Signal not neriodic if multiplied by decaying exponential?

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SUMMARY

The discussion clarifies that the signal x(t) = e^{jt} is periodic, while the signal x(t) = e^{-t}e^{jt} is not periodic due to the presence of the decaying exponential. The decay causes the complex exponential to converge toward zero, preventing the existence of a real period T such that x(t) = x(t+nT). Additionally, the signal x(t) = e^{-t}e^{jt} is not considered time-limited, as it never actually reaches zero.

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The signal [tex]x(t) = e^{jt}[/tex] is periodic.

The signal [tex]x(t) = e^{-t}e^{jt}[/tex] is not periodic.

The decaying exponential makes the complex exponential decay and converge toward zero, but why is it not periodic?

Also, this signal is not considered time limited since it never actually reaches zero. Right?
 
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SpaceDomain said:
The signal [tex]x(t) = e^{jt}[/tex] is periodic.

The signal [tex]x(t) = e^{-t}e^{jt}[/tex] is not periodic.

The decaying exponential makes the complex exponential decay and converge toward zero, but why is it not periodic?

Because you can't find a real T such that x(t)=x(t+nT).
 
Neat. I was thinking about this qualitatively and forgot about the required condition you mentioned.

Thanks.

Is what I said about it not being time limited correct?
 

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