Fourier Transform Theorems
• Addition Theorem
• Shift Theorem
• Convolution Theorem
• Similarity Theorem
• Rayleigh’s Theorem
• Differentiation Theorem
Addition Theorem
F { f + g} = F + G
Proof:
F { f + g}(s) =
=
Z
∞
−∞
f (t)e
Z
∞
−∞
− j2πst
= F(s) + G(s)
[ f (t) + g(t)] e− j2πst dt
dt +
Z
∞
−∞
g(t)e− j2πst dt
Shift Theorem
Proof:
F { f (t − t0)}(s) = e− j2πst0 F(s)
F { f (t − t0)}(s) =
Z
∞
−∞
f (t − t0)e− j2πst dt
Multiplying the r.h.s. by e j2πst0 e− j2πst0 = 1 yields:
=
Z
∞
−∞
= e
F { f (t − t0)}(s)
f (t − t0)e− j2πst e j2πst0 e− j2πst0 dt
− j2πst0
Z
∞
−∞
f (t − t0)e− j2πs(t−t0)dt.
Substituting u = t − t0 and du = dt yields:
F { f (t − t0)}(s) = e
− j2πst0
Z
∞
−∞
f (u)e− j2πsudu
= e− j2πst0 F(s).
Shift Theorem Example
F {sin (2π (t + 1/4))} (s)
=e
=
=
=
=
=
=
j2πs
4
F {sin(2πt)}
j
e · [δ(s + 1) − δ(s − 1)]
2
i
h
jπs
j jπs
e 2 δ(s + 1) − e 2 δ(s − 1)
2h
i
jπ(+1)
j jπ(−1)
e 2 δ(s + 1) − e 2 δ(s − 1)
2
j
[− jδ(s + 1) − jδ(s − 1)]
2
1
[δ(s + 1) + δ(s − 1)]
2
F {cos(2πs)}
jπs
2
Shift Theorem (variation)
−1
F
Proof:
F
−1
{F(s − s0)}(t) = e j2πs0 t f (t)
{F(s − s0)}(t) =
Z
∞
−∞
F(s − s0)e j2πst ds
Multiplying the r.h.s. by e j2πs0 t e− j2πs0 t = 1 yields:
−1
F
=
Z
∞
−∞
=e
{F(s − s0)}(t)
F(s − s0)e j2πst e j2πs0 t e− j2πs0 t ds
j2πs0 t
Z
∞
−∞
F(s − s0)e j2π(s−s0)t ds.
Substituting u = s − s0 and du = ds yields:
F
−1
{F(s − s0)}(t) = e
j2πs0 t
Z
∞
−∞
F(u)e j2πut du
= e j2πs0 t f (t).
Convolution Theorem
F { f ∗ g} = F · G
Proof:
=
Z ∞ Z
−∞
∞
−∞
F { f ∗ g}(s)
f (u)g(t − u)du e− j2πst dt
Changing the order of integration:
F { f ∗ g}(s)
Z ∞
Z ∞
g(t − u)e− j2πst dt du
f (u)
=
−∞
−∞
By the Shift Theorem, we recognize that
Z ∞
g(t − u)e− j2πst dt = e− j2πsuG(s)
so that
−∞
F { f ∗ g}(s) =
Z
∞
f (u)e− j2πsuG(s)du
−∞
= G(s)
Z
∞
−∞
f (u)e− j2πsudu
= G(s) · F(s)
Convolution Theorem Example
The pulse, Π, is defined as:
1 if |t| ≤ 12
Π(t) =
0 otherwise.
The triangular pulse, Λ, is defined as:
1 − |t| if |t| ≤ 1
Λ(t) =
0
otherwise.
It is straightforward to show that Λ = Π ∗ Π.
Using this fact, we can compute F {Λ}:
F {Λ}(s) = F {Π ∗ Π}(s)
= F {Π}(s) · F {Π}(s)
sin(πs) sin(πs)
=
·
πs
πs
sin2(πs)
=
.
2
2
πs
Convolution Theorem (variation)
F
−1
Proof:
F
=
Z ∞ Z
−∞
∞
−∞
{F ∗ G} = f · g
−1
{F ∗ G}(t)
F(u)G(s − u)du e j2πst ds
Changing the order of integration:
F −1{F ∗ G}(t)
Z ∞
Z ∞
G(s − u)e j2πst ds du
F(u)
=
−∞
−∞
By the Shift Theorem, we recognize that
Z ∞
G(s − u)e j2πst ds = e j2πtug(t)
so that
F
−1
−∞
{F ∗ G}(t) =
Z
∞
F(u)e j2πtug(t)du
−∞ Z
∞
= g(t)
−∞
F(u)e j2πtudu
= g(t) · f (t)
Similarity Theorem
Proof:
1 s
F { f (at)} (s) = F
|a| a
F { f (at)}(s) =
There are two cases.
Z
∞
f (at)e− j2πst dt
−∞
• a > 0. Multiplying the integral by a/|a| = 1
and the exponent by a/a = 1 yields:
Z
1 ∞
f (at)e− j2π(s/a)at adt
F { f (at)}(s) =
|a| −∞
We now make the substitution u = at and
du = adt:
Z
1 ∞
F { f (at)}(s) =
f (u)e− j2π(s/a)udu
|a| −∞
1 s
=
F
|a| a
Similarity Theorem (contd.)
• a < 0. Multiplying the integral by |a|/|a| = 1
and the exponent by −|a|/a = 1 and using
the fact that a = −|a| yields:
F { f (at)}(s) =
1 t=∞
f (−|a|t)e− j2π(s/a)(−|a|t)|a|dt
|a| t=−∞
We now make the substitution u = −|a|t and
du = −|a|dt:
Z
1 u=−∞
f (u)e− j2π(s/a)udu
F { f (at)}(s) = −
|a| u=∞
Z
1 ∞
f (u)e− j2π(s/a)udu
=
|a| −∞
1 s
=
F
|a| a
Z
Similarity Theorem Example
Let’s compute, G(s), the Fourier transform of:
−t 2/9
g(t) = e
.
We know that the Fourier transform of a Gaussian:
−π t 2
f (t) = e
is a Gaussian:
−π s2
.
F(s) = e
We also know that :
1 s
F { f (at)}(s) = F
.
|a| a
We need to write g(t) in the form f (at):
g(t) = f (at) = e
Let a = 3√1 π :
g(t) = e
−t 2/9
=e
1 t
−π 3√
π
2
−π(at)2
=f
.
It follows that:
√ −π(3√πs)2
.
G(s) = 3 π e
1
√ t .
3 π
Rayleigh’s Theorem
Z
∞
−∞
Proof:
Z
| f (t)|2dt =
∞
−∞
2
| f (t)| dt =
Z
∞
−∞
Z
∞
−∞
|F(s)|2ds
f ∗(t) f (t)dt
Substituting {F −1{F}}∗(t) for f ∗(t):
∗
Z ∞ Z ∞
F(s)e j2πst ds f (t) dt
=
−∞
−∞
Z ∞ Z ∞
=
F ∗(s)e− j2πst ds f (t) dt
−∞
−∞
Rayleigh’s Theorem (contd.)
Changing the order of integration:
Z t=∞ Z s=∞
=
F ∗(s)e− j2πst ds f (t) dt
t=−∞
s=−∞
Z t=∞
Z s=∞
e− j2πst f (t) dt ds
=
F ∗(s)
=
=
Zs=−∞
∞
t=−∞
F ∗(s)F(s) ds
Z−∞
∞
−∞
|F(s)|2 ds
Differentiation Theorem
F { f 0}(s) = j2πsF(s)
Proof:
f =F
Therefore
f (t) =
Z
∞
−1
{F}
F(s)e j2πst ds
−∞
Differentiating both sides with respect to t:
f 0(t) =
or
Z
∞
j2πsF(s)e j2πst ds
−∞
f 0 = F −1{ j2πsF(s)}
Taking the Fourier transform of both sides:
F { f 0}(s) = F {F
−1
{ j2πsF(s)}}
= j2πsF(s)
Differentiation Theorem Example
F
d sin(2πt)
(s)
dt
= j2πsF {sin(2πt)} (s)
j
= j2πs · [δ(s + 1) − δ(s − 1)]
2
= −πs [δ(s + 1) − δ(s − 1)]
= π [sδ(s − 1) − sδ(s + 1)]
= π [(+1)δ(s − 1) − (−1)δ(s + 1)]
= π [δ(s − 1) + δ(s + 1)]
= πF {2 cos(2πt)}(s)
= F {2π cos(2πt)}(s)