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

S uch a plot called t he f ourier s pectrum o f fk

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Unformatted text preview: mediately w ithout t he f ormality o f finding t he Fourier coefficients. We have intentionally chosen this trivial example t o i ntroduce t he r eader g ently t o t he n ew concept of t he d iscrete-time Fourier series a nd i ts periodic n ature. T he F ourier series is a way o f e xpressing a periodic signal f lk] i n t erms of exponentials o f t he form e jrnok a nd i ts harmonics. T he r esult in Eq. (10.15) is merely a s tatement o f t he (obvious) fact t hat s in 0.l1rk c an b e e xpressed as a s um o f t wo exponentials e jO . bk a nd e - jO . hk . B ecause o f t he p eriodicity p roperty of t he d iscrete-time exponentials e jrnok, t he Fourier series components c an b e s elected in a ny r ange o f l ength No = 20 ( or 0 = 211&quot;). For example, if we select t he frequency range 0 ::; 0 &lt; 211&quot; ( or 0 ::; r &lt; 20), we o btain t he Fourier series as o -4 - 16 4 28 16 32 36 k-- (b) It -It Fig. 1 0.2 n -- P eriodic s ampled g ate p ulse a nd i ts Fourier s pectrum. • E xample 1 0.2 F ind t he d iscrete-time Fourier series for t he p eriodic sampled g ate f unction shown i n F ig.1O.2a. In t his c ase No = 32 a nd 0 0 = ~ = f6. T herefore f lk] = L D rejrfok (10.17) r =&lt;32&gt; w here Dr 1 = 32 L f lk]e- jrfok (10.18) k =&lt;32&gt; f l k] = sin 0.l1rk = ~ (e jO. 1&quot;k _ e- j 1. 9d ) 2j (10.16) T his series is equivalent t o t hat in Eq. (10.15) because, as seen in Sec. 8.2, t he t wo exponentials e j 1. 9&quot;k a nd e - jO . hk a re identical. We could have selected t he s pectrum over a ny o ther r ange o f w idth 0 = 211&quot; i n Figs. 1O.lb a nd c as a valid discrete-time Fourier series. T he r eader m ay s atisfy himself by proving t hat s uch a s pectrum s tarting a nywhere ( and o f w idth 0 = 211&quot;) is e quivalent t o t he same two components on t he r ight-hand side o f E q. (10.15). • F or o ur convenience, we s hall choose t he i nterval - 16 ::; k ::; 15 for t he s ummation (10.18), a lthough a ny o ther i nterval o f t he s ame w idth (32 points) would give t he s ame result. 15 Dr = ;2 L f lk]e-jrfok k =-16 Now f lk] = 1 for - 4 ::; k ::; 4 a nd is zero for all o ther values o f k. T herefore L e -jr.!!..k 4 Dr = -1 32 16 k =-4 10 Fourier Analysis of Discrete-Time Signals 624 10.2 Aperiodic Signal R epresentation by Fourier Integral flkl T his is a geometric progression w ith a c ommon r atio e -jfBr. T herefore,[see Sec. (8.7-4)J [_jU1u: =(~) e _.JUJu: [e e _).~ e 16 J _ }'.!hO= 16 - - e}. JUJu:] -N N 0 (aJ s in (~) (&quot;&quot;1'1 ( 1) sin(4.5rrlo) 32 .rTllnll!. .... 16 (10.19) s in (O.5rrlo) T his s pectrum ( with i ts p eriodic extension) is d epicted i n Fig. 1 O.2b.t o .... .. e j~] 16 s in (4.~;r) 1 = ( 3 2) = 16 16 625 • C omputer E xample C 10.1 Do E xample 10.2 using MATLAB. N O=32;k=0:NO-1; f =[ones(1,5} z eros(1,23) o nes(1,4)]; F r=1/32*fft(f); r =k; s tem ( k,Fr},grid 0 1111]11 11 ..... . .(IILIll!. .... .. II rnII!! o -N N ( b) F ig. 1 0.3 G eneration o f a p eriodic signal b y p eriodic extension o f a s ignal f[kJ. T hus, t he Fourier series representing I No [k] will also represent I [k] in t he limit No --+ 0 0. T he e xponential Fourier series for INo[k] is given by INo[k] = L D rejrflok r . _ 211&quot...
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