This preview shows page 1. Sign up to view the full content.
Unformatted text preview: es c orresponding t o r = 0, 2,
a nd 4 I II E xample 10.9 a re i dentical t o t he s amples c orresponding t o r = 0, 1, a nd 2 i n
~x~ple 10.8. T he D FT s pectrum i n F ig. 1O.12b c ontains a ll t he t hree s amples a ppearing
I II F Ig. 1O.11b p lus 3 m ore s amples i n between. Clearly, t he z ero p adding a llows us a b etter
asses~ment o f t he D TFT. B ut e ven i n t his case, t he v alleys o f t he F (n) a re m issed b y t his
( 6polllt) D FT.
O bserve t hat F s = F1 a nd F.4 *
.
 <::) = Z=f[k]ejr(~)k C omputer E xample C 10.3
Use M ATLAB t o d o E xample 10.8 t o y ield 32 s ample v alues o f D TFT
B ecause t he s ignal l ength is 3, we n eed t o p ad 29 zeros t o t he s ignal. . • k =O I5 I i ii I 0 I  i I i I pr~perty. F Igure 1 0.12b s hows t he p lots o f • Fr E xercise E lO.4
A 3point signal I [k] is specified by 1[0] = 2, f [I] = f(1] = 1, a nd I[k] = 0 for all
other k. Show t hat t he O FT o f t his signal is Fo = 4, H = 1, a nd F2 = 1. F ind F (fl), t he
O TFT o f t his signal, and verify t hat t he O FT is e qual t o t he samples o f t he O TFT a t intervals o f
wo = 27r/No = 27r/3. \ l
tc,. tc,. E xercise E lO.S
Show t hat t he 8point O FT of t he signal f[k] in Exercise ElO.4 is Fo = 4, F I = 3.4142, F2 = 2, F3 = 0.5858, F4 = 0, Fs Observe the conjugate symmetry o f Fr a bout r = N o/2 = 4. = 0.5858, F6 = 2, F7 = 3.4142 \l P ractical C hoice o f No T he value of No is d etermined by t he desired resolution
no. However, t here is a nother consideration in selecting a value of No. I f we are
using t he F FT a lgorithm t o c ompute t he D FT, t hen for efficient computation of 650 10 Fourier Analysis o f Discrete Time Signals f [ k]
w. ¥ " IF(Q)I for signal (O.8)k u [k] ( a) 8point truncated signa1
( O.8)k u [kJ o 2 8 k~ 10.6 Signal processing Using D FT a nd F FT 651 The DFT of the windowed signal is obtained using the F FT algorithm. Figure 1O.13a shows
the windowed signal and Fig. 1O.13b shows the corresponding 8point DFT. The dotted
curve shows the D TFT amplitude IF(!1)\ of the complete (untruncated) signal (O.8)ku[k]
for comparison. The unbroken oscillating curve is the plot of the D TFT amplitude for
the truncated (8point) signal. The oscillations are because of the Gibbs phenomenon
arising from the rectangular window. Observe the interesting fact t hat t he 8 DFT values
computed from t he truncated signal are exactly equal to the 8 samples of the D TFT of
the complete signal. 8 point D Ff for signal (O.8)ku[k] F ig. 1 0.13 DFT computations for f [k] = (0.8)ku[k] using an 8point rectangular window.
D FT, No should be a power of 2. Hence, we o ften p ad sufficient numbers o f zeros
t o ensure this requirement. Effect o f Signal t runcation
So f~r we have considered only finite length sequences. For such sequences, we
c an readIly find the Nopoint D FT, where No is a t l east equal t o t he l ength of the
sequence. How do we h andle signals o f infinite length? I t is p ractically impossible t o
p rocess infinite length sequences, because they generally require a n infinite number
o f comp.utations....
View
Full
Document
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

Click to edit the document details