I'm trying to find the DTFT of $$(-1)^n \cdot \frac{\sin (\pi n/2)}{\pi n}$$ I know the DTFT of $\frac{\sin \pi n/2}{\pi n}$ = a box function of amplitude 1, cutoff $\pi/2$. And I know that multiplication in time domain = convolution in frequency, but I'm still not sure how to piece everything together.

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    $\begingroup$ Hint exp(j pi n) $\endgroup$ – user28715 Apr 24 '18 at 0:36
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    $\begingroup$ Not trying to nitpick, but itsn't the DTFT, the discrete-time fourier transform? The equation is discrete, isn't it? $\endgroup$ – Ben Apr 24 '18 at 2:49
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    $\begingroup$ @Ben, usually when "$n$" is used as the argument, it is meant to be an integer value. if the "$n$" were replaced with "$t$", maybe that would be interpreted to be continuous-time. but then you would have to do something about the $(-1)^t$ factor. for integer $n$, the expression $(-1)^n$ is unambiguous. but $(-1)^t$ is a little bit ambiguous. is it $(e^{j\pi})^t = e^{j \pi t}$? or is it $(e^{-j\pi})^t = e^{-j \pi t}$? they ain't the same thing for non-integer values of $t$. so i think the equation is meant to be a discrete-time equation. $\endgroup$ – robert bristow-johnson Apr 24 '18 at 11:21
  • $\begingroup$ Hint: find out which n the sinc function is non-zero and how these n relate to (-1)^n $\endgroup$ – Hilmar Apr 24 '18 at 17:19

Considering @Stanley Pawlukiewicz's hint, you should take into account that $e^{~j \pi} = -1$, thus $(-1)^n = e^{~j \pi n}$, for any integer $n$.

Therefore, time-domain multiplication by $(-1)^n$ corresponds to a frequency-domain shift by $\pi$ radians in digital frequency.

Recall the Fourier transform's theorem: $\mathscr{F}\{x[n]~e^{j \omega_0 n}\} = X(\omega - \omega_0)$ where $\mathscr{F}\{\}$ is, in this case, the Discrete-Time Fourier Transform (DTFT).

Now, since

$$\mathscr{F}\left\{\frac{\sin (\pi n/2)}{\pi n}\right\} = \sum\limits_{k=-\infty}^{+\infty} \Pi\left(\frac{\omega + 2k\pi}{\pi}\right)$$

then, answering to your question:

$$\mathscr{F}\left\{(-1)^n \cdot \frac{\sin (\pi n/2)}{\pi n}\right\} = \sum\limits_{k=-\infty}^{+\infty}\Pi\left(\frac{\omega - \pi + 2k\pi}{\pi}\right)$$

which is a box function of amplitude $1$ and width $\pi$, centered on $\omega = \pi$. The box function or rect function is

$$\Pi(u) \triangleq \begin{cases} 0 & \mbox{if } |u| > \frac{1}{2} \\ \frac{1}{2} & \mbox{if } |u| = \frac{1}{2} \\ 1 & \mbox{if } |u| < \frac{1}{2}. \\ \end{cases}$$

Since the DTFT is periodic in $\omega$ with period $2\pi$, there is a box centered on $\omega = (2k-1)\pi$, for every integer $k$.

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  • $\begingroup$ and $$(-1)^n = e^{j \pi n}$$ is just as true as $$(-1)^n = e^{-j \pi n}$$ for integer $n$, but they are not equivalent for non-integer $n$. $\endgroup$ – robert bristow-johnson Apr 24 '18 at 20:06
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    $\begingroup$ The DTFT is periodic in $\omega$ with period $2\pi$, there is no problem with using $e^{~j \pi} = -1$ or $e^{-j \pi} = -1$, which are equivalent. The result is the same whether we apply a frequency shift of $\pi$ or $-\pi$. $\endgroup$ – Luis M Gato Apr 24 '18 at 20:11
  • $\begingroup$ yes. you are correct, Luis, if $n \in \mathbb{Z}$. that's all i wanted to point out. $\endgroup$ – robert bristow-johnson Apr 24 '18 at 23:03

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