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

# Similarly t he component of f requency 3 is 6e j te

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: oreover, we m ay c ompute D n = a n /2 u sing E q. (3.66b), w hich r equires i ntegration o ver a h alf p eriod only. Similarly, w hen j (t) h as a n o dd s ymmetry, an = 0, a nd D n = - jb n /2 is i maginary ( positive o r n egative). H ence, L Dn c an o nly b e 0 o r ±11' / 2. M oreover, we m ay c ompute D n = - jb n /2 u sing E q. ( 3.67b), which r equires i ntegration o ver a h alf p eriod only. N ote, h owever, t hat i n t he e xponential c ase, we a re u sing t he s ymmetry p roperty i ndirectly b y f inding t he t rigonometric coefficients. We c annot a pply i t d irectly t o f ind D n f rom eq. (3.71). 6 E xercise E 3.9 The exponential Fourier spectra of a certain periodic signal f (t) are shown in Fig. 3.17. Determine and sketch the trigonometric Fourier spectra of f (t) by inspection of Fig. 3.17. Now write the (compact) trigonometric Fourier series for f (t). Answer: n wo = 211' 6 f (t) = 4 + 6 cos (3t - ~) + 2 cos (6t -:,f) + 4 cos (9t - ~) E xercise E 3.10 Find the exponential Fourier series and sketch the corresponding Fourier spectrum Dn vs. w for the full-wave rectified sine wave depicted in Fig. 3.18. Answer: f (t) = ~ ~ _ 1_e1 2nt 2 11' ~ 1 -4n 3 214 S ignal R epresentation b y O rthogonal S ets 3.5 215 E xponential F ourier S eries t L D, -9 -6 -1tI6 -1tI4 6 3 . .• ! -1tI2 ,- 9 .1 F ig. 3 .17 Fourier spectra for t he signal in Exercise E3.9. __ ___ A ............... 1+ Sint ~ Fig. 3 .19 (a) A clipped sinusoid cos wot (b) t he d istortion component fd(t) of t he signal in (a). ?t~ For a real f (t), ID-nl = IDnl. T herefore F ig. 3 .18 A full-wave rectified sine wave in Exercise E3.10. 00 P f = D02 + 2 L ID nl 2 ( 3.83b) n =l 3 .5-2 Parseval's Theorem • T he t rigonometric F ourier s eries o f a p eriodic s ignal f (t) is given b y 00 f (t) = Co + L C n cos (nwot + lin) n =l E very t erm o n t he r ight-hand s ide o f t his e quation is a power signal. M oreo:er, a ll t he F ourier c omponents o n t he r ight-hand s ide a re o rthogonal o ver o ne p eriod. H ence, t he p ower o f f (t) is e qual t o t he p ower o f t he s um ~f a ll t he s inusoidal c omponents o n t he r ight-hand side. We a lready d emonstrated I II E xample 1.2 t hat t he p ower o f t he s um o f sinusoids is e qual t o t he s um of t he p owers of all t he 2 s inusoids. M oreover, t he p ower of a sinusoid of a mplitude C n is C n / 2 r~gardless o f t he v alues o f i ts f requency p hase, a nd t he p ower o f a d c t erm Co is Co . T hus, t he p ower o f f (t) is given b y Pf 2 1 ~ = Co + 2 L ..&quot;Cn (3.82) 2 E xample 3 .9 T he i nput signal t o a n audio amplifier of gain 100 is given by x (t) = 0.1 cos wot. Hence, t he o utput is a sinusoid 10 cos wot. However, t he amplifier, being nonlinear a t higher amplitude levels, clips all amplitudes beyond ± 8 volts as shown in Fig. 3.19a. We shall determine t he harmonic distortion incurred in this operation. The o utput y(t) is t he clipped signal in Fig. 3.19a. The distortion signal Yd(t), shown in Fig. 3.19b, is t he difference between t he u ndistorted sinuso...
View Full Document

Ask a homework question - tutors are online