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 2


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, $y[0] \overset{\text{def}}{=} \int_{-\infty}^\infty x(t)\operatorname{sinc}\left(\frac{t}{T}\right) \mathrm dt$ 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:


  • $\begingroup$ 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. $\endgroup$
    – bp7070
    Mar 26, 2019 at 17:42

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