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HW4solutionbookproblems

HW4solutionbookproblems - 2125(Q We may write a[n...

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Unformatted text preview: 2125 (Q) We may write a:[n] as .~1M*[email protected]" Zoo (1/3) "( ([email protected]+: Z(1/3)k( (-1/4)" k u[n—k+3] : LE ‘0 r 26% J. h '9 a 13—3 r391) “W33 + 5 fl Q/WBJ H a ,NMMA I? >‘L! @Q g i / By conmder each summation in the above equafion separately, we may show that (124/11)3n, 7 n g 4 Mn] ={ ’C (1/4)"(1/11) + —3(1/4)n + 3(g§§)(1/3)n, n > —3 fl (b) Now consider the convolution ylln] = [(1/3)”u[nll * [(1/4)"U[n + 31]- We may show that _ 0, n < —3 91M ‘ { —3(1/4)n + 3(256)(1/3)"Aa " Z *3 ' k—v 8\ Also, consider the convolution 3121”] = [(3)"u[-n ~ 1]] * [(1/4)"u{n + 3]] - We may show that . (124/11)3", n g —4 312M: { (1/4)n(1/11), n 2 ~ -"1'l Clearly, y1[n] + yzln] = y[n] obtained in the previous part. l h -’ k ‘ (if) Z (5%) {zormizo :0 ( Q {fill fir n 3'3 30 Mul‘lrplg L26 ' l " 3 _ atelSQMl’the HEM-fl , 1 “’ ‘73 first Sag)? ‘twm H? 79”“ g g? ‘ .5 U it’ll “(3&4 20h gubvwwa‘lbq °§"%W.€?‘3¢ “Kg 21 , E3 k nuk 72W}: 7. “Buflfl {*1} flimflgj ~0 Rm”; '3? 353“” Ass-lat) n—\<+93"O Egg—4 szfi . ’ Izrl‘ 4% +0-44 fining h‘fl'l‘rfi L9 «3‘ 52) Ifn>kut )4 L9 2%)? \2 = w)" :m 2 P)“ 7, \2 ll‘v‘gfi Q‘m'ri’ ’ H F54» fl ‘ ‘L_§7- M h ‘3 ,l t? S? _L :: —(’¢:)n§l-57:)—r *‘lfzmlilizi m .an '7’) i [w H i J n y - ..(« \ l?“ n I; l; (L?) [”3ng H ‘1) a l 3.21. Using the Fourier series synthesis eq. (3.38), $(t) = aleflZW/Tfl + a_le—j(21r/T)t + a58j5(21r/T)t + a_56—j5(27r/T)t jej(21r/8)t _ je—j(27r/8)t + 2ej5(27r/8)t + 28~j5(27r/8)t ' = —2sin(§t) + 4cos(§;17—rt) —-2 cos(Z—t — 7r/2) + 4cos(§47—rt). 3.25. (a) The nonzero FS coefficients of :I:(t) are (11 = a_1 = 1/2. We : 27]: = 2]: : Li‘i'r‘ (b) The nonzero FS coefficients of :c(t) are I); = 12:1 = 1129') V2 - flirt “J“Wt first, .34 rrt 5”) If“ ‘ 63 W E (’03 Lifi‘ix‘ € *9 2 if z i ' e {d s eds g ‘ :i‘ rim 3 947%) y , 4-, v ‘ (c) ,Using the multiplication property, we know that if Jé?%mai\% _ €Q§tfi0§¥€ - oo zm = mun/(t) +13 at = Z asz @ (38, >15 >13” lit—~00 CV :0; ~ ho oug¢[email protected] (a ;\ , gang Therefore, 1 @Ck : 2 , 56‘ air) 1 ck=ak=kbk=~—.6[k—2]——.6[k+2]. Cwaimfifif 4] 43 ‘ é; eggs; .Jge‘? $4 This implies that the nonzero Fourier series coefficients of 2(t) are (:2 : c_2 ‘2 (1/4j). (d) We have \ z(t) = sini‘ifl?os(‘lfi)= ésinfi HT) Therefore, the nonzero Fourier series coefficients of z(t) are 62 : c-2 : (1 /4j ). 3.27. Using the Fourier series synthesis eq. (3‘14) $[n] = a0+a2ej2(27r/N)n+a_2e—j2(27r/N)n+a4ej4(27r/N)n+a_4e—j4(27r/N)n : 2+26j7r/6ej(47r/5)n + 2e—j7r/66—j(41r/5)n + ej‘1r/3ej(87r/5)n +e—j7r/3e‘j(87r/5)n = 2 + 4cos[(47rn/5) + 7r/6] + 2cos 87m/5) + 7r 3 = 2 + 4sin[(41rn/5) + 27r/3] {gifl ' n - mfg, loinr iieesrevrlliihdgge \ a-” Om -= :3 v 2cos(?-gh _fi‘/3) \\\ . 7- 2 3m Q Qgh’ffifb) 3.37. The frequency response of the system may be easily shown to be ‘49an ) i I w _ 1 ____1_ H =2): :1 ‘ ”:7 we] ) ‘ 1—%e—J‘w 1~2e‘j“" ( E. 22 \ “E (a) The Fourier series coefficients of x[n] are ‘ .. 1 f 11 [C qk-‘RZ XV] Q \ J 781nk Z \ ak — Z’ or a . F: {N5 ‘ Lt We Also, N z: 4. Therefore, the Fourier series coefficients of y[n] are ,1" ""'“‘”§»~\,,M :00 2-2-5: a? r 2 1‘: st . ~ “we ‘ - ‘ {‘4 ,2. 23:3 \\ , . 1 1 1 » »b '3; Es _ 2k1r/N _ _ _ / 4w «file» bk _ akme; ) 4 1— Ame/2 1 — 2e—j7rk/2] ' W V (I ' ‘ a ’ ' , 2 3%: 2’2 q ~ i Z“ .2 1;: .. (b) In this case, the Fourier series coefficients of w[n] are k‘ (a if e "”" f :«s i ak = l[1+ 2COS(k7T/3)]a for allk. W} 6 43%” wwwwe.MWWWMWWWWW r Also, N = 6. Therefore, the Fourier series coefficients of y[n] are ‘ j2k7r/N 1' 1 1 bk = akH(e )= E5-[1 + 2cos(k7r/3)] —-—————1 _ %e—j7rk/3 — —————1 _ 2e_j1rk/3 . E, —\< I} ~l< 3.38. The frequency response of the system may be evaluated as “(2} 1 Z .. 2, SE . . i . ~ = t; 3 " .2 HM”) = —623“ — e“ + 1 + e‘J“ + (2‘29“. F ° "i For a:[n], N = 4 and we = 1r/2. The PS coefficients of the input m[n] are (1;; = 21;" for all n. (0‘5 Silbu’m chew) Therefore, the FS coefficients of the output are NQTKCQ ’) 2i “”2 ' ”J W" blc = awake) «.— 41[1 — ej’m/2 +e—W/2]. ~E’; ” + e - ~ <— a) A —\ Io ...
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