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Thread: Laplace Transformation help

  1. June 23rd 2008, 04:36 PM #1
    bsbooker
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    Laplace Transformation help

    y"+3y'+2y=u2(t) and y(0)=0 and y'(0)=1

    y"-y=20δ(t-3) and y(0)=0 and y'(0)=0

    I am having a bit of trouble with these problems. Can anyone help?

    Ben
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  2. June 23rd 2008, 05:00 PM #2
    TheEmptySet
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    Quote Originally Posted by bsbooker View Post
    y"+3y'+2y=u2(t) and y(0)=0 and y'(0)=1

    y"-y=20δ(t-3) and y(0)=0 and y'(0)=0

    I am having a bit of trouble with these problems. Can anyone help?

    Ben
    The transform of the first is(I'm guessing u_2(t) is the unit step function)

    s^2Y-s(0)-1+3sY+2Y=\frac{1}{s}

    The transform of the 2nd is

    s^2Y-Y=20e^{-3s}

    Can you take it from here?

    Good luck.
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  3. June 23rd 2008, 06:45 PM #3
    bsbooker
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    thanks

    Quote Originally Posted by TheEmptySet View Post
    The transform of the first is(I'm guessing u_2(t) is the unit step function)

    s^2Y-s(0)-1+3sY+2Y=\frac{1}{s}

    The transform of the 2nd is

    s^2Y-Y=20e^{-3s}

    Can you take it from here?

    Good luck.

    I am having most trouble with next step.
    Thanks
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  4. June 23rd 2008, 07:02 PM #4
    TheEmptySet
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    Alright here is the first one

    s^2Y-s(0)-1+3sY+2Y=\frac{1}{s}

    First we need to isolate Y

    s^2Y-s(0)-1+3sY+2Y=\frac{1}{s} \iff (s^2+3s+2)Y=s + 1 +\frac{1}{s}

    Y =\frac{s+1}{(s^2+3s+2)}+\frac{1}{s(s^2+3s+2)}

    Factoring and reducing like terms we get

    Y=\frac{s+1}{(s+1)(s+2)}+\frac{1}{s(s+1)(s+2)}=\fr ac{1}{s+2}+\frac{1}{s(s+1)(s+2)}

    Depending on what theorems you know there are many ways to take the inverse Laplace transform, but lets do partial fractions on the 2nd fraction
    (Work not shown)
    \frac{1}{s(s+1)(s+2)}=\frac{1/2}{s}+\frac{-1}{(s+1)}+\frac{1/2}{s+2}

    Putting this into the above and collecting like terms we get

    Y=\frac{1}{2}\left( \frac{1}{s} \right)-\frac{1}{s+1}+\frac{3}{2}\left( \frac{1}{s+2}\right)

    Now taking the inverse transform we get

    y(t)=\frac{1}{2}-e^{-t}+\frac{3}{2}e^{-2t}

    Try the 2nd one and show us what you have done!!

    Good luck.
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  5. June 25th 2008, 03:59 PM #5
    bsbooker
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    Thanks, here is what I came up with

    Thanks,

    Y(t)=cost +[20(sin(t-3)]u(t-3)

    is it ok to leave a u in it?

    Ben
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  6. June 25th 2008, 06:57 PM #6
    bsbooker
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    Laplace correction?

    For the first one,

    Should it be 1/2 instead of 3/2?
    Ben
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  7. June 25th 2008, 07:02 PM #7
    TheEmptySet
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    Quote Originally Posted by bsbooker View Post
    Thanks,

    Y(t)=cost +[20(sin(t-3)]u(t-3)

    is it ok to leave a u in it?

    Ben
    If you take two derivatives and plug in back in you can check your work. However, this solution doesn't check. It doesn't satify the inital condition aswell.

    s^2Y-Y=20e^{-3s} \iff (s^2-1)Y=20e^{-3s} \iff Y =20e^{-3s}\frac{1}{(s-1)(s+1)}

    By partial fractions or this trick

    \frac{1}{(s-1)(s+1)}=\frac{\frac{1}{2}s-\frac{1}{2}s+\frac{1}{2}+\frac{1}{2}}{(s-1)(s+1)}=\frac{\frac{1}{2}s+\frac{1}{2}}{(s+1)(s-1)}+\frac{-\frac{1}{2}s+\frac{1}{2}}{(s+1)(s-1)}

    \frac{\frac{1}{2}(s+1)}{(s+1)(s-1)}+\frac{-\frac{1}{2}(s-1)}{(s+1)(s-1)}=\frac{\frac{1}{2}}{(s+1)}-\frac{\frac{1}{2}}{(s-1)}

    Now we get

    Y=\frac{10e^{-3s}}{(s+1)}-\frac{10e^{-3s}}{(s-1)}

    Now taking the inverse Laplace transform we get

    y(t)=10e^{-(t-3)}\mathcal{U}(t-3)-10e^{(t-3)}\mathcal{U}(t-3)
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  8. June 25th 2008, 09:11 PM #8
    bsbooker
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    thanks

    Thanks again.

    for the first problem, where does the 3/2 come from?

    Ben
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  9. June 25th 2008, 09:32 PM #9
    Chris L T521
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    Quote Originally Posted by bsbooker View Post
    Thanks again.

    for the first problem, where does the 3/2 come from?

    Ben
    Quote Originally Posted by TheEmptySet View Post
    Alright here is the first one

    s^2Y-s(0)-1+3sY+2Y=\frac{1}{s}

    First we need to isolate Y

    s^2Y-s(0)-1+3sY+2Y=\frac{1}{s} \iff (s^2+3s+2)Y=s + 1 +\frac{1}{s}

    Y =\frac{s+1}{(s^2+3s+2)}+\frac{1}{s(s^2+3s+2)}

    Factoring and reducing like terms we get

    Y=\frac{s+1}{(s+1)(s+2)}+\frac{1}{s(s+1)(s+2)}={\c olor{red}\frac{1}{s+2}}+\frac{1}{s(s+1)(s+2)}

    Depending on what theorems you know there are many ways to take the inverse Laplace transform, but lets do partial fractions on the 2nd fraction
    (Work not shown)
    \frac{1}{s(s+1)(s+2)}=\frac{1/2}{s}+\frac{-1}{(s+1)}+\frac{1/2}{s+2}

    Putting this into the above and collecting like terms we get

    Y=\frac{1}{2}\left( \frac{1}{s} \right)-\frac{1}{s+1}+\frac{3}{2}\left( \frac{1}{s+2}\right)

    Now taking the inverse transform we get

    y(t)=\frac{1}{2}-e^{-t}+\frac{3}{2}e^{-2t}

    Try the 2nd one and show us what you have done!!

    Good luck.
    Don't forget about the other term!! {\color{red}\frac{1}{s+2}}+\underbrace{\frac{1/2}{s+2}}_{\text{from PFD}}=\frac{3}{2}\left(\frac{1}{s+2}\right)

    --Chris
    Last edited by Chris L T521; June 26th 2008 at 07:44 PM. Reason: typo... =(
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  10. June 26th 2008, 04:19 PM #10
    bsbooker
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    thanks

    Thank you
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  11. June 26th 2008, 05:45 PM #11
    bsbooker
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    what if

    What if I set either y(0)=1 or y'(0)=1? How does that change the result?
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