Fairly simple inverse laplace check

Discussion in 'Homework Help' started by ihaveaquestion, Oct 12, 2009.

  1. ihaveaquestion

    Thread Starter Active Member

    May 1, 2009
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  2. silvrstring

    Active Member

    Mar 27, 2008
    159
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    I hate to bum you out, but I think you're going to have to do them again. When you see e^(-as) in the transform, it means there is a shift involved in the time domain. So instead of ending up with f(t)u(t), you're going to end up with f(t-a)H(t-a).

    Review the second shift theorem, and you'll see what I mean.
    laplace{f(t-a)H(t-a)} = e^(-as)F(s)

    Take care,
    silvrstring
     
  3. silvrstring

    Active Member

    Mar 27, 2008
    159
    0
    okay. It's also called the delay theorem. In the time domain, you'll end up with a piecewise function. I've attached problem c. I'm pretty sure it's right. It should be enough to guide you for other problems. You'll also want to start off differently in problem b. Separate those numerator coefficients. It looks like you should end up with a 3 part piecewise function.

    Mark44 and Ratch are great for help with this stuff too.

    Take care,
    silvrstring
     
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