# Laplace transform

1. May 28, 2008

### jesuslovesu

[SOLVED] Laplace transform

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

Find the inverse laplace transform of $$\frac{e^{-2s} }{s^2 + s - 2}$$

2. Relevant equations

3. The attempt at a solution

I'm able to do about half of the problem
using partial fractions, I've found
$$\frac{e^{-2s}}{3(s+1)}+\frac{e^{-2s}}{3(s-2)}$$

I can find the inverse Laplace transform of the latter part of that expression $$1/3u_2(t) e^{t-2}$$ unfortunately, I don't know how I can modify the first part so that it's shifted by -2. Anyone know what I should do?

2. May 28, 2008

### DavidWhitbeck

Use calculus of residues (have you taken complex analysis yet?)

3. May 28, 2008

### dirk_mec1

Look in the wiki page about laplace transformations under 'time shifting'.

4. May 28, 2008

### jesuslovesu

Unfortunately I haven't studied complex analysis.
I made an error in my original post (and I can't edit it for some reason)
It's now worse than before
$$s^2 + s - 2 = (s+2)(s-1)$$
$$-\frac{e^{-2s}}{3(s+2)} + \frac{e^{-2s}}{3(s-1)}$$

I see the time shift equation on wikipedia, but since the 's' is in the denominator I don't see how to shift either so that they match up with $$u_2(t)$$

5. May 28, 2008

### DavidWhitbeck

Hum, you seem to be misreading it, do this for your two terms separately

Step 1-- cover the exponential up with your hand. What's left is your 1/(s+2) (or 1/(s-1) for the other term) with your 3 and your signs of course. Call that F(s).

Step 2-- look up on a table the f(t) such that L[f]=F(s) from Step 1. Write that f(t) down.

Step 3-- now look on that exponential you were ignoring before (exp(-2s)) that's telling you that your answer will not be f(t), but instead be f(t-2)H(t-2) where H is the unit step function. So write that down.

Now repeat for the other term and then add the two expressions you get from Step 3 together.

6. May 28, 2008

### jesuslovesu

got it now thanks, those steps are really handy