Are Shifted Sinc Functions Orthogonal for Any Real Value of x_0?

[mnb96]

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

\int_{-\infty}^{+\infty}sinc(x)sinc(x-x_0)dx
Observing this quantity can be obtained by evaluating the Fourier transform at zero, we have:

\mathcal{F}\{ sinc(x)sinc(x-x_0) \}(0)

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

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

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

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

= \frac{sin(\pi x_0)}{\pi x_0} =

= sinc(x_0)

Now, this quantity is 0 iff x_0 is a non-zero integer!
Is this the correct result?
Aren't two sinc functions supposed to be orthogonal for any x_0 real?

 

[emanuel_hr]

I think it is correct, you have shown that:

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

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

\left \langle sinc(x),sinc(x) \right \rangle = \int_{-\infty}^{+\infty}sinc(x)^{2}dx

 

[mnb96]

Thanks!
that means that the set of sinc functions:

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

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.

 

[JSuarez]

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 f\left(t\right), such that its Fourier transform F\left(\omega\right) is such that F\left(\omega\right)=0, \left|\omega\right|>B, then:

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

If T&gt; \pi/B. 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"

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.

 

[mnb96]

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:

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

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

The orthonormal basis for the bandlimited functions is then: \{ sinc(x-n) | n \in \mathbb{Z} \}

 

[JSuarez]

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