Re: Fourier Duality

Ok, thank you very much for the help, I just want to make sure I am
understanding this correctly.

So, is this saying that if I have a time domain signal x(t) whos FT I know
to be X(w),
I know that the freq domain signal with a graph that is the same as my
original time
domain graph is simply 2*pi*X(-w).

That is if I have say a sinc fctn in my time domain and I know its FT if I
multiply that FT
by 2*pi I will get a sinc fctn in the freq domain (ignoring the -w part
because sinc is
and even fctn?) And furthermore if I divide that original time domain sinc
fctn by
2*pi I will end up with a box graph in the time domain that is similar to
the freq domain
box graph I got by taking the FT of my sinc fctn?

Oih, I hope my understanding is correct (and that I actually typed what I
mean!)
Thank you all for helping me wrap my head around all of this.

-matt



"mobi" <mobien@xxxxxxxxx> wrote in message
news:1141116532.304943.146370@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
The fomula for fourier analysis containts exp(-jwt), while for
synthesis we have exp(jwt). Analysis means finding the transform and
synthesis means making the original signal back. The only difference is
that in synthesis we divide by 2*pi. Now there is not much difference
between exp(-jwt) and exp(jwt) except a sign difference. Which means
that if x(t) leads to X(w), then X(t) must lead to x(-w), intuitively,
as both are complex exponentials with only a sign change.
Duality is a property. We donot have any desired to create properties
they are just there to be discovered. So with fourier transformed
someone figured out that hey
if I know x(t) <-> X(w), and if X(t) is given then i can easily find
x(w). Why? For the reason explained above. For example we might know
what is the fourier transform of a sinc function (it is a box
function), now what is the fourier of a box function? Duality helps
here. It is a smart way of finding the transform, a shortcut way, when
you already know the fourier, and the function on the other side
appears, you donot need to compute the fourier again.

I hope it helped.
~Mobien



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