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A very basic laplace transform problem (picture included)

  1. Sep 4, 2011 #1
    Hi..new here. I'm having trouble finishing off a very basic laplace transform problem. I'd greatly appreciate it if someone could show me the way.

    The problem is #5 from this picture: http://www.flickr.com/photos/67208468@N08/6111476312/



    What I did was use the Euler's formula for sin t and I got to this: 1/i * (-s2e-pi*s+i+e-pi*s)/s2+1

    The answer is (1+e-pi*s)/s2+1

    I always kinda get confused typing math/equations, so I apologize if what i described wasn't clear. Thanks for any help.
     
  2. jcsd
  3. Sep 4, 2011 #2

    lanedance

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    i would probably just write it as a combination of sin, say u(t) is the unit step function
    [tex]f(t) = u(t) sin(t) - u(t-\pi)sin(t) [/tex]

    most laplace transform tables will have entries for all these terms, or do you need to perform the integral yourself?
     
  4. Sep 4, 2011 #3
    We haven't been taught the step function stuff yet, so I've never seen that equation before but I'm sure the professor meant for us to solve this without tables. It's just been one lecture so far, so we haven't seen the tables you're talking about either.

    We have to do the integral, yes. Thanks for the response.
     
  5. Sep 4, 2011 #4

    lanedance

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    ok so you just need to perform

    [tex]
    F(s) = L\{f(t)\} = \int_0^{\infty}e^{-st}f(t)dt = \int_0^{\pi}e^{-st}sin(t)dt
    [/tex]

    can you show your steps for the integration and we'll try and find where things are falling over
     
  6. Sep 4, 2011 #5
    Last edited by a moderator: Apr 26, 2017
  7. Sep 4, 2011 #6
    You may have figured it out by now, but in case you haven't this might help a little. The case you have is a little bit tricky, and you have to solve it similarly to the situation e^t*sint back in calc 2.

    F(t) = [itex]\int[/itex][itex]^{\pi}_{0}[/itex]e-stsin(t)dt = [itex]\frac{-1}{s}[/itex]e-stsin(t) + [itex]\frac{1}{s}[/itex][itex]\int[/itex][itex]^{\pi}_{0}[/itex]e-stcos(t)dt
    In that we used integration by parts with u = sin(t) and dv = e-st. Now, we see that the last term is [itex]\int^{\pi}_{0}[/itex]e-stcos(t)dt. If we tried to integrate that we would come back to our original equation:
    F2(t) = [itex]\int[/itex][itex]^{\pi}_{0}[/itex]e-stcos(t)dt = [itex]\frac{-1}{s}[/itex]e-stcos(t) - [itex]\frac{1}{s}[/itex][itex]\int[/itex][itex]^{\pi}_{0}[/itex]e-stsin(t)dt

    From Here we can make a substitution:

    F(t) = [itex]\int[/itex][itex]^{\pi}_{0}[/itex]e-stsin(t)dt = [itex]\frac{-1}{s}[/itex]e-stsin(t) + [itex]\frac{1}{s}[/itex]([itex]\frac{-1}{s}[/itex]e-stcos(t) - [itex]\frac{1}{s}[/itex][itex]\int[/itex][itex]^{\pi}_{0}[/itex]e-stsin(t)dt)

    (1+[itex]\frac{1}{s^{2}}[/itex])[itex]\int^{\pi}_{0}[/itex]e-stsin(t)dt = [itex]\frac{-1}{s}[/itex]e-stsin(t) - [itex]\frac{1}{s^{2}}[/itex]e-stcos(t))

    [itex]\int^{\pi}_{0}[/itex]e-stsin(t)dt = ([itex]1/(1+\frac{1}{s^{2}})[/itex])([itex]\frac{-1}{s}[/itex]e-stsin(t) - [itex]\frac{1}{s^{2}}[/itex]e-stcos(t))

    You'll have to include the bounds and solve, of course, but that's the general breakdown of how you'd want to tackle that type of problem. Good luck!
     
  8. Sep 4, 2011 #7
    BestMilke, appreciate it a lot. If no one can do it using the Euler's formula, then I'll do it the way you did.
     
  9. Sep 4, 2011 #8

    Ray Vickson

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    Use brackets around the denominator: (1+ exp(-s*pi))/(s^2+1). I got the correct answer, so I cannot tell what you did wrong.

    RGV
     
  10. Sep 4, 2011 #9
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