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

1068 a nd 1069 observe t hat t here are no e lements

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Unformatted text preview: ion o f I [k]. B y a ppending s ufficient n umber o f z eros t o I [k], a s i llustrated i n F ig. 1O.lOc, we c an i ncrease t he p eriod No a s m uch a s w e wish, t hereby i ncreasing t he n umber o f p oints o f t he D FT. R ecall t hat t he n umber o f p oints o f b oth I [k] a s well a s F r a re i dentical (No). W e s hall r ework e xample 10.8 u sing z ero p adding t o i ncrease t he n umber o f s amples o f D TFT. • E xample 1 0.9 0 0 E xample 10.8 by padding three zeros t o t[k]. Figure 1O.12a shows t he zero padded I [k], a nd t he corresponding INo[k]. Now I [k] is a 6-point sequence. Hence, 2 F(f2) = L I[k]e- jflk = 3 +2e-jfl+3e-j2fl = e -jfl[2+3ejfl+3e-jfl] = e - jfl (2+6 cos f2) N o=6, a nd k=O T he amplitude and phase s pectra are given by a nd from Eq. (10.69), we o btain 5 Fr 1F(f2) 1= 12 + 6 cos f21 LF(f2) = - f2 { 1 T-f2 =L f[k]e - jr( t )k = 3 + 2e - jrf + 3e - jr "1f k=O when (2 + 6 cos f2) > 0 Therefore when ( 2+6 cos f2) < 0 Figure 1O.11b shows IF(f2)1 a nd LF(f2) ( dotted). Observe t hat O FT values are exactly equal t o O TFT values a t t he sampling frequencies; there is no approximation. This is always true of O FT of a finite length f[k]. However, if I [k] is obtained by truncating or windowing a longer sequence, we shall see t hat t he O FT gives only approximate sample values of the O TFT. Fo = 3 +2+ 3 = 8 I n t he s ame way, we find • 648 10 Fourier Analysis of Discrete-Time Signals N O=32;k=0:N-l; f =[3 2 3 z eros(1,N0-3)]; F r=fft(f) r =k; s ubplot(3,1,1) s tem ( k,f); x label('k');ylabel('f[k]'); s ubplot(3,1,2) s tem(r,abs(Fr» x label( ' r') ; ylabel( ' Fr'); s Ubplot(3,1,3); s tem(r,angle(Fr» x label('r');ylabel('angle F r');grid o n; f [k) • --<i -4 -2 0 I 2 34 5 6 7 -4 k --+ -2Jt - 4. ( a) T -2n: T I q 34 :ix "3 4. "3 649 10.6 Signal processing Using D FT a nd F FT 5 (; Q --+ 2. ( b) <::) • E xample 1 0.10 F or t he s ignal f [k] i n E xample 10.8, we n eed a f requency r esolution = tr/6 for a r easonable v iew o f F (n). D etermine t he n umber o f z eros n eeded t o b e p added, a nd w rite t he e xpression for c omputing t he D FT. F or a r esolution o f = tr / 6, t he p added l ength No o f t he s ignal i s no no F ig. 1 0.12 D FT o f a s ignal w ith z ero p adding. No = 2tr = no F2 = 3 + 2 e- j :lf + 3 e-j ¥ = e ji \'< +3e-j~ = 12 11 Fr = Z=J[k]e-jr(~)k F4 = 3 + 2e-j~ + 3 e-j \" = e -ji j tr/6 T he p added l ength r equired i s 12. T herefore w e n eed t o p ad 9 z eros t o f[k]. I n t his c ase F3 = 3 + 2e-j~ + 3 e-j2". = 4 F s = 3+2e- ~ k =O = 5eji H owever, t he l ast 9 s amples o f f [k] a re zero. T herefore, t he a bove e quation r educes t o T he m agnitUde a nd a ngles o f Fr a re s hown i n t he T able b elow 2 r 0 IFrJ 8 LFr 0 11 2 5 11 3 I 1 I4 I 4 1 151 F* a s e xpected f r om t he c onjugate s ymmetry . 2 IFrJ a nd L Fr . O bserve t hat we now h ave a 6~olllt D FT, w hich provides 6 s amples o f t he D TFT s paced a t t he f requency i nterval o f tr / 3 (Ill co~trast t o 2tr/3 s pacing i n E xample 10.8). T he s ampl...
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