# 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?

• Leaving $|\delta|^2$ undefined does not imply that its integral is infinite. The reason for the absence of a definition is that there is no consistent way to define it. If you take the approach of defining the Dirac distribution as a limit of unit area functions with the support approaching 0, then the square of that function simply can converge against anything you desire. Therefore the square is not implied by the defining property of the Dirac distribution. – Jazzmaniac Aug 17 '17 at 10:15
• Specifically, you can define such a sequence of functions that converges to $|\delta|$ so that the area of its point-wise square converges to 0, or any other non-negative number. – Jazzmaniac Aug 17 '17 at 10:20
• Yes, I understood my mistake. But solving Parseval's equation ∫∞−∞|x(t)|2dt=∫∞−∞|X(f)|2df, Energy of unit impulse signal Eω=∫∞−∞|1|2df = ∞. Is this approach of defining the energy to be infinite is fine? – Hendry Newman Aug 17 '17 at 10:24
• @Jazzmaniac what happens if you define the object $\delta(t)^2$ itself as the limit of the normal functions, (just as the usual $\delta(t)$ is defined so) such as: $\delta(t)^2 = \lim_{{\Delta} \rightarrow 0} F_{\Delta}(t)$ where $F_{\Delta} (t) = p_{\Delta}(t) p_{\Delta}(t) = p_{\delta}(t)^2$... So at least then we would not be asking ourselves about the properties of the square of a generalized function. Rather it's, itself, independent of the $\delta(t)$, defined directly... would that help ? – Fat32 Aug 17 '17 at 15:42
• A simple statement to remember : "dirac delta only makes sense under an integral sign" – percusse Aug 18 '17 at 9:52

## 4 Answers

[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. 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...

Some related posts:

• When you say this product is neither energy nor power do you mean in the sense of the function $x(t) = t u(t)$ which is neither energy, nor power signal, (as it has infinite energy and infinite average power) ? Or do you mean in the sense that it's undefined ( or not decidable ) ? – Fat32 Aug 17 '17 at 19:15
• I was expending my answer, and @endolith (in a nice move), did some edits and my corrections got lost (because of me on Chrome). So, until I find energy to redo: First, unless I find a way for a unique and well-grounded Dirac square definition, useful to DSP, for which energy/power character can be assessed, it is not (IMO), in the traditional mundane way. Second: if not unique or decidable, I should choose to add a "not power/energy axiom". Such a product is too wild for my mind at the present time (like $i\times i$ in the past) – Laurent Duval Aug 18 '17 at 21:50

$\delta(x)$ doesn't really exist at all for any particular $x$. Like Laurent Duval said, Dirac is not an $\mathbb{R}\to\mathbb{R}$ function, rather the whole mapping $$\backslash f \mapsto f(a) \equiv \int_\mathbb{R}\!\!\mathrm{d}t\: f(t) \cdot \delta(t-a)"$$ is a functional, mapping functions to values of the function evaluated at some particular point. Arguably, it would make sense to reflect that with a dedicated symbol, like $$\int\!\!\!\!\delta_a\mathrm{d}t\: f(t).$$ (The reason it makes sense to write $\delta$ as if it was an $\mathbb{R}\to\mathbb{R}$ function is that any square integrable function $g$ gives rise to a functional in a similar way, namely $$\gamma : L^2(\mathbb{R})\to\mathbb{R}, \quad \gamma(f) = \int_\mathbb{R}\!\!\mathrm{d}t\:f(t)\cdot g(t).$$ That's in fact just the $L^2$ scalar product between $f$ and $g$; the function space $L^2$ is a Hilbert space The benefit of the Dirac-delta notation is that it allows you to write superpositions of such real-function functionals and Dirac functionals, for example the high-pass impulse-respone $$\delta(t) - \sqrt{\frac{\omega_0}{2\pi}}\cdot\exp(-t^2\cdot\tfrac{\omega_0^2}2).$$ That's a function you can never actually implement in practice, only approximate, but it captures the concept of a high-pass filter which isn't really concerned with the impulse-response as such, but by the result of folding it with actual real-world signals, and it is the folding that provides the integral which defines the $\delta$'s meaning.)

So, because $\delta$ is not a function, there's no reason to believe it could make any sense to write $|\delta(t)|^2$ since in that expression the delta does not occur exactly once beneath an integral running over its variable. Even if you did write an integral around it, it would always have two deltas with the same parameter in it, and that's not defined.

Summary: you're right, Dirac is not a signal, neither power nor energy.

• Your latex formatting seems to be broken. Did you use a latex formula editor/converter or is this intentional? – Jazzmaniac Aug 17 '17 at 15:11
• @Jazzmaniac the formatting was fine in my browser, but I had accidentally used a couple of Unicode ℝ instead of LaTeX \mathbb{R}, perhaps MathJax doesn't reliably support that. Does it render ok now? – leftaroundabout Aug 17 '17 at 15:18

You're right that the square of a Dirac delta impulse is undefined, so energy and power cannot be defined in the usual way for signals containing Dirac impulses.

However, in analogy with discrete-time signals, it is common to define energy and power of a signal consisting of Dirac impulses in the following way. If a signal $x(t)$ is given by

$$x(t)=\sum_{n=-\infty}^{\infty}a_n\delta(t-t_n)\tag{1}$$

then its energy can be defined as

$$E_x=\sum_{n=-\infty}^{\infty}|a_n|^2\tag{2}$$

and its power can be defined by

$$P_x=\lim_{T\rightarrow\infty}\frac{1}{2T}\sum_{n:\;|t_n|<T}|a_n|^2\tag{3}$$

Using definitions $(2)$ and $(3)$, a signal consisting of Dirac impulses can either be an energy signal ($E_x<\infty$), or a power signal ($E_x\rightarrow\infty$, $P_x<\infty$), or neither of the two (both $(2)$ and $(3)$ do not exist).

• The generalisation that squares are not defined for any distribution is not accurate, at least not without additional qualification of the term distribution. – Jazzmaniac Aug 17 '17 at 10:16
• @Jazzmaniac: I removed the reference to general distributions because it doesn't add anything useful to the answer. By the way, that claim is taken from engineering literature, which might indeed be inaccurate when it comes to details about distributions. – Matt L. Aug 17 '17 at 19:07

You have got nice answers here but I'm trying to explain it in a simple way: An impulse is any signal that is entirely zero except for a short blip of arbitrary shape. For example, an impulse to a microwave transmitter may have to be in picosecond range because the electronics respond in nanoseconds. In comparison, a volcano that erupts for years may be a perfectly good impulse to geological changes that take millennia. Mathematicians don't like to be limited by any particular system, and commonly use the term impulse to mean a signal that is short enough to be an impulse for any system! That is a signal that is infinitesimally narrow and again mathematicians define an impulse as: 1. signal that is infinitesimally brief 2. The pulse occurring at time zero and 3. Pulse must have an area of one [By Steven W. Smith]