Unformatted text preview: t Fig. 3.9a shows t hat t he a verage value (dc) o f f (t) is zero, so t hat ao = O.
Also Using t his fact, we c an express t he series in (3.61) as
f (t) = iti ::; 2 At
{ 2 A(1  t) 1/2 T he d etailed evaluation o f t he a bove integrals shows t hat b oth have a value o f zero. Therefore ~
an = 0 (3.62a) n even
n o dd bn = 1 1/2 1 3/2 2 At sin n7l:t d t +  1/2 On = {O71: for all n n # 3 ,7,11,15, . .. T he d etailed evaluation o f t hese integrals yields = 3 ,7,11,15,··· tBecause cos ( x ± 71:) =  cos x , we could have chosen the phase 7r or  7r. In fact, cos ( x ± N7r)
 cos x for any odd integral value of N . Therefore, the phase can be chosen as ± N7r where N
any convenient odd integer. 2 A(1  t) sin n7rt d t
1/2 ~
IS b n 8 A.
= n 2 7r 2 S In ( ~)
2 n even
= {OBA
~
BA ~ n = 1, 5, 9, 13, . ..
n = 3, 7, 11, 15, . .. (3.62b) 3 Signal R epresentation by Orthogonal S ets 198 t Trigonometric Fourier Series 199 t erms only a nd t he series for any o dd p eriodic function f (t) consists of sine t erms
only. Moreover, because of s ymmetry (even or o dd), t he i nformation of one period
of f (t) is i mplicit in only h alf t he p eriod, as seen in Figs. 3.8a a nd 3.9a. In these
cases, knowing t he s ignal over a h alf p eriod a nd w hat k ind of s ymmetry (even or
odd) is present, we can d etermine t he s ignal waveform over a complete period. For
this reason, t he F ourier coefficients in these cases c an b e c omputed by integrating
over only h alf t he p eriod r ather t han a c omplete period. To prove t his r esult, recall
t hat SA 1t2 C. 3.4 1 ao =  SA 9it To
( b) T 2 jTO/2 bn = To 11
2 o (c)  It 2
F ig_ 3 .9 21
41 To Therefore
(3.63) ± sin k t = cos ( kt ' f 90°)
Using this identity, Eq. (3.63) can be expressed as :~ [cos (7ft  90°) + ~ cos (37ft + 90°) + +49 cos (77ft + 90°) + ...J
~ (3.64) The Effect of Symmetry T he F ourier series for t he p eriodic signal in Fig. 3.7b (Example 3.3) consists
of sine a nd c osine t erms, b ut t he series for t he s ignal f (t) i n Fig. 3.8a (Example
3.4) consists o f cosine t erms only, a nd t he series for the signal f (t) in Fig. 3.9a
( Example 3.5) consists of sine t erms only. T his o bservation is no accident. We c an
show t hat t he F ourier series of any even periodic function f (t) consists of cosine f(t)sinnwotdt (3.65c) TO 2
/ f(t)dt (3.66a) TO 2
/ f(t) cos nwot dt (3.66b) 0 =0 (3.66c) Similarly, if f (t) is an o dd f unction of t , t hen f (t) cos nwot is a n o dd fUllction of t
a nd j (t) s in nwot is a n even function of t. T herefore = an = 0 41 bn = To 90°) In this series all the even harmonics are missing. The phases of odd harmonics alternate
from _90° to 90 0 • Figure 3.9 shows amplitude and phase spectra for f (t).
• 3.41 bn ao ~ cos (57ft  (3.65b) 0 an =  To In order to plot Fourier spectra, the series must be converted into compact trigonometric
form as in Eq. (3.54). In this case, sine terms are readily converted into cosine terms with
a suitable phase shift. For example, j(t) cos nwot dt  To/2 ao =  + . .. ] (3.65a) Recall also t hat cos nwot is a n even function a nd sin nwot is an ~dd f unction of
t. I f f (t) is a n even functi...
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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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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