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 $-\pi\leq\omega<\pi$.


1 Answer 1


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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