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Unformatted text preview: l s pectra a re a g raphical r epresentation o f coefficients
D n a s a f unction o f w. E xistence o f t he spectrum a t w = - nwa i s m erely a n i ndication o f t he f act t hat a n exponential component e -jnwot exists i n t he series. We
know t hat [see E q. (3.72)J a s inusoid of frequency nwa c an b e e xpressed i n t erms o f
a p air o f e xponentials e jnwot a nd e -jnwot.
E quation ( 3.77) s hows t he close c onnection b etween t he t rigonometric s pectra
(C n a nd lin) w ith e xponential s pectra (lDnl a nd LDn). T he d c c omponents Do a nd
Co a re i dentical i n b oth s pectra. M oreover, t he e xponential a mplitude s pectrum
IDnl is h alf o f t he t rigonometric a mplitude s pectrum C n for n ~ 1. T he e xponential a ngle s pectrum LDn is i dentical t o t he t rigonometric p hase s pectrum lin for
n > O. W e c an t herefore p roduce t he e xponential s pectra m erely b y i nspection o f
t rigonometric s pectra, a nd v ice versa. T he following e xample d emonstrates t his
• - 12 00 . .... E xample 3 .7
T he trigonometric Fourier spectra of a certain periodic signal f (t) are shown in Fig.
3.15a. By inspecting these spectra, sketch the corresponding exponential Fourier spectra
and verify your results analytically. S pectra for Example 3.7. The trigonometric spectral components exist a t frequencies 0 ,3,6, a nd 9. T he exponential spectral components exist a t 0 ,3,6,9 a nd - 3, - 6, -9. Consider first t he a mplitude
spectrum. The dc component remains unchanged; t hat is, Do = Co = 16. Now jDnj is
an even function of w and IDnl = ID-nl = C n/2. T hus, all t he remaining spectrum IDnl
for positive n is half the trigonometric amplitude spectrum C n, a nd t he s pectrum IDnl for
negative n is a reflection about t he vertical axis of t he s pectrum for positive n , as shown
in Fig. 3.15b.
T he angle spectrum L Dn = I)n for positive n a nd is - I)n for negative n , as depicted
in Fig. 3.15b. We shall now verify t hat b oth s ets of spectra represent t he same signal.
Signal f (t) whose trigonometric s pectra are shown in Fig. 3.15a, has four spectral
components of frequencies 0, 3, 6, a nd 9. T he dc component is 16. T he amplitude and
the phase of t he component of frequency 3 are 12 a nd - ~, respectively. Therefore, this
component can be expressed as 12 cos (3t - ~). Proceeding in this manner, we can write
the Fourier series for f (t) as f (t) = 16 + 12cos (3t -~) + Bcos (6t - I) + 4cos (9t -~) Consider now t he exponential spectra in Fig. 3.15b. They contain components of
frequencies (dc), ± 3, ±6, a nd ±9. T he dc component is Do = 16. T he component
j3t (frequency 3) has magnitUde 6 a nd angle - ~. Therefore, this component strength is
6 e- j t, and i t can be expressed as ( 6e- J t)e J3 '. Similarly, t he component of f requency-3
is (6e J t)e- J3t . Proceeding in this manner, j (t), t he signal corresponding t o t he s pectra
in Fig. 3.15b, is ° j (t) = 16 + [ 6e-Jt eJ3t + 6 eJt e - J3t ] + [4e-J~ eJ6t + 4ej~ e - J6t ] + [ 2e-Jt eJ9t + 2 eJt e - J9t ]
= 16 + 6 [ ej(3t-tl + e -J(3t-tl] + 4 [eJ(6t-~l + e-J(6t-~l]...
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