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742 Chapter 23 Fourier series
that is, all the Fourier coefficients,b n
, are zero. Finally we can gather together all our
results and writedown the Fourier series representation of f (t):
f(t)=2+ 8 π cost− 8
3π cos3t + 8
5π cos5t−···
In this example we see that there are no sine terms at all. In fact, whenever a function
is even its Fourier series will possess no sine terms. To see this we note thatb n
can be
found from
b n
= 2 T
∫ T/2
−T/2
f(t)sin 2nπt
T
dt
Since f (t) is even and sin 2nπt is odd, the product f (t)sin 2nπt is odd also. Now the
T
T
integral of an odd function on an interval which is symmetrical about the vertical axis
wasshowninSection23.3tobezero.Hencewhenever f (t)isevenwecanimmediately
assumeb n
= 0 for alln.
Correspondingly, when a function is odd its Fourier series will contain no cosine or
constant terms.This isbecause the product
f(t)cos 2nπt
T
isodd alsoand so the integral
a n
= 2 T
∫ T/2
−T/2
f(t)cos 2nπt
T
dt
will equal zero. We conclude that when f (t) is odd,a n
= 0 for alln. These facts can
oftenbeusedtosavetimeandeffort.KnowingthefunctioninExample23.17waseven
before we started the Fourier analysis, we could have assumed that theb n
would all be
zero.
Example23.18 FindtheFourierseriesrepresentationofthesawtoothwaveformdescribedatthebeginning
ofthissection (see Figure 23.12).
Solution This function is defined by f (t) =t, −π <t < π, and has periodT = 2π. It is an odd
function and hencea n
= 0 foralln. To find theb n
wemustevaluate
b n
= 1 π
= 1 π
∫ π
−π
tsinntdt
{[ −tcosnt
n
] π
−π
= 1 {−πcosnπ − πcosnπ}
nπ
∫ π
}
cosnt
+ dt
−π n
since the lastintegral vanishes. Therefore,
b n
=− 2 n cosnπ=−2 n (−1)n