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I would like to know why we are not calculating the negative energy of a signal.For example for $sin$ wave there is both an positive and negative phase.But while we take the energy of the signal we are only summing up the square of its instantaneous values.

Energy of an Analog Signal is :$\int_{-\infty}^{\infty} s^2(t)dt$

Here are we actually neglecting the negative energy of the signal or whether there isn't any negative energy associated with an signal?Could anyone help me.

Source is:http://1drv.ms/1zLxlzg (p.17)

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  • $\begingroup$ The square of the signal is always positive, hence the sum (integral) of the square of the signal will also be positive. How do you propose to calculate negative energy? $\endgroup$ – theo Nov 22 '14 at 8:00
  • $\begingroup$ @theo:But doesn't the negative effect of energy vanishes when we take it's sqaure? $\endgroup$ – justin Nov 22 '14 at 8:02
  • $\begingroup$ Basically, life's too short to let negative energy influence you. $\endgroup$ – theo Nov 27 '14 at 14:28
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The energy density in an electric field $E$ is proportional to $E^2$. See this article on Wikipedia for a proof if you want the gory details.

That's why when you're calculating the total energy you integrate $E^2$, as you say in your question. This also explains why there is no negative energy associated with the negative half of the wave cycle, because the negative values of $E$ square to a positive value of $E^2$ so the energy is always positive.

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The negative energy doesn't exist. The negative symbol indicates that the direction of the signal is opposite. The signal may be voltage signal or current signal. Suppose the $R=1\: \mathrm{\Omega}$, as the energy is $\frac{V^2}{R}t$ or $I^2Rt$. Give $dt$ and $R=1$. The differential of the energy is $V^2dt$ or $I^2dt$. Instead both V and I with $s(t)$,the result is $s^2(t)dt$. the formula indicates the truth.No matter the signal flow the Resistance by the path or the opposite path, the Resistance will produce heat.not the negative heat.

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  • $\begingroup$ Yes, negative energies DO exist, but in bound states. Of course, it's not the case in the situation indicated in the question. $\endgroup$ – Sofia Nov 22 '14 at 11:57
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We neglect NOTHING, when we calculate the energy. You have to distinguish between the AMPLITUDE of the signal, given by your formula s(t), and the ENERGY of the signal. By the way, negative energies do exist, e.g. the electrons in an atom have negative energy - to escape from the nuclear attraction and become free to run away from the atom, one has to supply to them additional energy. But this is not your case with a signal moving (I guess) in the space free of fields.

Now, your question is analog to "energy of a signal". So, see my answer at

Energy of a signal

It is said there that what we can detect is actually the flux of the signal impinging on our apparatuses. The formula of the flux is given, and it shows why we come to the square of which you ask.

Good luck!

Sofia

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  • $\begingroup$ :Whether a signal having negative amplitude(for example sin wave in it's 2nd half has negative amplitude and then reaches to zero) would possess negative energy? $\endgroup$ – justin Nov 29 '14 at 4:34

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