1
$\begingroup$

Today I learned, instantaneous power is

$$ p_x(t) = |x(t)|^2 $$

and I've known, energy is

$$ E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt $$

What I also thought was, $|x(t)|^2$ is instantaneous energy, and that $P(t) = dE(t)/dt$, but these are inconsistent. What I can't figure out:

  1. Why is $|x(t)|^2$ defined to be instantaneous power?
  2. If $|x(t)|^2$ is instantaneous power, is there instantaneous energy, and what is it, in continuous and discrete? If not, why not?
  3. The rest of this post isn't optional reading.

Questions clarification

We can name $x$ whose $|x|^2$ is power, energy, or neither - physically. $P=IV$ from electronics, for example, hence isn't justification, as one can define $x(t) = i(t) v(t)$. The question then is, why is $|x|^2$ chosen to be power - is it just a choice of words, or is there physical justification for the notion of "energy" upon any general signal as an abstraction?

Some responses that don't answer the question include:

  • "The Parseval-Plancherel's relation, $\|x\|^2 = \|\hat x\|^2$, is a statement of conservation of energy." The statement is that of an equality of a mathematical operation upon functions in transformed domains. Calling it "energy" doesn't make it one. The concept of energy of a signal doesn't subsume the existence of the Fourier transform in the first place.
  • $P=IV$, again, or any other specific physical case that claims itself sufficient, without further explanation, to address questions 1 or 2 in context of generalizing to an abstraction.

Concerning 3, now the rest of this post is somewhat optional, at risk of getting the wrong idea about things read so far.

My thoughts

$(2)$ implies that energy is exclusively an aggregate quantity, and that instantaneous energy cannot exist. This is obviously false: we have kinetic and potential energy for mechanical systems defined at any point in time, but also electrical.

Instantaneous energy example

  1. $1C$ charge $2m$ away from infinite flat electrically charged plate of $F_e(t) = 3V/m$ has potential energy of $E_p(t) = (3 (J/C)/m) \cdot (1C) \cdot (2m) = 6J$, for all $t$.
  2. Repeat with $F_e(t) = 3 \cos(2\pi t) V/m$, we get $E_p(t) = 6\cos(2\pi t) J$. Both can convert to kinetic, and both are measurable at each point in time.

I can see that, in a circuit, since current and "voltage" are easier to measure, and conversion to mechanical energy is best expressed as a process over time, it makes sense to say "the" signal is $i(t)$, which unambiguously translates to power being $\propto i^2(t)$, and to work as integration over time. Yet, "signal" in general is context-agnostic, and could very well be $x(t) = i(t) v(t)$, making $|x(t)|^2$ no longer instantaneous power - so this is no justification.

Why instantaneous energy should be $|x(t)|^2$

Because spring energy is $\propto x^2$.

My argument is from, of all things, Fourier theory. Virtually all of signal processing is built upon it, and the Fourier transform is nothing but sinusoids.

Some have said on this network, energy's computed via $|x[n]|^2$ because it's "just a definition". While technically true, it overlooks an elephant in the room: what causes sine waves, physically? Why does a jerked string, or an electromagnetic wave, oscillate specifically sinusoidally, and not in terms of any other periodic motion, e.g. cycloid? Because of restoring force.

If there's a sine, there's a restoring force. A sine directly arises from equations of motion applied to a spring, whose force is given by $F= k x$ - and I'm not just talking differential equations, I'm talking how computers themselves evaluate sines:

$$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... $$

Begin at position $x$. The block is subjected to $F = kx$, at that time instant. Future position is current position plus time under acceleration, or $x + \int\int(F/m)dt dt$, or $x - (k/m) \int\int x dt dt$. For simplicity, suppose we're at $x=t=0$, and $k=m=1$, then we obtain $x - \frac{x^3}{3!}$. Yet, while traversing to this new position, our acceleration was changing in accordance to the change in position, $-\frac{x^3}{3!}$, so we repeat semi-recursively to arrive at $+\frac{x^5}{5!}$, and so on.

Obviously, the above is nonsense. Yet I think there's a way to do it legit while preserving the intuition I'm trying to convey, but I've not figured it out. Regardless, the end result is correct.

It's a fact that string and EM energies are defined instantaneously in terms of amplitude squared, and both these phenomena have well-known restoring forces - what I'm saying is it makes sense to generalize this to all wave phenomena, and since Fourier theory builds on waves, instantaneous energy should be $|x(t)|^2$.

$\endgroup$
11
  • $\begingroup$ This question is being discussed on Meta. $\endgroup$ Commented Mar 1, 2023 at 13:13
  • $\begingroup$ The energy of a signal $x(t)$ is not necessarily the energy in physics which is measured in Joules. For example, if you were to accept the definition power $|x(t)|^2$, you could always define $y(t)=|x(t)|^2$ and Parseval's theorem still holds for the "energy" of $y(t)$, no conflict. Other answers are not wrong, neither is yours because they do not share the same premise. $\endgroup$
    – AlexTP
    Commented Mar 1, 2023 at 14:29
  • $\begingroup$ you are welcome. So if you agree that the answer can be simply short, why did you have to resort to all these other kinds of stuff? I think the downvotes on your answer are mainly due to its low quality. Don't mix unnecessary terms, use the terminology correctly and try to make simple arguments next time. $\endgroup$
    – AlexTP
    Commented Mar 1, 2023 at 15:14
  • 1
    $\begingroup$ No, your answer is not an alternate or a deeper understanding. It is essentially my comment mixed with unnecessary kinds of stuff and, therefore, makes it incomprehensible. This is also the first of two reasons I am presenting my comment as news to you. The second reason is that this is not the first time you mix concepts from other fields that do not help. You can argue whatever you want, but the quality of your answer is low, at least for me, and I am not surprised by the downvotes. Take this as my subjective opinion, as implied by being in the comment section. $\endgroup$
    – AlexTP
    Commented Mar 1, 2023 at 16:17
  • 1
    $\begingroup$ It's too long now, but listen, for your "first", maybe, however even if this was true, it does not change the quality of your answer, try to produce good answers and people will appreciate them (if you have been through the peer review of scientific publications, you will know that this is true). For your "second", if you want something intuitive, choose my comment instead of your answer. This is to tell you that no need to be discouraged or angry by downvotes or criticism. Sometimes you have good points I acknowledge. But "If you can't explain it simply, you don't understand it well enough". $\endgroup$
    – AlexTP
    Commented Mar 1, 2023 at 17:07

2 Answers 2

5
$\begingroup$

Consider

$$ \begin{align} E_x(t) &= \int\limits_{-\infty}^{t} p_x(u) \, \mathrm{d}u \\ \\ &= \int\limits_{-\infty}^{t} \big|x(u) \big|^2 \, \mathrm{d}u \\ \end{align} $$

For a capacitor or some other energy storage device, there is an instantaneous stored energy that can be defined.

$\endgroup$
4
  • $\begingroup$ I think you've not fully absorbed my Q&A. This example is a "signal" in same sense a car's velocity is. $\endgroup$ Commented Feb 25, 2023 at 17:12
  • 2
    $\begingroup$ Anything can be a signal. You have even have a signal that means power and I do that in my pitch detection algorithms. But if your signal represents what we call "action" (as a function of time), then the square of the signal is "power", and the aggregate of that is "energy". $\endgroup$ Commented Feb 25, 2023 at 17:16
  • $\begingroup$ Yes, anything can be a signal, but not every signal has non-zero instantaneous energy, and your example isn't a counterexample to my OP. $\endgroup$ Commented Feb 25, 2023 at 17:57
  • $\begingroup$ I welcome your calls for undeletion, but if you actually care about this Q&A, I ask for your response. Here's the most relevant bits; elaborating further, acceleration would serve the role of power in your case: $E(t) = \frac{1}{2}m v(t)^2 = \frac{1}{2}m(\int_{-\infty}^t a(u) du)^2$. That there's a units or a square mismatch is conceptually immaterial, I can name $x$ such that $x^2$ physically is energy, power, or neither. I've not had the time to include every popular case. $\endgroup$ Commented Mar 1, 2023 at 13:10
0
$\begingroup$

Answer summary

The answer isn't written or meant to be a proof but a perspective with intuition some can appreciate - which is also the purpose of the question. The points are resolved as follows:

  1. $|x|^2$ is indeed best defined as power, and there is physical justification that generalizes to an abstraction. Specifically, it's justified if the choice of framework is Fourier transform. That's because the FT's basis is sines, and power and energy of physical sines obey the $|x|^2$ definitions with $x$ being amplitude.

    • Infinitesimal energy, e.g. $|x(t)|^2 dt$, is abstracted nicely with the interpretation of $dt$ as a local constant of proportionality, and a "mass" element (whose abstraction is "inertia", which one could conceptualize for a general physical unit in application).
    • Fourier is, in large part, less of a choice and more of an emergent descriptor, due to its characterization of Linear and Time-Invariant systems. Put differently, its statements apply even if our framework of choice differs.
    • "Choice of framework" is a matter of usefulness, in that it could be that, the most useful transform lacks the relevant properties that Fourier has, hence if we defined "energy" per Fourier anyway, it wouldn't be useful.
    • Put together, calling $|x|^2$ power is both useful and physically meaningful in generalization.
  2. Per 1, instantaneous energy as an abstraction stays zero. However, as acknowledged in the question and elaborated in this answer, this implies neither of: (1) there aren't signals with non-zero instantaneous physical energy; (2) there aren't signals whose instantaneous physical energy is $|x|^2$.

Below is "rest of the answer".


It's power.

I spent several hours to conclude this, and figured it's a waste to not post this in case someone has the same thought process. Non-flushed-out answer cause time. Other answers are welcome.


It bugged me that the implication is that energy is necessarily an aggregate quantity, yet we can obviously measure e.g. instantaneous kinetic energy. Wrestling with calculus and foundations of waves from restoring force, one finds that the energy of a differential element is indeed given by $|x(t)|^2$ (see below), but any finite realization requires aggregation, hence "instantaneous energy" is zero. And, $E(t)$, as derived from $\int P(t)dt$, is implicitly aggregating from some reference.

And that, a signal in abstract and general definition obeys the same logic per standard continuum, which subsumes uniformity and ordinality, plus, built on Fourier which is sines, put that all together and fix my lazy English... it's power. -- Now, to tackle the discrete case, if we insist on sample-by-sample, one could argue each sample is indeed energy, in that we aggregate over the sampling period with local uniformity assumption.

Some examples:

  1. $E_k = \frac{1}{2}mv^2$. Thing is, $m$ is an aggregate quantity - the total mass of an object. If we tried to find the "instantaneous energy" of a ball, at each point of a ball, we'd have to integrate over a non-zero interval to not end up with zero mass.
  2. For a string, the energy of a mass element is $\propto y^2$. Same story, either the mass is zero, or we're aggregating. We must recall that a signal spans a continuum and has a domain, so if $y(x)$ is vertical displacement vs horizontal distance, we can't say there's instantaneous energy "at" $x$, because there can't be non-zero mass at $x$, only over $\Delta x > 0$.
  3. Circuit current is defined as time-rate passage of charge through a cross-section; if we argue that at some point in a circuit there's charge, and $E = qV$, again $q$ is an aggregation. "Electrons are sizeless" is mathematically resolved as a Dirac delta.

Now I'd like to revisit:

  1. $E(t) = \frac{1}{2}m v^2(t)$. Yes, that's meaningfully instantaneous kinetic energy, and $v(t)$ can very well be "signal". But it's not a signal that represents a wave. $x(t) = i(t) v(t)$ (circuits) is also a signal, but only $i(t)$ and $v(t)$ are physically waves. This adds to the obviousness, the definition is arbitrary. But surely, if we are to choose, we best choose meaningfully - and repeating my OP, since signal processing is built upon Fourier, which is sines, the definition is best rooted in waves.

Also, I've certainly not contested $E = \sum |x|^2$, only conflated $E[n] = |x[n]|^2$. There's great benefit to defining energy as $\sum |x|^2$, mainly via Parseval-Plancherel's theorem.

Re: some points in OP

  1. Amplitude squared in string energy is indeed over the entire string, and in EM waves it's yet still energy density. Above example 1 would integrate over mass density; the alt would be a Dirac delta, but a ball isn't sizeless, so while there are options in terms of mathematical arbitration, only one maps to reality: aggregation over a non-Delta/non-infinite continuum can only shrink with a shrinking interval, converging to $0$ "at a point".

  2. "Instantaneous energy for signals?" - physically, for some signals (and for those signals "total energy" by aggregation is meaningless, as it's already total energy). But in physically-agnostic sense, the most useful abstraction is zero.

'differential element' energy clarification

There's of course no measure of a differential quantity without a differential, so $dt$ there is implicit and was omitted on purpose. But I should clarify.

I say "it's energy" here in a same sense we say $i^2(t)$ is energy in a circuit, but it can't be taken out of context of my post otherwise it's wrong. Following a derivation of energy of vibrating string, we have

enter image description here

Energy of entire string is computed by aggregating over mass differentials. For each such differential (or rather, infinitesimal), we have energy proportional to square of amplitude, so "it's energy" in same sense we'd say $v^2(t)$ is energy in kinetic energy, dropping mass.

The abstract generalization is in form of $|x(\cdot)|^2 d\cdot$, where $d\cdot$ is the abstract equivalent of "mass" over the input domain of the signal. There is and isn't a problem with treating the differential as a constant of proportionality, since it shares the input domain with the signal, but I'm focused on the "isn't".

$\endgroup$
1
  • $\begingroup$ For "historic" reasons one can explore at Meta, I've resubmitted the answer with a significant edit, which could become a separate answer with some work. If you still believe it flawed, I'll take the votes. $\endgroup$ Commented Mar 2, 2023 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.