Hi,
I need to find the inverse Laplace transform of
$\displaystyle F(s) = s^{-2} \tanh(sT/2)$.
I suspect I have to use the period definition for Laplace transforms but I can't see how to make it fit. Any tips?
Hi,
I need to find the inverse Laplace transform of
$\displaystyle F(s) = s^{-2} \tanh(sT/2)$.
I suspect I have to use the period definition for Laplace transforms but I can't see how to make it fit. Any tips?
I haven't used the residue theorem for a while, I know. The Bromwich integral I know by definition, but I don't think we're expected to use it (and haven't been taught how).
Anyway, I'll be keen to try whatever you're thinking, or if you know a way of doing it without these things, that's fine too.
You are 'lucky' because in this case You don't need to know sophisticated tools like Bromwitch integral, residue theorem and so on but only to have 'little imagination' ...
The LT to be inverted is...
$\displaystyle F(s)= \frac{1}{s^{2}}\cdot \tanh \frac{sT}{2}$ (1)
Now, taking into account that...
a) for $\displaystyle |s|>0$ the function $\displaystyle \tanh (*)$ can be written as...
$\displaystyle \tanh \frac{sT}{2} = \frac{1 - e^{-sT}}{1+e^{-sT}} = (1-e^{-sT})\cdot (1 - e^{-sT} + e^{-2sT} - \dots)$ (2)
b) is...
$\displaystyle \mathcal{L}^{-1} \{\frac{1}{s^{2}}\cdot (1-e^{-sT}) \}=t \cdot H(t) -(t-T)\cdot H(t-T)$ (3)
... that is the function represented here...
... is easy enough verify with the aid of (2) that the inverse LT we are searching is the function (3) 'periodized with alternating signs' , i.e. the periodic function represented here...
So your 'intuition' was exact: we have a periodic function ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Hi. Just for fun:
$\displaystyle \mathcal{L}^{-1}\left\{\frac{1}{s^2}\tanh(sT/2)\right\}=\frac{1}{2\pi i}\mathop\int\limits_{c-i\infty}^{c+i\infty}e^{st}\frac{1}{s^2}\tanh(sT/2)ds$
and with some effort, we can show a closed contour pinned at the line $\displaystyle c+i\gamma$ and allowed to expand without bounds, reduces to the Bromwich integral and sum of the enclosed residues:
$\displaystyle \mathcal{L}^{-1}\left\{\frac{1}{s^2}\tanh(sT/2)\right\}=\mathop\text{Res}\limits_{z=0}\left\{e^ {st}\frac{1}{s^2}\tanh(sT/2)\right\}$$\displaystyle +\sum_{n=0}^{\infty}\mathop\text{Res}\limits_{z=z_ n}\left\{e^{st}\frac{1}{s^2}\tanh(sT/2)\right\},\quad z_n=\pm\frac{(2n+1)\pi i}{T}$
$\displaystyle =\frac{T}{2}-\frac{4T}{\pi^2}\sum_{n=0}^{\infty}\frac{1}{(2n+1) ^2}\cos\left[\frac{(2n+1)\pi x}{T}\right]$
and I suspect that is the Fourier series of the answer obtained by Chisigma and suggested by Mr. F.