Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sinc functions orthogonality

  1. Feb 1, 2010 #1
    I'd like to prove the orthogonality of two "shifted" Sinc functions, but I can't find the mistake.
    Here is my attempt:

    Observing this quantity can be obtained by evaluating the Fourier transform at zero, we have:

    [tex]\mathcal{F}\{ sinc(x)sinc(x-x_0) \}(0)[/tex]

    and using the convolution theorem and the shift theorem (for the second sinc), we get:

    [tex](rect(\omega)\otimes e^{-i2\pi\omega x_0}rect(\omega))(0) = [/tex]

    [tex]= \int_{-1/2}^{+1/2} e^{-i2\pi\omega x_0}d\omega = [/tex]

    [tex]= \left[ \frac{sin(2\pi\omega x_0)}{2\pi x_0} \right]_{-1/2}^{1/2} = [/tex]

    [tex]= \frac{sin(\pi x_0)}{\pi x_0} = [/tex]

    [tex]= sinc(x_0) [/tex]

    Now, this quantity is 0 iff [itex]x_0[/itex] is a non-zero integer!
    Is this the correct result?
    Aren't two sinc functions supposed to be orthogonal for any [itex]x_0[/itex] real?
  2. jcsd
  3. Feb 1, 2010 #2
    I think it is correct, you have shown that:

    [tex] \int_{-\infty}^{+\infty}sinc(x)sinc(x-x_0)dx = \delta (x-x_{0})[/tex]

    For [tex]x_{0}=0[/tex] you have the inner product of the same vector(there's no shift) i.e.

    [tex] \left \langle sinc(x),sinc(x) \right \rangle = \int_{-\infty}^{+\infty}sinc(x)^{2}dx[/tex]
  4. Feb 1, 2010 #3
    that means that the set of sinc functions:

    [tex]\{ sinc(x-n) | n \in \mathbb{Z}^{+}_{0} \}[/tex]

    is an orthonormal basis....but basis for what???
    what functions can be represented as a linear combinations of sinc(x-n) ?

    Basically what confuses me, is that in any pair sinc(x-n) and sinc(x+n) the functions are not orthogonal, and I don't know if they are are redundant or not.
  5. Feb 1, 2010 #4
    For the bandlimited[\B] functions, the ones that have a Fourier transform with a bounded support. This is the Shannon-Whittaker sampling theorem: given a function [tex]f\left(t\right)[/tex], such that its Fourier transform [tex]F\left(\omega\right)[/tex] is such that [tex]F\left(\omega\right)=0, \left|\omega\right|>B[/tex], then:

    f\left(t\right)=\sum_{n=-\infty}^{n=+\infty}f\left(nT\right){\rm sinc}\left(\frac{t-nT}{T}\right)

    If [tex] T> \pi/B[/tex]. This is the called the Nyquist condition; if it's not satisfied, then you have what is called aliasing, where the highest frequency components of the function "appear" as low-frequency ones. For more details, see:

    http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem" [Broken]

    Notice that sinc(x-n) and sinc(x+n) are obtained from each other through a reflection on the y-axis, so you should not expect that they are orthogonal.
    Last edited by a moderator: May 4, 2017
  6. Feb 2, 2010 #5
    Sorry, my calculations were wrong.
    It is possible to prove that:

    \int_{-\infty}^{+\infty}sinc(x+x_0)sinc(x-x_0)dx = sinc(2x_0)

    Again, assuming [itex]x_0[/itex] is a non-zero integer, [itex]sinc(2x_0)=0[/itex], so those functions, despite being the reflection of each other, are orthogonal.

    The orthonormal basis for the bandlimited functions is then: [itex] \{ sinc(x-n) | n \in \mathbb{Z} \} [/itex]
  7. Feb 2, 2010 #6
    My apologies, I studied sampling a long time ago and I didn't remember if they were orthogonal.:redface: The rest is correct, thought.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook