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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?

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    Hint: what happens when you substitute $t-(1/2)$ for $t$ in that summation? – 2017-01-19
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    I'll get t-1 =n. However, how do I know I can make that substitution? – 2017-01-19
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    I don't see how that whole summation becomes $t-1=n$ when you replace $t$ in it everywhere by $t-(1/2)$. But, anyway, what could possibly stop you from making that substitution? The beauty of functions is that you can replace their variables by anything you like, so long as you are consistent about it. – 2017-01-19
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    Are you still here? – 2017-01-20
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    @GerryMyerson Very sorry for the delay; I was away on travel! I have figured out a solution and would be glad to post it if this could be reopened. Thank you. – 2017-01-23
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    You can now post your solution. – 2017-01-26
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    ((Comment deleted, which was addressing the "period" aspect but not the "fundamental period" question. Sorry for the noise.)) – 2017-01-26
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    @DanielFischer I have added the solution. Please consider removing the downvote. Thank you. – 2017-01-26
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    Removing other people's downvotes is not an ability moderators have. – 2017-01-26
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    @GerryMyerson please see new answer – 2017-01-29

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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 $T = \frac{1}{2}$.

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    That's the idea – you have shown that one of the terms in the expansion at $t$ goes over to one of the terms in the expansion at $t+(1/2)$. Now if you can show that works for all the terms in the expansion, and not just for the one with $n=0$.... – 2017-01-29
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    @GerryMyerson I have added the general term – 2017-01-29