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Root mean square (RMS) of a sinc function

  1. Jan 29, 2009 #1
    Given that the root mean square (RMS) of a sine function is as follows:

    RMS of (a*sin([tex]\omega[/tex]*r) = a / [tex]\sqrt{}2[/tex]

    Let a = 1/[tex]\omega[/tex]


    RMS of ((1/[tex]\omega[/tex])*sin([tex]\omega[/tex]*r)) = 1 / ([tex]\omega[/tex]*[tex]\sqrt{}2[/tex])

    But for sinc([tex]\omega[/tex]*x), what is formula for the RMS?
  2. jcsd
  3. Feb 1, 2009 #2
    The rms value of sinc is zero.

    I'll define rms as
    [tex]f_{rms} = \sqrt{ \lim_{a \to \infty} \frac{1}{2a} \int_{-a}^{a} f(x)^2 \,dx }.[/tex]
    Let's find the rms for sinc(x). The integral here is
    [tex]\int_{-a}^{a} \frac{\sin^2 x}{x^2} \,dx[/tex]
    [tex]= \int_{-a}^{-1} \frac{\sin^2 x}{x^2} \,dx + \int_{-1}^{1} \frac{\sin^2 x}{x^2} \,dx + \int_{1}^{a} \frac{\sin^2 x}{x^2} \,dx[/tex]
    [tex]\le \int_{-a}^{-1} \frac{1}{x^2} \,dx + \int_{-1}^{1} 1 \,dx + \int_{1}^{a} \frac{1}{x^2} \,dx[/tex]
    [tex]= 4 - \frac2a < 4.[/tex]

    Thus, the rms value of sinc is
    [tex]\sqrt{ \lim_{a \to \infty} \frac{1}{2a} \int_{-a}^{a} \frac{\sin^2 x}{x^2} \,dx } \le \sqrt{ \lim_{a \to \infty} \frac{1}{2a} 4 } = 0.[/tex]
  4. Feb 2, 2009 #3
  5. Jun 3, 2009 #4
    I understand the case when a approaches infinity, but what about the case where "a" is a finite number?
  6. Jun 3, 2009 #5


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    The sinc function is not periodic, in other words it's period is infinity. That's why adriank found it's RMS value to be zero. Think of it this way, it's got a finite total energy that is spread over an infinite time interval, the result is an RMS value of zero.

    In many ways it doesn't really make a lot of sense to even talk about the RMS value of an aperiodic function like that. What exactly is your application?
  7. Jun 3, 2009 #6
    I want to find a good approximation for the following function:
    a = k * sinc( pi k r) = sin (pi k r) / (pi r)

    One approximation would be the following
    a = 1 / (pi r)

    But I think a better approximation is based on RMS. My feeling is that the sinc funciton above is a periodic function, though with decreasing amplitude as a function of r and therefore can be approximated using RMS.
  8. Jun 3, 2009 #7
    Well, the rms value of sin is [itex]1/\sqrt{2}[/itex], so you might guess
    [tex]a = \frac{1}{\sqrt{2} \lvert \pi r \rvert}.[/tex]
    It doesn't seem very useful though. It's not square-integrable, for instance.
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