0.‘ INF)! 0.3 0.2 0.1 o 7 5 no 15 an 25 —-> F Figure 4.3: x(l) Figure 4.4: _—F-——"‘ (a) ' 7 'Inl :(n) = Zeke I E: Notethatifck = e::.,then 7 _ 7 . Zellzllelflznl = I"... k=0 n=0 - - 8) p=—n 0, paé-n - Since c1, = %[efl:—. +e_‘:"] +-213 [elk—i -e--:'.] We have :(n) = 46(n + 1) + 46(n - 1) — 4j6(n +3) +4j6(n - 3), —3 Si; 5 5 (b) c - c-éc—éc-Oc--[Ec-——‘/§c-c-O 0- 51-212-213-14- 215— 2,6-7— 7 4 :(n) = Snag-FL i=0 = —2—-¢4+¢4 —¢c —¢c = s/S[sinfl+sinfl em 2 4 . (c) 4 'II-b‘ :(n) = Zeal—I'— k=—3 2:. _1_-"- 1 2:. 1 :12. 1 1'."_- 1 _J__-'=-- = 2+e¢ +e ¢ +—e' +-e = +-e¢ +—e ¢ 2 2 4 4 - 2+2o""+ 1r3+1cos3‘l2 — cs4 cos2 2 4 f.— (8) ’0‘) = u(")-u(n-—6) co X(w) = Z z(h)e"""” n=—oo 5 = ze—jwn 1-e'j6'” = l—e'j'” (b) ’ :(n) = 2"u(—n) o X(w) = » z 2"e"""‘ > n=-oo °° w = MEX-2‘) _ 2 - 2-ej'” (c) I 1 2(a) = (3)"u(n+4) co 1 . x(w) = 2(3)?!“ n=—4 °° 1 . . = (:(Z)me-JWM)44CJ4W m=0 44ej4w = l—fie'j'” (d) 2(11) = a"sinwonu(n),|a|<l . °° ejwon _ e-jwon _ X w =‘ a"[ , ]e‘-""” ( ) a; 2] \- 1 °° _ n 1 °° , n. _ _ -J(W-Wo) __ -J(W+Wo) - I ~ I _ 1 1 _ 1 — a} 1 — ae'j(W'Wo) 1 _ ae-j(W+Wo) __ asinwoe'j'” _ 1 — 2acoswoe‘j'” + a’e'jz'” (e) 2(n) = |alnsinwon,|a|<1 W W Note that z |::(n)| = z la|"|sinwon| fl=-OO n=—oo Suppose that we = 2%, so that. lsinwonl ‘: 1. C” II [V] E i 1 8 Z M" nz-oo fl=—m 99 Therefore, the fourier transform does not exist. (1') 1n £(n)={ 2‘(§)x [71'54 0, otherwise 4 X(w) = Z z(n)e'j""' n=-4 4 : [Mgr] n=-4 2ej4w 1— e‘j'” a _% [_4ej4w + 4e-j4w _3ejaw +e—j3u'a _ 292w + 2e—j2w _ ejw +e-jw] 26j4w W + j [4sin4w + 3sin3w + 2sin2w +-sinw] (s) Q) X(w) = Z z(n)e‘j"’" n=-oo —2e1'"” - ej'” + ej'” + 2e'j2'” —2j [2sin2w + sinw] I _ (h) ' ‘ _ A(2M+1—l|, IISM zoo—{0’ n) M Z z(n)e'j""' n=-Al X(w) M = A Z (2M+1—-ln|)e-J"”" n=-Al M . (2M + l)A + A 2(2M + I — lc)(e‘j'"" + a“) k=l . M ‘ (2M + l)A + 2A 2(2M + l - k)coswk k=l (a) X1(w) = Z:(n)e‘j"’" n chw+cjw + 1+e-jw +e—j2w 1 + 2cosw + 2cos2w (b) X2(w) Z 22(n)e‘j"’" cj4w +cj2w + 1+e—j2w+ e-jqw 1 + 2cos2w + 2cos4w X3(w) Z 33(n)e‘j"’" chw-+ cj3w + 1 + e43.» + e—jGw 1 + 2cos3w + 2cos6w (d) X2(w) = X1(2w) and X3(w) = X1(3w). Refer to fig 4.7 (e) If _ :(f), ;—‘ an integer “(7') s { 0, otherwise Then, = Z zg(n)e'j"'" X g(w) ‘ mg an integer z z(n)e'j'""" £(kw) (a) X1(w) Z 2(2n + l)e'j‘"“ 11 Z 2(k)e—jwk/2ejw/2 k X<%)ej'"“ ejail? l - (Rim/2 (b) 2 2(n + 2)e'“/2e‘j‘"" 1| _ Z 2(k)e-jk(w+jr/2)ej2w k X2(w) —X(w+%)e”‘” (C) I X3(w) = Zz(—2n)e‘j‘”" II I H A a- V 0 I L. If E \ N v u E I J, ‘ (d) X400) 2 ‘;_(ejo.3rn +e—j0.37n)z(n)e—jwn 1| 1 - - '(w-0.3r)n -j(w+0.3!)n - 2(n) e 1 + e 1 = -;-[X(w — 0.3«) + X(w + 0.31)} (e) X5(w) =. X(w) [X(w)e"j“’] = X2(w)e'j“" (f) X6(w) = X(w)X(—w) l (l — ae‘j‘")(l — aej”) l (l - 2acosw + a2) 4.23 (a) Y1(w) = z" y1(n)c“j'”" = 2,”, even :(n)e"""" The fourier transform Y1(w) can easily be obtained by combining the results of (b) and (c). (b) y3(n) = 2(2n) Y2(w) = Zy2(n)e‘j‘”" Z z(2n)e"“"" fl = Z :|:(m)e'j"""/2 m = In?) Refer to fig 4.8. (c) _ :(n/Z), 1: even yam) — { 0, otherwise no») 293(n)¢'j"'" Z z(n/2)e'j“’" u even 2 :(m)c'j2Wh‘ ;(2w) —rc —n/2 0 75/2 1: 311/2 27? Figure 4.8: We now return to part.(a). Note that. yl (n) may be expressed as -y1<">={3?‘"”* 2:32“ Hence, Y1(w) = Y2(2w). Refer to fig 4.9. - Figure 4.9: 4.4 ' (3) Mn) - 2-11; X(w)ej'”"dw II S’l *‘ l——_I N or: II ‘L. E 3 R, E I B d! 0 .1. E 3 a. .__‘5_._. 5 3 3.5 _ 0 0.5 1 1.5 2 2. '20 0.5 1 1.5 2 25 3 3.5 —>' Figure 4.47: — 1 sinawn ' In — 1m 8 sun8 - 13in" c 1r — 1m 8n 0.941; (b) Let h1(n) = 25:51; Then, _ 2, M S % H1(w)f{0, fi<lwl<1r and Mn) = h1(n)cos-74:n 4m Refer to fig 4.48. IxMImn) 1mm 1») 4 8 3 0 g; i4 E 1 2 L °o 1 2 a 2 oo 1 2 a 4 "('an IXMI'VW) 1.5 ‘2 10 h 8 C 2 0.50 1 I: 3 ‘ 0° 1 2 3 2 Figure 4.48: 4.51 / (a) 11(2) = Refer to fig 4.49. (b) H(1) = 0.014.319“) 0.028cos(%n + 1342") 4.5? V (a) H(z) = bofii—Z-i. Refer to fig 4.50. . l‘o'sr Since the pole is inside the unit circle and the _ l (b) For a = 0.5,!» = —o.6. H(z) = bofifiT‘T' Figure 4.50: filter is causal, it is also stable. Refer to fig 4.51. (c) I 1 + 0.52“1 = ,— H(‘) b“ 1 — 0.5z-1 =2- + cosw 2 b0 5 z - cosw => |H(w)|2 = 147 Figure 4.51: The maximum occurs at w = 0. Hence, I l o. o u ulna H(w)lw=0 ll ‘0 v on M H p... =>bo ll H- “Iv-a (d) Refer to fig 4.52. (e) Refer to fig 4.53. obviously, this is a highpass filter. By selectin b = —1 the f highpass filter is improved. 8 , requency response or the 3:95: (a) (1—0)2 " 1+az—2acosw 1 4ra-1--a2 2 — — — 2 => coswl 20 H H (b) 1- a 2 (l + cosw)2 + sinzw (l - acosw)2 + azsinzw (1 - a)2 2(1+ cosw) 2 1 + 02 - Zacosw 1 20 2 - — = - 2 => coswz . 02 By comparing the results‘of (a)’and (b), we find that coswz > coswl and, hence w: < w; Therefore, the second filter has a. smaller 3dB bandwidth. ' mam»? IHMI 1‘5 2 Figure 4.52: Figure 4.53: 149 2.5 3.5 3.5 ...
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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_hw3sol - (a) _ 24(1) = Ae‘“‘u(t), a>0...

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