       Re: Fourier Transform

• To: mathgroup at smc.vnet.net
• Subject: [mg93584] Re: Fourier Transform
• From: "Nasser Abbasi" <nma at 12000.org>
• Date: Sun, 16 Nov 2008 07:03:41 -0500 (EST)
• References: <gf8rgp\$poq\$1@smc.vnet.net> <gfbur0\$43e\$1@smc.vnet.net> <gfgqhc\$dkv\$1@smc.vnet.net> <gfh3cl\$g1l\$1@smc.vnet.net> <gfimic\$g0f\$1@smc.vnet.net>
• Reply-to: "Nasser Abbasi" <nma at 12000.org>

```"Nikolaus Rath" <Nikolaus at rath.org> wrote in message
news:gfimic\$g0f\$1 at smc.vnet.net...
>

>
> But what is the point of that? There is nothing special about the
> integral in a Fourier transform, so any functionality that helps to
> evaluate integrals is best added to Integrate[].
>
>   -Nikolaus
>

my 2 cents on this good question is the following:

Using FourierTransform and using the equivlant integeration command is not
always the same.

-------------------
In:= func = Cos[w0 t];
Integrate[func* Exp[(-I)*w*t],   {t, -Infinity, Infinity}]

Integrate::idiv: Integral of E^(-I t w) Cos[t w0] does not converge on
{-\[Infinity],\[Infinity]}.

FourierTransform[func, t, w]
Sqrt[Pi/2] DiracDelta[w-w0]+Sqrt[Pi/2] DiracDelta[w+w0]
------------------------

Cos[] is not square integrable (its average power is not zero, the average
power of cos[] is 1/2)
So, Cos[] does not have a FourierTransform (one of the conditions to have
FourierTransform is for the function to be square integrable). So, for
non-square integrable functions, FourierTransform does not exist.

Yet, you see that FourierTransform does generate a FourierTransform for the
Cos[].

Since there are many useful functions which have infinite energy, but we
want to find its fourier transform,  a more generalized definition is used
for the integral which uses a dirac delta function to allow one to integrate
such functions, and that is why sometimes FourierTransform gives different
answer than direct use of the Intgeration function.

reference
Delta Functions: An Introduction to Generalised Functions by R. F. Hoskins

Nasser

```

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