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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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  • $\begingroup$ Hint: what happens when you substitute $t-(1/2)$ for $t$ in that summation? $\endgroup$ Commented Jan 19, 2017 at 8:11
  • $\begingroup$ I'll get t-1 =n. However, how do I know I can make that substitution? $\endgroup$
    – Jonathan
    Commented Jan 19, 2017 at 8:27
  • $\begingroup$ 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. $\endgroup$ Commented Jan 19, 2017 at 11:48
  • $\begingroup$ Are you still here? $\endgroup$ Commented Jan 20, 2017 at 21:52
  • $\begingroup$ @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. $\endgroup$
    – Jonathan
    Commented Jan 23, 2017 at 19:17

1 Answer 1

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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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  • $\begingroup$ 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 $n=0$.... $\endgroup$ Commented Jan 29, 2017 at 21:53
  • $\begingroup$ @GerryMyerson I have added the general term $\endgroup$
    – Jonathan
    Commented Jan 29, 2017 at 23:44

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