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I'm new to DSP and I'm struggling to solve a basic problem. I think I've solved it but I feel that something's off in my reasoning.

The problem involves investigating a signal of the form $e^{jk\pi n}$, where $n$ is an integer and $k$ is real. How do I confirm if this signal is periodic? I tried dividing the frequency by $2\pi$ as my classmate suggested but the how does the result $\frac{k}{2\pi}$ confirm whether a signal is periodic or not? $k$ isn't required to be an integer, so how does $\frac{k}{2\pi}$ being rational mean the signal is periodic? Even if the signal is periodic, how do I find the fundamental frequency? There's a formula given in Oppenheim that involves converting the exponential to an $e^{jm\frac{2\pi}{N}n}$ form but I don't quite understand how this would help find the fundamental period.

The second part of the problem involves finding the absolute value of $e^{jk\pi n}$. This should be straightforward enough - the absolute value is 1 because of Euler's formula, but I'm not missing out on anything, right?

The third part of the problem is finding out a value of $n$ to make the signal take on any value within its range. Now since the signal is a complex exponential, the only real values it takes on are $1$ and $-1$. But since the frequency depends on $k$, a real number, wouldn't we find a non-integer value of $n$ for the vast majority of cases?

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2 Answers 2

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You have a sequence of complex numbers

$$x[n]=e^{j\pi an},\quad a\in\mathbb{R},\quad n\in\mathbb{Z}\tag{1}$$

$x[n]$ is periodic with period $N$ if $x[n]=x[n+N]$ is satisfied for all $n$. For the given sequence, that means

$$e^{j\pi an}=e^{j\pi a(n+N)}=e^{j\pi an}e^{j\pi aN}\tag{2}$$

For $(2)$ to be satisfied we require $e^{j\pi aN}=1$, which is equivalent to

$$\pi aN=2\pi m,\qquad m\in\mathbb{Z}\tag{3}$$

Consequently,

$$a=2\frac{m}{N}\tag{4}$$

and

$$x[n]=e^{j2\pi n \frac{m}{N}}$$

If $m$ and $N$ in $(4)$ are coprime, then $N$ is the (smallest) period of $x[n]$.

As for the last question, if I understand correctly, it should be about the value of $a$ (not $n$) for which $x[n]$ assumes all possible values satisfying $|x[n]|=1$. This is the case if $x[n]$ is not periodic. I.e., any $a$ that does not satisfy Eq. $(4)$ results in $x[n]$ taking on all possible values with unity magnitude as $n$ is varied.

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A signal $x[n]$ is periodic with period $P$ if $$ x[n] = x[n+P]$$ for all $n$.

That means that some "sinusoidal" signals which are periodic when $n$ is real-valued and continuous are not necessarily periodic when $n$ is an integer.

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