2
$\begingroup$

I am trying to derive a simple expression for root-mean-square bandwidth $B_{\rm rms}$ for a signal that has flat spectral density over its entire bandwidth $B$. The expression for the $B_{\rm rms}$ is

$$ B_{\rm rms} = \sqrt{\frac{\displaystyle\int_{-B/2}^{B/2} \lvert H(\omega)\rvert^2 \omega^2\, \mathrm{d}\omega}{\displaystyle\int_{-B/2}^{B/2} \lvert H(\omega)\rvert^2\, \mathrm{d}\omega}} $$

where $\lvert H(\omega)\rvert^2 = A$ is the constant PSD over $B_\rm h$.

I read it at some sources that this gets simplified as follows: $B_{\rm rms} = \frac{B}{\sqrt{3}}$.

However, when I integrate this I do not see how I can get this expression:

$$ B_{\rm rms} = \sqrt{\frac{\displaystyle\int_{-B/2}^{B/2} A \omega^2\, \mathrm{d}\omega}{\displaystyle\int\limits_{-B/2}^{B/2} A\, \mathrm{d}\omega}} = \sqrt{\frac{\displaystyle\frac{1}{3}\cdot\frac{2B^3}{8}}{\displaystyle\frac{2B}{2}}} = \frac{B}{\sqrt{12}} .$$

$\endgroup$
  • $\begingroup$ sometimes this is called the "spectral centroid" $\endgroup$ – robert bristow-johnson Aug 17 '16 at 19:33
2
$\begingroup$

Your formula for the RMS bandwidth makes sense for (perfectly band-limited) low pass signals, i.e., for signals with a spectrum centered around $f_0=0$. The bandwidth of low pass signals is defined as the support of their spectrum at positive frequencies. So your integration limits must be $-B$ and $B$. This results in

$$B^2_{\rm rms}=\frac{\displaystyle\int_{-B}^{B}f^2df}{\displaystyle\int_{-B}^Bdf}=\frac{2B^3/3}{2B}=\frac{B^2}{3}\tag{1}\\$$

which is the expression that you're looking for.

Note that the RMS bandwidth is also defined for signals that are not ideally band-limited. In that case you have to integrate from $-\infty$ to $\infty$, as pointed out in MBaz's answer.

If you have a (not necessarily perfectly band-limited) band pass signal centered around $f_0\neq 0$, you must use the following formula for the RMS bandwidth:

$$B^2_{\rm rms}=\frac{4\displaystyle\int_{0}^{\infty}\lvert H(f)\rvert^2(f-f_0)^2df}{\displaystyle\int_{0}^{\infty}\lvert H(f)\rvert^2df}\tag{2}\\$$

If $\lvert H(f)\rvert$ is perfectly band-limited and if it is constant ($\lvert H(f)\rvert=c$) in the interval $[f_0-B/2,f_0+B/2]$, its RMS bandwidth is

$$B^2_{\rm rms}=\frac{4c^2\displaystyle\int_{f_0-B/2}^{f_0+B/2}(f-f_0)^2df}{c^2\displaystyle\int_{f_0-B/2}^{f_0+B/2}df}=\frac{4c^2\displaystyle\int_{-B/2}^{B/2}f^2df}{c^2B}=\frac{B^2}{3}\tag{3}\\$$

which is the same result as for the corresponding low pass signal.

Note the factor $4$ in Eq. $(2)$, which is caused by the fact that the RMS bandwidth (without the correction factor) measures the "width" (variance) of the function $|H(f)|^2$. For low pass signals this width is measured across the whole spectrum (positive and negative frequencies), whereas for band pass signals, this width is measured only at positive frequencies. If this width is the same for a low pass signal and for a band pass signal, then the bandwidth of the band pass signal must be twice the bandwidth of the corresponding low pass signal, because bandwidth is measured at positive frequencies only. This is the reason why the squared bandwidth in the case of a band pass signal needs a factor $4$ to comply with the standard definition of bandwidth for low pass and band pass signals.

$\endgroup$
  • $\begingroup$ Thanks. But, if my signal is instead bandpass, is there any approximation I can use? $\endgroup$ – r2d2 Aug 17 '16 at 13:15
  • $\begingroup$ Isn't integrating from $-B$ to $B$ integrating over double the bandwidth instead of $B$ i.e. $-B/2$ to $B/2$ (or $0$ to $B$) ? With the change of $\omega$ to $f$, shouldn't there be a $2\pi$ factor on the boundaries and in the expression ? $\endgroup$ – Gilles Aug 17 '16 at 13:17
  • $\begingroup$ @user164568: I've added the formula for band pass signals to my answer. $\endgroup$ – Matt L. Aug 17 '16 at 17:28
  • $\begingroup$ @Gilles: You have to integrate over $[-B,B]$ for low pass signals, not over $[-B/2,B/2]$, because the spectrum of low pass signals extends over $[-B,B]$. You can integrate over $[0,B]$ if you do it in the numerator as well as in the denominator, because that's just the same as integrating over $[-B,B]$. $\endgroup$ – Matt L. Aug 17 '16 at 17:31
  • $\begingroup$ @MattL.: You're right, I overlooked the first sentence in your answer. Thanks. $\endgroup$ – Gilles Aug 17 '16 at 17:51
1
$\begingroup$

Note that your definition for RMS bandwidth is not really standard; see for instance the definition here, where the RMS bandwidth of a signal with spectrum $S(f)$ is given as $$B_{\rm rms}^2 = \frac{\displaystyle\int_{-\infty}^\infty f^2\lvert S(f)\rvert^2}{\displaystyle\int_{-\infty}^\infty \lvert S(f)\rvert^2}.$$ Note that the integration limits are more general, and it uses $f$ (measured in Hz) instead of $\omega$ (measured in rad/s).

Let's say that the signal's conventional bandwidth is $W$; that is, its spectrum is zero iff $\lvert f\rvert>W$. Then, the integral limits are $-W$ and $W$ and @MattL.'s result follows immediately: $$B_{\rm rms}^2=\frac{W^2}{3}.$$

In your question, your integral limits imply that $W=B/2$; that is, you're saying that the signal's spectrum is zero iff $\lvert f\rvert>B/2$ (ignoring for the moment the fact that your definition uses $\omega$). So, your RMS bandwidth is $$B_{\rm rms}^2 = \frac{(B/2)^2}{3} = \frac{B^2}{12},$$ which is the result you obtained.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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