Sinc functions orthogonality

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712
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Hello,
I'd like to prove the orthogonality of two "shifted" Sinc functions, but I can't find the mistake.
Here is my attempt:

[tex]\int_{-\infty}^{+\infty}sinc(x)sinc(x-x_0)dx[/tex]
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?
 
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]
 
712
5
Thanks!
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.
 
402
1
is an orthonormal basis....but basis for what???
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:

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

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]

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.
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:
712
5
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.
Sorry, my calculations were wrong.
It is possible to prove that:

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

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]
 
402
1
My apologies, I studied sampling a long time ago and I didn't remember if they were orthogonal.:redface: The rest is correct, thought.
 

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