# Sinc functions orthogonality

1. Feb 1, 2010

### 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?

2. Feb 1, 2010

### 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.

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

3. Feb 1, 2010

### 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.

4. Feb 1, 2010

### JSuarez

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:

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

If $$T> \pi/B$$. 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" [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
5. Feb 2, 2010

### mnb96

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 $x_0$ 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: $\{ sinc(x-n) | n \in \mathbb{Z} \}$

6. Feb 2, 2010

### JSuarez

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