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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 $(2)$?

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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

$$E_p=\int_{-\infty}^{\infty}|p(t)|^2dt\tag{1}$$

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

$$\tilde{p}(t)=\frac{1}{\sqrt{E_p}}p(t)\tag{2}$$

with energy

$$\int_{-\infty}^{\infty}|\tilde{p}(t)|^2dt=\frac{1}{E_p}\int_{-\infty}^{\infty}|p(t)|^2dt=1\tag{3}$$

as expected.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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