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 ontinuousTime 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 {ea'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 22 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 ighthand 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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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