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I'm having DSP for the first time, and after some classes I got confused about the following:

Suppose I have a signal which its fourier transform in a frequency band $[ \omega_1,\omega_2] $ is just a delta function with a peak somewhere in that band . What is the energy of the signal in that specific frequency band?

It seems that the answer depends on the integral

$$ \int_{\omega_1}^{\omega_2}|\delta (\omega-\omega^*)|^2.d\omega $$

with $\omega_1 < \omega^* < \omega_2$.

I'd guess this is 1 (maybe 0...), but I don't know how to compute it.

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\begin{align} &\int_{\omega_1}^{\omega_2}|\delta (\omega-\omega^*)|^2.d\omega \\ &= \int_{\omega_1}^{\omega_2}|\delta (\omega-\omega^*)|.|\delta (\omega-\omega^*)|.d\omega \\ &= \int_{\omega_1}^{\omega_2}\delta (\omega-\omega^*).\delta (\omega-\omega^*).d\omega \\ &= \int_{\omega_1}^{\omega_2}\delta (\omega^*-\omega^*).d\omega \\ &= \int_{\omega_1}^{\omega_2}\delta (0).d\omega \\ &= \delta (0).\Delta \omega \end{align} Which is undefined! See this for better clarification: https://en.wikipedia.org/wiki/Distribution_(mathematics)#Adding_and_multiplying_distributions

What you have to understand is that $\delta$ is not a function, it is a distribution. What you learn in introductory DSP courses is just an "engineering trick" so that things will work the way they should (you will understand what that means once you study sampling and analog to digital conversion).

EDIT: Sorry I misinterpretation the question. (Thanks Jazzmaniac for pointing that out)

As far as I know the $\delta$ "function" in the continuous domain is not in $L^2$, it isn't a square-integrable function.

Think about your question in the time domain. If you have 2 symmetric $\delta$'s in the frequency domain, you will have a INFINITLY lasting sinusoid. It's energy is infinite, i.e, you can't associate a value to it (undefined, remember that $\infty$ is not a number). If you have a single $\delta(f)$, centered at zero, you will have a complex exponential in the time domain, and it's energy is also undefined (unbounded since $|e^x|^2=1$.)

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  • $\begingroup$ Wow. "Defined" to be 1? That was unexpected... Is there any definition about its power on that same frequency band? $\endgroup$ – JLagana Nov 12 '15 at 16:51
  • $\begingroup$ To answer these questions, usually you can think of the dirac function as the limit of a window that has length $1/a$ and amplitude $a$, when $a$ approaches $\infty$. $\endgroup$ – Daniel Severo Nov 12 '15 at 16:54
  • $\begingroup$ Okay, I'll try that. Thank you for the help! $\endgroup$ – JLagana Nov 12 '15 at 16:55
  • $\begingroup$ The energy of the Dirac delta is NOT defined to be 1. That would be in vast disagreement with the unitarity of the Fourier transform. If you want the Dirac delta to have energy 1 then its Fourier dual $\exp(i\omega t)$ would have to also have unit energy. The latter obviously leads to great inconsistencies. If anything then the energy is infinite. $\endgroup$ – Jazzmaniac Nov 12 '15 at 23:46
  • $\begingroup$ You are right, when I answered I had in my mind the discrete domain. The $\delta$ (continuous) is not in $L^2$ as far as I know, so you can't calculate it's energy. $\endgroup$ – Daniel Severo Nov 13 '15 at 0:26

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