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I think almost everything is in the title.

In an exercise, a DTFT is given : $$X(e^{j\Omega}) = \sin(\Omega) + \cos(\Omega/2)$$ The period of this DTFT is $4\pi$. Is that possible? I mean, the definition of the DTFT shows that it is $2\pi$-periodic $$X(e^{j\Omega}) = \sum_{n=-\infty}^\infty x[n]e^{-k\Omega n}.$$

I don't know if a $4\pi$-periodic DTFT has any sense. I'm really confused about this.

Thanks in advance,

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  • $\begingroup$ It doesn't look like that map is well defined since it depends on how you represent the number, since $\exp(j\Omega)=\exp(j\Omega+2j\pi)$ but $X$ gives two different values. Is there a restriction on $\Omega$? or should it just be $X(\Omega)$? $\endgroup$
    – snulty
    May 18, 2015 at 18:08
  • $\begingroup$ Also en.wikipedia.org/wiki/Discrete-time_Fourier_transform $\endgroup$
    – snulty
    May 18, 2015 at 18:12
  • $\begingroup$ No restriction given on $\Omega$ , $\Omega$ is here defined as $\omega/T_s$ where $T_s$ is the sampling period of the discrete time signal $x[n]$. Defining the DTFT as $X(e^{j\Omega})$ instead of $X(\Omega)$ is just a small trick to show explicitly the $2\pi$-periodicity, but there is no difference with the notation you suggested. $\endgroup$
    – anpar
    May 18, 2015 at 18:18
  • $\begingroup$ When you evaluate the function, isn't the domain of $\Omega$ restricted to $[-\pi,\pi)$ radians per sample? Outside of this domain are replicas, so I don't see how it could be, as you say, $4\pi$-periodic. $\endgroup$ May 18, 2015 at 18:29
  • $\begingroup$ I totally agree that with this restriction on $\Omega$, my problem disappears. But the exercise does not give this restriction... But does this DTFT has any sense without the restriction you just give? $\endgroup$
    – anpar
    May 18, 2015 at 18:34

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