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

Yet t he fourier transform accomplishes i t routinely

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Unformatted text preview: ered a t t he origin, as shown in Fig. 4.Sa t .(x) = F or t he f requency-domain case the system response to e jwt is H (w)e jwt f it) = - 1 211" Joo F(w )e iwt dw 1 foo F(w)H(w)e . shows J(t) as a sum of everlasting exponential components Jw t dw yet) is a sum of responses to exponential components - 00 {~_ 21xl Ixl 2: ~ Ixl < ~ (4.21) T he p ulse in Fig. 4.Sb is t.(~). Observe t hat here, as for t he g ate pulse, t he d enominator T o f t he a rgument o f t .( ~) i ndicates t he pulse width. ;k(X) - 00 a nd y(t) = 211" (4.20) T a nd 2 ( b) ~2" 2 x_ <aJ T he f requency-domain view sees a system in terms of its frequency response (system r esponse t o various sinusoidal components). I t views a signal as a s um of various sinusoidal components. Transmission of a signal t hrough a (linear) system is viewed as transmission of various sinusoidal components of t he signal t hrough t he s ystem. F ig. 4.8 A triangle pulse. t At x = 0, we require rect (x) = 0.5, because the inverse Fourier transform of a discontinuous signal converges to the mean of its two values at the discontinuity. 246 4 Continuous-Time S ignal Analysis: T he F ourier Transform 4.2 Transforms o f S ome Useful Functions a rgument , w= 'If. 6 = 11" w hen w = 'If. 247 T herefore, t he first zero o f t his function occurs a t E xercise E 4.1 Sketch: (a) rect(~) ( b) 6 (TIi) (c) s ince;W) ( d) sine (t)reet (-t,,). 'V • ' -"'x-+- E xample 4 .2 Find the Fourier transform of J (t) = rect Since rect(~) = 1 for (Fig. 4.10a). It I <~, and since it is zero for It I >~, F(w) = sine U·) (a) j T/2 e -jw'dt - T/2 e,f) (b) 14lt "3 Therefore F ig. 4 .9 A sine pulse. rect I nterpolation Function sine (x) T he f unction s in x /x is t he "sine over a rgument" f unction d enoted b y s inc (x). t T his f unction p lays a n i mportant role i n s ignal processing. I t is also known as t he f iltering o r i nterpolating f unction. W e define sin x (4.22) s inc(x) = - x I nspection o f E q. (4.22) shows t hat U:) = rsinc (~r) ( 4.23) Recall t hat s inc(x) = 0 when x = ±n11". Hence, s inc(T) = 0 when T = ±n11"; t hat is, when w = ± 2~", (n = 1 ,2,3, . .. ), as depicted in Fig. 4.10b. The Fourier transform F(w) shown in Fig. 4.10b exhibits positive and negative values. A negative amplitude can be considered as a positive amplitude with a phase of -11" or 11". We use this observation to plot the amplitude spectrum IF(w)1 = I sinc(T)1 (Fig. 4.10c) and the phase spectrum LF(w) (Fig. 4.lOd). The phase spectrum, which is required to be an odd function of w, may be drawn in several other ways because a negative sign can be accounted for by a phase of ±n11", where n is any odd integer. All such representations are equivalent. • 1. s inc (x) i s a n even function o f x. 2. sinc (x) = 0 w hen sin x = 0 e xcept a t x = 0, w here i t a ppears i ndeterminate. T his m eans t hat s inc x = 0 for x = ±11", ±211", ±311", . .. . ..--':'j-_ f (t) (aJ (bJ 3. Usin...
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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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