Chapter7_8_practice_problems_v2

Chapter7_8_practice_problems_v2 - T15. tail Figure STE-E...

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Unformatted text preview: T15. tail Figure STE-E Ehfl-Wfi the eulput signal :-i|11:r.1;rll.m Hue]. {L1} Uheerve that F'wa is the same :15 Mfu} with the ErL-queney epeetrum inverted. that i5. 11H: high Ithuullties are altil'ted tu leerer frequencies and vine veree. Time, the scrambler in Figure F?.T’—51'twe1't5 the irEqutrncy EPEELF'IJm- Te get hack the WEE-"141] 3?!“th MW}. 'I'l'fl HEM LE- itl‘I'EETL the spectrum FIT-er} lance. again. This can lee dene he passing the scrambled signal ylii] through the same eeramhler. 13—11. {e} m 13:“ Kim} = [mrflijfi-mld; and : film ERIE—jth! Changing the erder e!" tiHTerentintien end integratien yields g 2.1;” d-I—ilzr{ijfl"j”l} = f:[-jt55[t}]e_j“* dt T herel'ere F -jtm[t] i=e % 1 —ut e “[t] 4:: jm+fl d 1 ‘J' —n£ "F = HJEE flit] 1:? dte<jw+ej fiw+uj3 l te'“u{t}e=lr _ 3&1. The signal elj} = ElnEII‘zflli'Il'fl is sampled by e mt-tenguler pulse seque:1ee 11TH} wheee period is ri. the en that the fundamental frequency (which is also the sampling Frequency} is 25L! He. Henee1 w, = EIJEhr. The Fourier series fer will is given by Use ei' Eqs. {3.6m yields 09 = {In C" = 3?; sin [%], tiiet is1 Cu = 11.2, C1 = 0.3511 C: = [lit—131 Cg = BEBE. C" = 13.093, 05 = I]1 - r r Consequently fit} = J[i}py[t} = [12 m{ti+fl.3?4e{li eesfiflihrHEIfliiii fit} (:05 lUflElnHfllfl! mfit} ees lEflU‘lTl‘l“ end Elfin] = U.2X[w} 4- EAST [me - fiflfln] + Kite + fidihri] +0451 [Xee — lflflfl‘lT] + xiii: + lflflflwjl +e.1e1 [XIIw —1semr}+ Xe; + lfiflflwii + - -r In the present case Xij = [Li'lEI-E- reethfi'fifl]. The Spectrum EM} is shown in Figure SEE-1. Clbeerve that the spectrum consists of Kim} repeating periodically at the interval of seen redfs {250 He}. Heneer there is no overlap between cycles. and Kim] een be reeevered by using, en ideal lewpsse fiher ef bendwidth itifl H's. An ideal Inwpess filter nl' unit gain {and bandwidth Hi0 I12} will ellew the Iirst term on the 11!: right-side of the above equation to pass fully and 5111'!me all the other terms. Henee the eutput nit] is yit} = 0.23:8} Because the spectrum The] has a zero value in the band from lfll] te 150 He. we can use an ideal lee-1mm filter of bandwidth B H: where lflfl -:: B < 154}. But if B > 150 Hz, the filter will pick up the unwanted spectral enmpenents item the next cycle. end the output will be distorted. Figure 53.2-1 es-s. j, = we: Uni = “mill 5 Liz = m in] I = 8 He La- less that Li's : 1e [[1 Henna, tiiifi lreqneiiql.I is net ellesed end |fn| = I = 3 Hi. (In Jr n :2 Hz Ifnl = |12 - flfli = 3 Hz Eel I =‘2U H3 lisl = nu — 2e} = e H: [n] I =22 H2. Iiil = we — 2e] = 2 Hz ie] I = 32 Hit |h| = |se —4eJ = s H: 3.143. a «=1- Fulfill: 19')” The spectrum Lima” shows that. most. of Lhu aigna'l energy is mnmtrnterl within the hand of ! Hz. it can In.- aha-m1 that EIIEIES'Ffi £1313:ng i3 cuntained within the band mi 1 Hz. II we use 90% energy trim-rim. {ur bandwidth. sampling mm: of 2 H3 is fidflquata. However, [or u. htttu: uppnzm'maLiun. {higher energy bandwidth trim-inn], we may go m I, = 4 Hz. Theoretically. of amuse. f. = an. Hi. Sgt-é: 7.2-4. (a) “all Infill = % I: e'w'flal'wdw=2_:_: 1:0 awn-mm; I — _ l 'wfl— 1;“ =fiinwfltt‘tfl]=fl_ t_tn I ‘ {mm—=an I" w,“ —_-m[£—Lu} “3'”th 1'1 rm ...
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This note was uploaded on 02/27/2008 for the course BME 513 taught by Professor Yen during the Fall '07 term at USC.

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Chapter7_8_practice_problems_v2 - T15. tail Figure STE-E...

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