# Ideal sampling using sinc funcion

Let $$x(t)$$ be a bandwidth limited signal such as $$\forall |\omega|>\frac{\pi}{T} : X^F(\omega)=0$$ while $$X^F(\omega)$$ is the signal's Fourier Transform.

Let us denote $$y[n]=\int_{-\infty}^{\infty}x(\tau)sinc(\frac{\tau-nT}{T}) d\tau$$

I noticed that my professor said that because the signal is bandwidth limited and the width of the $$sinc$$ function is $$T$$ and it's orthogonal property, and with regard to ideal sampling and reconstruction we can say that $$y[n]=T\cdot x[nT]$$.

I understand that we can treat $$sinc$$ as a delta function somehow but I'm can't find the right mathematical explenation which had lead to this result.

I'd happy for some insights.

## 2 Answers

Your professor is using sloppy notation. When one writes the sample value $$x[\cdot]$$, it is generally accepted that the argument is an integer (which $$nT$$ might not be) and we relate this to the continuous-time signal $$x(t)$$ via $$x[n] = x(nT)$$. So, what your professor should have said is that $$y[n]$$ has value $$T\cdot x[n] = T\cdot x(nT)$$, and not that $$y[n] = T\cdot x[nT]$$ (even if $$T$$ happens to be an integer). An outline of a proof that $$y[n] = T\cdot x[n] = T\cdot x(nT)$$ is given below.

If $$x(t)$$ is a continuous-time signal bandlimited to $$\left(-\frac{1}{2T}, \frac{1}{2T}\right)$$, that is, its Fourier transform $$X(f)$$ has the property that $$X(f) = 0$$ for all $$|f| \geq \frac{1}{2T}$$,then it can be reconstructed from its samples spaced $$T$$ seconds apart. So, let's consider the signal $$x(t)\operatorname{sinc}\left(\frac{t}{T}\right)$$ where $$\operatorname{sinc}(t) = \begin{cases}\frac{\sin(\pi t)}{\pi t},& t \neq 0,\\1, & t = 0.\end{cases}$$ The Fourier transform of $$\operatorname{sinc}(t)$$ is $$\operatorname{rect}(f)$$ and so the Fourier transform of $$\operatorname{sinc}\left(\frac{t}{T}\right)$$ is $$T\operatorname{rect}(fT)$$, and the Fourier transform of $$x(t)\operatorname{sinc}\left(\frac{t}{T}\right)$$ is the convolution $$X(f)\circledast T\operatorname{rect}(fT)$$ of the Fourier transforms $$X(f)$$ and $$T\operatorname{rect}(fT)$$ of $$x(t)$$ and $$\operatorname{sinc}\left(\frac{t}{T}\right)$$ respectively. Now, in principle, the convolution of a function and a rectangular pulse is easy to compute but I refuse to write out the answer because it is not needed: what we need to find is the value of $$\int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t}{T}\right) \mathrm dt$$ which, if you think about it a bit, is actually the value of the Fourier transform of $$x(t)\operatorname{sinc}\left(\frac{t}{T}\right)$$ evaluated at $$f=0$$. That is, \begin{align} \int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t}{T}\right) \mathrm dt &= X(f)\circledast T\operatorname{rect}(fT)\bigg |_{f=0}\\ &= T\int_{-\infty}^\infty X(\lambda)\operatorname{rect}((0-\lambda) T) \, \mathrm d\lambda\\ &= T\int_{-\infty}^\infty X(\lambda)\operatorname{rect}(\lambda T) \, \mathrm d\lambda & \scriptstyle\text{rect is an even function}\\ &= T\int_{-\frac{1}{2T}}^{\frac{1}{2T}}X(\lambda) \, \mathrm d\lambda &\scriptstyle\text{rect has finite support}\\ &= T\int_{-\infty}^\infty X(\lambda) \, \mathrm d\lambda &\scriptstyle{X(\lambda)=0 ~\text{for}~|\lambda| > \frac{1}{2T}}\\ &= T \cdot x(0). \end{align} Thus, $$y \overset{\text{def}}{=} \int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t}{T}\right) \mathrm dt$$ equals $$T\cdot x(0)$$. I leave it as an exercise for the reader to work out the Fourier transform of $$\operatorname{sinc}\left(\frac{t-nT}{T}\right)$$ and show that $$y[n] \overset{\text{def}}{=} \int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t-nT}{T}\right) \mathrm dt$$ works out to be $$T\cdot x(nT)$$ and so if we define the samples of $$x(t)$$ as $$x[n] = x(nT)$$, then we have that $$y[n] = T\cdot x[n]$$ for all $$n$$.

I'm not sure I understood your question, but I think you want to know why sometimes the discrete $$\mathrm{sinc}$$ function is treated as a Kronecker delta.

In signal processing, the function is defined as

$$\mathrm{sinc}(x)=\frac{\sin(\pi x)}{\pi x}$$

When we move to the discrete domain, this becomes $$\mathrm{sinc}[n]=\frac{\sin[\pi n]}{\pi n}$$

But we should notice that $$\sin[n\pi]=0 \ \forall n\in\mathbb{Z}$$

Therefore, $$\mathrm{sinc}[n]$$ is $$0$$ everywhere... but hold on. When $$n=0$$, the denominator equals $$0$$ as well. That's not determined, but we redefine our discrete function in a way that it looks like its continuous partner. In the continuous case, due to the fact that both left and right limits tend to $$1$$ when $$x\to0$$, we say that $$\mathrm{sinc}(x)=1$$. We take this reasoning to the discrete case, and so we are left with a function that equals $$0$$ everywhere except at $$n=0$$, where it equals $$1$$. Does this sound familiar? Indeed, this is exactly what the Kronecker delta looks like. So:

$$\mathrm{sinc}[n]=\delta[n]$$

• Thanks, I do understand why $sinc$ is related this way to the delta function. My question is why $y[n]=T\cdot x[nT]$. If we denote $h(t)=sinc(\frac{t}{T})$ we can say that $y[n]=\int_{-\infty}^{\infty}x(\tau)sinc(\frac{\tau-nT}{T}) d\tau=(x*h)(t)|_{t=nT}$. If $h(t)$ was $\delta (t)$ I can understand the result (without the $T$ factor). I think that somehow, because $sinc$ behaves like $\delta$ we get this result, but I don't really get it. – bp7070 Mar 26 at 17:42