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Background to my question: I'm taking a signals and systems course where we are using the Dirac delta function, but since it's an engineering course the explanation of what it actually is has been very hand-wavy with just enough for us to be able to do transforms with it.

It was first explained to me that the Dirac delta function is an impulse of infinite height and infinitely small width, so that the area under the impulse ends up being equal to 1. Then we started solving for the Fourier transforms of functions and the area under these impulses has taken many different values. Now if the area under the impulse is no longer 1, is the width no longer infinitely narrow so that it extends past its specific point on the frequency axis? Or is the height greater than infinity? Or something else?

Jyrki Lahtonen
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Austin
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    The idea that the dirac delta is an infinitely thin spike with infinite height is just a heuristic. Really the dirac delta is defined by what it does to functions: $\int_{\text{all space}} f(x)\delta(x)dx = f(0)$. –  Nov 23 '15 at 03:29
  • So changing the area from 1 to another value doesn't affect anything else? – Austin Nov 23 '15 at 03:31
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    No, changing area from $1$ to $a$ changes $\delta$ to $a\delta$. – A.S. Nov 23 '15 at 04:02
  • But I mean that its height and width are still considered infinitely tall and narrow? – Austin Nov 23 '15 at 04:07
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    Yes, but there ratio of infinities of tallness is $a$ with the same infinitesimal width. – A.S. Nov 23 '15 at 04:16
  • Height greater than infinity. – Anixx May 13 '20 at 21:03

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The problem is Dirac delta "function" is not a function at all! No function can have infinity at one point and zero everywhere else. Even if you somehow make sense of delta function being a function, it is certainly not continuous and definitely not differentiable as a function.

More precisely Dirac delta function is a Distribution (or a generalized function). One way of making sense of this distribution is treating at as a limit of smooth functions $$ \delta(x) = \lim_{n\to \infty} \sqrt{\frac{n}{\pi}} e^{-nx^2} := \lim_{n\to \infty} f_n(x) $$ As you can see $\int_{-\infty}^\infty f_n(x)dx = 1$ for all $n$, all $f_n$ are infinitely differentiable. The limit is a Gaussian which is getting sharper and narrower. You can check that for any smooth function $g(x)$ $$\lim_{n\to \infty}\int_{-\infty}^\infty f_n(x)g(x)dx=g(0), \quad \lim_{n\to \infty}\int_{-\infty}^\infty f'_n(x)g(x)dx=-g'(0) $$ There are lots of other ways to represent Dirac delta function as a limit of functions.

Hamed
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