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Consider discrete time sinusoids of the form $$x[n] = \cos(\omega n) \ ,$$ where $n$ is an integer. What frequency range of $\omega$ would constitute all the possible sinusoids? I'm thinking that $0\leq\omega<\pi$ would suffice, but my teacher tells me it's $-\pi\leq\omega<\pi$.

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assuming $\omega$ is real, if the form of $x[n]$ is how you've defined it above, your thinking is correct. doesn't matter if it's $+\omega$ or $-\omega$, the sinusoid behaves exactly the same. if instead, the sinusoid had form

$$ x[n] = e^{j \omega n} \ ,$$

then your teacher would be correct.

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  • $\begingroup$ Thanks robert bristow-johnson. If we changed $\cos$ to $\sin$, does there exist a $\pi$ range of $\omega$ such that would constitute all the sinusoids? $0\leq\omega<\pi$ wouldn't work for $\sin$, because $\sin(-\pi/2 n)$ is not included. $\endgroup$
    – J. Sanders
    Commented Jan 23, 2016 at 6:48
  • $\begingroup$ it's a polarity change. you would have generality back again if your question was posed as $$ x[n] = A \sin(\omega n) $$ or $$ x[n] = A \cos(\omega n) $$ but it doesn't really matter. $\endgroup$ Commented Jan 23, 2016 at 7:15
  • $\begingroup$ You're saying that $\sin(-\pi/2 n)$ is just a sign change from $\sin(\pi/2 n)$? If you don't mind me asking - why doesn't the sign matter? $\endgroup$
    – J. Sanders
    Commented Jan 23, 2016 at 7:24
  • $\begingroup$ it might. but it's basically the same as a scaler. or it can be viewed as just a delay of 1/2 cycle. $\endgroup$ Commented Jan 23, 2016 at 7:25

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