Chapter 9
Lecture 13 - Fourier Series
We consider the expansion of the function f (x) of the form
nπx nπx a0 +
an cos
+ bn sin
= S(x)
2
L
L
∞
f (x) ∼
(9.1)
n=1
where
an =
bn =
1
L
1
L
L
f (x) cos
nπx L
−L
L
f (x) sin
nπx −L
L
dx
a0
1
=
2
2L
L
f (x) dx = average value of f .
−L
dx
(9.2)
Note:
nπx nπλ
2L
(x + λ) = cos
provided
= 2π, λ =
L
L
L
n
nπx nπ
(x + 2L) = sin
. Thus each of the terms
and similarly sin
L
L
of the Fourier Series S(x) on the RHS of (10.1) is a periodic function
having a period 2L. As a result the function S(x) is also periodic.
1. Note that cos
nπ
How does this relate to f (x) which may not be periodic?
The function S(x) represented by the series is known as the periodic
extension of f on [−L, L].
61
Lecture 13 - Fourier Series
2. If f (or its periodic extension) is discontinuous at a point x0 then S(x)
converges to the average value of f across the discontinuity.
S(x0 ) =
1
−
f (x+
0 ) + f (x0 )
2
(9.3)
−π < x < 0
0≤x≤π
(9.4)
Example 9.1
f (x) =
62
0
x
L=π
a0 =
an =
=
=
=
=
a2m+1 =
bn =
=
=
=
1
π
1
π
1
π
π
−π
π
1
f (x) dx =
π
π
x dx =
0
π
2
(9.5)
f (x) cos(nx) dx
−π
π
x cos(nx) dx
0
⎧
⎫
π
π
⎨
⎬
sin(nx) 1
1
x
−
1.
sin(nx)
dx
⎭
π⎩
n 0 n
0
π 1 π sin
1
(nπ)
+ 2 cos(nx)
π
n
n
0
1 n
1 2
3 4
(−1)n − 1
n − 1 −2 0 −2 0
2
(−1)
πn
2
−
m = 0, 1, 2, . . .
π(2m + 1)2
π
1
f (x) sin(nx) dx
π
1
π
(9.6)
(9.7)
−π
π
x sin(nx) dx
0
⎫
⎧
π
π
⎬
⎨
1
cos(nx) 1
+
1.
cos(nx)
dx
−x
⎭
π⎩
n 0 n
0
π cos(nπ) 0. cos 0
1
1
−π
+
+ 2 sin(nx)
π
n
n
n
0
= (−1)n+1 /n
∞
a0 an cos(nx) + bn sin(nx)
+
f (x) =
2
n=1
∞
∞
sin(nx)
2 cos (2m + 1)x
π
−
+
(−1)n+1
=
4 π
(2m + 1)2
n
m=0
(9.8)
(9.9)
n=1
63
Lecture 13 - Fourier Series
9.1
It can be useful to shift the interval of integration from [−L, L] to [c, c + 2L]
Since the periodic extension
fe (x)is periodic
with period 2L (as are the
nπx nπx basis functions cos
and sin
).
L
L
1
L
an =
1
L
bn =
L
f (x) cos
nπx L
−L
L
f (x) sin
nπx −L
L
1
dx =
L
dx =
1
L
c+2L
fe (x) cos
c
c+2L
fe (x) sin
c
nπx L
nπx L
dx (9.10)
dx. (9.11)
Example 9.2 Previous Example:
f (x) =
0 −π < x < 0
x
0≤x≤π
(9.12)
0
π < x < 2π
x − 2π 2π ≤ x ≤ 3π
(9.13)
On [π, 3π]
fe (x) =
an =
1
π
=
1
π
=
64
1
π
3π
π
3π
2π
π
0
fe (x) cos(nx) dx
(x − 2π) cos(nx) dx
t cos(nt) dt.
t = x − 2π dx = dt
x = t + 2π x = π ⇒ t = −π
(9.14)
x = 3π ⇒ t = π
since
cos n(t + 2π) = cos t
9.2. COMPLEX FORM OF FOURIER SERIES
9.2
Complex Form of Fourier Series
nπx nπx a0 +
an cos
+ bn sin
2
L
L
∞
f (x) =
cos
nπx L
n=1
=
e
nπx
nπx ei( nπx
−i( nπx
L ) − e
L )
+ e−i( L )
; sin
=
2
L
2i
∞
b nπx
nπx
nπx
an i( nπx )
n
+
e L + e−i( L ) +
ei( L ) − e−i( L )
2
2i
n=1
∞
an − ibn
nπx
nπx
an + ibn
+
ei( L ) +
e−i( L ) (9.15)
2
2
i( nπx
L )
Therefore f (x) =
a0
2
=
a0
2
n=1
↑
=
↑
↑
cn
c0
∞
cn ei(
nπx
L
c−n
)
n=−∞
an − ibn
cn =
2
=
=
1
2L
1
2L
L
−L
L
nπx nπx f (x) cos
− i sin
dx (9.16)
L
L
f (x)e−i(
nπx
L
) dx
b−n = −bn
(9.17)
−L
Therefore
f (x) =
∞
cn ei(
nπx
L
)
(9.18)
n=−∞
cn =
1
2L
L
f (x)e−i(
nπx
L
) dx.
(9.19)
−L
65
Lecture 13 - Fourier Series
Example 9.3
f (x) =
cn =
−1 −π ≤ x < 0
L=π
1
0<x<π
⎧
⎫
0
π
⎨
⎬
1
− e−inx dx + e−inx dx
⎭
2π ⎩
0
−L
0
e−inx −π
=
=
i −2 + e+inπ + e−inπ =
2πn
(−in)
+
(9.21)
π
e−inx 0
1
2π
−
(9.20)
(−in)
(9.22)
0
n even
(2/iπn) n odd
(9.23)
Therefore
f (x) =
66
2
ei (2n+1)x .
πi(2n + 1)
n=−∞
∞
(9.24)