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Homework Help: Easy Laplace

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data

    I just want to verify that I am dong this correctly and if not, where my misconceptions are. I keep getting a different answer so I am going to go through this step by step:

    After a some steps in a controls problem, I end up with a function in the Laplace domain:

    [tex]\Theta(s) = \frac{12}{\frac{6}{5}s^2 +96}\qquad(1)[/tex]


    2. Relevant equations

    From table, I know that

    [tex]\mathcal L[Be^{-at}\sin(\omega t)]=\frac{B\omega}{(s+a)^2 + \omega^2}\qquad(2)[/tex]



    3. The attempt at a solution

    So the idea is to make my equation (1) take the form of the known equation (2):

    First, I factored the denominator so that s2 has a coefficient of 1:

    [tex]\frac{12}{\frac{6}{5}s^2 +96}=\frac{1}{\frac{6}{5}}\cdot \frac{12}{s^2+80}[/tex]


    Comparing the denominators of each we have:

    [tex](s+a)^2 + \omega^2 = s^2 +80[/tex]

    [tex]\Rightarrow s^2 + 2as + a^2 + \omega^2 =s^2 +80[/tex]

    [tex]\therefore a = 0 \qquad\qquad \omega^2 = 80[/tex]

    So we now have to compare the numerators:

    [tex]12 = B*\omega[/tex]

    [tex]\Rightarrow B = \frac{12}{\omega}[/tex]

    [tex]\Rightarrow B=\frac{12}{\sqrt{80}}[/tex]


    Thus, we can write:

    [tex]\Theta(s) =\frac{5}{6}\cdot\frac{(\frac{12}{\sqrt{80}})*\sqrt{80}}{(s+0)^2+ \sqrt{80}^2}[/tex]


    So, in the time domain, we have:

    [tex]\mathcal L^{-1} [\Theta(s)]=\theta(t) = \frac{10}{\sqrt{80}}\cdot\sin(\sqrt{80}t)[/tex]

    or

    [tex]\frac{\sqrt{5}}{2}\sin(4\sqrt5 t)[/tex]

    I have gotten this result a couple of times, so I think I am correct, but my friend got something a little different.

    Are there any blatant errors I am making?

    Thanks!
     
  2. jcsd
  3. Feb 7, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Well, I used Maxima to find the Laplace transform of sqrt(5)*sin(4*sqrt(5)*t)/2. I got 10/(s^2+80). As that is the same function you started out with, I don't think you are doing anything wrong.
     
  4. Feb 7, 2010 #3
    Okie dokie! Thanks!
     
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