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Re: Problem with function


....and to make the result the same as the one Bob Hanlon provided you make
the replacement

Derivative[1][DiracDelta][\[Omega]]  -> 0

in the result I gave before.

This is allowed because, although the derivative of a Dirac delata function
is a positive impulse immediately followed by a negative impulse, the effect
of these impulses cancels when they are used to weight the integrand in in
any sufficiently smooth integral

Steve Luttrell
West Malvern, UK

"Steve Luttrell" <steve_usenet at _removemefirst_luttrell.org.uk> wrote in
message news:c8rulh$8m6$1 at smc.vnet.net...
> Define your function.
>
> f[t_] := (UnitStep[t] - UnitStep[t - 1])*t +
> (UnitStep[t - 1] - UnitStep[t - 3])*(3/2 - t/2)
>
> Fourier transform your function.
>
> Expand[FourierTransform[f[t], t, \[Omega]]]
>
> which gives
>
>
> -(1/(Sqrt[2*Pi]*\[Omega]^2)) + (3*E^(I*\[Omega]))/
> (2*Sqrt[2*Pi]*\[Omega]^2) - E^(3*I*\[Omega])/
> (2*Sqrt[2*Pi]*\[Omega]^2) - I*Sqrt[Pi/2]*
> Derivative[1][DiracDelta][\[Omega]] +
> (3/2)*I*E^(I*\[Omega])*Sqrt[Pi/2]*
> Derivative[1][DiracDelta][\[Omega]] -
> (1/2)*I*E^(3*I*\[Omega])*Sqrt[Pi/2]*
> Derivative[1][DiracDelta][\[Omega]]
>
> Steve Luttrell
>
> "DJkapi" <djkapi at poczta.onet.pl> wrote in message
> news:c8ptmu$kne$1 at smc.vnet.net...
> > How to compute in Mathematica  fourier transform  of function:
> >
> >             t     ,0=< t =<1
> > g(t)=
> >             1.5-0.5t , 1< t =<3
> >
> > to tell the truth i dont even know how to make Mathematica print that
> > function.
> >
> > Regards,
> > DJKapi
> >
> >
>


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