How to verify the function $$ \frac{e^{j10\pi n}}{10} $$ is a period/aperiodic function ?

My attempt:

$\omega_o=10\pi\implies f_o=\omega_o/2\pi=5$, which is a rational number. So the given function is periodic. The fundamental period, $N=2\pi\frac{m}{\omega_o}=2\pi\frac{5}{10\pi}=1$

Is it the proper way to approach the problem?

  • 2
    $\begingroup$ You'll find that the signal, as written, is equal to $\frac{1}{10}$ for all $n$. Is it possible that you have a typo? $\endgroup$ – Jason R Mar 8 '17 at 12:17
  • 1
    $\begingroup$ I suppose I would say it's periodic, as it does meet that condition. I'm not aware of any formal classification for whether a DC signal is periodic; this is engineering, after all. Such arguments are never productive. :) $\endgroup$ – Jason R Mar 8 '17 at 12:48
  • 2
    $\begingroup$ @DanBoschen a discrete time constant signal, $x[n]=K$ for all $n$, can be considered to be periodic with any period $N$ that satify $x[n+N] = x[n]$ and the minimum of such $N$ is $N=1$. No formal problems here. A continuous time DC is periodic with any $T$ and since by definition the period is to be selected as the minimum of such numbers $T$ such that $x(t+T) = x(t)$ , then for a continuous time DC there may be some objections to whether to classify DC as a periodic signal or not. $\endgroup$ – Fat32 Mar 8 '17 at 14:05
  • 1
    $\begingroup$ @Fat32 Thanks that is helpful and I think you answered the OP (you should post your response below). So interesting point that a continuous DC signal would be periodic with t=0 (or at least a limit at t goes to 0), meaning it's periodic rate is also DC! $\endgroup$ – Dan Boschen Mar 8 '17 at 14:12
  • 1
    $\begingroup$ @DanBoschen The way I prefer to put it is, a DC signal is periodic but it doesn't have a fundamental period, defined as the smallest $T>0$ such that $x(t)=x(t+T)$. $\endgroup$ – MBaz Mar 8 '17 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.