How is FT of $ \delta $(t) equal to 1 ? Normal FT gives the result $\infty$. Can someone please explain? I did the normal integration and substituted the limits.

Is it because $ \delta $(t) is a unit impulse function so as it's height is large it's width is very small so no matter what the FT will always be equal to 1 ? (I'm just trying to figure out the logic)


1 Answer 1


If follows directly from the definition of the Dirac delta distribution. It is defined so that

$$\int_\mathbb{R} \delta(x) f(x) dx := f(0)$$

for any test function $f(x)$. In other words, the Dirac distribution is the generator of the linear functional that extracts a single function value.

With this definition the Fourier transform of the Dirac distribution is simply: $$\int_\mathbb{R} \delta(t) \exp(-2\pi i \omega t) dt=\exp(-2\pi i \omega\cdot 0)=\exp(0)=1$$

It doesn't make much sense to say the Dirac distribution is infinitely high or infinitely narrow. Just use the definition given above and apply it. If you really need to, you can understand the distribution as the limit of a sequence of certain functions, but it's not a function itself. And specifically it doesn't have a graph.

  • $\begingroup$ You do need $f(x)$ to be continuous at $0$. $\endgroup$ Aug 10, 2014 at 16:14
  • $\begingroup$ @DilipSarwate, I see that you need continuity if you define it using measure theory or even weak $C^\infty$ if you come from general distribution theory. But I don't see the requirement of f being continuous at 0 if you just define the linear functional $\delta(f):=f(0)$ without further context. Do you see this differently? $\endgroup$
    – Jazzmaniac
    Aug 10, 2014 at 16:40
  • $\begingroup$ If you want to use the Dirac delta in Fourier transform applications, which is the context here, then it is perhaps best to include the continuity requirement without insisting on the non-necessity of including such a requirement. You might want to raise this question on math.SE if you want a complete answer as to whether the more general definition that you are using is adequate for use in the context of Fourier transforms. $\endgroup$ Aug 11, 2014 at 0:40
  • $\begingroup$ @DilipSarwate, this is a good example on, just at a pragmatic level, that people from a background discipline of mathematics will not see things the same way as people from an engineering or physics background. the continuity requirement for $f(0)$ is necessary for the "nascent delta" definition (which uses a limit and what engineers often see) of the dirac delta. but from the "distribution" definition, no such requirement is needed. but then $\delta(t)$ is not a function in the strict mathematical sense, so viewing $f(t) \cdot \delta(t)$ by itself (not inside an integral) makes no sense. $\endgroup$ Oct 3, 2014 at 2:18
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    $\begingroup$ Robert, please don't change notation just because you like it better. The use of $\exp()$ has nothing to do with the availablility of $e$ as a symbol. I strongly prefer the function notation for several reasons and I also find it easier to read. $\endgroup$
    – Jazzmaniac
    Oct 3, 2014 at 9:36

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