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Another Laplace Transform problem, need region of convergence help

  1. Nov 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Find L[x(t)], where $$ x(t) = tu(t) + 3e^{-1}u(-t) $$

    Also determine the region of convergence


    2. Relevant equations


    Laplace properties, Laplace table:

    L[te-at = 1/(s+a)2

    L[u(t)] = 1/s

    L[t] = 1/s2



    3. The attempt at a solution

    I don't really know what to do with this as my table doesn't give the product of these two.

    Do you just combine them like this?:

    tu(t) ---> (1/s)(1/s2)

    3e-3u(-t) ---> wait...

    I don't even know if my table says anything about a transform for u(-t), could it be -1/s?

    I was going to do this and then add them to each other. Also need help with RoC
     
  2. jcsd
  3. Nov 15, 2014 #2

    NascentOxygen

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    Staff: Mentor

    No. You make use of this rule:

    L t.f(t) = -F'(s)

    so if you have a function f(t) whose Laplace transform you know to be F(s)
    then multiplying that function by t results in a Laplace transform which can be calculated by you as the derivative of F(s) multiplied by -1

    F'(s) means the derivative of F(s). So you need to know how to differentiate F(s) with respect to s.
     
    Last edited: Nov 15, 2014
  4. Nov 15, 2014 #3

    NascentOxygen

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    Staff: Mentor

    Check that e-1
    If it's really e-1 then that's just a constant, it's a number.

    I don't know anything about u(-t)

    Paging rude man ☎ ....
     
  5. Nov 16, 2014 #4
    Yeah that was actually -t and I read it too fast.


    So since I'm converting it to s-domain does that mean I integrate it instead (so it becomes F(s) now)?
     
  6. Nov 16, 2014 #5

    NascentOxygen

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    No. You start with F(s) and by differentiating F(s) you're calculating F'(s) which is what you need.
     
  7. Nov 16, 2014 #6

    NascentOxygen

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    Staff: Mentor

  8. Nov 16, 2014 #7
    Ok for that the derivative is tδ(t) + u(t)? And that's it at least for that right? Still not sure what RoC is
     
  9. Nov 16, 2014 #8

    NascentOxygen

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    We're looking at only the t.u(t) term right now.

    What is the Laplace Transform of u(t)?
     
    Last edited: Nov 16, 2014
  10. Nov 16, 2014 #9

    rude man

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    Homework Helper
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    This is a math problem more than an engineering problem. The usual application of the Laplace transform is to solve a linear differential equation with constant coefficients and with given initial conditions.

    This problem on the other hand is purely math and probably purely useless, but here goes:

    The mathematically correct Laplace transform is L{f(t)} = integral from -∞ to +∞ of f(t)exp(-st)dt.
    In the real problems to which I referred, the transform is integral from 0 to +∞ of the same integrand.
    Thus, taking your expression, the fact that it includes a u(-t) forces you to integrate from -∞ rather than zero.
    In other words, and we've gone thru all this before, pick you limits to accord with the u function's argument.

    As to convergence, look at the given time function. The region of convergence is simply the region where the time function does not blow up to ∞. The "region" is the region in the complex s plane, with x axis = σ = Re{s} and y axis = Im{s} = jw.

    Example:
    f(t) = exp(-at) u(t)
    F(s) = integral fro 0 to infinity of exp(-at)exp(-st)dt
    = integral from 0 to infinity of exp[-(s+a)t]dt
    = (-1/(s+a)[exp(-(s+a)t evaluated from t=0 to infinity.

    Now, you can see that, since 0<t<∞, the expression (s+a) must be positive or you get infinity for evaluating the integral between its limits.
    But s = σ + jw
    So σ must be > -a
    and the region of convergence is the region to the right of σ = -a in the s plane, sine there σ> -a.

    The u(t) term is handled similarly.

    Look at the attached for more info.
     

    Attached Files:

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