Talk:Dirac delta function/Archive 2

Archive 1

Archive 2

The frequency domain

I would like to see some specifics on the Dirac Delta in the frequency domain. The mere substitution of "x" by "f" in the current definition formulae would be misleading. This clarification is important since some other articles depend on it. For instance, the statement The Fourier transform of a Dirac comb is also a Dirac comb here.MaskedAce (talk) 03:31, 30 September 2012 (UTC)

The Fourier transform is already covered in detail in the relevant section. It's not clear what you specifically believe is inadequate in the present treatment. Sławomir Biały (talk) 21:24, 30 September 2012 (UTC)

In the equation that follows the expression "The inverse Fourier transform of the tempered distribution f(ξ) = 1 is the delta function. Formally, this is expressed": When I naively try to do some calculus I obtain $\int _{f=-\infty }^{\infty }e^{2\pi ift}df={\biggl [}{\frac {e^{2\pi ift}}{2\pi it}}{\biggr ]}_{f=-\infty }^{\infty }={\biggl [}{\frac {\cos {2\pi ft}+i\sin {2\pi ft}}{2\pi it}}{\biggr ]}_{f=-\infty }^{\infty }={\biggl [}{\frac {\sin {2\pi ft}}{2\pi t}}{\biggr ]}_{f=-\infty }^{\infty }=\lim _{f\to \infty }{\frac {2\sin {2\pi ft}}{2\pi t}}={\Bigg \{}{\begin{matrix}\lim _{f\to \infty }2f&{\textrm {as}}&t\to 0\\\in [-1,1]/(\pi t)&{\textrm {if}}&t\neq 0\end{matrix}}$

Is the Dirac pulse amplitude defined to be $2\infty$ so to speak? My question is if there is a reason that the definition doesn't contain a compensating factor of 1/2. I simply just don't understand and I believe that the article would benefit much if someone had any idea on how to explain the above in a pedagogical sense. Geo39geo (talk) 13:23, 12 February 2018 (UTC)

The value of the function $\textstyle {\frac {2\sin {2\pi ft}}{2\pi t}}$ at 0 is roughly the height of the hump; but did you think about the width of the hump? The integral of this function (over the whole real line) is equal to 1 (rather than 2), since $\textstyle \int _{0}^{\infty }{\frac {\sin at}{t}}\,dt={\frac {\pi }{2}}$ for all $a>0.$ Boris Tsirelson (talk) 18:53, 12 February 2018 (UTC)

Aha ok, I get it! This makes sense. Thank you very much for the succinct and fast reply. Geo39geo (talk) 22:10, 13 February 2018 (UTC)