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Unformatted text preview: rithm. k=O Vm = no 4>[r] = L 4>[r] (10.11) r=<No> N oI L k =O f [k]ejmrlok (10.7) tIr we let flk] = N oh a nd Dr = Fr , Eqs. (1O.8) a nd (1O.9) a re identical t o Eqs. (5.18b) a nd
(5.18a), respectively. 10 F ourier Analysis o f D iscrete Time Signals 620 10.1 Periodic Signal R epresentation by DiscreteTime Fourier Series ( DTFS) 621
I [k] where r = <No> i ndicates s ummation over a ny No c onsecutive values of r . T his
follows because t he r ighthand side of Eq. (10.11) is t he s um o f all t he No c onsecutive
values of .plr]. B ecause .plr] is periodic, this s um m ust b e t he s ame r egardless of
where we s tart t he first term. Now e jrflok is Noperiodic because Therefore, if I lk] is Noperiodic, I lk]ejrflok is also Noperiodic. Hence, from Eq.
(10.9) i t follows t hat 'Or is also Noperiodic, as is v rejrflok. Now, because o f p roperty (10.11), we c an express Eqs. (10.8) a nd (10.9) as
I lk] = r 0.5 l: v rejrflok l: I lk]ejrflok (10.12) ( b) r =<No> a nd
1
V r = (10.13)  2lt I f we p lot 'Or for all values of r ( rather t han o nly 0 :::; r :::; No  1), t hen
t he s pectrum 'Or is Noperiodic. Moreover, Eq. (10.12) shows t hat I lk] c an b e
synthesized by n ot only t he No e xponentials corresponding t o 0 :::; r :::; No  1,
b ut b y any successive No e xponentials in t his s pectrum, s tarting a t a ny value of r
(positive o r n egative). For this reason, i t is c ustomary t o s how t he s pectrum 'Or for
all values of r ( not j ust over t he i nterval 0 :::; r :::; No  1). Y et we m ust r emember
that to s ynthesize I lk] [ rom t his spectrum, we n eed to a dd o nly No consecutive components.
T he s pectral c omponents 'Or a re s eparated b y t he f requency n o = ; ;., a nd t here
a re a t otal o f No c omponents r epeating p eriodically along t he n axis. T hus, o n t he
frequency scale n , 'Or r epeats every 211" intervals. E quations (10.12) a nd (10.13)
show t hat b oth I lk] a nd i ts s pectrum 'Or a re periodic a nd b oth h ave e xactly t he
s ame number of components (No) over one period. T he p eriod o f I lk] is No a nd
t hat o f 'Or is 211" r adians.
E quation (10.13) shows t hat 'Or is complex in general, a nd V r is t he c onjugate
of 'Or if I lk] is real. T hus  20 ~ 2 lt ~ 10 No k=<N >
o Q 10 30 20 10 0 LVr ~ 2 ~ ~ 10 2lt 10 (c) ~ 2 2 lt Q 1 F ig. 1 0.1 Discretetime sinusoid sin 0.11I"k and its Fourier spectra. r =<20> where the sum is performed over any 20 consecutive values of T . We shall select the range
 10:5 r < 10 (values of r from  10 to 9). This choice corresponds to synthesizing I[k]
using the spectral components in the fundamental frequency range ( 11" :5 fl < 11"). Thus,
9 I[k] = so t hat t he a mplitude s pectrum IVrl is a n even f unction, a nd L Vr is a n o dd f unction
of T (or n ). All these concepts will b e clarified by t he e xamples t o follow. l : VrejO.brk
r =lO where, according to Eq. (10.13),
E xample 10.1
Find the discretetime Fourier series (DTFS) for I[k] = sin 0.11I"k (Fig. 1O.1a). Sketch
the a...
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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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