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Unformatted text preview: 13) is satisfied. Otherwise, there is no guarantee. We have seen in Example
4.1 t hat t he F ourier transform does n ot exist for a n exponentially growing signal
(which violates this condition). Although this condition is sufficient, i t is n ot necessary for t he e xistence of t he Fourier transform of a signal. For example, t he signal
(sin a t)lt v iolates condition (4.13), b ut does have a Fourier transform. Any signal
t hat c an b e g enerated in practice satisfies t he Dirichlet conditions a nd t herefore has
a Fourier transform. Thus, t he physical existence of a signal is a sufficient condition
for t he e xistence of its transform. n Ax xI n (a) I/(t)1 dt 241 [ ;;/lJ Dn .
I I I I ..... I XI Because lejwtl = 1, from Eq. (4.8a), we o btain IF(w)1 ::; 4.1 = l xn F (x)dx Xl In t he case of discrete loading (Fig. 4.5a), t he load exists only a t t he n discrete
points. At o ther p oints there is no load. O n t he other hand, in the continuously
loaded case, t he load exists a t every point, b ut a t any specific point x , t he load
is zero. T he load over a small interval A x, however, is [F(nAx)] A x (Fig. 4.5b).
Thus, even though t he load a t a p oint x is zero, t he relative load a t t hat p oint is
F (x).
An exactly analogous situation exists in t he case of a signal spectrum. When
I (t) is periodic, t he s pectrum is discrete, a nd I (t) can be expressed as a sum of
discrete exponentials with finite amplitudes: I (t) = L Dnejnwot
For a n aperiodic signal, t he s pectrum becomes continuous; t hat is, the spectrum
exists for every value of w , b ut t he a mplitude of each component in the spectrum
is zero. T he meaningful measure here is n ot t he a mplitude of a component of some
frequency b ut t he s pectral density per unit bandwidth. From Eq. (4.6b) it is clear
t hat I (t) is synthesized by adding exponentials of t he form ejnll. wt , in which t he
c ontribution by a nyone e xponential component is zero. B ut t he contribution by
exponentials in a n infinitesimal band Aw l ocated a t w = n Aw is .l.. F (n A w ) Aw ,
a nd t he a ddition o f all these components yields I (t) in t he i ntegrall;rm: 242 4 ContinuousTime Signal Analysis: T he Fourier Transform f (t)= lim 1 00
1
'""' F (nilw)e(jnt:>.w)tilw=_
211" ~w~O 211" n =oo
L...J 1 00 F (w)eiwtdw (4.15)  00 T he c ontribution by components within a b and dw is frrF(w) dw = F (w) d F, where
d F is t he b andwidth in hertz. Clearly, F (w) is t he s pectral d ensity p er u nit b andwidth (in h ertz). I t also follows t hat even if t he a mplitude of a nyone c omponent is
zero, t he r elative amount of a component of frequency w is F (w). A lthough F(w) is
a s pectral density, in practice it is c ustomarily called the s pectrum of f (t) r ather
t han t he s pectral density of f (t). Deferring t o t his convention, we shall call F(w)
t he Fourier s pectrum (or Fourier transform) of f (t). t  a
F ig. 4 .6 4.1 Aperiodic Signal R epresentation by Fourier Integral 243 in t he n otation introduced in C hapter 2. T hus, t o b e consistent with t he previous
notatio...
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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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