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

# 22 s incx x i nspection o f e q 422 shows t hat u

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Unformatted text preview: ngle everlasting exponential e jwt w ith w = woo Therefore, t he s pectrum consists of a single component a t frequency w = Woo F rom Eq. (4.26a) i t follows t hat dw = - 2 ~ e - jwot (4.25a) (4.25b) T his result shows t hat t he spectrum of a constant signal f (t) = 1 is a n impulse 2~6(w), as i llustrated in Fig. 4.12. T he result [Eq. (4.25b)J also could have been anticipated on qualitative grounds. Recall t hat t he Fourier transform of f (t) is a s pectral representation of f (t) in terms of everlasting exponential components of the form e jwt . Now, t o represent a constant signal tTo compute bandwidth, we must consider the spectrum only for positive values of w. See discussion on p. 212. • = 2~6(w + wo) E xample 4 .6 F ind t he Fourier transforms of t he everlasting sinusoid cos wot. Recall the Euler formula (4.26b) • t The Jwt constant multiplier 2rr in the spectrum [F(w) = 2~6(w)J may be a bit puzzling. Since 1= e with w = 0, it appears that the Fourier transform of f (t) = 1 should be an impulse of strength unity rather than 2~. Recall, however, t hat in the Fourier transform f (t) is synthesized not by exponentials of amplitude F (ntl.w)tl.w, b ut of amplitude 1 /2rr times F (ntl.w)tl.w, as seen from Eq. (4.6b). Had we used variable F (hertz) instead of w, the spectrum would have been a unit impulse. 250 4 C ontinuous-Time S ignal A nalysis: T he F ourier T ransform 4 .3 r t 251 S ome P roperties o f t he F ourier T ransform u ( I) (b) (a) F ig. 4 .13 A cosine signal and its Fourier spectrum. .................. cos wot = o 1I"[6(w + wo) + 6(w - wo)] • E xample 4 .7 F ind t he F ourier transform of t he u nit step function u( t). Trying t o find t he Fourier transform of u (t) by direct integration leads t o a n indeterminate result, because 1 00 = jw u (t)e- , dt 1 00 = - 00 jw e - ' dt J E xercise E 4.2 Show t hat t he Fourier transform of the sign function sgn(t) depicted in Fig. 4.15a is 2 jjw. Hint: Note that sgn(t) shifted vertically by 1 yields 2u(t) 'V. = aim e -a'u(t) l _O 6. E xercise E 4.3 Show t hat the inverse Fourier transform of F(w) illustrated in Fig. 4.15b is f (t) = Sketch f (t). 'V. 6. E xercise E 4.4 = aim F {e-a'u(t)} = a_O _ 1_._ l lim _O a + JW W_] U(w) = lim [_ __ _ j _ _ a a _O a 2 + w 2 a2 + w 2 a _O (4.28b) o - - - - - - 1 -1 2 T he function a /(a 2 + w 2 ) has interesting properties. First, the a rea u nder this function (Fig. 4.14b) is 7 r regardless of the value of a 1 00 - 00 a -2--2 a +w dw = t an -1 w a 100 1 = 1I"6(w) + JW -:- t __ (a) (b) F ig. 4 .15 = 11" - 00 Second, when a - > 0, t his function approaches zero for all w t 0, a nd all its area (11") is concentrated a t a single point w = 0. Clearly, as a - > 0, t his function approaches an impulse of s trength 11". t T hus U(w) = sgn (t) [_a_] +.2.. +w jw a2 sine (wot). (4.28a) Expressing t he r ight-hand side in terms of its real and imaginary p arts yields = lim ~ 1I"[o(w + wo)e- j9 + o(w - wo)e j9 ]. Hint: cos (wot + 0) = ~[ej(wot+9) + e- j (w o'+9)] \7 Show t hat cos (wot + 0) a nd U(w) D erivation of t he F ourie...
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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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