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Unformatted text preview: he components in this band of l lF (in hertz)
2
IS 1F (w ) 1 llF. T he total signal energy is t he sum of energies of all such bands and
is indicated by t he a rea under IF(w Was in Eq. (4.63). Therefore, IF(w)1 2 is t he
e nergy s pectral d ensity (per unit bandwidth in hertz).
For real signals, F(w) a nd F (  w) are conjugates, and IF(w)1 2 is a n even function of w because
IF(w)12 = F(w)F*(w) = F (w)F(w) w2 t ::.Ej=IF(w)1 2 dw
11" w, 00  00 (4.65) Therefore, Eq. (4.63) can be expressed ast dt Here we used t he fact t hat f *(t), being t he conjugate of f (t), can be expressed as
t he c onjugate of t he r ighthand side of Eq. (4.8b). Now, interchanging t he o rder of
integration yields 1
= 211" l lw 2 ; = l lF Hz (4.63) (4.67) t I? Eq: (4.66) i t is ?Ssumed t hat F(w) does not c ontain a n impulse a t w = O. I f s uch a n impulse
eXISts, It s hould b e m tegrated s eparately with a mUltiplying factor of 1/211" r ather t han l/rr. 4 2 76 Continuous Time S ignal Analysis: T he F ourier T ransform • E xample 4 .16
F ind the energy of signal jet) = ea'u(t). Determine the frequency W ( rad/s) so
t hat t he energy contributed by t he spectral components of all the frequencies below W is
95% of the signal energy E j .
We have 4.7 Application t o C ommunications: A mplitude M odulation Suppression of all t he s pectral c omponents o f j et) b eyond t he e ssential bandwidth r esults in a signal j et), which is a close a pproximation o f j et). I f we use t he
95% criterion for t he e ssential b andwidth, t he e nergy of t he e rror ( the difference)
j et)  jet) is 5% o f E f . Energy Spectral Density From Autocorrelation Function
C orrelation of a function j et) w ith i tself is its a utocorrelation f unction
,pf(t), which, for a real j et), is given by [see Eq. (3.32)] We c an verify t his result by Parseval's theorem. For this signal
1
F (w)=. ,pJ(t) = Jw+a a nd 1 = r1r 0 1 00 00 Ej ! F(w) I d w
2 1
= ;: 0 1
1
+ a2 dw = ~ t an 1 W ;; 1 0 = 21a T he band w = 0 to w = W contains 95% of the signa! energy, t hat is, 0 .95/2a.
Therefore, from E q. (4.67) with W 1 = 0 and W 2 = W , we obtain
0.95 _ 2a  .rr!.lW ~ = 2ra tan1 ~IW = 2ra tan1 W
..
..
+
r
r
0 w2 a2 a a 0 or
0.95rr = t an1 W = =} W = 12.706a r ad/s
2
a
This result indicates t hat the spectral components of jet) in the band from 0 (dc) to
1 2.706a r ad/s ( 2.02a Hz) contribute 95% o f t he t otal signa! energy; all t he remaining
spectral components (in the band from 1 2.706a r ad/s to 00) contribute only 5% o f the
signal energy.
£::, • E xercise E 4.13
Use Parseval's theorem to show that the energy of the signal
2a J(t) = t2 + a 2
is ~. Hint: Find F (w) using pair 3 and the symmetry property. 1: j (x)j(x  t) dx (4.68a) Also, from Eq. (3.31) w ith get) = j et), i t follows t hat 00 w2 277 'V T he Essential B andwidth o f a Signal
S pectra o f m ost o f t he s ignals e xtend t o infinity. However, b ecause t he e nergy
o f a ny p ractical s ignal is finite, t he s ignal s pectrum m ust a pproach 0 a s w ...
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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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