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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\$ That exponential isn't periodic unless it has a compex number in the power. Non-periodic x anything = non-periodic (I believe). \$\endgroup\$
    – Andy aka
    Commented Jan 19, 2017 at 8:19
  • \$\begingroup\$ However, this does have a period. \$\endgroup\$
    – user91567
    Commented Jan 19, 2017 at 8:29

1 Answer 1

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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.

enter image description here

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  • \$\begingroup\$ But how would I mathematically derive this? \$\endgroup\$
    – user91567
    Commented Jan 19, 2017 at 16:47

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