Laplace transforms and resonance

1. Sep 3, 2010

joriarty

1. The problem statement, all variables and given/known data

[PLAIN]http://img231.imageshack.us/img231/3904/q2a.png [Broken]

3. The attempt at a solution

Below is a screenshot of my Maple worksheet. I am currently stuck on part (a). I have taken the Laplace transform of both sides of the equation, solved for Z(s), then converted it Heaviside form, yet Maple cannot compute the inverse transformation! So, I've done something wrong, but I cannot figure out what! Could I have a few pointers please? (bear in mind this is coursework, so it's really only appropriate that you guide me in the right direction rather than explicitly giving any steps towards the solution)

Note that there's something funny with the rendering of the sum notation. In the middle of the "∑" you'll see a "K" - this should be above the ∑ symbol because the sum is from k = 0..K.

Thanks!

[PLAIN]http://img3.imageshack.us/img3/9652/q2b.png [Broken]

Last edited by a moderator: May 4, 2017
2. Sep 3, 2010

ystael

I can't help you with your problem, but I am chortling with joy because I have finally seen someone else (besides me and Graham/Knuth/Patashnik) use the beautiful Euler math fonts in public.

3. Sep 3, 2010

gabbagabbahey

I haven't used Maple in a long time, so I can't tell you why it is writing the $K$ inline with the summation symbol. But it looks as though the reason it isn't computing the inverse Laplace transform is because you haven't told it to assume anything about $d$ and $v$. I'd assume that both are positive constants.

You really shouldn't need to use Maple for this though, since the Laplace transform (and its inverse) are linear, you should know that if $$Z(s)=\sum_{k} G_k(s)[/itex], then [tex]z(t)=\sum_{k} g_k(t)[/itex]. As long as you know how to take the inverse transform of [tex]\frac{e^{-kds/v}}{(s+1)^2}$$ , you shouldn't have any trouble finding $z(t)$.

4. Sep 3, 2010

joriarty

Thanks for your reply. I'm not sure the inverse laplace transform of $$\frac{e^{-kds/v}}{(s+1)^2}$$ actually exists. I have told Maple to assume that k > 0 and v > 0, but the result is no different.

In my course notes (which have been prepared as a complete set of notes in lieu of a textbook) we are given a small table of Laplace transformations, which is presumably all we would require for the course. The closest equation in this table is e-cs/s ---> H(t - c)*f(t - c), which doesn't really match at all. This leads me to assume that my lecturer doesn't expect us to have to compute the inverse laplace of such a function, therefore, I am wrong!

However if it does exist - I understand what you are saying. Could I just take the inverse laplace transform of the piecewise function and get another piecewise function back? (rather than converting to Heaviside first) Of course the inverse laplace of 0 is 0 so my result would be a piecewise function of the sum from k = 0 to K of the inverse laplace of $$\frac{e^{-kds/v}}{(s+1)^2}$$, where kd/v >= 0, and 0 elsewhere (hope that makes sense, I don't really know LaTeX)

I'm thinking I have made a mistake earlier on in my working, but I can't for the life of me figure out what it is. Perhaps I should ask my lecturer, as he wrote the problem. (I come here sometimes to avoid asking too many questions and annoying my lecturers )

5. Sep 3, 2010

gabbagabbahey

You should have a table of several rules, like the time-shifting rule:

$$\mathcal{L}\left[H(t-c)f(t-c)\right]=e^{-cs}\mathcal{L}\left[f(t)\right]$$

or the frequency shift rule:

$$\mathcal{L}\left[e^{at}f(t)\right]=F(s-a)$$

Begin by using those. For starters, applying the 2nd rule:

\begin{aligned}\mathcal{L}^{-1}\left[\frac{e^{-kds/v}}{(s+1)^2}\right] &= e^{-t}\mathcal{L}^{-1}\left[\frac{e^{-kd(s-1)/v}}{s^2}\right] \\ &= e^{\frac{kd}{v}-t}\mathcal{L}^{-1}\left[\frac{e^{-kds/v}}{s^2}\right]\end{aligned}

Now apply the 1st rule (I assume you know what the inverse tranform of $1/s^2$ is?)...

6. Sep 5, 2010

joriarty

Thanks for your help - I think I have it now. Perhaps the recent seismic activity jolted by brain a bit! Does this make sense to you?

[PLAIN]http://img844.imageshack.us/img844/3443/eq3.png [Broken]

And the amplitude dies away nicely as well, more quickly at higher velocities as expected (if multiple plots are drawn).

[PLAIN]http://img228.imageshack.us/img228/865/screenshot20100906at121.png [Broken]

Interesting that Maple couldn't solve this yet a few manipulations by hand could. Just goes to show that computers can't do everything!

Last edited by a moderator: May 4, 2017
7. Sep 5, 2010

hunt_mat

Taking the laplace transform of your equation results in:
$$s^{2}\mathcal{L}(z)+2s\mathcal{L}(z)+\mathcal{L}(z)=\sum_{n=0}^{K}e^{-\frac{nsd}{v}}$$
The RHS can be summed using a geometric progression formula, from there it is a matter of doing a little algebra and using information about the laplace transform. I would advise you to do this using a pen and a piece of paper...

8. Sep 5, 2010

hunt_mat

I will even give you a hint on the geometric series, the common factor is:
$$e^{-\frac{sd}{v}}$$

9. Sep 5, 2010

joriarty

Hunt_mat, thanks for your reply but I'm not sure exactly what you are trying to say, I'm working on taking an inverse laplace transform. Are you saying my original Laplace transformation done in Maple is wrong?

Turns out though that WolframAlpha can compute this inverse laplace where Maple can't. Of course I'm not going to just accept this answer, I still need to work this through myself and actually understand what's going on. http://www.wolframalpha.com/input/?i=inverse+Laplace+transform+(exp(-k*d*s/v))/(s+1)^2". One problem. Our solutions are identical except where WolframAlpha has (t-dk/v) and I just have t, and I can't see where this has happened!

Edit: I see where I went wrong. My application of the shift theorem (what you called the time shifting rule) was wrong. Looks like I've finally got this bit sorted!

Last edited by a moderator: Apr 25, 2017