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Math Help - More Laplace transforms

  1. #1
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    More Laplace transforms

    This problem looks a lot like the one I posted yesterday, but I am still not sure that I have the hang of how to do these.

    Using only the definition fo the Laplace transform, and showing your work, find the laplace transform of the following:

    e^(-t)sin((pi)t)

    Thanks in advance for any help!
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  2. #2
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    Quote Originally Posted by Hollysti View Post
    This problem looks a lot like the one I posted yesterday, but I am still not sure that I have the hang of how to do these.

    Using only the definition fo the Laplace transform, and showing your work, find the laplace transform of the following:

    e^(-t)sin((pi)t)

    Thanks in advance for any help!
    You want:

    L[e^(-t)sin((pi.t)] = integral_{t=0 to infty} e^{-t - st} sin(pi.t) dt

    .................= Im[integral_{t=0 to infty} e^{(-1-s+pi.i)t} dt]

    .................= Im 1/(1+s-pi.i) = pi/[(1+s)^2 + pi^2]

    If you need more explanation just ask (looking at this again it does seem a bit
    cryptic).

    RonL
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  3. #3
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    what do you mean by Im and pi.i?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Hollysti View Post
    what do you mean by Im and pi.i?
    Im means "the imaginary part." For example:
    Im(a + i*b) = b where i^2 = -1

    pi.i simply is the number (pi) times i where i^2 = -1.

    -Dan
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  5. #5
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Hollysti View Post
    This problem looks a lot like the one I posted yesterday, but I am still not sure that I have the hang of how to do these.

    Using only the definition fo the Laplace transform, and showing your work, find the laplace transform of the following:

    e^(-t)sin((pi)t)

    Thanks in advance for any help!
    L{e^(-t)(-sin(pit))} = INT{0,inf} e^(-st)*e^(-t)(sin(pi*t)) dt
    = INT{0,inf} e^[-(s + 1)t]sin(pi*t) dt

    For this, we need to use by-parts integration. I'm going to do it the quick way by using a table:

    Code:
          u = e^[-(s + 1)t]     |      dv = sin(pi*t) dt       
                                |
      -(s + 1)e^[-(s + 1)t]     |    -(1/pi)cos(pi*t)
                                |
     (s + 1)^2e^[-(s + 1)t]     |  -(1/pi^2)sin(pi*t)
    Now, we multiply each function on each downward-sloping diagnal, adding each diagnal while switching the signs of each diagnal from + to -, starting with +. For now I'm going to ignore the limits of integration.

    INT e^[-(s + 1)t]sin(pi*t) dt = (-(1/pi)cos(pi*t))e^[-(s + 1)t] - (-(1/pi^2)sin(pi*t))(-(s + 1)e^[-(s + 1)t]) + INT (s + 1)^2e^[-(s + 1)t](-(1/pi^2)sin(pi*t)) dt

    INT e^[-(s + 1)t]sin(pi*t) dt = -(1/pi)cos(pi*t)*e^[-(s + 1)t] - ((s + 1)/pi^2)sin(pi*t)e^[-(s + 1)t] - ((s + 1)/pi)^2*INT e^[-(s + 1)t]sin(pi*t) dt

    [((s + 1)/pi)^2 + 1]*INT e^[-(s + 1)t]sin(pi*t) dt = -(1/pi)cos(pi*t)*e^[-(s + 1)t] - ((s + 1)/pi^2)sin(pi*t)e^[-(s + 1)t]

    Plugging back in the limits of integration {0,inf}:
    -(1/pi)cos(pi*t)*e^[-(s + 1)t] - ((s + 1)/pi^2)sin(pi*t)e^[-(s + 1)t]
    [0 + 0] - [-1/pi - 0] = 1/pi

    [((s + 1)/pi)^2 + 1]*INT e^[-(s + 1)t]sint dt = 1/pi
    INT e^[-(s + 1)t]sin(pi*t) dt = 1/{pi[((s + 1)/pi)^2 + 1]} = pi/[(s + 1)^2 + pi^2]
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  6. #6
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    You want:

    L[e^(-t)sin((pi.t)] = integral_{t=0 to infty} e^{-t - st} sin(pi.t) dt

    .................= Im[integral_{t=0 to infty} e^{(-1-s+pi.i)t} dt]

    .................= Im 1/(1+s-pi.i) = pi/[(1+s)^2 + pi^2]

    If you need more explanation just ask (looking at this again it does seem a bit
    cryptic).

    RonL
    This looks simpler than what I did, but I have no clue what you've done.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by ecMathGeek View Post
    This looks simpler than what I did, but I have no clue what you've done.
    As exp(i.x) = cos(x)+i sin(x), and it is easier to integrate a term like
    exp((a+i.b)x than exp(a.x) sin(x) or exp(a.x) cos(x), and the real and
    imaginary parts do not mix during integration it is often more convienient
    to use:

    integral f(x) sin(x) dx = Im [integral f(x) exp(i.x) dx]

    or:

    integral f(x) cos(x) dx = Re [integral f(x) exp(i.x) dx]

    In the case of Laplace transforms the technique make things very much
    neater.

    RonL
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  8. #8
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Im 1/(1+s-pi.i) = pi/[(1+s)^2 + pi^2]
    RonL
    Can you show me how Im 1/(1 + s - p*i) = pi/[(1 + 2)^2 + pi^2]?

    It appears you rationalized the denominator, but it seems you've done more than that because the numerator would not be pi if that were the case.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by ecMathGeek View Post
    Can you show me how Im 1/(1 + s - p*i) = pi/[(1 + 2)^2 + pi^2]?

    It appears you rationalized the denominator, but it seems you've done more than that because the numerator would not be pi if that were the case.
    Inside the Im(.) multiply top and bottom by (1+s+ pi*i)

    Im 1/(1 + s - p*i) = Im (1+s + pi*i)/[(1+s - pi*i)(1+s + pi*s)]

    ....................... = Im (1+s + pi*i)/[(1+s)^2 +pi^2]

    ....................... = Im {(1+s)/[(1+s)^2 +pi^2] + pi*i/[(1+s)^2 +pi^2]}

    ....................... = pi/[(1+s)^2 +pi^2] (taking the imaginary part)

    RonL
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