Integrating e^-|t|: Signals & Systems Demystified Help

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Discussion Overview

The discussion revolves around the integration of the function involving the sine and exponential terms, specifically the integral \(\int_{-\infty}^{\infty}{\sin(2t)e^{-|t|}e^{-j2\pi ft}} dt\). The context is related to signals and systems, focusing on techniques for handling absolute values in integrals and integration by parts.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant, Sam, expresses difficulty in integrating the function due to the absolute value and attempts to break it down into two integrals over different intervals.
  • Another participant asserts that Sam's expansion is correct without providing further details.
  • A third participant suggests using integration by parts on each decomposed integral, implying that this method may yield a satisfactory result.
  • Sam later confirms that the suggested method worked for them.

Areas of Agreement / Disagreement

The discussion shows some agreement on the correctness of the initial expansion, but the effectiveness of the integration method remains contingent on individual approaches, as evidenced by Sam's initial struggle and subsequent success.

Contextual Notes

There may be limitations related to the assumptions made about the convergence of the integrals and the handling of the absolute value, which are not fully explored in the discussion.

sam.green
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I am working through Signals and Systems Demystified on my own. I need to integrate:

[tex] \int_{-\infty}^{\infty}{sin(2t)e^{-|t|}e^{-j2\pi ft}} dt[/tex]

I first went about dealing with the absolute value sign by using the following

[tex] \int_{-\infty}^{\infty} e^{-|t|} dt = \int_{-\infty}^{0} e^{t} dt + \int_{0}^{\infty} e^{-t} dt[/tex]

Going along this route seems to not work and makes me think the expansion is incorrect. Can anyone give me a pointer?

Thanks,
Sam
 
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The expansion is correct.
 
Use the decomposition you have now and do integration by parts twice (on each decomposed integral). I think you'll be pleasantly surprised.
 
That worked. Thanks!
 

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