| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1661.1 | addendum | 18889::THIBAULT | | Tue Sep 08 1992 15:20 | 3 |
| forgot to add that the / was used to denote the integral. pt
/
/
|
| 1661.2 | voila | AMCCXN::BERGH | Peter Bergh, DTN 523-3007 | Tue Sep 08 1992 17:48 | 35 |
| <<< Note 1661.0 by 18889::THIBAULT >>>
-< Help needed with Calculus-Fourier transforms >-
<< /
<< If s(t)= |0 -T <t< T
<< |1 - -
<< \
I assume that you mean that the function is 1 in the interval from -T to +T and
that it's zero otherwise. If not, we run into distributions and it gets a
whole lot hairier.
<< Show its Fourier transform is given by
<< s(f)=2T {sin 2#ft}
<< {--------} -oo<f<+oo
<< {2#ft }
We know that the Fourier transform is given by integral of s(t)*exp(-2#jtf)
over the whole real line with respect to t. Since s(t) is zero outside
(-T,+T) (and 1 inside the interval), the integral reduces to one from -T to +T
of exp(-2#jtf).
But the integral of exp(zx) with respect to x is exp(xz)/z, so the Fourier
transform we're after is (exp(-2#jTf) - exp(2#jTf))/(-2#jf). This simplifies
to (exp(2#jTf) - exp(-2#jTf))/(2#jf).
Recalling that sin(z) is (exp(jz)-exp(-jz))/2j, we find that the Fourier
transform we're after is sin(2#jTf)/(#f), or, equivalently,
(2Tsin(2#jTf)/(2#fT) -- the result you were after. (Note that the variable
of integration (t) can not appear in the definite integral; I have assumed
that 2#ft in your desired result should in fact be 2#fT.)
Interestingly enough, the Fourier transform of s(f) does not exist in the
Lebesgue sense!
|
| 1661.3 | That is what I get | 19584::ABBASI | Have you spelled checked today? | Tue Sep 08 1992 17:59 | 60 |
| s(t) is
------------+ +------------...
| |
| |
+-----------------+
-T 0 T
let w= 2Pi f
+oo -T oo
/ / /
s(f) = | s(t) e^-jwt = | e^-jwt + | e^-jwt
/ / /
-oo -oo +T
|-T |oo
= 1/(-2 w) e^- jwt | + 1/(- jw) e^- jwt |
|-oo |T
= 1/(- jw) lim k->oo { e^ jwT - e^ jw(K) + e^- jw(K) - e^- jwT}
= 1/ jw lim k->oo { e^- jwT - e^ jwT +e^ jwK - e^- jwK }
= 1/w { e^- jwT - e^ jwT + lim k->oo e^ jwK - e^- jwk }
---------------- ----------------
j j
= 1/w { -2 sin(wT) + lim k->oo (2 sin wK) }
humm... I think you got your time function the other way round,
I think it should be
/
If s(t)= |1 -T <t< T
|0 - -
\
not
/
If s(t)= |0 -T <t< T
|1 - -
\
in this case we get
+oo 0 T
/ / /
s(f) = | s(t) e^-jwt = | e^-jwt + | e^-jwt
/ / /
-oo -T 0
= -1/jw { 1- e^jwT + e^-jwT - 1} = 2 SIN wT
--------
W
/nasser
|
| 1661.4 | | 3D::ROTH | Geometry is the real life! | Wed Sep 09 1992 11:37 | 16 |
| >I assume that you mean that the function is 1 in the interval from -T to +T and
>that it's zero otherwise. If not, we run into distributions and it gets a
>whole lot hairier.
But an engineer would simply add an appropriate DC term and leave it at that.
(insert smiley face)
>Interestingly enough, the Fourier transform of s(f) does not exist in the
>Lebesgue sense!
While s(f) is not in L1, it is in L2 and has an inverse transform there.
This does restrict the space of functions that can be handled (at least
for mathematicians and probabilists, engineers and physicists can safely
ignore it.)
- Jim
|
| 1661.5 | | 9055::BERGH | Peter Bergh, DTN 523-3007 | Wed Sep 09 1992 12:45 | 13 |
| <<< Note 1661.4 by 3D::ROTH "Geometry is the real life!" >>>
<< >Interestingly enough, the Fourier transform of s(f) does not exist in the
<< >Lebesgue sense!
<< While s(f) is not in L1, it is in L2 and has an inverse transform there.
<< This does restrict the space of functions that can be handled (at least
<< for mathematicians and probabilists, engineers and physicists can safely
<< ignore it.)
Fourier-transform theory is best done in L2, in that F is a norm-preserving
(Parseval's formula) function from L2 to L2. My remark was meant as a factoid,
not as an interesting fact with deep implications.
|