to 1
How Compute Some of the Nonstandard,
Extended, or Generalized Fourier Transforms
Recall: F-transform exists as a normal function if
(a) x(t) is piecewise continuous
(b)
|x(t)|dt
< (absolute integrability)
We will look at a few cases where the absolute integrability condition is violated,
but yet it is still convenient to talk about an extended Fourier transform.
(1) x(t) = ej0 t F {ej0t } = 2 ( 0 ) ?
Note:
j0 t
|dt
|e
= A normal Fourier transform does NOT exist.
Let xT (t) = ej0 t [u(t + T ) u(t T )], 0 < T < . Then
T
|xT (t)|dt =
1dt = 2T <
T
Can compute F {xT (t)}.
XT ( ) = F {xT (t)}
T
ej0 t ejt dt
=
T
T
ej (0 )tdt
=
T
1
ej (0 )t |T T
j ( 0 )
1
=
ej (0 )T ej (0 )T
j ( 0 )
=
2
ej (0 )T ej (0 )T
2T
=
T ( 0 )
2j
sin((0 )T )
.
= 2T Sa((0 )T)
= 2T Sa(( 0)T) because Sa(.) is even
( 0)T
x
= 2T sinc(
) because Sa(x) = sinc( )
Fact: lim T Sa(( 0 )T)) = ( 0 ). Note that the factor of is due to the
T
easily veried fact that, T > 0
T Sa(T)d =
sin(T)
d =
Thus,
ej0 t lim 2T Sa(( 0)T)) = 2 ( 0 )
T
Note: We can also arrive at this conclusion from the other direction:
F
1
1
{ ( 0 ) } =
2
( 0 )ejtd =
1 j0 t
e
( 0 )
2
1 j0 t
e
2
3
2
?
j
We note that |sgn(t)|dt = . This time, instead of using a nite window, we
use an exponential decay term: sgn(t) = lim , sgn(t)ea|t| a > 0
a0
2
We note that for every a > 0, |sgn(t)ea|t| |dt = < . We further check that
a
sgn(t)ea|t| = eat u(t) eat u(t)
(2) sgn(t)
We now take Fourier transforms of both sides and take the limit as a decreases to
zero:
1
1
a0 2
F {sgn(t)ea|t|} =
a + j a j
j
11
+ sgn(t)
22
1
12
1
F {u(t)} = (2 ( ) + ( )) = ( ) +
2
2 j
j
(3) u(t) =
(4) Note: Writing 1 = u(t) + u(t) gives a consistent result because taking Fourier
1
1
transforms of both sides yields 2 ( ) = ( ) +
+ ( )
= 2 ( ), which
j
j
is what we had previously computed in class. [Note: we used the fact that is even.]
This concludes our exercise of justifying some of these nonstandard Fourier transforms. You are NOT responsible for the derivations. You will be called upon to use
the extended Fourier transforms presented in the textbook.
4
We have formulas that compute the Fourier transform and the inverse Fourier transform. The following is a sketch of a proof that if
X ( ) =
x(t)ejtdt, then x(t) =
1
2
X ( )ejt d
That is, the proof outlines why the inverse transform really is an inverse.
From previous pages,
T
j
e
ej d = limT 2T Sa(T ) = 2 ( )
d = lim
T
1
2
T
jt
X ( )e
1
d =
2
1
=
2
x( )ej d ejt d
x( )
ej(t ) d d
2 (t )
1
2
2
= x(t)
=
x( ) (t )d
Note: To do a more correct proof requires at least Math 451.