**ab initio**

Well, either a) the signal must be finite in length, (i.e., at some point for t < t_0, x(t)==0 and for t > t_end, x(t) ==0). Here's a link to a university lecture that discusses energy in signals (http://ocw.usu.edu/Electrical_and_Computer_Engineering/Signals_and_Systems/5_10node10.html). Here they require a finite length signal.

Or b) you can talk about the signal's energy per unit time. I believe this is the more typical case. Here, Parseval's theorem is useful for showing that the energy in the signal is the same whether you calculate it in time domain or Fourier domain (http://en.wikipedia.org/wiki/Parseval%27s_theorem). One might do this by considering the energy of some periodic signal. One integrates over the period of the signal.

The energy in the signal is bounded if the energy in the Fourier coefficients decrease sufficiently fast with increasing wavenumber. In the case of a band-limited signal, this is the case, because all Fourier coefficients are zero beyond the Nyquist frequency.

Cheers

**mikeaj**

It's also been a while for me, but something like a sinusoid over infinite time does have infinite energy. (Think of a voltage signal like that operating for eternity—you'd need an infinite amount of energy from some source to keep applying that voltage across a resistor or something.) That's why these are analyzed in terms of power, so divide by the time of integration. If you're not integrating over infinite time and you're not integrating something that's going to infinity in nasty ways, then the integral will be finite.

Look at power spectral density, etc. and see if there is nonzero content at frequencies above Nyquist. Or just filter those frequencies out, depending on the application.

Amazing how the fundamentals slip out your head after some time. You guys more or less covered it for me. Something along the lines of a certain class of energy signals that fall into category of power signals which can still be analyzed using fourier integration without blowing up, if I recall correctly. Gets you that delta functions as a consequence, as a sort of a limit.

Anyways, this was a bit off topic. The phrase "all real signals are finite energy" triggered the question for me.