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"0- - r=<No> No (10.20) where 1 0.2 Aperiodic Signal representation by Fourier I ntegral I n Sec. 10.1 we succeeded in representing periodic signals as a s um of (everlasting) exponentials. In this section we e xtend this representation t o a periodic signals.
T he procedure is identical conceptually t o t hat in C hapter 4 used for continuoustime signals.
Applying a limiting process, we now show t hat aperiodic signals I[k] c an be
expressed as a continuous sum (integral) of everlasting exponentials. To represent
a n aperiodic signal I[k] such as t he one illustrated in Fig. 1O.3a by everlasting
exponential signals, let us construct a new periodic signal I No [k] formed b y r epeating
t he signal I[k] every No u nits, as shown in Fig. 10.3b. T he p eriod No is made long
enough t o avoid overlap between t he r epeating cycles (No 2: 2N + 1). T he periodic
signal I No [k] can be represented by a n e xponential Fourier series. I f we l et No --+ 0 0
, t he signal I[k] r epeats after a n infinite interval, a nd therefore
lim INo[k] = J[k] N o--too t In t his e xample we have used t he s ame e quations as t hose for O FT i n E xample C5.2, w ith a
m inor difference. I n t he p resent e xample, t he v alues o f J[k] a t k = 4 a nd - 4 a re t aken a s 1,
whereas in E xample 5.3 t hese values a re 0.5. T his is t he r eason for t he s light difference i n s pectra
i n Fig. 1O.2b a nd F ig. 5.16d. Unlike continuous-time signals, discrete-time signals c an h ave n o
d iscontinuity. ~ L No Dr = k =-oo 00 I[k]e-jrflok (10.21) T he limits for t he s um o n t he r ight-hand side of Eq. (10.21) should be from - N
t o N . B ut because I[k] = 0 for Ikl > N , i t does not m atter if t he limits are taken
from - 00 t o 0 0.
I t is i nteresting t o see how t he n ature of t he s pectrum changes as No increases.
To understand this fascinating behavior, let us define F (n), a continuous function
of n, as
00 F (n) = L I[k]e- jflk (10.22) k =-oo From this definition a nd Eq. (10.21), we have
Dr = 1 - F(rno)
No (10.23) This result shows t hat t he Fourier coefficients D r a re ( l/No) times t he samples o f
F (n) t aken every no r ad/s.t Therefore, ( l/No)F(n) is t he envelope for t he coefficients D r. We now let No --+ 0 0 by doubling No repeatedly. Doubling No halves t he
f undamental frequency no so t he s pacing between successive spectral components
(harmonics) is halved, a nd t here a re now twice as many components (samples) in
t For t he sake o f s implicity we assume D r a nd t herefore F (f!) t o b e r eal. T he a rgument, however,
is also v alid for complex D r [or F (f!)]. 626 10 Fourier Analysis of Discrete-Time Signals t he s pectrum. At the same time, by doubling No, t he envelope of t he coefficients
'Or is halved, as seen from Eq. (10.23). I f we c ontinue this process o f doubling
No repeatedly, t he n umber of components doubles in each step; t he s pectrum progressively becomes denser while its magnitude 'Or becomes smaller. Note, however,
t hat t he re...
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