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?


2 Answers 2


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

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




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

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.