# Inverse Laplace Transform - Step Function?

1. Dec 16, 2007

### 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!!!

2. Dec 17, 2007

### 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.

3. Dec 17, 2007

### 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.

4. Dec 17, 2007

### 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.

5. Dec 17, 2007

### Ben Niehoff

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

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

6. Dec 17, 2007

### jmg498

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

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