# How can I interpret “energy” in signals?

I am learning about various signal processing methods in my university course, and I can't seem to grasp what 'energy' in signals represent. I mean, I know that it is the integral of the absolute value of energy squared (side question: why is absolute value needed?), but I don't know the meaning of what this energy is. Is there an intuitive way to think about these energies? Thanks a lot.

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Since this is signal processing, might you be asking about the energy spectral density? –  Unreasonable Sin Oct 26 '10 at 16:04
+1. In my answer I try to explain the implied physical constant one assumes. –  Américo Tavares Oct 26 '10 at 17:50

The absolute value is probably because signal amplitudes are often represented by phasors, especially in the context of electromagnetic radiation or in the context of processing via Fourier analysis. And since it uses complex numbers, and the square of a complex number $z^2$ is not the same as its 'absolute value squared' $|z|^2$ (simplest case, take $z = -i$, the imaginary unit, then the former is -1 and the latter 1$), in the formla it is customary to write an absolute value. That we associate energy with the square of the wave amplitude is due to the mechanical model of small amplitude waves of d'Alembert. Essentially the wave medium is thought of as a bunch of springs coupled together. For springs, with Hooke's law we have that the mechanical energy is proportional to the square of the amplitude of displacement. The analogy that the displacement of the media due to the wave should be well approximated by a simple spring leads us to take the infinitesimal mechanical energy associated to an infinitesimal chunck of the medium to be the square of the wave amplitude. So adding them all up you get an integral. - Link to this one in FAQs on "how to write a good answer". Each aspect of question answered clearly in its own paragraph. Intuition conveyed. Economical but clear explanation with minimal use of formulas. – T.. Oct 26 '10 at 16:35 +1 for your nice answer. – Américo Tavares Oct 26 '10 at 17:50 Nice answer. Just for curiosity: Why did you take$-i$instead of$i$as your example for$z^2 \ne |z|^2$? – j.p. Oct 27 '10 at 7:53 @jug: purely an editing issue. I was typing$\sqrt{-1}$, thought better of it, and accidentally left the minus sign still intact. – Willie Wong Oct 27 '10 at 9:25 Thanks for solving this mystery, I really started to wonder about why$-i$is simpler than$i$;-). – j.p. Oct 27 '10 at 11:57 I try to give an explanation in the time and frequency domains, as normaly used in Telecomunication Systems. If you have a physical process described in the time by an oscillating signal$s(t)$such as an electrical voltage or current, the description in the frequency domain is given by its spectrum $$\overrightarrow{S}(\omega )=\frac{1}{% 2\pi }\int_{-\infty }^{\infty }s(t)e^{-i\omega t}dt$$ The signal is $$% s(t)=\int_{-\infty }^{\infty }\overrightarrow{S}(\omega )e^{-i\omega t}d\omega ,$$ where$\omega =2\pi f=\frac{2\pi }{T}$is the angular frequency. The function$\overrightarrow{S}(\omega )$is the Fourier Transform of$s(t)$and$s(t)$is the inverse transform. Note: a definition of the pair$(s(t),\overrightarrow{S}(\omega ))$with different constant factors is possible. The "energetic activity" during the infinitesimal time interval$% \delta t$is$\delta E=s^{2}(t)\delta t$(excluding a constant with physical dimensions, see note below). The instantaneous power is $$p(t)=\underset{\delta t\rightarrow 0% }{\lim }\frac{\delta E}{\delta t}=s^{2}(t).$$ Therefore the "energy" of the full process is $$E=\int_{-\infty }^{\infty }s^{2}(t)dt=2\pi \int_{-\infty }^{\infty }S^{2}(\omega )d\omega ,$$ where$S(\omega )=\left\vert \overrightarrow{S}(\omega )\right\vert $and$\overrightarrow{S}(\omega )=S(\omega )e^{i\psi (\omega )}$is the distribution or spectrum with the properties$S(-\omega )=S(\omega )$,$\psi (-\omega )=\psi (\omega ).$The equality between both integrals is known as the Parseval's Theorem. Note on the physical dimensions. e.g. in electrical circuits the product of a voltage$V$by a current$I$is a power (energy/time). By the Ohm's law, in d.c. circuits,$V=RI$and$VI=RI^2=\dfrac{1}{R}V^2$; in a.c. circuits$| \overrightarrow{V}|| \overrightarrow{I}|=R |\overrightarrow{I}|^2$. "The power spectral density (PSD) (...) describes how the power of a signal or time series is distributed with frequency. Here power can be the actual physical power, or more often, for convenience with abstract signals, can be defined as the squared value of the signal, that is, as the actual power dissipated in a load if the signal were a voltage applied to it." - This is a fairly good answer, though I was surprised not to see the words "Fourier transform" or "power spectrum"! – Noldorin Oct 27 '10 at 11:20 Answer edited to include them. – Américo Tavares Oct 27 '10 at 12:58 Yeah, I'm amazed Américo wrote something neat that does not even mention Fourier! +1. – Ｊ. Ｍ. Oct 27 '10 at 13:00 The electrical power in a resistive circuit is given by$P = I V$. Frequently one has a fixed resistance (or more in general, impedance), so that voltage and current are variable; taking voltage as independent variable, we have$P = V^2/R$. This gives the instantaneous power (energy over time unit), and hence also gives the average power (or energy over some finite time interval) if the voltage is constant (DC current). When the voltage varies peridocally in time (for example, AC current), the above equation applies to each instant of time ($ P(t) = V(t)^2/R \$) , so that, for getting the average power over a period one must integrate (see here). That's why AC energy is typically measured in RMS units: a 120V power outlet gives a sinuosid of approx. 170V peak voltage: that sinusoid has the same energy as a constant 120V voltage.