Signal Processing and Linear Systems-B.P.Lathi copy

# 1 e quation lo31 involves summing a n infinite n

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Unformatted text preview: e r eader may confirm these values from from Eq. (10.44). Fr = L f[k]e - jr( 1 f)k = 3 + 2e - j r 1f + 3e - jr¥- k=O T herefore Fo = 3 + 2 + 3 = 8 "h "4. ' .0\ F1 =3+2e _1 3 + 3e _1 3=3+ ( - 1-Jv3)+ F2 = 3 + 2 e _ "!1!. J3 "8. + 3 e _1 3 .0\ = 3 + ( - 1 + J"v 3 ) + (3 .3v'3) = (1 "2v'3) (32 - "3v'3) (12 - .v'3) 2 - 2+ J -2- - J - 2- 2+ J = J = ej1!. 3 = e _ j1!. 3 646 10 F ourier A nalysis o f D iscrete- Time S ignals 6 47 T he O FT i n this example has too few points t o give a reasonable p icture of t he O TFT. T he p eak of O TFT appearing between t he second a nd t he t hird sample (between r = 1 a nd 2), for instance, is missed by t he O FT. T he two valleys of t he O TFT are also missed. We definitely need more points in t he O FT for a n acceptable resolution. This goal is accomplished by z ero p adding, explained below. • f [kl • • 1 0.6 S ignal p rocessing U sing D FT a nd F FT • Use o f Zero Padding -<; --4 -2 -3 -2 k -+ -2>< - 4. - 2. i -1 0 1234567 3 3 1 q 2>< T ( a) 4. T T he D FT y ields t he s amples o f D TFT a t t he f requency i ntervals o f n o = { l-+ 2. ( b) L F(Q) F ig. 1 0.11 C omputation of O FT of a signal I[k]. T he magnitudes and angles of Fr a re shown in t he Table below. r 8 LFr IT 21 1I 1 I 0 IFrl 0 t l-fl Observe t hat F2 = F { ( the conjugate symmetry property). Because I[k] is a 3-point sequence, Fr (its O FT) is also a 3-point sequence, which repeats periodically. Figure 1O.11b shows Fr a nd L Fr . Recall t hat t he O FT gives only t he sample values of F(f2). We want t o know if O FT has enough samples t o give a reasonably good idea of O TFT. T he O TFT for I[k] is given by [Eq. (10.31)] 21T/No, w here n o is t he f requency resolution. S eeing t he D TFT t hrough D FT is like viewing F (n) t hrough a p icket fence. O nly t he s pectral c omponents a t t he s ampled f requencies (which a re i ntegral m ultiples o f n o) will b e v isible. B ut f requency c omponents l ying i n b etween will b e h idden b ehind t he p icket fence. I f D FT h as t oo few p oints, m ajor p eaks a nd v alleys o f F (n) e xisting b etween t he D FT p oints ( sampled f requencies) will n ot b e s een, t hus g iving a n e rroneous v iew o f t he s pectrum F (n). T his is precisely t he c ase in E xample 10.8. A ctually, u sing t he i nterpolation f ormula, i t is possible t o c ompute a ny n umber o f v alues o f D TFT f rom t he D FT. B ut h aving t o u se t he i nterpolation f ormula r eally d efeats t he p urpose o f D FT. W e t herefore s eek t o r educe n o s o t hat t he n umber o f s amples is i ncreased for a b etter v iew o f t he D TFT. B ecause n o = 21T/No, we c an r educe n o b y i ncreasing No, t he l ength o f I [k]. F or a f inite l ength s equence, t he o nly w ay t o i ncrease No is b y a ppending s ufficient n umber o f z ero v alued p oints t o J[k]. T his p rocedure o f z ero p adding is d epicted i n F ig. 1O.lOc. R ecall t hat No is t he p eriod o f INo[kJ, w hich is f ormed b y p eriodic r epetit...
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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.

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