1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Hint exp(j pi n) $\endgroup$ – Stanley Pawlukiewicz Apr 24 '18 at 0:36
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
2
$\begingroup$

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 $k$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.