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

# A glance a t fig 39a shows t hat t he a verage value

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: on of t, t hen f (t) cos nwot is also a n even function a nd f (t) s in nwot is a n o dd f unction of t (see Sec. 1.5-1). Therefore, use of Eqs. (1.33a) and (1.33b) yields A triangular periodic signal and its Fourier spectra. 8 . 1. 1. 1. f (t) = - A [sm 7 ft- -sm 37ft + - sm 57 ft- -sm 77f t 7f2 9 25 49 f(t)dt - To/2 2 jTO/2 e. f(t) = O/2 - To/2 an = - To t j TO 2 / (3.67a) j(t) sin nwot dt (3.67b) 0 Observe t hat, b ecause of symmetry, t he i ntegration required to c ompute t he coefficients need be performed over only h alf t he p eriod. I f a p eriodic signal j (t), s hifted by half t he p eriod, remains unchanged except for a s ign-that is, if f (t - :!f) = - f(t) t he s ignal is s aid t o h ave a h alf-wave s ymmetry. We c an demonst.rate t hat in a signal w ith a half-wave symmetry, all t he e ven-numbered harmonics vanish (see P rob. 3.4-7). T he s ignal in Fig. 3.9a is a clear example of such a symmetry. T hb half-wave s ymmetry is also present in t he signal in Fig. 3.8a b ut in a s\lbtle form. T he half-wave s ymmetry b ecomes obvious, however, when we s ubtract t he de component 0.5 from t his signal. Note t hat t his signal has a dc component 0.5 a nd only o dd h armonics. 3 Signal R epresentation by Orthogonal Sets 200 (a) F ig. 3 .10 P eriodic s ignals f or E xercise E 3.6. l' n en Answers: (a) f it) = ~ - ~ (cos 7rt - :t cos 27rt + = .!. + : ;; [cos (7rt - 7r) + 3 ( b) 3 .4-2 " ~ cos 37rt - i·t Ll. E xercise E 3.7 D etermine whether t he s ignal f (t)=cos (~t+300) + s in (~t+45°) is periodic. I f i t is p eriodic, find t he f undamental frequency a nd t he p eriod. W hat h armonics a re p resent in f it)? .!:. cos 27rt + .!:. cos (37rt - 1f) + . .. J 4 9 271" V2 T o= - - =-71" (3V2) 3 16 cos 41ft + . .. ) 9 2; [sin 1ft - ! sin 27rt + ~ sin 37rt - :t sin 47rt + . .. J = ¥ [ cos(1ft-900) + ! cos (27rt + 9 0°) + ! cos(37rt-900)+ . .. J t n umber of which ~, a nd t a re integral multiples is Moreover, 3 (i) ~, 4(~) = ~, a nd 7 (6) = t . T herefore t he f undamental frequency is ~. T he t hree frequencies in t he s pectrum a re t he t hird, fourth, a nd seventh harmonics. Observe t hat t he f undamental frequency component is a bsent in this Fourier series. T he signal h (t) is n ot periodic because t he r atio of two frequencies in t he s pectrum is 2/71", which is n ot a rational number. T he signal h (t) is periodic because t he r atio of frequencies 3 V2 a nd 6 V2 is 1 /2, a r ational number. T he g reatest common divisor of 3 V2 a nd 6 V2 is 3V2. Therefore t he f undamental frequency Wo = 3V2, a nd t he p eriod Ll. E xercise E 3.6 F ind t he c ompact t rigonometric Fourier series for t he p eriodic signals depi~ted in Fi~s. 3 .lOa a nd 3.10b. S ketch t heir a mplitude a nd p hase spectra. Allow t o take o n negatIve values If b = 0 so t hat t he p hase s pectrum c an be eliminated. Hint: Use Eqs. (3.66) a nd (3.67) for symmetry. ~l 3.4 Trigonometric Fourier Series Answer: P eriodic w ith Wo = ft a nd p eriod To = 151f. T he fifth a nd s ixth h armonics. \l f it) = \l Determinin...
View Full Document

## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

Ask a homework question - tutors are online