- #1
Niles
- 1,866
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Homework Statement
Hi all.
I have found the Laplace transform of the following piecewise function:
[tex]
f(x) = \left\{ {\begin{array}{*{20}c}
{0\,\,\,\,{\rm{for}}\,\,\,\,x < 0} \\
{x\,\,\,\,{\rm{for}}\,\,x \in (0;1)} \\
{0\,\,\,\,{\rm{for}}\,\,\,\,x > 1} \\
\end{array}} \right.
[/tex]
I get the Laplace transform to be
[tex]
\overline f (s) = \frac{{ - \left( { - 1 + e^{ - s} + e^{ - s} s} \right)}}{{s^2 }}.
[/tex]
Now I wish to find the inverse Laplace transform, and this is given by the Bromwich integral:
[tex]
f(x) = - \frac{1}{{2\pi i}}\left[ {\int\limits_{\lambda - i\infty }^{\lambda + i\infty } {\frac{{ - e^{st} }}{{s^2 }}ds + \int\limits_{\lambda - i\infty }^{\lambda + i\infty } {\frac{{e^{s(t - 1)} }}{{s^2 }}ds + } \int\limits_{\lambda - i\infty }^{\lambda + i\infty } {\frac{{e^{s(t - 1)} }}{s}ds} } } \right].
[/tex]
Doing these integrals, I end up with f(x) being x for x<1 and 0 for x>1, which is not correct. I have gone through the calculations many times, and I do not believe I have made an error. Here's my reasonings:
1) For the first integral, I close the line by a semi-circle in the left half-plane, and this integral gives me -x.
2) For the second and third integral, I close the line by a semi-circle in the right half-plane for x<1, and this gives me 0, since there are no poles.
For x>1 I close the line in the left halfplane, and I get x.So, as stated earlier, I end up with f(x) being x for x<1 and 0 for x>1, and my reasoning has been 100% valid so far. Can you tell me what is wrong here?
Niles.