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
f (t) = - A [sm 7 ft- -sm 37ft + - sm 57 ft- -sm 77f t
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"
T o= - - =-71"
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...
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