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Suppose a signal with two-sided power spectral density $N/2$ is passed through an ideal LPF having cutoff frequncy $B\,\mathrm{Hz}$. The output power is equal to $N/2 \times 2B = NB$ Watts. Why not $NB/2$?

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    $\begingroup$ The question in the title is backwards. Let us ask instead, if negative frequencies cannot exist in practice, why do we even define a two-sided power spectral density? It is possible to deny the existence of negative frequencies and carry out all calculations using non-negative frequencies only, in which case the one-sided power spectral density would be $N$, and we have to take special precautions if the PSD has an impulse at $f=0$. See e.g. here for a discussion of the issues. $\endgroup$ – Dilip Sarwate Oct 22 '12 at 13:23
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Negative frequencies do exist. The vector base for Fourier transforms (and therefore power spectral density) is the set of complex exponentials $e^{j\omega t}$ where the frequency $\omega$ can be positive or negative.

A real sinusoid can be decomposed into complex exponentials:

$$\cos(\omega t) = \frac{e^{j\omega t} + e^{-j \omega t}}{2}$$

There you have your practical signal (the cosine wave) expressed in terms of the Fourier vector base (complex exponentials). Both sides of the spectrum must be considered in order to calculate power.

This is similar for any other real signal. In general, for real signals, the power spectrum will be symmetric and you must consider power in both sides of the spectrum.

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    $\begingroup$ "Negative frequencies do exist" but its only a conceptual idea arising from complex math. The relation to compute power from spectral density has to account for the underlying math notation and hence we consider both sides of spectrum. As @Juancho said for real signals, the spectrum will always be symmetric. $\endgroup$ – sauravrt Oct 22 '12 at 12:58
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    $\begingroup$ @sauravrt Yes, but all mathematical models are just "conceptual ideas". Generally the model that is the simplest and most consistent is considered to be the best, and in this case that is the model with negative frequencies. $\endgroup$ – Jim Clay Oct 22 '12 at 18:20
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Negative frequencies just represent the clockwise rotating phasors. Whereas , positive frequencies represent the counter-clockwise rotating phasors . In complex mathematics , $cos \omega t= (e^{j\omega t} + e^{-j\omega t}) /2$ . Now if you draw the phasor diagram of the RHS equation , there will be two phasors in it , $e^{j\omega t}$ rotating in counter clockwise direction . and $e^{-j\omega t}$ rotating in clockwise manner . The value of $cos \omega t$ , at any instant $t$ , is nothing but , the sum of projection of each of these phasors on x-axis , whole divided by $2$. I hope this clears up what exactly Negative frequencies represent!

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  • $\begingroup$ But, $j$ is an imaginary number and so does not exist either. Thus, you are explaining a nonexistent concept (negative frequencies) by saying that negative frequencies exist if we assume that another nonexistent object (the square root of $-1$) exists! Everybody knows that $-1$ has no square root and that negative frequencies do not exist! $$ $$ This answer does not add anything beyond what juancho has already said. $\endgroup$ – Dilip Sarwate Apr 8 '15 at 13:51
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The negative is comparatively concept .. this means when your reference signal has a main component @ fo there are also other components that leading and lagging to fo .. so the leading and lagging produce the (+) and (-) components on the right and left sides if fo respectively ..

I hope this may clear your confusion about negative frequencies..

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