Fourier Transform Exercises
Fourier Transform Exercises
Madas
FOURIER
TRANSFORM
Created by T. Madas
Created by T. Madas
∫
1
• F f ( x ) = fˆ ( k ) = f ( x ) e−i kx dx
2π −∞
∫
−1 1
fˆ k = f ( x ) = f ( k ) ei kx dk
( )
• F
2π −∞
Useful Results
• F f ′ ( x ) = i k fˆ ( k )
d
• F x f ( x ) = i fˆ ( k )
dk
Shift Results
• F f ( x + c ) = ei kc fˆ ( k )
• F −1 fˆ ( k + c ) = e−i cx f ( x )
Convolution Theorem
F {[ f ∗ g ] ( x )} = 2π F f ( x ) F g ( x )
∞
where [ f ∗ g ]( x ) =
∫ −∞
f ( x − y ) g ( y ) dy
Parseval’s Theorem
∞ ∞ ∞ ∞
∫ ∫ ∫ ∫
2 2
h ( y ) g ( y ) dy = hˆ ( k ) gˆ ( k ) dk or h( y) dy = hˆ ( k ) dk
−∞ −∞ −∞ −∞
Created by T. Madas
Created by T. Madas
FINDING FOURIER
TRANSFORMS
and
INVERSES
Created by T. Madas
Created by T. Madas
Question 1
f ( x ) = e− ax , x > 0 ,
a − ik
fˆ ( k ) =
(a 2
+ k2 ) 2π
Created by T. Madas
Created by T. Madas
Question 2
1 x < 1a
2
f ( x) = ,
0 1
x > a
2
2 a
fˆ ( k ) =
k 2π
( )
sin 1 ka =
2 2π
( )
sinc 1 ka
2
Created by T. Madas
Created by T. Madas
Question 3
1 0 ≤ x ≤ 2
f ( x) = .
0 otherwise
2
fˆ ( k ) = e−ik sinc k
π
Question 4
1 x ≤ω
f ( x) = ω
0 x >ω
2
fˆ ( k ) = sinc ω
π
Created by T. Madas
Created by T. Madas
Question 5
The function f ( x ) is defined in terms of the positive constant a , by
x
1 − x ≤a
f ( x) = a
0 x >a
2 1 a
F f ( x ) = fˆ ( k ) = 1 − cos ( ak ) =
π ak 2 2π
( )
sinc 2 1 ka
2
Created by T. Madas
Created by T. Madas
Question 6
1
mx x≤
m
f ( x) = .
0 1
x >
m
i 2
fˆ ( k ) =
k π ( )
m ( )
cos k − sinc k
m
Created by T. Madas
Created by T. Madas
Question 7
f ( x ) = x e−2 x , x > 0 .
1
MM3-D , fˆ ( k ) =
( 2 + ik ) 2 2π
Created by T. Madas
Created by T. Madas
Question 8
The triangle function Λ n ( x ) is defined as
1
n2 ( n + x ) −n< x <0
Λn ( x ) = 1
n2 ( n − x ) 0< x<n
0 otherwise
1
2π
( )
sinc2 1 kn .
2
proof
Created by T. Madas
Created by T. Madas
Question 9
The function f is defined by
−a x
f ( x) = e ,
2 a
F e = fˆ ( k ) =
−a x
π a + k2
2
Created by T. Madas
Created by T. Madas
Question 10
The function f is defined by
1
f ( x) = , x ≠ 0.
x
∫
sin ax
standard results about dx .
0 x
−ε x
b) By introducing the converging factor e and letting ε → 0 , invert the
answer of part (a) to obtain f .
1 π
MM3-E , F = fˆ ( k ) = − i sign ( k )
x 2
Created by T. Madas
Created by T. Madas
Question 11
The impulse function δ ( x ) is defined by
∞ x = 0
δ ( x) =
0 x ≠ 0
a) Determine
i. … F δ ( x ) .
1 1 −i ka 1
F δ ( x ) = , F δ ( x − a ) = e , F −1 δ ( k ) = ,
2π 2π 2π
F [1] = 2π δ ( k ) , F −1 [1] = 2π δ ( x )
Created by T. Madas
Created by T. Madas
Question 12
The signum function sign ( x ) is defined by
1 x>0
sign ( x ) =
−1 x<0
−ε x
By introducing the converging factor e and letting ε → 0 , determine the Fourier
transform of sign ( x ) .
i 1
F sign ( x ) = −
k π
Created by T. Madas
Created by T. Madas
Question 13
The Unit function U ( x ) is defined by
U( x) = 1 .
−ε x
By introducing the converging factor e and letting ε → 0 , determine the Fourier
transform of U ( x ) .
1 ε
You may assume that δ ( t ) = lim 2 2 .
π ε →0 ε + t
F U ( x ) = 2π δ ( k )
Created by T. Madas
Created by T. Madas
Question 14
The Unit function U ( x ) is defined by
U ( x ) = 1.
−ε k
By introducing the converging factor e and letting ε → 0 , find F −1 U ( k ) .
1 ε
You may assume that δ ( t ) = lim 2 2 .
π ε →0 ε + t
F −1 U ( k ) = 2π δ ( x )
Created by T. Madas
Created by T. Madas
Question 15
The function g ( x ) has Fourier transform given by
gˆ ( k ) = −i sign ( k ) .
−ε k
By introducing the converging factor e and letting ε → 0 , find F −1 ĝ ( k ) .
2 1
MM3-B , F −1 ĝ ( k ) =
π x
Created by T. Madas
Created by T. Madas
Question 16
The Heaviside function H ( x ) is defined by
1 x≥0
H( x) =
0 x<0
By introducing the converging factor e−ε x and letting ε → 0 , determine the Fourier
transform of H ( x ) .
1 ε
You may assume that δ ( t ) = lim 2 2 .
π ε →0 ε + t
1 i
F [ H( x)] = πδ ( k ) − k
2π
Created by T. Madas
Created by T. Madas
Question 17
The impulse function δ ( x ) is defined by
∞ x = 0
δ ( x) =
0 x ≠ 0
F [1] = 2π δ ( k )
Created by T. Madas
Created by T. Madas
Question 18
The function f is defined by
1 x>0
f ( x ) = sign( x) =
−1 x<0
−ε x
a) By introducing the converging factor e and letting ε → 0 , find the Fourier
transform of f .
−ε x
b) By introducing the converging factor e and letting ε → 0 , find the Fourier
transform of g ( x ) = 1 .
1 ε
You may assume that δ ( t ) = lim 2 2 .
π ε →0 ε + t
1 x≥0
H( x) =
0 x<0
i 1 1 i
F [sign( x)] = − , F [1] = 2π δ ( k ) , F [ H( x)] = πδ ( k ) − k
k π 2π
Created by T. Madas
Created by T. Madas
Question 19
The Fourier transforms of the functions f ( x ) and g ( x ) are
1
fˆ ( k ) = δ ( k ) and gˆ ( k ) = ,
ik
Find simplified expressions for f ( x ) and g ( x ) , and use them to show that
1 1
F [ H( x) ] = π δ ( k ) + i k ,
2π
1
f ( x) = , g ( x ) = 1 π sgn ( x )
2π 2
Created by T. Madas
Created by T. Madas
Question 20
The function f is defined by
sin ax
f ( x) = , a >0.
x
π
k <a
2
sin ax
F = π
x k =a
8
0 k >a
Created by T. Madas
Created by T. Madas
Question 21
Given that l is a non zero constant, show that
F
( )
exp − x2
l 2 = fˆ k =
( )
1 k 2l 2
exp − .
l π 2π 4
proof
Created by T. Madas
Created by T. Madas
Question 22
The Gaussian function f ( x ) is defined by
2
f ( x ) = A e−α x ,
k2
MM3-A , F A e −α x = fˆ ( k ) =
2 A
exp −
2α 4α
Created by T. Madas
Created by T. Madas
Question 23
The function f is defined by
1
f ( x) = ,
x + a2
2
1 ˆ π e−a k
F 2 = f (k ) =
x + a 2 2 a
Created by T. Madas
Created by T. Madas
Question 24
The function f is defined by
2
f ( x ) = x e− x , x ∈ ℝ .
2
F x e− x = 1 k 2 e 4
2 −1k
4
Created by T. Madas
Created by T. Madas
Question 25
The function f is defined by
x
f ( x) = ,
x + a2
2
−a k
x ˆ π e sign k
MM3-F , F 2 2
= f (k ) = − i
x +a 2 a
Created by T. Madas
Created by T. Madas
Question 26
Find the inverse Fourier transform of
2 2
σ t
gˆ ( k ) = e− k ,
x2
F −1 e− k σ t
2 2 1
exp −
4tσ 2
=
2t σ
proof
Created by T. Madas
Created by T. Madas
Question 27
The Fourier transform fˆ ( k ) , of function f ( x ) is
2 a
fˆ ( k ) = ,
π a + k2
2
ax
f ( x) = e
Created by T. Madas
Created by T. Madas
Question 28
The function f is defined by
1
f ( x) = ,
2 2
(x 2
+a )
where a is a positive constant.
−a k
1 ˆ π (1 + a k ) e
MM3-C , F 2
= f (k ) =
a3
2
x + a
2
( )
8
Created by T. Madas
Created by T. Madas
VARIOUS PROBLEMS
on
FOURIER
TRANSFORMS
Created by T. Madas
Created by T. Madas
Question 1
Find the Fourier transform of an arbitrary function f ( x ) if
i. f ( x ) is even.
ii. f ( x ) is odd.
∞ ∞
∫ ∫
2 2
fˆ ( k ) = f ( x ) cos kx dx , fˆ ( k ) = −i f ( x ) sin kx dx
π 0 π 0
Question 2
Use the definition of the Fourier transform, of an absolutely integrable function f ( x ) ,
to show that
F f ′ ( x ) = i k F f ( x ) .
proof
Created by T. Madas
Created by T. Madas
Question 3
The Fourier transform of an absolutely integrable function f ( x ) , is denoted by fˆ ( k ) .
Show that
d
F x f ( x ) = i fˆ ( k ) .
dk
proof
Question 4
Given that c is a constant show that
F f ( x + c ) = ei kc F f ( x ) .
proof
Created by T. Madas
Created by T. Madas
Question 5
Given that c is a constant show that
F −1 fˆ ( k + c ) = ei cx f ( x ) ,
where fˆ ( k ) ≡ F f ( x )
proof
Created by T. Madas
Created by T. Madas
Question 6
Given that c is a constant prove the validity of the two shift theorems
a) F f ( x + c ) = ei kc F f ( x ) .
b) F −1 fˆ ( k + c ) = ei cx f ( x ) .
Note that fˆ ( k ) ≡ F f ( x ) .
proof
Created by T. Madas
Created by T. Madas
Question 7
The convolution [ f ∗ g ] ( x ) , of two functions f ( x ) and g ( x ) is defined as
∞
[ f ∗ g ]( x ) =
∫
−∞
f ( x − y ) g ( y ) dy .
Show that
F {[ f ∗ g ] ( x )} = 2π F f ( x ) F g ( x ) = 2π fˆ ( k ) gˆ ( k ) .
proof
Created by T. Madas
Created by T. Madas
Question 8
It is given that c is a constant and fˆ ( k ) ≡ F f ( x ) .
F −1 fˆ ( k + c ) = ei cx f ( x ) .
− k −a 2
F −1 e ( ) ,
− k −a 2 1 − 14 x2
F −1 e ( ) = e [ cos ax + isin ax ]
2
Created by T. Madas
Created by T. Madas
Question 9
The convolution theorem for two functions f ( x ) and g ( x ) asserts that
F {[ f ∗ g ] ( x )} = 2π F f ( x ) F g ( x ) ,
where
∞
[ f ∗ g ]( x ) =
∫ −∞
f ( x − y ) g ( y ) dy .
∞ ∞
∫ ∫
2 2
h( y) dy = hˆ ( k ) dk .
−∞ −∞
∫
1
dx .
0 x + a2
2
−a x 2 a
You may assume that if f ( x ) = e , then fˆ ( k ) = .
π a + k2
2
π
4a3
Created by T. Madas
Created by T. Madas
Question 10
The convolution [ f ∗ g ] ( x ) , of two functions f ( x ) and g ( x ) is defined as
∞
[ f ∗ g ]( x ) =
∫−∞
f ( x − y ) g ( y ) dy .
a) Show that
F {[ f ∗ g ] ( x )} = 2π F f ( x ) F g ( x ) = 2π fˆ ( k ) gˆ ( k ) .
∞ ∞
∫ −∞
h ( y ) g ( y ) dy =
∫ −∞
hˆ ( k ) gˆ ( k ) dk .
∫(
1
dx .
0
x2 + a2 )( x2 + b2 )
−a x 2 a
You may assume that if f ( x ) = e , then fˆ ( k ) = .
π a + k2 2
π
2ab ( a + b )
Created by T. Madas
Created by T. Madas
APPLICATIONS
of
FOURIER
TRANSFORMS
Created by T. Madas
Created by T. Madas
Question 1
The function ϕ = ϕ ( x, y ) satisfies Laplace’s equation in Cartesian coordinates
∂ 2ϕ ∂ 2ϕ
+ = 0.
∂x 2 ∂y 2
Use Fourier transforms to convert the above partial differential equation into an
ordinary differential equation for ϕˆ ( k , y ) , where ϕˆ ( k , y ) is the Fourier transform of
ϕ ( x, y ) with respect to x .
d 2ϕˆ
2
− k 2ϕˆ = 0
dx
Created by T. Madas
Created by T. Madas
Question 2
The function ϕ = ϕ ( x, y ) satisfies Laplace’s equation in Cartesian coordinates,
∂ 2ϕ ∂ 2ϕ
+ = 0,
∂x 2 ∂y 2
• ϕ ( x, y ) → 0 as x2 + y 2 → ∞
1 x <1
• ϕ ( x,0 ) = 2
0 x >1
∫
1 1 − ky
ϕ ( x, y ) = e sin k cos kx dk ,
π 0 k
MM4-B , ϕ ( ±1, 0 ) = 1
4
Created by T. Madas
Created by T. Madas
Question 3
The Airy function Ai ( x ) satisfies the differential equation
d2y
− xy = 0 .
dx 2
∫
1
Ai ( x ) =
π 0
(3 )
cos 1 t 3 + xt dt ,
d
You may assume that F x f ( x ) = i
dk
{
F f ( x ) . }
proof
Created by T. Madas
Created by T. Madas
Question 4
The function ψ = ψ ( x, y ) satisfies Laplace’s equation in Cartesian coordinates,
∂ 2ψ ∂ 2ψ
+ =0,
∂x 2 ∂y 2
• ψ ( x, y ) → 0 as x2 + y 2 → ∞
Use Fourier transforms to convert the above partial differential equation into an
ordinary differential equation and hence show that
1 y
ψ ( x, y ) = 2 .
π x + y2
MM4-C , proof
Created by T. Madas
Created by T. Madas
Question 5
The function u = u ( x, t ) satisfies the partial differential equation
∂u 1 ∂ 3u
+ = 0.
∂t 3 ∂x3
• u ( x, t ) → 0 as x → ∞
Use Fourier transforms to convert the above partial differential equation into an
ordinary differential equation and hence show that
1 x
u ( x, t ) = Ai 1 ,
1
t3 t3
∫
1
Ai ( x ) = cos 1 k 3 + kx dk .
π 3
0
proof
Created by T. Madas
Created by T. Madas
Question 6
The function ϕ = ϕ ( x, y ) satisfies Laplace’s equation in Cartesian coordinates,
∂ 2ϕ ∂ 2ϕ
+ = 0,
∂x 2 ∂y 2
• ϕ ( x, y ) → 0 as x2 + y 2 → ∞
∂
• ϕ ( x,0 ) = 0
∂x
Use Fourier transforms to convert the above partial differential equation into an
ordinary differential equation and hence show that
y +1
ϕ ( x, y ) = 2
.
2
x + ( y + 1)
proof
Created by T. Madas
Created by T. Madas
Question 7
The function Φ = Φ ( x, y ) satisfies Laplace’s equation in Cartesian coordinates,
∂ 2Φ ∂ 2Φ
+ = 0,
∂x 2 ∂y 2
• Φ ( x, y ) → 0 as x2 + y 2 → ∞
Use Fourier transforms to find the solution of the above partial differential equation
and hence show that
−1
1 y2
δ ( x ) = lim 1 + 2 .
α →0 πα α
proof
Created by T. Madas
Created by T. Madas
Question 8
The function y = y ( x ) satisfies the differential equation
dy
+ λ y = f ( x) ,
dx
∞
y ( x) =
∫
0
eλt f ( x − t ) dt .
proof
Created by T. Madas
Created by T. Madas
Question 9
The function ϕ = ϕ ( x, y ) satisfies Laplace’s equation in Cartesian coordinates,
∂ 2ϕ ∂ 2ϕ
+ = 0,
∂x 2 ∂y 2
• ϕ ( x, y ) → 0 as x2 + y 2 → ∞
Use Fourier transforms to convert the above partial differential equation into an
ordinary differential equation and hence show that
∞
f ( x − u)
∫
y
ϕ ( x, y ) = du .
π −∞ u2 + y2
MM4-E , proof
Created by T. Madas
Created by T. Madas
Question 10
The function ϕ = ϕ ( x, y ) satisfies Laplace’s equation in Cartesian coordinates,
∂ 2ϕ ∂ 2ϕ
+ = 0,
∂x 2 ∂y 2
∂ ∂
• ϕ ( x,0 ) = f ( x )
∂y ∂x
• ϕ ( x, y ) → 0 as x2 + y 2 → ∞
Use Fourier transforms to convert the above partial differential equation into an
ordinary differential equation and hence show that
∞
f (u )
∫
1
ϕ ( x, 0 ) = du .
π −∞ x −u
proof
[ solution overleaf ]
Created by T. Madas
Created by T. Madas
Created by T. Madas
Created by T. Madas
Question 11
The function ϕ = ϕ ( x, y ) satisfies Laplace’s equation in Cartesian coordinates,
∂ 2ϕ ∂ 2ϕ
+ = 0 , −∞ < x < ∞ , y ≥ 0 .
∂x 2 ∂y 2
ϕ ( x, y ) → 0 as x2 + y 2 → ∞
1 1 x
ϕ ( x, y ) = + arctan .
2 π y
1 1
F [ H( x) ] = π δ ( k ) + i k .
2π
proof
Created by T. Madas
Created by T. Madas
Question 12
The function u = u ( x, y ) satisfies Laplace’s equation in Cartesian coordinates,
∂ 2u ∂ 2u
+ =0, −∞ < x < ∞ , 0 < y < 1.
∂x 2 ∂y 2
u ( x,1) = f ( x )
where f ( − x ) = f ( x ) and f ( x ) → 0 as x → ∞
∫
2 fˆ ( k ) cos kx sinh ky
u ( x, y ) = dk , fˆ ( k ) = F f ( x ) .
π sinh k
−∞
sin π y
u ( x, y ) = .
2 [ cosh π x + cos π y ]
proof
Created by T. Madas
Created by T. Madas
Question 13
The function ψ = ψ ( x, y ) satisfies Laplace’s equation in Cartesian coordinates,
∂ 2ψ ∂ 2ψ
+ =0,
∂x 2 ∂y 2
ψ ( x, y ) → 0 as x2 + y 2 → ∞
c) Use Fourier transforms to convert the above partial differential equation into
an ordinary differential equation and hence show that
∫
y f (u )
ψ ( x, y ) = du .
π
−∞
( x − u )2 + y 2
ii. … f ( x ) = sgn x
iii. … f ( x ) = H ( x )
2 x 1 1 x
ψ ( x, y ) = 1 , ψ ( x, y ) = arctan , ψ ( x, y ) = + arctan
π y 2 π y
[ solution overleaf ]
Created by T. Madas
Created by T. Madas
Created by T. Madas
Created by T. Madas
Question 14
The function θ = θ ( x, t ) satisfies the heat equation in one spatial dimension,
∂ 2θ 1 ∂θ
2
= , −∞ < x < ∞ , t ≥ 0 ,
∂x σ 2 ∂t
Given further that θ ( x,0 ) = f ( x ) , use Fourier transforms to convert the above partial
differential equation into an ordinary differential equation and hence show that
∞
u2
∫
1
θ ( x, t ) = f ( x − u ) exp
4tσ 2
du .
2σ π t −∞
proof
Created by T. Madas
Created by T. Madas
Question 15
The function u = u ( x, y ) satisfies Laplace’s equation in Cartesian coordinates,
∂ 2u ∂ 2u
+ =0,
∂x 2 ∂y 2
u ( 0, y ) = 0
u ( x, y ) → 0 as x2 + y 2 → ∞
u ( x,0 ) = f ( x ) , f ( 0 ) = 0 , f ( x ) → 0 as x → ∞
∫
y 1 1
u ( x, y ) = f ( w) 2
− 2
dw .
π y 2 + ( x − w ) y 2 + ( x + w )
0
proof
[ solution overleaf ]
Created by T. Madas
Created by T. Madas
Created by T. Madas
Created by T. Madas
Question 16
The function T = T ( x, t ) satisfies the heat equation in one spatial dimension,
∂ 2θ 1 ∂θ
2
= , x ≥ 0 , t ≥ 0,
∂x σ ∂t
• T ( 0, t ) = 0
• T ( x, t ) → 0 as x → ∞
Use Fourier transforms to convert the above partial differential equation into an
ordinary differential equation and hence show that
∫
1 ( x − u )2
T ( x, t ) = f ( u ) exp du .
4πσ t 4tσ
−∞
2
1 4k a
You may assume that F e ax =
2
e .
2a
proof
Created by T. Madas
Created by T. Madas
Question 17
The function f = f ( x ) satisfies the integral equation
∫
f (t ) 1
2
dt = 2
,
−∞ (x −t) +1 x +4
where f ( x ) → 0 as x → ∞
Use Fourier transforms to find the solution of the above integral equation.
1 1 π −a k
You may assume that F 2 = e .
x + a 2 a 2
1
f ( x) =
(
2π 1 + x 2 )
Created by T. Madas
Created by T. Madas
Question 18
The function f = f ( x ) satisfies the integral equation
∫
1
f ( x − u ) f ( u ) du = ,
−∞ 1 + x2
where f ( x ) → 0 as x → ∞
Use Fourier transforms to find the solution of the above integral equation.
∫
cos kx
dx = 1 π e .
k
2
x +1 2
0
2
f ( x) =
(1 + 4 x2 ) π
Created by T. Madas
Created by T. Madas
Question 19
The function f = f ( x ) satisfies the integral equation
∫
− 12 x 2 − x −u
e =1 e f ( u ) du ,
2
−∞
where f ( x ) → 0 as x → ∞
Use Fourier transforms to find the solution of the above integral equation.
2
1 4k a
F e ax =
2
• e .
2a
• F e
ax = 2 a
.
π a + k2
2
− 12 x 2
( )
f ( x ) = 2 − x2 e
Created by T. Madas