Chapter 2
Fourier Series and Transforms
2.1 Fourier Series
Let f (x) be an integrable functin on [−L, L]. Then the fourier co-ecients are dened as
an =
1
L
bn =
1
L
ˆ
L
−L
ˆ L
−L
nπx f (x) cos
f (x) sin
L
nπx L
dx
n ≥ 0,
dx
n > 0.
(2.1)
The claim is that the function f then can be written as
nπx nπx a0 an cos
+
+ bn sin
.
2
L
L
∞
f (x) =
n=1
The rhs is called the fourier series of f . Using fourier trick, one can obtain the expressions
for some m > 0 and
for an and bn . Multiplying both sides of the expression by cos mπx
L
integrating one gets
ˆ
=
L
−L
ˆ L
−L
= 0+
f (x) cos
mπx L
dx
ˆ L
ˆ L
∞ mπx mπx mπx nπx nπx a0
an
cos
cos
cos
dx +
cos
dx + bn
sin
dx
2
L
L
L
L
L
−L
−L
n=1
∞
[an Lδm,n + 0] = Lam
n=1
This proves the formula given by Equation 2.1.
Example 22. Let f (x) = x on [−π, π]. Find fourier series.
8
The function is odd, hence for all n
ˆ
1
an =
π
and
1
bn =
π
Thus the fourier series is
ˆ
π
−π
π
−π
x cos(nx)dx = 0
x sin(nx)dx =
2
(−1)n+1
n
1
f (x) = 2 sin (x) − sin (2x) + · · ·
2
Figure 2.1: Fourier series with 1 term, 3 terms and 100 terms
Two points can be noted from the example
1. The convergence is not uniform, some points converge faster than other points.
2. At x = ±π the fourier series does not converge to f .
Example 23. Let f (x) = 1 − |x| /L where x ∈ [−L, L]. This is a trianglular function. The
function is even, thus fourier series will contain only cos terms.
1
bn =
L
and for n > 0,
an =
1
L
=
1
L
ˆ
L
f (x) cos
−L
ˆ 0 −L
ˆ
L
−L
f (x) sin
nπx nπx L
dx = 0.
dx
L
ˆ
nπx nπx x
x
1 L
1+
1−
cos
dx +
cos
dx
L
L
L 0
L
L
2
=
(1 − (−1)n )
2
n π2
4
=
odd n
2
n π2
= 0
even n
9
Simillarly, a0 = 1. The fourier series is
πx 1
1
4
3πx
+ ··· .
f (x) = + 2 cos
+ cos
2 π
L
9
L
Dirichlet Condition
Denition 24. A function f on [−L, L] is said to be satisfy Dirichlet conditions if
f has nite number of extrema,
f has nite number of discontinuities,
f is abosultely integrable,
f is bounded.
Theorem 25. Suppose a function f satises Dirichlet conditions. Then the fourier series of
f converges to f at points where f is continuous. The fourier series converges to the midpoint
of discontinuity at points where f is discontinuous.
Example 26. A square function f is dened as
0
f (x) =
1
x ∈ [−L, 0]
x ∈ [0, L]
The fourier coecients are
a0 = 1,
bn =
The fourier series is
an = 0
2
nπ
0
odd n
even n
πx 1 2
2
+ sin
+
sin
2 π
L
3π
3πx
L
+ ···
Clearly, at x = 0 series has a value 0.5 which is equal to [f (0+) + f (0−)] /2.
Theorem 27. If a function f on [−L, L] is square integrable, that is
ˆ
L
−L
|f (x)|2 dx
exists, then the fourier series of f converges to f almost everywhere.
10
Exponential Fourier Series
Substituting cos x = [ex + e−x ] /2 and sin x = [ex − e−x ] /2i, the fourier series of a function f
can be rewritten as
nπx nπx a0 an cos
+
+ bn sin
2
L
L
∞
f (x) =
n=1
nπx cn exp i
=
L
n=−∞
∞
where
Or,
⎧
1
⎪
⎨ 2 (an − ibn )
cn = 12 (an + ibn )
⎪
⎩ a0
2
1
cn =
2L
ˆ
n>0
n<0
n=0
nπx f (x) exp −i
L
−L
L
Typically, if the argument of f is a time variable, then ωn = nπ/L are called frequencies. The
fourier coecients are usually labelled by frequencies, that is cn is written as cωn .
Example 28.
Exponential fourier coecients for triangular function are given by
1
2
= 0
c0 =
c2n
2
(2n − 1)2 π 2
c2n−1 =
The following graph shows the fourier coecients as a function of frequencies.
CΩ
0.5
5 Π
3 Π
Π
Π
3Π
5Π
Ω
2.2 Fourier Transform
For an interval [−L, L], the fourier frequencies are given by ωn = nπ/L. When L → ∞,
the fourier frequencies become a continuous variable. To get this idea, redene the fourier
coecients
ˆ
cωn =
L
−L
f (x) exp (−iωn x)
11
Then the fourier series becomes
f (x) =
∞
1 cω exp (iωn x)
2L n=−∞ n
Now, let ∆ω = π/L, then
f (x)
=
→
∞
1 cω exp (iωn x) ∆ω
2π n=−∞ n
ˆ ∞
1
c(ω) exp (iωx) dω
2π −∞
as L → ∞
The last step is just the denition of Riemann integral. The function c(ω) is called as fourier
transform of f (x). It is denoted as c = F (f ). Dierent text books dene fourier transform
dierently by placing the constat factor 2π dierently. In this note, fourier transform is
dened as
c(ω) =
f (x) =
ˆ ∞
1
√
f (x) exp (−iωx) dx
2π −∞
ˆ ∞
1
√
c (ω) exp (iωx) dω.
2π −∞
Example 29. Let f (x) = 1 for |x| ≤ a, and is 0 otherwise. If g is the FT of f , then
g(ω) =
=
=
ˆ ∞
1
√
f (x) exp (−iωx) dx
2π −∞
ˆ a
1
√
exp (−iωx) dx
2π −a
2a sin (ωa)
√
2π ωa
fx
gΩ
1
1
4 Π
a
a
2 Π
x
Here is another expample that illustrates the uncertainty principle.
12
2Π
4Π
Ω
Example 30. Let f (x) =
g(ω) =
=
=
=
a/π exp[−ax2 ]. If g is the FT of f , then
ˆ ∞
1
√
f (x) exp (−iωx) dx
2π −∞
ˆ ∞
1
a
√
exp −ax2 − iωx dx
2π π −∞
ˆ
iω 2
a −ω2 /4a ∞
1
√
dx
exp −a x +
e
2a
2π π
−∞
1
ω2
√ exp −
4a
2π
The width of the gaussian function is dened by parameter a. Thus f (x) is broad if a is small.
And g(ω) is narrow and sharp for small a.
Here is a very useful theorem called Parseval theorem.
Theorem 31.
If
g
is the fourier transform of a square integrable function
ˆ
∞
−∞
|f (x)|2 dx =
ˆ
∞
−∞
f
then
|g(ω)|2 dω.
Uncertainty Relation
The uncertainty relation is built into the fourier transform. It may be interpreted dierently
depending on the situation in which it is applied. Here we state the theorem without proof.
Theorem 32.
Suppose
g
is a fourier transform of a square integrable function
normalized to unity, that is
ˆ
∞
−∞
. Let
ˆ
µf
=
σf2
=
µg =
σg2 =
then
σf σg ≥
|f (x)|2 dx = 1
∞
ˆ−∞
∞
ˆ−∞
∞
ˆ−∞
∞
−∞
x|f (x)|2 dx
(x − µf )2 |f (x)|2 dx
ω|f (ω)|2 dω
(ω − µω )2 |f (ω)|2 dω
1
2.
The quantities σ are called uncertainties.
13
f
that is
Example 33. Let f (x) = (a/π)1/4 exp[−ax2 /2]. Then
ˆ
∞
|f (x)|2 dx = 1
−∞
Clearly, µf = 0 and
σf2
ˆ
=
∞
−∞
x2 |f (x)|2 dx =
1
2a
If g is the FT of f , then
g(ω) =
Note that
ˆ
∞
−∞
Now µg = 0. Then
σg2 =
ˆ
1
πa
1/4
ω2
exp −
2a
|g(ω)|2 dω = 1
∞
−∞
ω 2 |g(ω)|2 dω =
Then,
σf σ g =
14
1
2
a
2
.