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.