On page 5 of this document I read that

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

Could anyone please explain why this is true?

  • $\begingroup$ since the DTFT maps an infinitely long discrete signal $x[n]$ to a periodic (with period $2 \pi$) continuous spectrum: $$ X(e^{j \omega}) = \sum\limits_{n=-\infty}^{+\infty} x[n] e^{-j \omega n} $$, you need to be clear which is the discrete signal and which is the periodic continuous function. it appears that neither is periodic. $\endgroup$ – robert bristow-johnson Nov 10 '15 at 6:32
  • $\begingroup$ Do you know the difference between the discrete-time Fourier transform (DTFT) and the Discrete Fourier transform (DFT)? This is important for your question. $\endgroup$ – Matt L. Nov 10 '15 at 7:54
  • $\begingroup$ @MattL.The output of discrete-time Fourier transform (DTFT) is continuous in frequency and periodic. discrete Fourier transform (DFT) can be seen as the sampled version (in frequency-domain) of the DTFT output... Correct me if i am wrong $\endgroup$ – Aadnan Farooq A Nov 10 '15 at 9:25
  • $\begingroup$ Yes, but that means that the DTFT of $\text{sinc}(k)$ can't be $\delta(k)$ because both are discrete. $\endgroup$ – Matt L. Nov 10 '15 at 10:09
  • $\begingroup$ After, you what is the valueof $\text{sinc}(k)$ ? $\endgroup$ – Yves Daoust Nov 10 '15 at 17:12

EDIT: After editing according to the comments, the real question is if and why the following equality holds:

$$\text{sinc}[k]\doteq\frac{\sin(k\pi)}{k\pi}=\delta[k],\quad k\in\mathbb{Z}\tag{a}$$

The answer is, yes, of course it holds. Since $\sin(k\pi)=0$ for any integer $k$, the value of the left-hand side of the above equation must be zero for any integer $k\neq 0$. The case $k=0$ is different because here also the denominator is zero. It can be shown by a limit argument that for $k=0$, the function $\sin(k\pi)/(k\pi)$ equals $1$. Another way to see it is to use the equation that you quoted in your comment:

$$\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{j\omega k}d\omega=\ldots=\text{sinc}[k]$$

If you evaluate this equation separately for $k=0$ you get


So we get a sequence which is $1$ for $k=0$ and zero for all other (integer) values of $k$. Consequently, this sequence can be rewritten using the discrete-time unit impulse $\delta[k]$, as shown in Equation (a).

(Below is the original answer explaining that $\delta[k]$ can't be the DTFT of any sequence.)

As mentioned in the comments, this can't be true. Be careful to distinguish between the Discrete-Time Fourier Transform (DTFT) and the Discrete Fourier Transform (DFT). The first is defined by

$$\begin{align}\text{DTFT:}\quad X(e^{j\omega})&=\sum_{n=-\infty}^{\infty}x[n]e^{-jn\omega}\\x[n]&=\frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\omega})e^{j\omega n}d\omega\end{align}\tag{1}$$

The DTFT $X(e^{j\omega})$ of a sequence $x[n]$ is a $2\pi$-periodic function that is continuous in $\omega$.

The (length $N$) DFT is defined by

$$\begin{align}\text{DFT:}\quad X[k]&=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}\\x[n]&=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\end{align}\tag{2}$$

Here, $X[k]$ is also a discrete sequence.

The discrete unit impulse $\delta[k]$ can be the DFT of some time domain sequence, but it cannot be the DTFT of any sequence (because it is itself a sequence, not a continuous function). To see which sequence $x[n]$ corresponds to the DFT $X[k]=\delta[k]$, simply use the formula given in $(2)$:

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}\delta[k]e^{j2\pi nk/N}=\frac{1}{N}\tag{3}$$

From $(3)$, it is the constant sequence $x[n]=1/N$ that corresponds to the unit impulse $X[k]=\delta[k]$.

  • $\begingroup$ Proof On Page 5 you can see the proof $sinc(k) =\delta(k)$. Above in Equation 1, if you take out $X(e^{j\omega})$ and the take the integration and apply upper and lower limit it will be $\frac{sin(wn)}{wn}$ which be equal to $sinc$ $\endgroup$ – Aadnan Farooq A Nov 10 '15 at 11:15
  • $\begingroup$ @AadnanFarooqA: From your confused question it sounded to me like you thought that the DTFT of $\text{sinc}(k)$ equals $\delta(k)$. So your real question is if the equality $\text{sinc}(k)=\delta(k)$ holds. This has absolutely nothing to do with the DTFT. $\endgroup$ – Matt L. Nov 10 '15 at 11:44
  • $\begingroup$ It was mentioned in the text that in DTFT the $sinc(k)$ holds the equality of $\delta(k)$. so i asked that $\endgroup$ – Aadnan Farooq A Nov 10 '15 at 11:47
  • $\begingroup$ @AadnanFarooqA: An equality can either be true or false, it can't be true "in DTFT" and false otherwise. Please see my updated answer. $\endgroup$ – Matt L. Nov 10 '15 at 11:56
  • $\begingroup$ @AadnanFarooqA: I'll edit your title and question because your title explicitly asks if the DTFT of sinc(k) equals $\delta(k)$, which is not what you want to know. Please be more careful when formulating titles and question, it can save a lot of time. $\endgroup$ – Matt L. Nov 10 '15 at 11:59

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