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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.
    Commented Jan 31, 2019 at 19:28
  • $\begingroup$ @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
    Commented Jan 31, 2019 at 19:47
  • $\begingroup$ @MBaz Yes, it looks like you're correct, it just doesn't have a fundamental or minimal period. $\endgroup$
    – Peter K.
    Commented Jan 31, 2019 at 20:00
  • $\begingroup$ @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
    Commented Feb 1, 2019 at 16:15
  • $\begingroup$ @abhishek Surely that’s true in discrete time too? If the fundamental period must be greater than zero? $\endgroup$
    – Peter K.
    Commented Feb 1, 2019 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
    Commented Feb 1, 2019 at 14:46

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