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Unformatted text preview: nd :Fo 1 = To = 78.125 Hz T hus, increasing N o from 200 t o 256 can improve t he frequency resolution from 100 t o
78.125 Hz while m aintaining t he s ame aliasing error ( T = 50l's). ( iii) C ombination o f t hese t wo o ptions
We could choose T = 451's a nd To = 11.5 ms so t hat :Fo = 86.96 Hz. • E xample 5 .6
Use D FT t o c ompute t he Fourier transform of e  2t u(t). P lot t he r esulting Fourier
s pectra.
We first d etermine T a nd To. T he Fourier transform of e  2t u(t) is l /(jw + 2). T his
lowpass signal is n ot b andlimited. In Sec. 4.6, we used t he energy criterion t o c ompute t he
e ssential b andwidth o f a signal. Here, we s hall present a simpler, b ut workable alternative
t o t he e nergy criterion. T he e ssential b andwidth o f a signal will b e t aken a s t hat frequency
where IF(w)1 r educes t o 1% of its p eak value (see footnote on p. 276). In this case, t he
p eak value occurs a t w = 0, w here IF(O)I = 0.5. Observe t hat • I F(w)l= ~""..!:.
yw 2 + 4
w 5.2 N umerical C omputation o f t he F ourier Transform: T he D FT 3 45 those over t he r ange r = 129 to 255. Thus, F127 = F 29, F126 = F'30, . .. F _l = F255.
'
I n a ddition, because o f t he p roperty of conjugate s ymmetry o f t he Fourier transform,
F r = F ;, i t follows t hat F l = F i, F 2 = F 2, ... , F  128 = Fi28' T hus, we need Fr only
over t he range r = 0 t o N o/2 (128 in this case).
Figure 5.15 shows t he p lots of IFrl a nd LFr a nd t he e xact m agnitude a nd p hase
s pectra ( depicted by continuous curves) for comparison. Note t he n early perfect agreement
between t he two sets of spectra. We have depicted t he p lot of only t he first 28 p oints r ather
t han all t he 128 p oint b ecause i t would have m ade t he figure very crowded, resulting in
loss o f clarity. T he p oints are a t t he intervals of l /To = 1 /4 Hz o r wo = 1.5708 r ad/s. T he
28 samples, therefore, exhibit t he p lots over t he r ange w = 0 t o w = 28(1.5708) "" 44 r ad/s
o r 7 Hz.
I n t his example, we knew F (w) b eforehand a nd hence could make intelligent choices
for B (or t he s ampling frequency :Fs ). I n p ractice, we g enerally do n ot know F (w) beforehand. In fact, t hat is t he very thing we a re trying t o d etermine. In such a case, we
m ust make an intelligent guess for B o r :Fs from circumstantial evidence. We should then
continue reducing t he value of T a nd r ecomputing t he t ransform until t he r esult stabilizes
w ithin t he desired number of significant digits. T he MATLAB program, which implements
t he D FT using t he F FT a lgorithm, is p resented in Example C5.1. •
0.5
0.4 w »2
0.3 Also, 1% of t he p eak value is 0.01 x 0.5 = 0.005. Hence, t he essential b andwidth B is a t
w = 2trB, where
1
IF(w)1 "" 2trB = 0.005 ~ B = 1t00 Hz
r 0.2
0.1 a nd from Eq. (5.25b), T<~
 2B = ~ = 0.015708
200 H ad we u sed 1% energy criterion to determine the essential bandwidth, following t he
p rocedure in E xample 4.16, we would have obtained B = 20.26 Hz, which is s omewhat
small~r t ha...
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 Spring '13
 Bayliss
 Signal Processing, The Land

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