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.


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 at 17:42

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