Why is $x(t)$ not periodic but $x[n]$ is?

Question

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.

Answer 1

You have shown*:

If $x(t)$ is periodic, then there is some $T>0$ such that $\dfrac{T^2+2tT}{16}$ is an integer for every real $t$.

*Edit: As pointed out by @S.H.W in the comments, this is not quite true. Rather, it should be

$x(t)$ is periodic if and only if there is some $T > 0$ such that at least one of $\dfrac{T^2+2tT}{16}$ or $\dfrac{T^2+2tT + 2t^2}{16}$ is an integer for every real $t.$

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.

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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.

Answer 2

Let $x(t) = \cos(\frac{\pi t^2}{8})$. If $x(t)$ is periodic with $T$ then there exists $T \gt 0$ such that $x(t) = x(t+T)$ for all $t \in \mathbb{R}$. So in this case we have $$\cos(\frac{\pi (t+T)^2}{8}) = \cos(\frac{\pi t^2}{8})$$If $t = 0$ then $\cos(\frac{\pi T^2}{8}) = 1$. Differentiating both sides and let $t = 0$ we have $$ T\sin(\frac{\pi T^2}{8}) = 0$$ It means $T = 0$ or $\sin(\frac{\pi T^2}{8}) = 0$. The first case isn't allowed so we conclude that $\sin(\frac{\pi T^2}{8}) = 0$. If we differentiate twice and again let $t = 0$ then $$-\frac{\pi}{16} (4 \sin(\frac{\pi T^2}{8}) + \pi T^2 \cos(\frac{\pi T^2}{8})) = 0$$ Combining the results leads to the $T = 0$ which isn't allowed according to $T \gt 0$. The motivation for using differentiation here is that $\frac{d}{dt}\cos(u(t)) = -u'(t)\sin(u(t))$ which helps us to get $T$ out of the $\cos$ function and reaches a contradiction. Of course, Brian's answer is much more elegant and it doesn't require derivative calculations.