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 $$\int \lvert \delta(f-f_0)\rvert^2df\quad?$$

  • $\begingroup$ 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 :) $\endgroup$ Commented Mar 28, 2017 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.
    Commented Mar 28, 2017 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. $\endgroup$ Commented Mar 28, 2017 at 12:13

1 Answer 1


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

  • $\begingroup$ 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
    Commented Mar 28, 2017 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
    Commented Mar 28, 2017 at 10:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.