Inverse Laplace Transform - Step Function?

[jmg498]

Given...

(2s$$^{2}$$+1)e$$^{-s}$$
------------------
(s - 1)(s$$^{2}$$ + 4s + 5)

Find the inverse Laplace Transform.

I am unsure where to start and am just looking for a little direction, not an answer. From past experience, I am assuming I will have to rewrite this in another form...perhaps by taking out the e$$^{-s}$$ term. I also believe I will need to complete the square on the quadratic in the denominator.

If someone could just point me in the right direction, that would be greatly appreciated! Thanks!

 

[Ben Niehoff]

First, use the properties of the Laplace transform to get rid of the $e^{-s}$. Then use partial fractions to do the rest.

 

[jmg498]

Ok, so I did that...here's my work...(note, I factored out the 1/10 that appears in both terms).

(e$$^{-s}$$/10)$$\frac{17s+5}{(s+2)^{2}+4s+5}$$+$$\frac{3}{s-1}$$

Then I completed the square on the quadratic in the denominator and rewrote it...

(e$$^{-s}$$/10)$$\frac{17s+5}{(s+2)^{2}+1}$$+$$\frac{3}{s-1}$$

Now, finding the inverse Laplace Transform of 3/(s-1) is very simple. So is finding the inverse transform of e$$^{-s}$$. But the term containing the 17s + 5 is confusing me.

 

[Ben Niehoff]

Try writing

$$\frac{17s}{(s+2)^2+1} + \frac{5}{(s+2)^2+1}$$

Also, your formulas will be easier to read if you write the entire thing in TeX instead of just bits and pieces.

 

[Ben Niehoff]

And you should be able to get rid of the $e^{-s}$ part by using this property:

$$\mathcal{L}\{f(t-a)\} = e^{-as}F(s)$$

 

[jmg498]

Thanks very much! This is only my second time using TeX so I apologize for that. Getting used to it.