Is the Dirac delta (impulse) signal a power signal or an energy signal?

I'm a beginner so sorry if this question is very fundamental. Dirac impulse has finite area i.e = 1. But I've heard that $|\delta(t)|^2$ is undefined. So area under $|\delta(t)|^2$ is also undefined and signal doesn't exist in all time $t$ so it cant be a power signal. So my guess is Neither Power nor Energy signal. Am I right?

Answer

[Added a reference on Schwartz's impossibility theorem for products of distribution]

The continuous Dirac delta $\delta$ is not considered a true function or signal, but a distribution. From its wikipedia page:

The delta function can also be defined in the sense of distributions exactly as above in the one-dimensional case.[25] However, despite widespread use in engineering contexts, (2) should be manipulated with care, since the product of distributions can only be defined under quite narrow circumstances.

It can be defined such that, for any function $f$ satisfying some important properties, and for $a\in \mathbb{R}$:

$$ \int f(t)\delta(t-a)dt=f(a).$$

From the best of my knowledge, those important properties are not satisfied by $\delta$, so one cannot directly replaced $f$ by $\delta$ and get a meaningful result. As far as know, the product of two Dirac distributions is not well defined, unless one talks about $n$-dimensional versions, or so-called "formal" manipulations as used in physics for instance, or more complicated maths. A short account of Simplified production of Dirac delta function identities is provided by Nicholas Wheeler. If one wants to dig deeper, I'd suggest The Colombeau theory of generalized functions, by Ta Ngoc Tri, 2005:

Soon after the introduction of his own theory, L. Schwartz published a paper in which he showed an impossibility result (see [Sch54]) about the product of two arbitrary distributions.

One result is Schwartz impossibility result. It (somehow) says that if one wants to encompass the derivative of continuously differentiable functions whle keeping Leibniz's rule of derivation, then one get $\delta^2(|x|)=0$.

However, from an informal point of view, sometimes used in DSP (and in physics), this "product" is, as far as I know, neither energy nor power. From a logical point of view though, if it does not exist, one could affect this "product" a lot of properties...

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