5. Fundamental Theorems We are now at a stage, where we can ...

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The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
1


5. Fundamental Theorems
We are now at a stage, where we can formulate properly (and proof) the fundamental
theorems of Fourier Theory:
5.1 Fourier Theorem
"Fourier Trafo and subsequent Back-Trafo reproduces the original function"
With the Fourier-Trafo of
f

x

,
F

s

=





e

2

ixs
f

x

dx
(

2

i
-trafo), one obtains
f

x

through the "Back-Trafo" (

2

i
-trafo):
f

x

=





e

2

i
xs
F

s

ds
=





ds
e
2

i
xs





dx
'
e

2

i
xs
f

x
'

=





dx
'





ds
e
2

i

x

x
'

s



x

x
'

f

x
'

=
f

x

, q.e.d.
Note: the tough work has been done by proving that the FT of the

-function is



s

=
1
!
math_theorems_1.odt
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
2

5.2 Scaling Theorem
with
g

x

=
f

ax

:
G

s

=
1

a

F

s
a

Proof:
G

s

=





e

2

i
xs
g

x

dx
=





e

2

i
xs
f

ax

dx
=
1
a





e

2

i
ax
s
a
f

ax

a
dx
=
1

a






e

2

i
u
s
a
f

u

du
=
1

a

F

s
a

i.e.:
stretching in the spatial/time-domain implies compressing in the frequency-domain,
and vice versa
5.3. Linearity Theorem
the linear combination
a
f

x


b
g

x

has as its FT the linear combination
a
F

s


b
G

s

Trivial, but nevertheless useful!
Example:


x

=


x




2x

;



s

=



s


1
2



s
2

=
sinc

s


1
2
sinc
2

s
2

math_theorems_2.odt
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
3

5.4 Shifting-Theorem
with
f

x


F

s

:
f

x

a


e

2

i
as
F

s

,
i.e. a translation of a in the time/spatial domain

multiplication with phase factor
e

2

i
as
in frequency domain
Proof:

e

2

i
xs
f

x

a

dx
=
e

2

i
a
s

e

2

i

x

a

s
f

x

a

dx
=
e

2

i
a
s
F

s

5.5 Modulation-Theorem
(the inverse of the Shifting-Theorem)
with
f

x


F

s

FT of
f

x

e
2

i

x
, i.e. of a oscillation with frequency

(amplitude-)modulated with
f

x

?
simple calculus gives

f

x

e
2

i

x
e

2

i
xs
dx
=

f

x

e

2

i

s



x
dx
=
F

s



, i.e. the FT
gets shifted by


or: FT of
f

x

cos

2


x

gives

1
2
[
F

s




F

s



]
note:
signal bandwidth around
carrier frequency
math_theorems_3.odt
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
4

5.6 Convolution-Theorem
with
f

x


F

s

g

x


G

s

h

x

=
f

g

x

H

s

=

e

2

i
xs
h

x

dx
=

dx
e

2

i
xs

dx
'
f

x
'

g

x

x
'

=

dx
'
f

x
'


dx
g

x

x
'

e

2

i
xs

e

2

i
x
'
s
G

s

=
G

s


dx
'
f

x
'

e

2

i
x
'
s
=
F

s

G

s

i.e. convolution in time/spatial domain

multiplication in frequency domain,
and vice versa!
Note:

"1"-element of convolution is


x



"1"-element of multiplication,
f

x

=
1

shifting- and modulation theorem are special case of convolution theorem:
multiplication with
e

2

i
as


convolution with


x

a


examples:


x


sinc

s



x

=




x


sinc
2

s



x



x

=


x


sinc

sinc

s

=
sinc

s


Note: the latter says: self-convolution of
sinc

x

reproduces
sinc

x

!
And from this

sinc
2

x

dx
=

sinc

x

sinc

a

x

dx

a
=
0
=
sinc

sinc

0

=
sinc

0

=
1
(proof promised above)
math_theorems_4.odt
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
5

Note:
convolution-theorem is a real heavy-weight, useful to derive pretty tough things
example: consider the "famous" relation
1
sin


x

=
1


k
=






1

k
x

k
(recall: envelope of


N

s

)
How to prove this relation?
Consider the identity


x



x

=


x


(hat-fct. cuts out central spike of comb).
In the Fourier-domain with the convolution theorem,
and using



s

=
sinc

s

and



s

=


s

this reads:






s

=
sinc



s

=



s

=
1
.
Evaluating the convolution gives
1
=
sinc



s

=





sinc

s
'


k
=






s

s
'

k

ds
'
=

k
=




sin



s

k




s

k

=

k
=




sin


s

cos


k




s

k

=
sin


s

1


k
=






1

k
s

k
, q.e.d
math_theorems_5.odt
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
6

alternative epression (e.g. Abramowitz & Steguhn page 75, 4.3.93)

sin


x

=

k
=






1

k
x

k
=
1
x


k
=
1



1

k

1
x

k

1
x

k

=
1
x


k
=
1



1

k
x

k

x

k

x

k


x

k

=
1
x

2
x

k
=
1



1

k
x
2

k
2

Similarly, from


x

=


x



x

, i.e.
1
=
sinc
2



s

, one gets:
1
sin
2


x

=
1

2

k
=




1

x

k

2


Or, from


x

=


ax



x

,
a

1
(narrower hat also cuts out just the central spike!):
in Fourier-space:

1
=
1
a
sinc


a







s
=

k
sin


a

x

k




x

k


=
sin


a
x


k
cos


a
k



x

k


cos


a
x


k
sin


a
k



x

k

, (and for
a
=
1
the result from) above
Note
:
infinite sum of periodically repeated,
sinc
-functions of width

1
interfere to unity,
independent of width!
math_theorems_6.odt
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
7

this allows to derive many interesting infinite sums:
e.g. for
a
=
2
, and with
sin


2
k

=
{
0,
k
=
2
n


1

n
,
k
=
2
n

1
,
cos


2
k

=
{


1

n
,
k
=
2
n
0,
k
=
2
n

1
, we get
1
=
1

sin


2
x


n


1

n
x

2
n

1

cos


2
x


n


1

n
x

2
n

1

for this, with
x
=
0
:

=

n
=






1

n
2
n

1
=
1

2

n
=
1



1

n

2
n

1


2
n

1


and with
x
=
1
2
:

=
sin


4


n


1

n
1
/
2

2
n

cos


4


n


1

n
1
/
2

2
n

1

=
1

2

n


1

n
[
1
2
n

1
/
2

1
2
n

1
/
2
]
=

2

2

n


1

n

4
n

1


4
n

1


etc.

math_theorems_7.odt
a=1.
a=0.65
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
8

5.7 Rayleigh's Theorem
(special case of convolution theorem)
with
F

s

=
FT

f

x








f

x


2
dx
=






F

s


2
ds
proof:





f

x

f
*

x

dx
=





f

x

f
*

x


FT
:
F

s
'


F
*


s
'

(convolution theorem)
e

2

i
x
s
'
dx

s
'
=
0
=
F

s
'


F
*


s
'


s
'
=
0

=

F

s

F
*

s

s
'

ds
'

s
'
=
0
=





F

s

F
*

s

ds
q.e.d.
or more general:
5.7a Parseval Theorem (power theorem):





f

x

g
*

x

dx
=





F

s

G
*

s

ds
(proof as above)
and in the special case that
f

x

,
g

x

real valued, i.e.
F

s

,
G

s

hermitean:





f

x

g

x

dx
=





F

s

G
*

s

ds
=
1
2





F

s

G
*

s

ds

1
2





F


s
'

G
*


s
'

ds
'






F
*

s

G

s

ds
=





1
2
[
F

s

G
*

s


F
*

x

G

s

]
ds
=






[
F

s

G
*

s

]
d
=





[

F

s


G

s



F

s


G

s

]
ds
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
9

physical interpretation:
mechanical
electrical
power =
force
×
velocity
voltage
×
current
impedance =
force / velocity
voltage / current
with power
P
=
f
g
, impedance
Z
=
f
g
, we have
P
=
Z
g
2
and hence: the total energy spent or consumed:





P

t

dt
=
Z





g
2

t

dt
=
Z






G

f


2
df
with
Z

G


f

2
the spectral
energy density.
math_theorems_9.odt
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
10

5.8 Autocorrelation-Theorem
(again a special case of the convolution-theorem)
The autocorrelation function
A

x

=
f
°
f
*
=





f

u

f
*

u

x

du
has as
its Fourier-Transform the power spectrum
FT

A

x


=
P

s

=

F

s


2
proof:
A

x

=
A


x

=
f




FT

F

s


f
*





FT

F
*


s


x

conv. theor.
F

s

F
*

s

=

F

s


2
so that:






F

s


2
e

2

i
xs
ds
=
A

x

, and

F

s


2
=





A

x

e

2

i
x
s
dx
normalized AC:
a

x

=
A

x

A

0

=
A

x



f

u


2
du
, with
A

0

=





f
*

u

f

u

x

du

x
=
0
=






f

u


2
du
=
Parzival







F

s


2
ds
,
so that
FT

a

x


=
1
A

0

FT

A

x


=

F

s


2


F

s


2
ds
=


s


normalized power spectrum
math_theorems_10.odt
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
11

works also
"in the limit"
, e.g.
sinusoidal signal with frequency


0

s

t

=
A
sin

2


t



, with
S

f

=
A
2
i



e

i
f

/



f
2


=
A
2
i
e

i
f

/

[


f






f



]
.
We know from above:
a

t

=
cos

2


t

, and hence suspect


f

=
FT

a

t


=
1
2





f
2


Note that

S

f


2
=
A
2
4
[


f






f



]
2
=
A
2
4
[



f




2




f




2

2


f





f




=
0
]
=
?
,
in fact: (see below)


T
/
2

T
/
2

s

t


2
dt
=
A
2
2
T

T



.
Thus first consider a truncated sinusoidal:
s
T

t

=
s

t




t
T

FT





S
T

f

=
S

f

*
T
sinc

fT

,
i.e.
S
T

f

=
S

f


T
sinc

Tf

=





A
2
i



e

i
f
'

/



f
'
2


T
sinc

T

f

f
'


df
'
=
A
2
i
e

i
f

/

T





[


f
'






f
'



]
sinc

T

f

f
'


df
'
=
A
2
i
e

i
f

/

T
[
sinc

T

f





sinc

T

f




]
math_theorems_11.odt
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
12

and therefore
P
T

f

=

S
T

f


2
=
A
2
T
2
4
[
sinc

T

f





sinc

T

f




]
2
and





P
T

f

df
=






s
T

t


2
dt
=
A
2


T
/
2

T
/
2
sin
2

2


t



dt

A
2
T
2
.
hence:

T

f

=
P
T

f


P
T

f

df

T
2
[
sinc

T

f





sinc

T

f




]
2
=
T
2
[
sinc
2

T

f





sinc
2

T

f





2
sinc

T

f




sinc

T

f




]

T


1
2
[


f






f



]

O

1
T

=
1
2





f
2


, q.e.d.
The latter follows, because
lim
T


T
sinc
2

Tx

=
lim


0
1

sinc
2

x





x

,
namely
lim
T


T



/
2


/
2
sinc
2

Tx

dx
=
lim
T




T

/
2

T

/
2
sinc
2

u

du
=





sinc
2

u

du
=
1
and
T
sinc

T

f




sinc

T

f




=
sin


T

f





T

f



sin


T

f





T

f




O

1
T


T


0
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
13

5.9 derivative-theorem
(again: special case of convolution theorem!)
from above:

f

x

=
f

x

1
2


f

x

1
2

=
f

2


x

,
so that
f
'

x

=
lim


0
f

x


/
2


f

x


/
2


=
lim


0
f



x
/


1
/
2



f



x
/


1
/
2



=
lim


0
1

[
f




2



1



]

x

=
f

x


lim


0
1

2


1

x



'

x

therefore:
FT

f
'

x


=

f
'

s

=
FT

f

x


FT


'

x


=

f

s

FT


'

x


What is
FT


'

x


?
FT


'

x


=
lim


0
2

2
FT



x




i

sin



s

=
lim


0
2
i
sin



s


=
2

i
s
.
Hence:

f
'

s

=
2

i
s

f

s

The FT of the derivative is the FT of the function mutliplied by
2

i
s
formal derivation:
with
f

x

=

e
2

i
x
s

f

s

ds
, we have
f
'

x

=
2

i
s

e
2

i
x
s

f

s

ds
=

e
2

i
s
2

i
s

f

s


FT

f
'

x


ds
math_theorems_13.odt
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
14

recall:
convolution with
H

x


integral of
f

x

=
F

x

FT


multiplication with

H

s

=
1
2

i
s

1
2

i
s

f

s

=

F

s

math_theorems_14.odt
The Fourier Transform and its Applications, Jürgen Stutzki, Sommersemester 2007
Page
15

5.10 derivative of convolution
with
h

x

=
[
f

g
]

x

=

f

u

g

x

u

du
=

g

u

f

x

u

du
we have
h
'

x

=

f

u

g
'

x

u

du
=
[
f

g
'
]

x

=

g

u

f
'

x

u

du
=
[
f
'

g
]

x

CAUTION
:
no
"product rule", i.e.
h
'

x


[
f
'

g
]

x


[
f

g
'
]

x

!!!!!!
in Fourier space: trivially

h
'

s

=
2

i
s

h

s

=
2

i
s

g

s


f

s

=

2

i
s

f

s



g

s

=

f

s


2

i
s

g

s


=

f
'

s


g

s

=

f

s


g
'

s

math_theorems_15.odt