Results 1 to 4 of 4

Math Help - Inverse Laplace Transform

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    7

    Inverse Laplace Transform

    Sorry if this is in the wrong section.

    Basically I am trying to find the inverse laplace transform of

    F(s)=\frac{s}{\left ( s+1 \right )^2+1}

    using the bromiwich integral and the standard bromwich contour. I therefore used the residue theorem to evaluate

    \frac{1}{2\pi i}\int _\gamma \frac{e^{zt} z}{\left ( z+1 \right )^2+1} dz

    and I found the answer to be

    \left ( \frac{1}{2}+\frac{i}{2}  \right )e^{\left ( -1+i \right )t}+\left ( \frac{1}{2}-\frac{i}{2}  \right )e^{\left ( -1-i \right )t}

    Which I believe is right, however when I try to show that the curved part of the contour goes to zero as R tends to infinity I find that it does not, and so my answer if not what I want.

    Now I have been looking at ways around this and jordans lemma keeps appearing, although I really have no idea how to apply it (or even if this is the right approach), could someone please explain it or point me in the right direction?

    Thanks,

    Lewis.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by MathMan170 View Post
    Sorry if this is in the wrong section.

    Basically I am trying to find the inverse laplace transform of

    F(s)=\frac{s}{\left ( s+1 \right )^2+1}

    using the bromiwich integral and the standard bromwich contour. I therefore used the residue theorem to evaluate

    \frac{1}{2\pi i}\int _\gamma \frac{e^{zt} z}{\left ( z+1 \right )^2+1} dz

    and I found the answer to be

    \left ( \frac{1}{2}+\frac{i}{2}  \right )e^{\left ( -1+i \right )t}+\left ( \frac{1}{2}-\frac{i}{2}  \right )e^{\left ( -1-i \right )t}

    Which I believe is right, however when I try to show that the curved part of the contour goes to zero as R tends to infinity I find that it does not, and so my answer if not what I want.

    Now I have been looking at ways around this and jordans lemma keeps appearing, although I really have no idea how to apply it (or even if this is the right approach), could someone please explain it or point me in the right direction?

    Thanks,

    Lewis.
    Hello Lewis and welcome to MHF.

    When you say the standard contour what do you mean? You need to integrate on a vertical line that is to the right of the real part of all of the poles of the integrand.

    One such vertical line is

    s=iz

    The imaginary axis.

    This gives the integral

    \frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{ize^{izt}}{(iz+1)^2+1}(idz)=  \frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{ze^{izt}}{(z-i-1)(z-i+1)}dz

    This is now an integral with z real (an integral on the real axis) edit: sorry this was a bad choice of variables

    Now we can apply the residue theorem on the semicircle above the real axis and the line segment on the real axis.

    So by the residue theorem we get

    \frac{(i+1)e^{i(i+1)t}}{2}-\frac{(i-1)e^{i(i-1)t}}{2}=-e^{-t}\sin(t)+e^{-t}\cos(t)

    Now the reason we wanted to write the integral this way

    \frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{ze^{izt}}{(z-i-1)(z-i+1)}dz

    is so we can apply Jordan's lemma

    We can identify

    f(z)=\frac{z}{(z-i-1)(z-i+1)}

    and note that

    \lim_{R \to \infty}f(Re^{i\theta})=0

    So by Jordan's Lemma Jordan's Lemma -- from Wolfram MathWorld

    \lim_{R\to \infty}\int e^{itz}f(z)dz=0

    and we are done.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,972
    Thanks
    1011
    Just rewrite this as

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

    and then use \displaystyle \mathcal{L}^{-1}\left\{\frac{s - a}{(s - a)^2 + \omega ^2}\right\} = e^{a\,t}\cos{(\omega \,t)} and \displaystyle \mathcal{L}^{-1}\left\{\frac{\omega}{(s - a)^2 + \omega ^2}\right\} = e^{a\,t}\sin{(\omega\,t)}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2011
    Posts
    7
    Quote Originally Posted by TheEmptySet View Post
    Hello Lewis and welcome to MHF.

    When you say the standard contour what do you mean? You need to integrate on a vertical line that is to the right of the real part of all of the poles of the integrand.

    One such vertical line is

    s=iz
    .
    .
    .
    \lim_{R \to \infty}f(Re^{i\theta})=0

    So by Jordan's Lemma Jordan's Lemma -- from Wolfram MathWorld

    \lim_{R\to \infty}\int e^{itz}f(z)dz=0

    and we are done.
    Thanks alot, it makes alot more sense now, I was just struggling with how I could apply it the integral I had, but I see how now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inverse of Laplace Transform
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: April 10th 2011, 02:52 PM
  2. Inverse Laplace Transform
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: August 27th 2009, 09:42 AM
  3. Inverse Laplace Transform
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 13th 2008, 04:57 PM
  4. Inverse Laplace Transform
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 21st 2008, 10:56 PM
  5. inverse laplace transform
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 29th 2008, 11:41 PM

Search Tags


/mathhelpforum @mathhelpforum