Deriving the Oversampling formula

Question

I am familiar with the Shanon Sampling theorem, which states that:

Let $f \in L_1(\mathbb{R})$ and $supp(\mathcal{F}f) \subseteq [-B,B]$,then $f(x)=\sum_{n \in \mathbb{Z}} f(\frac{n}{2B}) sinc(2B(x-\frac{n}{2B}))$

in the sense that the RHS converges to $f$ in $L_2(\mathbb{R})$.

This states that a function can be recoverd from the values $(f(\frac{k}{2B}))_{k \in \mathbb{Z}}$.

Now I heard that one can "oversample" to get better convergence. I got a hint on how to do the prove: Use a function $g$, such that $\mathcal{F}g(\xi)=1$ on $[-B,B]$ and
$\mathcal{F}g(\xi)=0$ on $|\xi| > B'$

My approach:

Now assume $supp(\mathcal{F}f) \subseteq [-B,B]$, and consider some $B'$>B.

One can write $f(x)=\int_{-B}^{B} \mathcal{F}f(\xi) e^{2 \pi i \xi x}d\xi$

Now if I choose a function $g$, such that

$g(\xi)=1 for \xi \in [-B,B]$

and

$g(\xi)=0 for |\xi|>B'$

the above equals to $=\int_{-B'}^{B'} \mathcal{F}f(\xi) e^{2 \pi i \xi x}g(\xi)dx $

If I now consider the Fourier series of f on $(-B',B')$ I further get $= \int_{-B'}^{B'}\sum_{n \in\mathbb{Z}} \frac{1}{2B'} f(-\frac{n}{2B'}) e^{\frac{2 \pi i n \xi}{2B'}} g(\xi) e^{2 \pi i x \xi} d \xi=$ $\sum_{n \in\mathbb{Z}} f(-\frac{n}{2B'}) \frac{1}{2B'}\mathcal{F}^{-1}(g)(x+\frac{n}{2B'})$

To sum it up, one expands $\mathcal{F}f$ as a Fourier series, recover Fourier coefficients and then do the inverse Fourier transform.

Now it seems to come down to calculating $g$. For that I also got a hint:

Assume $g$ has the form $g=\frac{1}{2a} \chi_{[-a,a]} \ast ... \ast \frac{1}{2a} \chi_{[-a,a]} \ast \chi_{[-B-ka,B+ka]}$, where $\chi_{[c,d]}$ denotes the indicator function on some intervall $[c,d]$

That's where I am stuck.

Alternatively if someone has another prove this would be fine too.

Answer 1

First let us use the hint and define $g$ precisely. Denote by $*$ the convolution operator. Then choose $g(\xi)$ to have the form $$ g(\xi) = \underbrace{\frac{1}{2a}\chi_{[-a,a]} * \dots * \frac{1}{2a}\chi_{[-a,a]}}_{k \text{ times}} * \chi_{[-B-ka,B+ka]}. $$ with $ka=\frac{B'-B}{2}$ such that the condition \begin{equation} g(\xi) = \begin{cases} \text{const}, &|\xi|\leq B\\ 0, &|\xi|>B'\end{cases} \tag{1}\label{gxi} \end{equation} is fulfilled (see explanation below).

Denote by $X_b(\xi)$ the Fourier transform of the rectangular function $\chi_{[-b,b]}(x)$. Since $\mathcal{F}^{-1}g(x) = \mathcal{F}g(-x)$ and by the Convolution Theorem,

$$ \mathcal{F}^{-1}g(x) = \mathcal{F}g(-x) =\left(\frac{X_a(-x)}{2a}\right)^{k} X_{B+ka}(-x) $$

The Fourier transform of the rectangular function may be computed to

\begin{align*} X_b(\xi)&=\mathcal{F}\chi_{[-b,b]}(\xi) =\int_{-b}^b \mathrm{e}^{i2\pi x\xi}\mathrm{d} x = \frac{\mathrm{e}^{i2\pi b\xi}-\mathrm{e}^{-i2\pi b\xi}}{i2\pi \xi}= \frac{\sin(2\pi b\xi)}{\pi \xi}. \end{align*}

Plugging $X_b$ into the expression for $\mathcal{F}^{-1}g(x)$, gives $$ \mathcal{F}^{-1}g(x) = \left(\frac{\sin(2\pi ax)}{2\pi ax}\right)^{k}\frac{\sin(\pi (B+B')x)}{\pi x}. $$

This function decays as $x^{-k-1}$, which is better than the original $x^{-1}$ from the sinc function.

Here is a freedom of choosing $k$ (or more generally the interpolation function in the interval $B<|\xi|<B'$). This freedom gives the opportunity to construct smooth functions which decay faster than the sinc function which results in an improved convergence speed.

Choice of the function $g$

Following the hint, we have $ g(\xi) = \underbrace{\frac{1}{2a}\chi_{[-a,a]} * \dots * \frac{1}{2a}\chi_{[-a,a]}}_{m \text{ times}} * \chi_{[-B-ka,B+ka]}, $ where $k,a,m$ are not specified. We will specify $k,a,m$ such that \eqref{gxi} is satisfied.

First, the convolution of $m$ indicator functions (rectangles) is a function, which is only non-zero in $|\xi|\leq am$ (e.g. the convolution of two rectangles is a triangle defined between $\xi=-2a$ to $\xi=2a$, see this explanation)

Next, the convolution of a function which is non-zero only in $|\xi|<am$ with a rectangle from $-(B+ka)$ to $B+ka$ will satisfy

• $g(\xi) = 0$ for $|\xi|>B+ka+am$
• $g(\xi) = \text{const}$ for $|\xi|<B+ka-am$

Comparing with \eqref{gxi}, it is evident that we should choose $B+ka+am=B'$ and $B+ka-am=B$. Solving the second equation for $am$ and plugging into the first gives $B+2ka = B'$ or, equivalently, $2ka = B'-B$ and also $m=k$.