Compute the IDTFT of the following signal:

$$X(\Omega)=\sum_{k=-\infty}^{+\infty}\left(u(\Omega+\pi)+u\left(\Omega+\frac{\pi}{4}\right)-u\left(\Omega-\frac{\pi}{4}\right)-u(\Omega-\pi)\right)\star \delta(\Omega-2k\pi)$$

Using the IDTFT definition, I obtain:

$$x[n]=\frac{1}{2\pi}\int_{-\pi}^\pi X(\Omega) e^{j\Omega n}d\Omega = \frac{1}{2\pi}\left(\int_{-\pi}^{-\frac{\pi}{4}} e^{j\Omega n}d\Omega + 2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} e^{j\Omega n}d\Omega + \int_{\frac{\pi}{4}}^{\pi} e^{j\Omega n}d\Omega\right) = \frac{1}{2\pi jn}\left( e^{-j\pi n/4} -e^{-j\pi n} +2e^{j\pi n/4} -2e^{-j\pi n/4} + e^{j\pi n} -e^{j\pi n/4} \right) = \frac{\sin(\pi n/4)}{\pi n}$$

However, from the transformation table we know that this $x[n]$ has a different transform:

$$X_d(\Omega)=\sum_{k=-\infty}^{+\infty}\left(u\left(\Omega+\frac{\pi}{4}\right)-u\left(\Omega-\frac{\pi}{4}\right)\right)\star \delta(\Omega-2k\pi)$$

which is clearly different from $X(\Omega)$. So, I am asking: what have I done wrong here in this IDTFT definition application?


You made a mistake in this last equality

$$x[n] = \frac{1}{2\pi jn}\left( e^{-j\pi n/4} -e^{-j\pi n} +2e^{j\pi n/4} -2e^{-j\pi n/4} + e^{j\pi n} -e^{j\pi n/4} \right) \neq \frac{\sin(\pi n/4)}{\pi n}$$

The right way to do this would be:

$$\begin{align} x(n)&=\frac{1}{2\pi jn}\left( e^{-j\pi n/4} -e^{-j\pi n} +2e^{j\pi n/4} -2e^{-j\pi n/4} + e^{j\pi n} -e^{j\pi n/4} \right) \\ &=\frac{1}{\pi n}\left( \frac{e^{-j\pi n/4}-e^{j\pi n/4}}{2j}+ \frac{e^{j\pi n} -e^{-j\pi n}}{2j} +\frac{2e^{j\pi n/4} -2e^{-j\pi n/4}}{2j} \right) \\ &=\frac{1}{\pi n}\left( -\sin\left(\pi n/4\right)+\sin\left(\pi n\right)+2\sin\left(\pi n/4\right) \right)\\ &=\frac{1}{\pi n}\left( \sin\left(\pi n\right)+\sin\left(\pi n/4\right) \right) \end{align}$$

Which makes sense as one can see that the DTFT is the sum of two rectangular windows, one of width $2\pi$ and one of width $\pi/2$, each corresponding to each $\mathrm{sinc}()$.


I've just noticed judging by your comment that the mistake you made was to think that

$$\frac{\sin(\pi n)}{\pi n} =0$$

As you already know, the numerator is $0$ for all $n$... except for $n=0$. In that case, the denominator also is zero, so you have an indetermination. The same happens in the continuous case. Remember that we assume


by taking the limit when $t\to0$. To respect the fact that the discrete $\mathrm{sinc}$ is a sampled version of the continuous one, then it equals $1$ at the origin too. So we can state that:

$$\frac{\sin(\pi n)}{\pi n} =\delta(n)$$

Notice that the DTFT you wrote can also be expressed (using the fact that convolution is distributive) as:

$$X(\Omega)=\sum_{k=-\infty}^{+\infty}\left(u(\Omega+\pi)-u(\Omega-\pi)\right)\star \delta(\Omega-2k\pi) + \sum_{k=-\infty}^{+\infty}u\left(\left(\Omega+\frac{\pi}{4}\right)-u\left(\Omega-\frac{\pi}{4}\right)\right)\star \delta(\Omega-2k\pi)$$

The first term is a window of width $2\pi$ that is $2\pi$-periodic... So it's basically $1 \ \forall \Omega$.

$$X(\Omega)=1 + \sum_{k=-\infty}^{+\infty}u\left(\left(\Omega+\frac{\pi}{4}\right)-u\left(\Omega-\frac{\pi}{4}\right)\right)\star \delta(\Omega-2k\pi)$$

Now it's easier to see that the IDTFT I got at the beginning of the question corresponds indeed to the given DTFT.

| improve this answer | |
  • $\begingroup$ I don't think so... You know, $sin(\pi n) = 0$ for all $n$ since we are in natural numbers... $\endgroup$ – Jason Jan 31 '18 at 17:39
  • $\begingroup$ @Jason Indeed, and in fact $\frac{\sin(\pi n)}{\pi n} = \delta(n)$. Its transform is $$\sum_{k=-\infty}^{+\infty}\left(u(\Omega+\pi)-u(\Omega-\pi)\right)\star \delta(\Omega-2k\pi)$$ If you watch closely, that is the same as writing $1 \ \forall \Omega$, which corresponds to the DTFT of the delta. Thus the solution given in the answer is correct. $\endgroup$ – Tendero Jan 31 '18 at 17:45
  • $\begingroup$ @Jason I've added some useful information to the answer. $\endgroup$ – Tendero Jan 31 '18 at 17:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.