I've been stuying signals and systems and I came across this problem.
By definition, $x(t)$ denotes continuous-time signal and $x[n]$ denotes discrete-time signal.
$x(t)$ is periodic if there exists a constant $T>0$ such that $x(t) = x(t+T)$ for all $t$ is a subset of real numbers.
$x[n]$ is periodic if there exists a constant $N>0$ such that $x[n] = x[n+N]$ for all $n$ is a subset of integers.
Then I came across this question: Why is $x(t)$ aperiodic?
$x(t) = \cos((\pi t^2)/8)$
The workings I made is as follows:
$x(t+T) = \cos((\pi(t+T)^2)/8$
Assume $x(t) = x(t+T)$
i.e $(\pi t^2)/8 + 2\pi k = (\pi(t+T)^2)/8$
$\Rightarrow t^2 + 16k = (t+T)^2
\Rightarrow 16k = T^2 + 2tT $
Considering $k$ is an integer, isn't this periodic? Please let me know if my calculation is wrong.
Apologies if I'm posting an irrelevant topic and thanks for your feedback.
Best Answer
You have shown*:
*Edit: As pointed out by @S.H.W in the comments, this is not quite true. Rather, it should be
Since $T \neq 0$, it should be fairly apparent that there will be some $t$ such that neither of those expressions yields an integer, showing that $x(t)$ is not periodic.
To prove it, note that, for each integer $k$, there is a unique real $t$ such that $\dfrac{T^2+2tT}{16} = k$ and at most two real numbers $t$ such that $\dfrac{T^2+2tT + 2t^2}{16} = k.$ Since there are countably many integers, there are countably many $t$ such that at least one of $\dfrac{T^2+2tT}{16}$ or $\dfrac{T^2+2tT+2t^2}{16}$ is an integer. Since there are uncountably many real numbers, there must be some real $t$ such that neither expression yields an integer.
As I mentioned above, this shows $x(t)$ is not periodic.
On the other hand, we could set e.g. $T=8$ to see that $\dfrac{T^2+2tT}{16}$ is an integer whenever $t$ is an integer, showing $x[n]$ is periodic.