# Deriving the Oversampling formula

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.

• How is $g$ defined from $B$ to $B'$ and from $-B$ to $B'$. As far as I understand it, $g$ should be a piece-wise linear function. But I a not really sure about it. Commented Aug 9, 2023 at 10:01
• Are you fine with using the Dirac distribution $\delta$? Commented Aug 10, 2023 at 17:41
• @AndreasLenz I would prefer to do it without Distributions if possible Commented Aug 10, 2023 at 17:44
• @AndreasLenz Or if you do it with distributions, I would be thankful if it is detailed, since I have some difficulties with fully understanding distributions Commented Aug 10, 2023 at 17:48
• Does my reply answer your question? If not, what is missing? Commented Aug 13, 2023 at 20:52

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 $$$$g(\xi) = \begin{cases} \text{const}, &|\xi|\leq B\\ 0, &|\xi|>B'\end{cases} \tag{1}\label{gxi}$$$$ 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|). 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 $$|\xi| with a rectangle from $$-(B+ka)$$ to $$B+ka$$ will satisfy

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

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 $$m=k$$.

• What is the reason behind defining $g$ in that way? Commented Aug 14, 2023 at 10:46
• @Walli I have added some more explanation for the choice of $g$ in my post. Does that make sense? Commented Aug 14, 2023 at 11:26
• Yes this helps. But I don't really understand why the convolution of $m$ rectangles of width $2a$ has width $2am$. Here's how I tried to understand it: The Fourier transform of $\frac{1}{2a}\chi_{[-a,a]}$ is $\frac{sin(ay)}{ay}$, thus the convolution of it with itself $m$-times should be $(\frac{sin(ay)}{ay})^m$. Now since $(\frac{sin(ay)}{ay})^m$ is a (decaying) periodic function, it has infinitely many zeros, thus I can't choose some $x$ such that $(\frac{sin(ay)}{ay})^m=0$ for $|\xi|>x$ Commented Aug 14, 2023 at 12:02
• @Walli I think it is easier to directly think what the convolution does instead of going over the Fourier transform. I have added a link (dsp.stackexchange.com/questions/66129/…) which explains the convolution between rectangles. Commented Aug 15, 2023 at 7:30