#18 (10/30/2023)

Fourier series/transform

  1. Fourier series (−π, π)
    Fourier coefficient
    Fourier series
    Fm = 1



    π

    −π 
    f(x) ei m xdx
    f(x) = m=∞

    m=−∞ 
    Fm ei m x
    (1)
  2. Fourier transform (−∞, ∞) Three definitions:

    Fourier transform
    Inverse Fourier transform
    F(ω) = 1





    −∞ 
    f(x) ei ωxdx
    f(x) =


    −∞ 
    F(ω) ei ωx dω
    F(ω) =


    −∞ 
    f(x) ei ωxdx
    f(x) = 1





    −∞ 
    F(ω) ei ωx dω
    F(ω) = 1








    −∞ 
    f(x) ei ωxdx
    f(x) = 1








    −∞ 
    F(ω) ei ωx dω
    (2)
Definition:
From now on, the Fourier transform of a (non-periodic) function, f(x), is defined as

F(f) = F(ω) =
^
f
 



−∞ 
f(x)eiωx dx.
(3)
The inverse Fourier transform is defined as

F−1(
^
f
 
) = f(x) ≡ 1





−∞ 
F(ω) eiωx dω.
(4)

Properties of Fourier transform


    (1) (Linearity)

    F(α f + β g) = α F(f) + β F(g)
    (5)
    where α and β are constants. This property comes from the linearity of the integral operator.
    (2)

    F(f(n)(x)) = (iω)n F(f(x))
    (6)
    (Proof)
    For n=1,

    F(f (x))
    =



    −∞ 
    f  (x) eiωx dx
    =

    f(x) eiωx

    −∞ 



    −∞ 
    f(x) (−iω) eiωx dx
    =
    iω


    −∞ 
    f(x) eiωx dx
    =
    iωF(ω),
    (7)
    where f(x)→ 0 as x→ ±∞ was used. This condition is required for the existence of Fourier transforms. Use mathematical induction for general n. It is seen that once transformed into the frequency domain, differentiation becomes multiplication by iω.
    (Example):
    u"" + u = w(x).
    (8)
    where w(x) is a given function defined over (−∞,∞). Fourier transforming both sides of Eq.(8) into the frequency domain yields

    (iω)4 U (ω) + U(ω) = W(ω),
    (9)
    where U(ω) and W(ω) are the Fourier transforms of u(x) and w(x), respectively. Equation (9) can be solved as

    U(ω) = W(ω)

    1 + ω4
    .
    (10)
    By using the inverse Fourier transform, one obtains

    u(x)
    =
    1





    −∞ 
    U(ω) eiωx dω
    =
    1





    −∞ 
    W(ω)

    1 + ω4
    eiωxdω
    =
    1





    −∞ 




    −∞ 
    w(y) ei y ω dy
    ei x ω

    1 + ω4
    dω
    =



    −∞ 

    1





    −∞ 
    eiω(xy)

    1 + ω4
    dω
    w(y) dy
    =



    −∞ 
    G(xy) w(y) dy,
    (11)
    where
    G(xy) ≡ 1





    −∞ 
    eiω(xy)

    1 + ω4
    dω.
    (12)
    The function, G(xy), is called the Green's function and is the inverse Fourier transform of [1/(1+ω4)].



    (3) (Fourier convolution)
    Fourier convolution between two functions, f(x) and g(x), is defined as
    f * g


    −∞ 
    f(xy) g(y) dy.
    (13)
    Note that
    f * g = g * f.
    (Proof)
    f * g
    =



    −∞ 
    f(xy) g(y) dy
    =

    −∞

     
    f(z) g(xz) (−dz)
    =



    −∞ 
    g(xy) f(y) dy
    =
    g * f.
    (14)
    where xyz was used in Eq.(14).
(Fourier convolution theorem)
F(f * g) = F(ω) G(ω).
(15)
(Proof)
f * g
=



−∞ 
f(xy) g(y) dy
=



−∞ 

1





−∞ 
F(ω) eiω(xy) dω
g(y) dy
=
1





−∞ 




−∞ 
g(y) eiωy dy
F(ω) eiωx dω
=
1





−∞ 
G(ω) F(ω) eiωx dω
=
F−1 (G(ω) F(ω)).
(16)
Convolution in the x domain is converted into ordinary multiplication in the frequency domain.
Example

u""(x) + u(x) = w(x),     −∞ < x < ∞
(17)
Fourier transform the both sides yields
(iω)4 U(ω) + U(ω) = W(ω),
(18)
where U(ω) and W(ω) are the Fourier transforms of u(x) and w(x), respectively. By solving Eq.(18), one obtains
U(ω) = 1

1+ω4
W(ω).
(19)
Therefore, using the Fourier convolution theorem, u(x) can be expressed as
u(x)
=
F−1
1

1+ω4

* w(x)
=
G(x)* w(x)
=



−∞ 
G(xy) w(y) dy,
(20)
where

G(x)
F−1
1

1+ω4

=
1





−∞ 
eiωx

1+ω4
dω
(21)
is called Green's function for Eq.(17).
Let's see if a computer can do the inverse Fourier transform:
Enter the function to be inverse Fourier transformed.
with respect to .

Enter 1/(1+w^4) and w in the boxes above and press the button. Verify that you get
G(x) = 1

2
e−|x|/√2 sin
π+2√2|x|

4

.
(22)




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On 29 Oct 2023, 07:49.