Inverse Laplace Transform

harbottle · #1 $Old$ November 6th, 2009, 06:24 PM

Hi,

I need to find the inverse Laplace transform of

$F(s) = s^{-2} \tanh(sT/2)$ .

I suspect I have to use the period definition for Laplace transforms but I can't see how to make it fit. Any tips?

mr fantastic · #2 $Old$ November 6th, 2009, 06:33 PM

Quote:

Originally Posted by harbottle $View Post$

Hi,

I need to find the inverse Laplace transform of

$F(s) = s^{-2} \tanh(nT/2)$ .

I suspect I have to use the period definition for Laplace transforms but I can't see how to make it fit. Any tips?

There is no s in your tanh function ....?

harbottle · #3 $Old$ November 6th, 2009, 07:11 PM

Sorry, typo'd

. Was thinking of periods while I was writing it down, and an n found its way in there somewhere

mr fantastic · #4 $Old$ November 7th, 2009, 02:59 AM

Quote:

Originally Posted by harbottle $View Post$

Hi,

I need to find the inverse Laplace transform of

$F(s) = s^{-2} \tanh(sT/2)$ .

I suspect I have to use the period definition for Laplace transforms but I can't see how to make it fit. Any tips?

Are you familiar with:

The Bromwich inversion integral.
The residue theorem.
Fourier series.

Because these are the things you would need to understand if you were to be able to follow the outline of the type of solution I have in mind and fill in the details.

harbottle · #5 $Old$ November 7th, 2009, 03:35 AM

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.

mr fantastic · #6 $Old$ November 7th, 2009, 03:39 AM

Quote:

Originally Posted by harbottle $View Post$

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.

There's no point my posting an outline of a solution if you have not learned the mathematics it requires. I might think of a less sophisticated way at some stage, or perhaps someone else will.

chisigma · #7 $Old$ November 7th, 2009, 07:19 PM

Quote:

Originally Posted by harbottle $View Post$

Hi,

I need to find the inverse Laplace transform of

$F(s) = s^{-2} \tanh(sT/2)$ .

I suspect I have to use the period definition for Laplace transforms but I can't see how to make it fit. Any tips?

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...

$F(s)= \frac{1}{s^{2}}\cdot \tanh \frac{sT}{2}$ (1)

Now, taking into account that...

a) for $|s|>0$ the function $\tanh (*)$ can be written as...

$\tanh \frac{sT}{2} = \frac{1 - e^{-sT}}{1+e^{-sT}} = (1-e^{-sT})\cdot (1 - e^{-sT} + e^{-2sT} - \dots)$ (2)

b) is...

$\mathcal{L}^{-1} \{\frac{1}{s^{2}}\cdot (1-e^{-sT}) \}=t \cdot H(t) -(t-T)\cdot H(t-T)$ (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

$\chi$ $\sigma$

mr fantastic · #8 $Old$ November 7th, 2009, 08:12 PM

Quote:

Originally Posted by chisigma $View Post$

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...

$F(s)= \frac{1}{s^{2}}\cdot \tanh \frac{sT}{2}$ (1)

Now, taking into account that...

a) for $|s|>0$ the function $\tanh (*)$ can be written as...

$\tanh \frac{sT}{2} = \frac{1 - e^{-sT}}{1+e^{-sT}} = (1-e^{-sT})\cdot (1 - e^{-sT} + e^{-2sT} - \dots)$ (2)

b) is...

$\mathcal{L}^{-1} \{\frac{1}{s^{2}}\cdot (1-e^{-sT}) \}=t \cdot H(t) -(t-T)\cdot H(t-T)$ (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

$\chi$ $\sigma$

Nice reply. Clearly I have no imagination ....

shawsend · #9 $Old$ November 8th, 2009, 07:02 AM

Hi. Just for fun:

$\mathcal{L}^{-1}\left\{\frac{1}{s^2}\tanh(sT/2)\right\}=\frac{1}{2\pi i}\mathop\int\limits_{c-i\infty}^{c+i\infty}e^{st}\frac{1}{s^2}\tanh(sT/2)ds$

and with some effort, we can show a closed contour pinned at the line $c+i\gamma$ and allowed to expand without bounds, reduces to the Bromwich integral and sum of the enclosed residues:

$\mathcal{L}^{-1}\left\{\frac{1}{s^2}\tanh(sT/2)\right\}=\mathop\text{Res}\limits_{z=0}\left\{e^{st}\frac{1}{s^2}\tanh(sT/2)\right\}$ $+\sum_{n=0}^{\infty}\mathop\text{Res}\limits_{z=z_n}\left\{e^{st}\frac{1}{s^2}\tanh(sT/2)\right\},\quad z_n=\pm\frac{(2n+1)\pi i}{T}$

$=\frac{T}{2}-\frac{4T}{\pi^2}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\cos\left[\frac{(2n+1)\pi x}{T}\right]$

and I suspect that is the Fourier series of the answer obtained by Chisigma and suggested by Mr. F.

CaptainBlack · #10 $Old$ November 8th, 2009, 09:01 AM

Quote:

Originally Posted by harbottle $View Post$

Hi,

I need to find the inverse Laplace transform of

$F(s) = s^{-2} \tanh(sT/2)$ .

I suspect I have to use the period definition for Laplace transforms but I can't see how to make it fit. Any tips?

You can also look this up in a table of LT and ILT, it is item 32.158 in the Laplace Transform chapter of the Schaum Mathematical Handbook.

(interestingly enough WolframAlpha does not know this!)

CB

harbottle · #11 $Old$ November 10th, 2009, 01:06 AM

Nice! That looks easy now with $\chi\sigma$ 's hint.

And I'll have a look at your solution too shawsend though as Mr.F suggested I might have some difficulties.

Thanks everyone!