# Band-limited function, Shannon-Nyquist sampling distance

1. Oct 16, 2015

### Incand

1. The problem statement, all variables and given/known data
If $\delta$ and $\Omega$ are two numbers with $0 < \frac{\pi}{\Omega} < \delta$ find a function $f\in L^2(\mathbf R )$ such that $\hat f(\omega)=0$ for $|\omega|> \Omega$ and $f(n\delta) = 0$ for $n \in \mathbf Z$, but $f\ne 0$ as an element of $L^2(\mathbf R )$. This shows that the Shannon-Nyquist sampling distance $\Delta t = \frac{\pi}{\omega_{\max}}$ is the largest possible.

2. Relevant equations
Fourier transform
$\frac{\sin ax}{x} \to \pi \mathscr{X}_a(\xi ) = \begin{cases} \pi \; \; |\xi | < a \\ 0 \; \; |\xi | > a \end{cases}$

3. The attempt at a solution
I Choose $f(t) = \frac{\sin \pi \delta t}{t}$ which seems to satisfy all the criteria since
$F.T.(\frac{\sin \pi \delta t}{t}) = \pi \mathscr{x}_{\pi /\delta } < \pi \mathscr{x}_{\pi /\Omega }$
Is this what I was supposed to do in the exercise? There's no answer so I'm not sure I proved everything I should. What does it mean that $f\ne 0$ as an element of $L^2(\mathbf R )$? That it's not the zero-function?

Last edited: Oct 16, 2015
2. Oct 16, 2015

### MisterX

I assume you meant "$\delta$ and $\Omega$ are two numbers...". Shouldn't it be $f(t) \propto \frac{\sin \pi t/\delta}{t}$ ? Even still it fails for $n=0$. But this could still be the answer they want. Yes, $f\neq 0$ means it's not the zero function (which is a trivial solution).

3. Oct 16, 2015

### Incand

Yes should've been $\delta$. Your version of $f(t)$ is of cause equally correct (and more general). Since they only ask me to find a solution I didn't bother with the constant. I think I take your post as confirmation that at least I probably got what the question asked for. I guess there may be a solution with $f(0)=0$ or atl where $\lim_{x\to 0} f(x) =0$ but I think they probably meant except that point.
I guess I could also choose some function that is zero past a point say $|x| > a$ and get other solutions than trigonometric functions and have convergence even without a $t$ in the denominator but I feel quite happy as it is!