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

1015 because as seen in sec 82 t he t wo exponentials

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Unformatted text preview: 0) k=-oo 10.2-1 As No - + 0, !1o becomes infinitesimal (!1 o - + 0). For t his reason i t will be appropriate t o replace !1o with a n infinitesimal notation L'l.!1: L'l.!1= 271" No L F(!1) = f[kJ (1O.25b) f[kJ. Therefore [F(r!1 o)!1o ] 271" J27r 271" F(!1) = F {f[k]} r=<No> In the limit as No r F(!1)e jkfl d!1 ~ where J27r indicates integration over any continuous interval of 271". T he s pectrum F(!1) is given by [Eq. (1O.22)J Using Eqs. (10.23) a nd (10.21), we c an express Eq. (10.20) as (1O.25a) 627 Aperiodic Signal Representation by Fourier Integral F(!1 + 271") = k=-oo L 00 f [kJe- j (fl+27r)k = f[kJe-jflke-j27rk = F(!1) (10.32) k=-oo Clearly, t he s pectrum F(!1) is a continuous a nd periodic function o f!1 w ith period 271". We must remember, however, t hat t o synthesize J[k], we need t o use the spectrum over a frequency interval o f only 271", s tarting a t any value of!1 [see Eq. (1O.30)J. 628 10 Fourier Analysis of Discrete-Time Signals 629 10.2 Aperiodic Signal R epresentation by Fourier Integral As a m atter o f convenience, we shall choose this interval t o be t he f undamental frequency range (-11", 11"). I t is, therefore, not necessary t o show discrete-time-signal spectra beyond t he f undamental range, although we often do so. f [kJ C onjugate S ymmetry o f F{!1) for real j [k] From Eq. (1O.31), we o btain (a) 00 F {-!1) = L j [k]e jflk 10 5 k-- k =-oo T he r ight-hand side of this equation is t he conjugate o f t he r ight-hand side of Eq. (1O.31) for real j [k]. Therefore, for real j [k], F{!1) a nd F { - !1) a re conjugates; t hat is, F {-!1) = F*{!1). Since F{!1) is generally complex, we have b oth a mplitude a nd angle (or phase) spectra F{!1) = IF{!1)le jLF (fl) (10.33) -3" - 2" 0 - It " 21t 3" Q __ Because o f c onjugate symmetry of F{!1), i t follows t hat IF{!1)1 = I F{ -!1)1 LF{!1) = - LF{ - !1) L F(Q) (1O.34a) (1O.34b) Therefore, t he a mplitude spectrum IF{!1)1 is a n even function of !1 a nd t he p hase spectrum LF{!1) is a n o dd function o f!1 for real j [k]. Linearity o f t he D TFT According t o Eq. (10.31), it follows t hat if F ig. 1 0.4 Exponential ' lurk] and its frequency spectra. and then (1O.35) E xistence o f t he D TFT Because l e-jkfll = 1, from Eq. (1O.31), i t follows t hat t he existence of F{!1) is g uaranteed if j [k] is absolutely summable; t hat is, exponentials (or sinusoids). T he Fourier s pectrum of a signal indicates t he relative amplitudes a nd p hases o f t he e xponentials (or sinusoids) required t o synthesize j [k]. I f J[k] is periodic, t hen i ts Fourier s pectrum has finite amplitudes a nd exists a t discrete frequencies (!1 0 a nd i ts multiples). Such a s pectrum is easy t o visualize, b ut t he s pectrum of a n a periodic signal is n ot e asy t o visualize because i t is continuous. T he physical meaning of t he continuous s pectrum is fully explained in Sec. 4.1-1. • E xample 1 0.3 Find the DTFT of j[k] = -yku[k]. 00 L Ij[k]1 < 0...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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