7.7 Fourier Transform
Theorems, Part II
We have already covered four theorems.
The following three theorems will be covered
in this section
• shift theorem
• derivative theorem
• convolution theorem
The autocorrelation theorem will be covered
under electronics. The derivative/
convolution theorem will not be covered.
7.7 : 1/9
Shift Theorem
The Fourier transform of a shifted variable is the transform of the
un-shifted variable multiplied by a complex harmonic.
Given,
then,
F (t ) ↔ Φ ( f )
F ( t ± t ′ ) ↔ Φ ( f ) e± i 2π ft ′
where t' is a constant.
Proof: Start with the integral for
the forward transform.
Multiply the integrand by
e+i2πft'e-i2πft'.
∞
∫
F ( t − t ′ ) e−i 2π ft dt
−∞
∞
=
∫
F ( t − t′) e
−i 2π f ( t −t ′ ) −i 2π ft ′
e
dt
−∞
Since dt = d(t-t') the previous
integral can be written in the
form of a forward transform.
=e
−i 2π ft ′
∞
∫
F ( t − t′) e
−∞
= e−i 2π ft′Φ ( f )
7.7 : 2/9
−i 2π f ( t −t ′ )
d ( t − t′)
Shift Theorem Example
2A
The Fourier transform of a single impulse
at f = f 0 with amplitude, A, can be
determined by the shift theorem.
A
First write the transform pair for the
single impulse function δ(f=0).
δ(f=0) ↔ C(t) = A
-f 0
Now write the transform of the impulse at
f 0 using the shift theorem.
δ(f-f 0=0) ↔ Aexp(-i2πf 0t) = Acos(2πf 0t)-iAsin(2πf 0t)
The temporal result can be written as a phase shifted cosine.
R = A2 + A2 = 2 A
θ = tan −1 ( − A A ) = −π 4
(
F ( t ) = 2 A cos 2π f 0t − π 4
7.7 : 3/9
)
f0
Derivative Theorem
F (t ) ↔ Φ ( f )
Given,
then,
dF ( t )
↔ i 2π f Φ ( f )
dt
dΦ ( f )
↔ −i 2π tF ( t )
df
The theorem can be extended to multiple levels of differentiation.
d n F (t )
dt n
↔ ( i 2π f ) Φ ( f )
n
Proof:
F (t ) =
∞
∫
d nΦ ( f )
df n
↔ ( −i 2π t ) F ( t )
Φ ( f ) ei 2π ft df
−∞
dF ( t )
= i 2π
dt
7.7 : 4/9
∞
∫
−∞
f Φ ( f ) ei 2π ft df
n
Derivative Theorem Example
What is the Fourier transform of the derivative of a cosine,
cos(2πt/t 0)?
(
)
( )
cos 2π t t 0 ↔ δδ + f 0
Given,
(
d cos 2π t t 0
then,
dt
) ↔ i2π f δδ +
( f 0)
where,
(
d cos 2π t t 0
dt
) = − 2π sin 2π t t 0
(
)
t0
i2πf 0
7.7 : 5/9
0
f0
( )
iπf 0
1/2
-f 0
-i2πf 0
( )
and i 2π f δδ + f 0 = −i 2π f 0δδ − f 0
→
-1/2
-f 0
f0
-iπf 0
Convolution Theorem
Consider the following two Fourier transform pairs.
F1 (t ) ↔ Φ1 ( f ) and
F2 (t ) ↔ Φ 2 ( f )
Let a new function, C(t) be the convolution of F1(t) and F2(t).
C (t ) ≡ F1 ⊗ F2 = ∫
+∞
F (t ′) F2 (t − t ′) dt ′
−∞ 1
The convolution theorem gives the following relationships
between the convolution and multiplication operators.
F1 (t ) ⊗ F2 (t ) ↔ Φ1 ( f ) ⋅ Φ 2 ( f )
F1 (t ) ⋅ F2 (t ) ↔ Φ1 ( f ) ⊗ Φ 2 ( f )
This is the most important and useful theorem - the convolution
of two functions in the time domain has a transform given by
multiplication in the frequency domain.
The theorem can eliminate the need to evaluate convolution
integrals.
7.7 : 6/9
Theorem Proof
Start by writing out the Fourier transform of the convolved function.
ΦC ( f ) = ∫
+∞
−∞
C (t )e−i 2π ft dt
Substitute the convolution integral for C(t).
ΦC ( f ) = ∫
+∞ +∞
∫
−∞ −∞
F1 (t ′) F2 (t − t ′) dt ′e −i 2π ft dt
Let t - t' = T, and dt = dT.
ΦC ( f ) = ∫
+∞ +∞
∫
−∞ _ ∞
F1 (t ′) F2 (T )e −i 2π f (t′+T ) dt ′dT
Finally, separate the double integral into a product of integrals.
ΦC ( f ) = ∫
+∞
−∞
F1 (t ′)e −i 2π ft′dt ′∫
Φ C ( f ) = Φ1 ( f ) ⋅ Φ 2 ( f )
7.7 : 7/9
+∞
−∞
F2 (T )e−i 2π fT dT
Properties of Convolution
Commutation:
Distribution:
Association:
F1 (t ) ⊗ F2 (t ) = F2 (t ) ⊗ F1 (t )
F1 (t ) ⊗ [ F2 (t ) + F3 (t ) ] = F1(t ) ⊗ F2 (t ) + F1(t ) ⊗ F3 (t )
F1 ( t ) ⊗ ⎣⎡ F2 ( t ) ⊗ F3 ( t ) ⎤⎦ = ⎡⎣ F1 ( t ) ⊗ F2 ( t ) ⎤⎦ ⊗ F3 ( t )
Combinations with multiplication: multiplication and convolution
do not commute!
⎡
⎤
⎢ F (t ) ⊗ F (t )⎥ ⋅ F (t )
⎣ 1
2 ⎦ 3
↔ Φ ( f ) ⋅ ⎡⎢Φ ( f ) ⊗Φ ( f )⎤⎥
1
⎡
⎤
⎡
⎢ F (t ) ⊗ F (t )⎥ ⋅ F (t ) ↔ ⎢Φ ( f
⎣ 1
2 ⎦ 3
⎣ 1
⎤
⎡
⎤ ⎡
⎡
⎢ F (t ) ⊗ F (t )⎥ ⋅ ⎢ F (t ) ⊗ F (t ) ⎥ ↔ ⎢Φ ( f
4 ⎦ ⎣ 1
2 ⎦ ⎣ 3
⎣ 1
⎣
2
3
⎦
) ⋅Φ ( f )⎤⎥ ⊗Φ ( f )
2
⎦
3
) ⋅Φ ( f )⎤⎥ ⊗ ⎡⎢Φ ( f ) ⋅Φ ( f )⎤⎥
4 ⎦
2
⎦ ⎣ 3
A more complicated example:
[ F1(t ) ⊗ F2 (t ) + F3 (t )i F4 (t )] ⋅ F5 (t ) ↔ [Φ1( f ) ⋅ Φ 2 ( f ) + Φ3 ( f ) ⊗ Φ 4 ( f )] ⊗ Φ5 ( f )
7.7 : 8/9
Convolution Examples
Gaussian: The convolution of two Gaussian functions is a
Gaussian function,
gauss
( ) ⊗ gauss ( )
t10
t20
⎛ 0
= gauss ⎜ t3 =
⎜
⎝
( ) +( )
t10
2
t20
2
⎞
⎟⎟
⎠
where the variance of the two convolved functions adds to give
the variance of the result. In some cases it is more convenient
to use the FWHM, Γ32 = Γ12 + Γ22.
Impulse: The convolution of an impulse with any other function,
replicates the function and centers it at the position of the
impulse. For this reason, the impulse function is called a
replicator.
Comb: The convolution of a comb with any function with a size
smaller than the comb spacing, replicates the function at the
center of each comb impulse. Thus a comb function is a
multiple copy replicator.
7.7 : 9/9