Duality in continous Fourier transform

Hi,
If X(w) is the Fourier Transform (FT) of x(t), I know that the FT of
dx(t)/dt is jwX(w). Using the duality property I obtained that the FT of
tx(t) is (2pi/j)(dX(-w)/dw), but I found in a table that the correct FT is
in fact jX(w). Could you help me to find the mistake here? 

I will be thankful if somebody could help me with that.


Luis Felipe

luispipe16 wrote:
> Hi,
> If X(w) is the Fourier Transform (FT) of x(t), I know that the FT of
> dx(t)/dt is jwX(w). Using the duality property I obtained that the FT of
> tx(t) is (2pi/j)(dX(-w)/dw), but I found in a table that the correct FT is
> in fact jX(w). Could you help me to find the mistake here? 
> 
> I will be thankful if somebody could help me with that.

Luis,

You are almost there, but where did the 2pi come from? Note that
1/j = -j, and that dX(-w)/dw = -dX(w)/dw, so (dX(-w)/dw)/j = jdX(w)/dw.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

Thanks for your answer. The duality property establishes that 

if  FT{x(t)}=X(w), then FT{X(t)}=(2pi)x(-w) ; 

Using this property, I obtained 

FT{tx(t)}= (2pi)jdX(w)/dw

I don't know how I could cancel the 2pi in order to get the correct
answer. 


Luis Felipe



>You are almost there, but where did the 2pi come from? Note that
>1/j = -j, and that dX(-w)/dw = -dX(w)/dw, so (dX(-w)/dw)/j = jdX(w)/dw.
>
>Jerry
>-- 
>Engineering is the art of making what you want from things you can get.
>�����������������������������������������������������������������������
>

luispipe16 wrote:
> Thanks for your answer. The duality property establishes that 
> 
> if  FT{x(t)}=X(w), then FT{X(t)}=(2pi)x(-w) ; 
> 
> Using this property, I obtained 
> 
> FT{tx(t)}= (2pi)jdX(w)/dw
> 
> I don't know how I could cancel the 2pi in order to get the correct
> answer. 
> 
> 
> Luis Felipe

Ah! *That* 2pi! It's a bookkeeping factor. Some writers put it on the 
direct transform, others on the inverse. Some use sqrt(2pi) on each one 
for symmetry's sake. Other authors leave it up to the user/reader to 
deal with it as he likes. I think that's where you are. Sleep easy!

Jerry
-- 
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������

On Oct 28, 7:12&#4294967295;am, "luispipe16" <luispip...@yahoo.com> wrote:
> Hi,
> If X(w) is the Fourier Transform (FT) of x(t), I know that the FT of
> dx(t)/dt is jwX(w). Using the duality property I obtained that the FT of
> tx(t) is (2pi/j)(dX(-w)/dw), but I found in a table that the correct FT is
> in fact jX(w). Could you help me to find the mistake here?
>
> I will be thankful if somebody could help me with that.
>

dunno what table you were looking at, but the FT pf t*x(t) is not
j*X(w) (which is the FT of j*x(t)).

BTW, to use the Duality Property of the continuous Fourier Transform
most easily and effectively, i would recommend the definition of the
FT that has non-radian frequency in it ("f" instead of "omega").  the
reason why is that the form for the forward and inverse transform is
the same, making what you do with duality clear.  there are no 1/
(2*pi) scaling constants to screw you up, the convolution theorem and
Parseval's theorem are simple and need no icky scaling factors, and
are the same form regardless of which is in the frequency domain and
which is in the time domain.  when i first had my Communications
Theory class (1977 or 78) and we used a textbook by A. Bruce Carlson,
i immediately saw the value of that convention and have been a
believer ever since.

r b-j

OK. Thank you guys!

PS: Oops! I wrote j*X(w) instead of jwX(w). So, the table is ok. 

>dunno what table you were looking at, but the FT pf t*x(t) is not
>j*X(w) (which is the FT of j*x(t)).

Let's say Fourier Transform (FT) of x(t) is X(ω) and duality property says
FT of X(t) is 2πx(-ω). 

We know FT of dx(t)/dt is jωX(ω). From above duality property, we can
get FT of tX(t) is 2πjdx(ω)/dω. However, from this property, we can't
get FT of tx(t), which may sometimes cause confusion.

Paul 
http://www.ecvale.com/index.php?main_page=user_blogs_info&upar=174124429&blgspar=8

>On Oct 28, 7:12=A0am, "luispipe16" <luispip...@yahoo.com> wrote:
>> Hi,
>> If X(w) is the Fourier Transform (FT) of x(t), I know that the FT of
>> dx(t)/dt is jwX(w). Using the duality property I obtained that the FT
of
>> tx(t) is (2pi/j)(dX(-w)/dw), but I found in a table that the correct FT
i=
>s
>> in fact jX(w). Could you help me to find the mistake here?
>>
>> I will be thankful if somebody could help me with that.
>>
>
>dunno what table you were looking at, but the FT pf t*x(t) is not
>j*X(w) (which is the FT of j*x(t)).
>
>BTW, to use the Duality Property of the continuous Fourier Transform
>most easily and effectively, i would recommend the definition of the
>FT that has non-radian frequency in it ("f" instead of "omega").  the
>reason why is that the form for the forward and inverse transform is
>the same, making what you do with duality clear.  there are no 1/
>(2*pi) scaling constants to screw you up, the convolution theorem and
>Parseval's theorem are simple and need no icky scaling factors, and
>are the same form regardless of which is in the frequency domain and
>which is in the time domain.  when i first had my Communications
>Theory class (1977 or 78) and we used a textbook by A. Bruce Carlson,
>i immediately saw the value of that convention and have been a
>believer ever since.
>
>r b-j
>

paulz wrote:
> Let's say Fourier Transform (FT) of x(t) is X(ω) and duality property says
> FT of X(t) is 2πx(-ω). 
> 
> We know FT of dx(t)/dt is jωX(ω). From above duality property, we can
> get FT of tX(t) is 2πjdx(ω)/dω. However, from this property, we can't
> get FT of tx(t), which may sometimes cause confusion.

Does everyone see those lovely omegas? I'd much rather write ω than w, 
but I've been concerned that it would come through as '?' or worse.

Jerry
-- 
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Jerry Avins wrote:
> paulz wrote:
> 
>> Let's say Fourier Transform (FT) of x(t) is X(ω) and duality property 
>> says
>> FT of X(t) is 2πx(-ω).
>> We know FT of dx(t)/dt is jωX(ω). From above duality property, we can
>> get FT of tX(t) is 2πjdx(ω)/dω. However, from this property, we can't
>> get FT of tx(t), which may sometimes cause confusion.
> 
> 
> Does everyone see those lovely omegas? I'd much rather write ω than w, 
> but I've been concerned that it would come through as '?' or worse.
> 
> Jerry

I see omega(ω). Is there a simple general way of getting those in a 
post. I did it this time with cut-n-paste.

On Oct 29, 9:33&#4294967295;am, Jerry Avins <j...@ieee.org> wrote:
> paulz wrote:
> > Let's say Fourier Transform (FT) of x(t) is X(&#4294967295;) and duality property says
> > FT of X(t) is 2&#4294967295;x(-&#4294967295;).
>
> > We know FT of dx(t)/dt is j&#4294967295;X(&#4294967295;). From above duality property, we can
> > get FT of tX(t) is 2&#4294967295;jdx(&#4294967295;)/d&#4294967295;. However, from this property, we can't
> > get FT of tx(t), which may sometimes cause confusion.
>
> Does everyone see those lovely omegas? I'd much rather write &#4294967295; than w,
> but I've been concerned that it would come through as '?' or worse.
>

i can see the omega &#4294967295; and pi &#4294967295; (reading with Google Groups), but i
don't know what key sequence to hit to get it.  here i just copied and
pasted.  maybe someday we'll be able to use LaTeX paste-up on USENET
to *really* present our meaning clearly.

i still think (paulz) that representing the FT with "f" instead of "&#4294967295;"
in it results in a much simpler representation of the Duality and
Convolution properties and Parseval's theorem.

r b-j