# In the RMS bandwidth equation, why do we have the energy of the signal in the denominator?

In the following equation: $$B_{\rm rms}^2 = \frac{\displaystyle\int_{-\infty}^\infty f^2\lvert G(f)\rvert^2df}{\displaystyle\int_{-\infty}^\infty \lvert G(f)\rvert^2df}.$$

It is not clear why do we divide the RMS of the signal by its energy, I tried looking for answers in several books but did not find any explination.

• To normalize it. Otherwise bandwidth will be dependent on the energy of the $g(t)$. It's like that for any average or mean. You gotta thing that is measured in units of fargs. What is the mean-square farg of all of the things? Commented Aug 1, 2023 at 19:10

$$|G(f)|^2$$ is a nonnegative function of $$f$$. Another nonnegative function that might be familiar to you is a probability density function (pdf). Is $$|G(f)|^2$$ a pdf? Not necessarily because the area under the $$|G(f)|^2$$ curve is not necessarily equal to $$1$$. Well, the area under the $$|G(f)|^2$$ curve is $$\int_{-\infty}^\infty \lvert G(f)\rvert^2 \,\mathrm df$$ and so $$\frac{|G(f)|^2}{\displaystyle \int_{-\infty}^\infty \lvert G(f)\rvert^2 \,\mathrm df}$$ is indeed a valid pdf of some random variable $$X$$. (It is also the energy special density of a unit-energy signal). Then, what is $$E[X^2]$$? it is given by \begin{align}E[X^2] &= \int_{-\infty}^\infty f^2\cdot \left[\frac{|G(f)|^2}{\displaystyle \int_{-\infty}^\infty \lvert G(f)\rvert^2 \,\mathrm df}\right] \,\mathrm df\\ &= \frac{\displaystyle \int_{-\infty}^\infty f^2{|G(f)|^2}\,\mathrm df}{\displaystyle \int_{-\infty}^\infty \lvert G(f)\rvert^2 \,\mathrm df}\\ &= B_{\text{rms}}^2~~!! \end{align} If we were to amplify the signal by a factor ot $$2$$, say, then its RMS bandwidth will not change, but its energy spectral density will increase. Thus, normalizing the energy spectral density to be that of a unit-energy signal gives us $$B_{\text{rms}}^2$$.