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

# Each of the following equations specifies an ltid

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Unformatted text preview: rithm. k=O Vm = no 4&gt;[r] = L 4&gt;[r] (10.11) r=&lt;No&gt; N o-I L k =O f [k]e-jmrlok (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 Discrete-Time Fourier Series ( DTFS) 621 I [k] where r = &lt;No&gt; i ndicates s ummation over a ny No c onsecutive values of r . T his follows because t he r ight-hand 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 No-periodic because Therefore, if I lk] is No-periodic, I lk]e-jrflok is also No-periodic. Hence, from Eq. (10.9) i t follows t hat 'Or is also No-periodic, 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]e-jrflok (10.12) ( b) r =&lt;No&gt; 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 No-periodic. 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&quot; 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&quot; 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=&lt;N &gt; o Q- -10 30 20 10 0 LVr ~ 2 -~ ~ 10 -2lt 10 (c) -~ 2 2 lt Q- 1 F ig. 1 0.1 Discrete-time sinusoid sin 0.11I&quot;k and its Fourier spectra. r =&lt;20&gt; where the sum is performed over any 20 consecutive values of T . We shall select the range - 10:5 r &lt; 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&quot; :5 fl &lt; 11&quot;). 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 discrete-time Fourier series (DTFS) for I[k] = sin 0.11I&quot;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.

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