Fourier Series for Functions with Period 2L
Suppose that we have a periodic function f (t) with arbitrary period
P = 2L, generalizing the special case P = 2π which we have already seen.
Then a simple re-scaling of the interval (−π, π ) to (− L, L) allows us to
write down the general Fourier series and Fourier coefficent formulas:
f (t) =
� π �
� π �
∞
a0
+
∑ an cos n t
+ bn sin n
t
2
n=1
L
L
(1)
with Fourier coefficients given by the general Fourier coefficent formulas
1 L
a0 =
f (t) dt,
L −L
�
1 L
π
an =
f (t) cos(n t) dt,
L −L
L
�
1 L
π
bn =
f (t) sin(n t) dt.
L −L
L
�
Note: The number L =
P
2
(2)
is called the half-period.
Example. Let f (t) be the period 2 function, which is defined on the window
[−1, 1) by f (t) = |t|. Compute the Fourier series of f (t).
The graph of f (t) below shows why this function is called either a tri­
angle wave or a continuous sawtooth function.
−2
2
−1
1
Figure 1: The period 2 triangle wave.
Solution. In this case the period is P = 2, so the half-period L = 1. This
means n πL = nπ and we compute the coefficients from the formulas in (2),
using integration by parts, as follows.
For n �= 0:
an =
1
1
� 1
−1
�
= 2
|t| cos(nπt) dt = 2
� 1
0
�1
�
t cos(nπt) dt
t sin(nπt) cos(nπt) �
2
+
= 2 2 ((−1)n − 1) =
nπ
n2 π 2 �0
n π
�
− n24π2
0
for n odd
for n even
Fourier Series for Functions with Period 2L
OCW 18.03SC
and for n = 0:
�
� 1
1 1
a0 =
|t| dt = 2
t dt = 1
1 −1
0
Since f (t) is an even function and sin(nπt) is odd, the sine coefficients
bn = 0. (We will justify this carefully in the next session. For now you can
compute the integrals for bn as an exercise and verify it in this case.)
Thus, the Fourier series for f (t) is
�
�
1
4
cos 3πt cos 5πt
1
4
f (t) = − 2 cos πt +
+
+··· = − 2
2
2
2 π
3
5
2 π
2
cos(nπt)
.
n2
n odd
∑
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