Fourier Analysis A Basic Introduction Johar M. Ashfaque 1
Fourier Series
Fourier series is a type of series representation which is well suited for the analysis of periodic functions. Consider a function f defined on R and with period 2π that is f (x + 2π) = f (x), x ∈ R. And assume further that f belongs to the vector space Z π 2 L (−π, π) = f : |f (x)|2 dx < ∞ . −π
To such a function, associate formally the series ∞
a0 X (an cos nx + bn sin nx), + 2 n=1
f∼ where an = and
1 π
π
Z
f (x) cos nx dx, n = 0, 1, 2, ... −π
1 bn = π
Z
π
f (x) sin nx dx, n = 1, 2, ... −π
The assumption that f belongs to 2
L (−π, π) = f :
Z
π
|f (x)|2 dx < ∞
−π
implies that the integrals defining an and bn are well defined. The series
∞
f∼
a0 X + (an cos nx + bn sin nx) 2 n=1
is called the Fourier series associated to f and an , bn are called Fourier coefficients. Calculation of the Fourier coefficients can often be simplified using the following • If f is 2π-periodic then Z
2π
Z
a+2π
f (x)dx, ∀a ∈ R.
f (x)dx = 0
a
• If f is even, that is f (x) = f (−x) for all x then Z a Z f (x)dx = 2 −a
a
f (x)dx, ∀a > 0.
0
• If f is odd, that is f (x) = −f (−x) for all x then Z a f (x)dx = 0, ∀a > 0. −a
1
If f is an even function, then x 7→ f (x) cos nx is even and x 7→ f (x) sin nx is odd; if f is odd, then x 7→ f (x) cos nx is odd and x 7→ f (x) sin nx is even. Theorem 1.1 If f is an even function then bn = 0 for all n and Z 2 π f (x) cos nx dx, n = 0, 1, 2, .... an = π 0 If f is odd then an = 0 for all n and bn =
2
2 π
π
Z
f (x) sin nx dx, n = 1, 2, .... 0
Fourier Series: Complex Form
Complex exponential functions give a convenient way to rewrite Fourier series. By considering Fourier series in complex form allows to work with one set of coefficients {cn }∞ n=−∞ where all the coefficients are given by a single formula. Then
an cos nx + bn sin nx
=
and
π
einx + e−inx 2 −π Z π inx 1 e − e−inx + f (x) sin nx dx π 2i −π Z π 1 f (x)e−inx dx einx 2π −π Z π 1 inx f (x)e dx e−inx 2π −π 1 π
=
Z
f (x) cos nx dx
1 a0 = 2 2π
Z
π
f (x)dx. −π
Let us now define the numbers cn =
1 2π
Z
π
f (x)e−inx dx, n ∈ Z.
−π
When speaking of Fourier series of f in complex form, one simply means the infinite series appearing in ∞ X
f∼
cn einx .
n=−∞
Note. There is an easy way to come from the Fourier coefficients with respect to sine functions and cosine functions to the coefficients for Fourier series in complex form. In fact c0 = and for n ∈ N cn =
an − ibn , 2
a0 2 c−n =
an + ibn . 2
On the other hand, if the Fourier coefficients in complex form are known then a0 = 2c0 ,
an = cn + c−n ,
2
bn = i(cn − c−n ).
3
Parseval’s Theorem
∞ Assume that the function f ∈ L2 (−π, π) has the Fourier coefficients {an }∞ n=0 , {bn }n=1 or in complex ∞ form {cn }n=−∞ . Then
1 2π
Z
π
|f (x)|2 dx =
−π
∞ ∞ X 1 1X |a0 |2 + (|an |2 + |bn |2 ) = |cn |2 . 4 2 n=1 n=−∞
As a consequence of Parseval’s theorem, note that if f ∈ L2 (−π, pi) then ∞ X
|cn |2 < ∞.
n=−∞
The converse also holds: if
∞ X
|cn |2 < ∞
n=−∞
then
∞ X
f (x) =
cn einx
−∞
define a function in L2 (−π, π).
4
Fourier Transform By Examples
Example 1. We determine the Fourier transform f (ω) ≡ F (f (x); ω) for f (x) = 1 for −a < x < a and f (x) = 0 otherwise.
Z f (ω)
a
dx e−iωx
= −a
= =
i −iaω (e − eiaω ) ω 2 sin(aω) ω
Example 2. We determine the Fourier transform f (ω) ≡ F (f (x); ω) for f (x) = e−a|x| with a > 0.
Z f (ω)
0
=
dx eax e−iωx +
−∞
= =
Z 0
1 1 + a − iω a + iω 2a a2 + ω 2
3
∞
dx e−ax e−iωx
Example 3. Using our result from Example 2, we would like to determine the Fourier transform of f (x) =
1 x2 + ω 2
without carrying out any further integration. We would also like to determine A(a) such that f (x) = Ae−a|x| leading to the representation of Dirac’s δ-function in the limit a → ∞. The integrals for Fourier transform and inverse transform differ by i → −i which is irrelevant for even functions. In addition the inverse Fourier transform includes an extra prefactor (2π)−1 . We have that 2aF and therefore
−1
1 ;x 2 a + ω2
1 F ;ω a2 + x2
=
= e−a|x|
2π −a|ω| π e = e−a|ω| 2a a
In the limit a → ∞: The function e−a|x| vanishes at x 6= 0 for a → ∞. Moreover, Z ∞ 2A dx A e−a|x| = =1 a −∞ for A =
a 2
and lim
a→∞
The Fourier transform of
a −a|x| 2e
a −a|x| e 2
is a2
a2 . + ω2
Taking the limit a → ∞ we have that f (ω) = 1 for f (x) = δ(x).
4
= δ(x).