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Homework Help: Laplace of y = -0.0563x + 0.7831

  1. Jul 18, 2012 #1
    I have a linear equation y = -0.0563x + 0.7831

    I am trying to find Laplace of this equation in Y(s)/ X(s) format. I do not know how to achieve it? Please help me.

    This equation describes the performance of a motor... torque vs angular speed (motor characteristic curve approximated to a linear equation for simplicity).

    It is important to have solution in Y(s)/X(s) format. I need to do a step and impulse response analysis on the system in matlab. Please give me some hints.
  2. jcsd
  3. Jul 20, 2012 #2


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    Staff: Mentor

    Welcome to the PF.

    Here is what Wolfram Alpha says:


    [tex]\frac{y}{s} = \frac{0.7831}{s} - \frac{0.0563}{s^2}[/tex]
  4. Jul 20, 2012 #3
    hello berkeman ,

    thanks for the solution. that link is really wonderful. i did not know this up to know.

    best regards,
  5. Jul 22, 2012 #4


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    Science Advisor

    You mean the Laplace transform?

    That's pretty straight forward, isn't it? The Laplace transform of any function, f(x) is
    [tex]\int_0^\infty e^{st}f(t)dt[/tex]

    Here, that is
    [tex]-0.0563\int_0^\infty te^{-st}dt+ 0.7831\int_0^\infty e^{-st}dt[/tex]
    The second integral is just
    [tex]\left[0.7831 (-\frac{1}{s}e^{-st}\right]_0^\infty= \frac{0.7831}{s}[/tex]

    The second integral can be done by parts, taking u= t, [itex]dv= e^{-st}dt[/itex] so that [itex]du= dt[/itex] and [itex]v= -(1/s)e^{-st}[/itex] so that
    [tex]-0.0563\int_0^\infty te^{-st}dt= -0.0563(\left[-(1/s)te^{-st}\right]_0^\infty+ \frac{1}{s}\int_0^\infty e^{-st}dt= -\frac{0.0563}{s^2}[/tex]
  6. Jul 22, 2012 #5
    Hello HallsofIvy,

    Thanks for providing mathematical proof of it. This helped me to understand the maths behind it. It has been quite a while away from Laplace and I had forgot the basics. I got it now !! Thanks again !!
  7. Jul 22, 2012 #6


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    Staff: Mentor

    Beauty Halls! :biggrin:
  8. Jul 24, 2012 #7

    rude man

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    Homework Helper
    Gold Member

    If y and x are functions of time:
    y(t) = a - bx(t)

    then the Laplace transform of that equation is
    Y(s) = a/s - bX(s).

    It seems that y is rotational speed and x is applied torque.

    I must confess I don't follow Halls and not Wolfram either.
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