# Finding the period of a continuous time signal

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?

• That exponential isn't periodic unless it has a compex number in the power. Non-periodic x anything = non-periodic (I believe). – Andy aka Jan 19 '17 at 8:19
• However, this does have a period. – Jonathan Jan 19 '17 at 8:29

## 1 Answer

As the sum goes from $-\infty$ it is periodic, with period 1. Sketch it and you'll see the trend. The function is plotted in Excel, below, starting at $t=0$ rather than $t=-\infty$ (!), so there's a small transient at the beginning of the plot. • But how would I mathematically derive this? – Jonathan Jan 19 '17 at 16:47