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I have this signal: $$ X(f)= 2\delta(f)+ \delta\left(f-\frac 1{T_0}\right)+\delta\left(f+\frac 1{T_0}\right)+\textrm{rect}\left(\frac{f-\frac 4{T_0}}{\frac 2{T_0}}\right)+\textrm{rect}\left(\frac{f+\frac 4{T_0}}{\frac 2{T_0}}\right)$$

I must calculate the energy. How can I find $$\int \lvert \delta(f-f_0)\rvert^2df\quad?$$

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  • $\begingroup$ Hint: strictly speaking, the value of $\delta(x)$ at $x=0$ is undefined. But the integral over it is not. You've got your question backwards :) $\endgroup$ – Marcus Müller Mar 28 '17 at 7:43
  • $\begingroup$ @MarcusMüller: You're right that the integral over $\delta(x)$ is well-defined, but here we're dealing with the integral of $\delta^2(x)$, which is undefined, because the square of a distribution is undefined. $\endgroup$ – Matt L. Mar 28 '17 at 11:55
  • $\begingroup$ @MattL that's true, but that's basically because this spectrum has discrete components, so it must be periodic and hence cannot have Energy as defined in the question. $\endgroup$ – Marcus Müller Mar 28 '17 at 12:13
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The signal, whose total energy you want to calculate, is periodic therefore it will have infinite energy...

To see that note, the following Fourier transform pair: $$ x(t) = \cos(2\pi f_0 t) \longleftrightarrow X(f) = 0.5 \delta(f + f_0) + 0.5 \delta(f - f_0) $$

And based on Parseval's relation, $$ E_x = \int_{-\infty}^{\infty} |x(t)|^2 dt = \int_{-\infty}^{\infty} |X(f)|^2 df $$

you can conlude that total energy is infinite

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  • $\begingroup$ many thanks. this signal will have over power, and the power is $1/T_0(\int |\delta(f)+\delta(f+f_0)+\delta(f-f_0)+rect+rect|^2df)$ and will have the same problem $\endgroup$ – Santo1991 Mar 28 '17 at 7:59
  • $\begingroup$ But it did not tell "How find $\int |\delta(f-f_0)|^2df?$". It is easy to understand the Fourier transform pair and the fact that a power signal has infinite energy. However, it cannot clearly be concluded from the given $E_x$ and the PT. Maybe more rigorous discussion on the integral itself? I can see this has been also mentioned in the comments. $\endgroup$ – msm Mar 28 '17 at 10:36

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