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Inverse Laplace transform of ex(-s)/(s+3)^3

  1. Dec 15, 2008 #1
    matlab:
    >> ilaplace(exp(-1*s)/(s+3)^3)

    ans =

    1/2*heaviside(t-1)*(t-1)^2*exp(-3*t+3)

    But I think there should be an additional exp(3) multiplied by all.

    =exp(-1*s)/(s+3)^3
    =exp(-1*(s+3)+3)/(s+3)^3
    Taking out exp(-3t). In above step I did (s+3)+3 because exp(-3t) says all variables should be shifted by 3 ..
    =[exp(-1*(s)+3)/(s)^3]*exp(-3t)
    Now I take care anything but Heaviside
    =[exp(-1*(s))]*exp(-3t)*[t^2/2]*exp(3)
    now I take care of Heaviside
    =exp(-3(t-1))*[(t-1)^2/2]*exp(3)

    I might have made some error while explaining what I was doing - just ignore it. But, the thing I am concerned about is the one in red. It gives me exp(3) in the end which matlab does not.

    Thanks
     
  2. jcsd
  3. Dec 15, 2008 #2

    gabbagabbahey

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    Re: ilaplace(exp(-1*s)/(s+3)^3)

    This makes no sense. Are you trying to apply a frequency-shift or a time-shift? The [itex]e^{-s}[/itex] should correspond to a time shift. The rule for such is:

    [tex]u(t-a)f(t-a)=\mathcal{L}^{-1}[e^{-as}F(s)][/tex]

    where [itex]f(t)=\mathcal{L}^{-1}[F(s)][/itex] and [itex]u(t-a)[/itex] is the Heaviside function.

    In this case, you would take [itex]a=3[/itex] and [tex]F(s)=\frac{1}{(s+3)^3}[/tex]. What does that make [itex]f(t)[/itex]? And so [itex]f(t-a)[/itex] is___?
     
  4. Dec 15, 2008 #3
    Re: ilaplace(exp(-1*s)/(s+3)^3)

    I was using the formula:

    ILaplace{F(s-3)} = ILaplace{F(s)}e^3t
    Yes exp(-as) corresponds to Heaviside but shouldn't it also be exp(-a(s-3)) if I want to take out e^3t (or use the above formula)?

    For F(s-3) e^3t makes it F(s) or f(t) .. (?)
     
  5. Dec 15, 2008 #4

    gabbagabbahey

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    Re: ilaplace(exp(-1*s)/(s+3)^3)

    Okay, so you are doing a frequency shift first and then a time-shift.

    So for the frequency shift you have:

    [tex]\mathcal{L}^{-1}[\frac{e^{-s}}{(s+3)^3}]=e^{-3t}\mathcal{L}^{-1}[\frac{e^{-(s-3)}}{s^3}][/tex]

    But when you apply the time-shift, you seem to be shifting the [itex]e^{-3t}[/itex]. Why would you do that? The [itex]e^{-3t}[/itex] is just a factor multiplying the stuff you are taking the inverse transform of; it shouldn't be affected.
     
    Last edited: Dec 15, 2008
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