Signal Processing and Linear Systems-B.P.Lathi copy

F or t his r eason we s hall b ypass t he t

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Unformatted text preview: mplitude and phase spectra. = is a rational number In this case the sinusoid sin O.111"k is periodic because and the period No is [see Eq. (8.9b)] • * to to l : sin 9 'Or = = m (211") = m (.111" = 20m ~) 0 fl T he smallest value of m t hat makes 20m an integer is m = 1. Therefore, the period No = 20, so t hat flo = i,: = 0.111", and from Eq. (10.12) 0.111"ke-jO.brk k=-lO 9 ~ 1 No 40 r- (10.14) a nd 4lt 10 1( = 20L..2'e k=-l0 J = 4~j [ t k=-l0 jO.l1rk -e ejO .1>rk(l-r - jo.17rk) l- t k=-l0 - jO.17rrk e e - iO .1>rk(1+ r l] 10 F ourier A nalysis o f D iscrete- Time S ignals 6 22 I n these sums, r t akes on all values between - 10 a nd 9. F rom Eq. (5.43) i t follows t hat t he first s um o n t he r ight-hand side is zero for all values o f r e xcept r = I , w hen t he s um is e qual t o No = 20. Similarly, t he second sum is zero for all values o f r e xcept r = - I, when i t is equal t o No = 20. T herefore 1 0.1 P eriodic S ignal R epresentation b y D iscrete-Time F ourier S eries ( DTFS) 6 23 E xercise E I0.1 From the spectrum in Fig. 10.1 write the Fourier series corresponding t o t he interval - 10 ?: r > - 30 (or -7r ?: ! l > -37r). Show t hat this Fourier is equivalent to t hat in E q. (10.15). \ l £; £; a nd E xercise E I0.2 Find the period and the DTFS for Ilk] = 4 cos 0.27rk + 6 sin 0.57rk and all o ther coefficients a re zero. T he c orresponding Fourier series is given by f lk] = sin 0.l1rk = -h (e jO. bk _ e- jO. bk ) (10.15) over the interval 0 Answer: ~ r ::; 19. Use Eq. (10.9) to compute D r. Here t he f undamental frequency 0 0 = O.11r, a nd t here a re o nly two nonzero components: - 1_ 1 D 1 - 2]- 2 e - j,,/2 f [kJ Therefore (a) L Dl = -~ a nd L D_l =~ - 28 - 32 -36 F igures a nd c shows t he s ketch o f D r for t he i nterval ( -10 ::; r < 10). According t o E q. (10.15a), t here a re o nly two components corresponding t o r = I a nd - 1. T he r emaining 18 coefficients are zero. T he r th c omponent D r is t he a mplitude o f t he frequency rOo = O .lnr. T herefore, t he frequency interval corresponding t o - 10 ::; r < 10 is -11" ::; 0 < 11", a s depicted in Figs. a nd c. T his s pectrum i n t he interval - 10 ::; r < 10 (or -11" ::; 0 < 11") is sufficient t o specify t he f requency-domain description (Fourier series), a nd we c an s ynthesize f lk] b y a dding t hese spectral components. Because o f t he p eriodicity p roperty discussed in Sec. 10.1-1, t he s pectrum D r is a p eriodic function o f r w ith p eriod No = 20. F or t his r eason, we r epeat t he s pectrum w ith p eriod No = 20 ( or 0 = 211"), a s illustrated in Figs. a nd c, which a re p eriodic extensions o f t he s pectrum i n t he r ange - 10 ::; r < 10. Observe t hat t he a mplitude s pectrum is a n even function a nd t he angle o r p hase s pectrum is a n o dd f unction o f r ( or 0 ) a s expected. T he r esult (10.15) is a t rigonometric identity, a nd c ould have been o btained im...
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