0
$\begingroup$

I would like to know how to calculate $(-1)^n$ and $(-1)^n(-1)^n$ being periodic and get an explanation for the functions sine, cosine, and tangent periods. I know that for a signal to be periodic there has to be such $T$, that satisfies: $$f(n)=f(n+T)$$ It may explain the reason why sine and cosine periods are $2\pi$, once $(−1)^n$ period is $2$, and why the tangent period is $\pi$, once $(−1)^n⋅(−1)^n$ has period $1$.

$\endgroup$
1
  • $\begingroup$ Why are you having questions. $\sin(n + 2\pi) = \sin n$ so $2\pi$ is a period of $\sin$ (but it might not be the smallest one) and $(-1)^{n+2} = (-1)^n$ so $2$ is a period of $f(n)= (-1)^n$. .... so what's your question? $\endgroup$ – fleablood Apr 24 '20 at 3:36
2
$\begingroup$

I know that for a signal to be periodic there has to be such T, that satisfies:

By that definition there may be multiple periods.

$\sin(n + 4,567,282\pi) = \sin n$ so $\sin n$ is periodic.

And $4,567,282\pi$ is a period of $\sin$ but in might not be the smallest.

Likewise if $f(n+T) = f(n)$ then $f(n+97T)=f(n)$ so $f$ is periodic but $97$ isn't the smallest period and, for all we know, $T$ might not be either.

Usually when saying a function is "periodic" we want to know what the smallest period is (which we usually say is "the" period).

If $T = 2k\pi$ then $\sin (n+T) = \sin n$ but the smallest such $T$ (greater than $0$) so that $\sin (n+T)=\sin n$ is $T= 2\pi$.

It is true that for any $n$ that $\sin (n + 2\pi) = \sin n$.

I won't prove this, but for any $T< 2\pi$ there will always be some $n$ where $\sin (n+T) \ne \sin n$. For example if $T = \pi$ then $\sin(n+\pi) =-\sin n$ and those aren't usually equal.

So why is $\tan$ different. Well, $\tan$ is a different function that $\sin$ so there is no reason it should be the same.

But notice: $\tan (n+\pi) = \frac {\sin (n + \pi)}{\cos (n+\pi)} = \frac {-\sin n}{-\cos n} = \frac {\sin n}{\cos n}$ so $\pi$ is a period, even though it isn't a period for $\sin$ or $\cos$.

$\endgroup$

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.