I have the following continuous time signal: $$x(t) = \sum_{n=-\infty}^{\infty}e^{-(2t-n)}u(2t-n)$$ where $u(t)$ is the unit function. I had previously determined that this signal was not periodic. However, it seems that it is. However, I'm not sure how I'd determine the fundamental period of such a function. Here is my work so far.
$$u(2t-n) = 1 \text{ for all } n \leq 2\lfloor t \rfloor \text{ and } 0 \text{ for all } n \geq 2\lfloor t\rfloor + 1$$ $$e^{-2t}*\frac{e^{\lfloor 2t \rfloor}}{1-e^{-1}}$$ But how do I proceed from here?
Let's consider slightly expanding out the equation. $$x(t) = \sum_{n=-\infty}^{\infty}e^{-(2t-n)}u(2t-n)$$ $$x(t) = ... + e^{2t-0}u(2t-0)+e^{2t-1}u(2t-1)+...$$ $$x(t) = x(t+T)$$ $$x(t+T) = ...+e^{2(t+T)-0}u(2(t+T)-0)+e^{2(t+T)-1}u(2(t+T)-1)+...$$ Here, I'm defining $T$ as the fundamental period. Now, the adjacent period would allow adjacent terms to be equal. By adjacent, I mean for $x(t) \text{ when n=0 } = x(t+T) \text{ when n = 1}$ . So, let's set those terms equal to each other. $$e^{2t-0}u(2t-0) = e^{2(t+T)-1}u(2(t+T)-1)$$ From this equation, we can set the following two terms equal to each other. $$2t = 2t + 2T -1$$ $$1 = 2T$$ $$T = \frac{1}{2}$$ **EDIT: **To show that this holds for all terms, we have the following $$e^{2t-n}u(2t-n) = e^{2(t+T)-(n+1)}u(2(t+T)-(n+1))$$ $$2t - n= 2t + 2T -n-1$$ And from here, you can see how $T = \frac{1}{2}$.