# Having difficulty finding the inverse laplace transform

1. Aug 14, 2004

### Theelectricchild

Having difficulty finding the inverse laplace transform!!

Hello everyone, I am really stuck on finding the inverse Laplace transform for this:

$$f(s)=\frac{5se^{-3s} - e^{-3s}}{s^{2}-4s+17}$$

Heres my reasoning: I feel that I should rewrite the denominator in some kind of form such as (s-2)^2 + 13, and note the similarity with some of the problems ive been doing before, however its that 13 that is bothering me! Its not something you can take the squareroot of--- and also in addtion, I tried factoring out e^-3s on top and splitting this into two equations, but its this denominator that I absolutely despise.

Any help with finding the right method would be greatly appreciated thank you!!!

Last edited: Aug 14, 2004
2. Aug 14, 2004

### arildno

I've forgotten just about every transform formula I knew (and I'm not in the mood rederiving them), but you may write $$13=(\sqrt{13})^{2}$$

if that helps..

3. Aug 14, 2004

### Theelectricchild

hmm i still am having difficulty---

4. Aug 14, 2004

### arildno

$$e^{-3s}=e^{-6}e^{-3(s-2)}$$

Then you would get one expression on the form:
$$k\frac{e^{-3w}}{w^{2}+a^{2}},k=-e^{-6},w=s-2,a=\sqrt{13}$$

Is this a familiar transform in w?

5. Aug 14, 2004

### Tom Mattson

Staff Emeritus
The exponentials are due to shifts in the t-domain:

L{f(t-T)}=e-sTF(s).

Just find the inverse transform of the rational functions of s, and then let the t-domain functions be delayed by the appropriate amount, given by the coefficient in the exponents of the exp functions.

6. Aug 14, 2004

### Theelectricchild

interesting let me see if i can get anywhere with that...

7. Aug 14, 2004

### Theelectricchild

Uh oh, you see, I cannot split the denominator into two linear factors like the book seems to be doing with most of the problems---- and because I cant, i dont know how to proceed like the examples do... Im sorry if im a bit slow but we just started Laplace and this is a challenge problem id like to know to prepare.

Last edited: Aug 14, 2004
8. Aug 15, 2004

### Theelectricchild

Well I cannot solve it but thanks for your help anyway...

9. Aug 16, 2004

### Tom Mattson

Staff Emeritus
You could factor the denominator into 2 linear factors, but the roots of that polynomial are complex. It would be better instead to complete the square in the denominator. The solution will be t-shifted, exponentially damped sines and cosines.