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

# Because ydt is a n e ven function of t a nd b ecause

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Unformatted text preview: ugh not necessary) t hat No be a power of 2, t hat is No = 2m , where m is a n integer. o C omputer E xample C 3.1 C ompute a nd p lot t he t rigonometric a nd e xponential Fourier s pectra for t he p eriodic signal in Fig. 3.7b (Example 3.3). 3 218 S ignal R epresentation b y O rthogonal S ets T he samples o f f (t) s tart a t t = 0 a nd t he l ast ( No-th) sample is a t t = To - T (the last sample is n ot a t t = To because the sample a t t = 0 is identical t o t he sample a t t = To, a nd t he n ext cycle begins a t t = To). At t he p oints of discontinuity, t he s ample value is t aken as the average of t he values of t he function on two sides of t he discontinuity. Thus, in t he present case, t he first sample (at t = 0) is n ot 1, b ut ( e-"/2 + 1 )/2 = 0.604. To determine No, we r equire t hat D n for n ~ N o/2 t o b e negligible. Because f (t) h as a j ump discontinuity, D n d ecays r ather slowly as l /n. Hence, choice of No = 200 is acceptable because the ( No/2)-nd (100th) h armonic is a bout 0.01 ( about 1%) of t he fundamental. However, we also require No t o b e power of 2. Hence, we shall take No = 256 = 28. We write a nd save a MATLAB file (or program) c31.m to compute a nd p lot the exponential Fourier coefficients. % ( c31.m) % M i s t he n umber o f c oefficients t o b e c omputed T O=pi;NO=256;T=TO/NO;M=10; t =O:T:T*(NO-l); t =t'; f =exp( - t/2) ; f(l) = 0.604; % f ft(f) i s t he F FT [ the s um o n t he r ight-hand s ide o f E q. ( 3.86)] D n=fft(f)/NO [ Dnangle,Dnmag]=cart2pol(real(Dn),imag(Dn»; k =O:length(Dn)-l;k=k'; s ubplot(211) , stem(k,Dnmag) s ubplot(212), s tem(k,Dnangle) 3.7 a ns= A mplitudes A ngles 0 .5043 0 0 .2446 - 75.9622 0 .1251 - 82.8719 0 .0837 - 85.2317 0 .0629 - 86.4175 0 .0503 - 87.1299 0 .0419 - 87.6048 0 .0359 - 87.9437 0 .0314 - 88.1977 0 .0279 - 88.3949 LT IC System Response to periodic Inputs A p eriodic s ignal c an b e e xpressed a s a s um o f e verlasting e xponentials (or sinusoids). W e a lso k now h ow t o find t he r esponse o f a n L TIC s ystem t o a n e verlasting e xponential. F rom t his i nformation we c an r eadily d etermine t he r esponse o f a n L TIC s ystem t o p eriodic i nputs. A p eriodic s ignal f (t) w ith p eriod To c an b e e xpressed a s a n e xponential F ourier s eries 00 f (t) = 2 1l' D nejnwot w o=To n =-oo e jwt " '-.r' i nput = =? H ( jw )e jwt '-v-' o utput T herefore, f rom l inearity p roperty 00 L n =-oo '--v-" input f (t) D nH(jnwo)ejnwot (3.87) n=-oo '-'- - - - " . . - - - r esponse y (t) T he r esponse yet) is o btained i n t he f orm o f a n e xponential F ourier s eries, a nd is t herefore a p eriodic s ignal o f t he s ame p eriod a s t hat o f t he i nput. W e s hall d emonstrate t he u tility o f t hese r esults b y t he following e xample. • E xample 3 .9 A full-wave rectifier (Fig. 3.20a) is used t o o btain a dc signal from a sinusoid sin t. T he rectified signal f (t), d epicted in Fig. 3.18, is applied t o t he i nput of a a low-pass R C filter, which...
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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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