My book says:

The width-1 NRZ pulse is $$ \mathrm{rect} (t) = \begin{cases} 1 , \qquad -1/2 \leq t \leq 1/2\\ 0, \qquad \mathrm{otherwise} \tag 1 \end{cases} $$ The unit-energy width-$T$ NRZ pulse is $$ \frac{1}{\sqrt T} \mathrm{rect}(\frac{t}{T}) \tag 2 $$

I need help who to derive the unit-energy.

With $\mathrm{rect}(t/T)$ I think it is the function $$ \mathrm{rect}(t/T) = \begin{cases} 1 , \qquad -T/2 \leq t \leq T/2\\ 0, \qquad \mathrm{otherwise} \end{cases} $$ The energy is definition as $ E = \int_{-\infty}^{\infty} \lvert x(t) \rvert^2 \, dt $, so $$ E = \int_{-T/2}^{T/2} \lvert \mathrm{rect}(t/T) \rvert^2 \, dt = \int_{-T/2}^{T/2} 1^2 \, dt = \frac{T}{2}-(-\frac{T}{2}) = T $$ And we want unit-energy so $E=1$, but I don't know how to proceed. How can I find $(2)$?


1 Answer 1


I think you misunderstand what the book is saying. There is nothing like a "unit-energy" that you can compute. There is, however, a "unit-energy [...] pulse", which is a pulse with energy equal to $1$. So if you have some pulse $p(t)$ with energy


and you want to normalize it such that its energy becomes unity, you simply have to scale it by $1/\sqrt{E_p}$, which gives you the corresponding unit-energy pulse


with energy


as expected.

  • $\begingroup$ Thank you, it clarified some things! But why is the scaling factor $1/\sqrt{E_p}$? I.e. can we derive $(2)$ somehow? I can't get a sense of $(2)$ by intuition. $\endgroup$
    – JDoeDoe
    Dec 31, 2021 at 12:39
  • 1
    $\begingroup$ @JDoeDoe: Any constant scaling factor $c$ applied to a function shows up as $c^2$ in the energy (because we integrate over the square of the function). So by scaling with $c$, the energy is multiplied by $c^2$. If I have a pulse with energy $E_p$ and I want to scale it such that its energy becomes $1$, I need to scale the pulse by $1/\sqrt{E_p}$ such that the energy gets scaled by $1/E_p$. $\endgroup$
    – Matt L.
    Dec 31, 2021 at 12:48

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.