are constant discrete time signals periodic?
example \begin{equation}
e^{i10\pi n}
\end{equation}
my proffesor says that this signal is aperiodic, in the discrete sense. but it seems wrong, because
unlike in the continuous case, i can calculate the smallest time period , which is 1.
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$\begingroup$ Welcome to SE.SP! As you say, $e^{\imath 10 \pi n}$ at integer $n$ is a constant. Are constants periodic? See this question and answer on SE.math. $\endgroup$ – Peter K.♦ Jan 31 at 19:28
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$\begingroup$ @PeterK. Since this is a discrete signal, I'd say it is periodic with period 1, as abhishek suspects. fourier.eng.hmc.edu/e101/lectures/Fundamental_Frequency/… $\endgroup$ – MBaz Jan 31 at 19:47
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$\begingroup$ @MBaz Yes, it looks like you're correct, it just doesn't have a fundamental or minimal period. $\endgroup$ – Peter K.♦ Jan 31 at 20:00
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$\begingroup$ @PeterK. the document you gave a reference to says that a constant continuous time signal has no fundamental or minimal period. The document says nothing about a constant discrete time signal. $\endgroup$ – abhishek Feb 1 at 16:15
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$\begingroup$ @abhishek Surely that’s true in discrete time too? If the fundamental period must be greater than zero? $\endgroup$ – Peter K.♦ Feb 1 at 20:56
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normally, "$n$" is the symbol we use here for discrete-time. if your professor said that:
$$\begin{align} x[n] &= e^{i10 \pi n} \\ &= e^{i 2 \pi (5n)} \\ \end{align}$$
is not periodic with a period of $1$ (assuming $n \in \mathbb{Z}$) or a period of $\frac15$ (assuming $n \in \mathbb{R}$), then your professor is mistaken.
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$\begingroup$ More generally, if the frequency is a rational multiple of $pi$, then the sequence is periodic. $10pi$ is a rational multiple of $pi$. $\endgroup$ – Juancho Feb 1 at 14:46
