ELEC 8501: The Fourier Transform and Its Applications
E. Lam
Handout #4
Mar 3, 2008
Some Properties of Fourier Transform
1
Addition Theorem
If g(x) ⊃ G(s) and h(x) ⊃ H(s), and a and b are some scalars, then
ag(x) + bh(x) ⊃ aG(s) + bH(s).
(1)
Proof.
F{ag(x) + bh(x)} =
= a
∞
∞
[ag(x) + bh(x)]e−j2πsx dx
∞
g(x)e−j2πsx dx + b
−∞ −∞
∞ ∞
−∞
−∞
= aG(s) + bH(s).
2
−∞
∞
h(x)e−j2πsx dx
−∞
Convolution Theorem
If g(x) ⊃ G(s) and h(x) ⊃ H(s), then
g(x) ∗ h(x) ⊃ G(s)H(s).
(2)
Proof.
g(u)h(x − u) du e−j2πsx dx
F{g(x) ∗ h(x)} =
−∞
−∞
∞
∞
−j2πsx
g(u)
h(x − u)e
dx du
=
−∞
−∞
∞
g(u)e−j2πus H(s) du
=
∞
∞
−∞
= G(s)H(s).
3
Similarity Theorem
If g(x) ⊃ G(s), then
g(ax) ⊃
1 s
G
.
|a|
a
1
(3)
Proof.
∞
−j2πsx
g(ax)e
dx =
−∞
=
∞
1
g(ax)e−j2π(s/a)(ax) d(ax)
|a| −∞
1 s
G
.
|a|
a
This means a contraction in one domain gives rise to expansion in another domain, and vice
versa.
4
Shift Theorem
If g(x) ⊃ G(s), then
Proof.
∞
−∞
g(x − a) ⊃ e−j2πas G(s).
−j2πsx
g(x − a)e
∞
dx =
−∞
(4)
g(x − a)e−j2πs(x−a) e−j2πsa d(x − a)
= e−j2πas G(s).
A shift in position in one domain gives rise to a phase change in another domain. In a similar
fashion, we can show that
(5)
ej2πax g(x) ⊃ G(s − a).
5
Modulation Theorem
If g(x) ⊃ G(s), then
g(x) cos(2πax) ⊃
Proof.
F{g(x) cos(2πax)} =
=
=
1
2
1
2
1
2
∞
−∞
∞
1
2
[G(s − a) + G(s + a)] .
g(x)ej2πax e−j2πsx dx +
g(x)e−j2π(s−a)x dx +
−∞
1
2
1
2
∞
(6)
g(x)e−j2πax e−j2πsx dx
−∞
∞
g(x)e−j2π(s+a)x dx
−∞
[G(s − a) + G(s + a)] .
Alternatively, if we make use of the fourier transform of a cosine and the convolution theorem
in Equation 2,
F{g(x) cos(2πax)} = F{g(x)} ∗ F{cos(2πax)}
= G(s) ∗ 12 [δ(s − a) + δ(s + a)]
=
1
2
[G(s − a) + G(s + a)] .
Multiplication with a cosine has the effect of shifting the spectrum to center on the frequency
of the cosine.
2
6
Time Reversal Theorem
If g(x) ⊃ G(s), then
g(−x) ⊃ G(−s).
(7)
Proof.
F{g(−x)} =
∞
g(−x)e−j2πsx dx
−∞
−∞
=
g(x̃)e−j2π(−s)x̃ (− dx̃)
∞
(x̃ = −x)
= G(−s).
7
Derivative Theorem
If g(x) ⊃ G(s), then
d
g(x) ⊃ j2πsG(s).
dx
(8)
Proof.
∞
G(s)ej2πsx ds
∞
d
d
g(x) =
G(s)ej2πsx ds
dx
dx
∞ −∞
d j2πsx e
G(s)
ds
=
dx
−∞
∞
[j2πsG(s)] ej2πsx ds
=
−∞
d
g(x)
= j2πsG(s).
F
dx
g(x) =
−∞
Hence
By extension, we have
8
dn
g(x) ⊃ (j2πs)n G(s).
dxn
(9)
g(x cos θ + y sin θ, −x sin θ + y cos θ) ⊃ G(fX cos θ + fY sin θ, −fX sin θ + fY cos θ).
(10)
2D rotation Theorem
If g(x, y) ⊃ G(fX , fY ), then
In words, that means an anti-clockwise rotation of a function by an angle θ implies that its Fourier
transform is also rotated anti-clockwise by the same angle.
Proof. We can define a new coordinate system (x̌, y̌), where
x̌
cos θ sin θ x
=
.
y̌
− sin θ cos θ y
3
(11)
This is shown in Figure 1. We can also express (x, y) in terms of the new coordinate system, where
x
cos θ − sin θ x̌
x̌ cos θ − y̌ sin θ
=
=
.
y
sin θ cos θ
y̌
x̌ sin θ + y̌ cos θ
Thus,
F{g(x̌, y̌)} =
∞
∞
−∞
−∞
∞ ∞
=
−∞
−∞
∞ ∞
=
−∞
−∞
∞ ∞
=
−∞
g(x̌, y̌)e−j2π(xfX +yfY ) dx dy
g(x̌, y̌)e−j2π(x̌fX cos θ−y̌fX sin θ+x̌fY
sin θ+y̌fY cos θ)
dx dy
g(x̌, y̌)e−j2π[x̌(fX cos θ+fY
sin θ)+y̌(−fX sin θ+fY cos θ)]
dx dy
g(x̌, y̌)e−j2π[x̌(fX cos θ+fY
sin θ)+y̌(−fX sin θ+fY cos θ)]
dx̌ dy̌
−∞
= G(fX cos θ + fY sin θ, −fX sin θ + fY cos θ).
In the derivation above, we have made use of the fact that
∂ x̌ ∂ y̌ cos θ − sin θ
∂x ∂x dx dy = dx dy.
dx̌ dy̌ = ∂ x̌ ∂ y̌ dx dy = sin θ cos θ ∂y ∂y One consequence of the two-dimensional rotation theorem is that if the 2D function is circularly
symmetric, its Fourier transform must also be circularly symmetric.
y
x̌
y̌
θ
x
Figure 1: Illustration of a rotation in coordinates.
4