# Inverse laplace tables, where am I going wrong?

1. Sep 16, 2011

### cdotter

1. The problem statement, all variables and given/known data
Find the inverse Laplace transform of $$\frac{2s+1}{s^2-2s+2}$$

2. Relevant equations

3. The attempt at a solution
\begin{align*} L^{-1}\{ \frac{2s+1}{s^2-2s+2} \} &= 2 L^{-1}\{ \frac{s}{(s-1)^2+1} \} + L^{-1}\{ \frac{1}{(s-1)^2+1} \} \\ &= 2 L^{-1}\{ \frac{s}{(s-1)^2+1} \} + e^t sin(t) \\ &= L^{-1}\{ \frac{2s+1-1}{(s-1)^2+1} \} + e^t sin(t) \\ &= L^{-1}\{ \frac{2s-1}{(s-1)^2+1} \} + L^{-1}\{ \frac{1}{(s-1)^2+1} \} + e^t sin(t) \\ &= 2e^t cos(t) + e^t sin(t) + L^{-1}\{ \frac{1}{(s-1)^2+1} \} \\ &= 2e^t cos(t) + e^t sin(t) + e^t sin(t) \end{align*}

My book has an answer of $2e^t cos(t) + 3e^t sin(t)$. Am I making a mistake in the algebra?

Last edited: Sep 16, 2011
2. Sep 16, 2011

### irtsrider

Instead of 2s+1-1 put 2s+2-2 then you will habe 2(s-1)+2 and the you will have the answer

3. Sep 16, 2011

### cdotter

Ok I see how that works, but what about my answer is wrong? What algebra mistake am I making?

4. Sep 16, 2011

### cdotter

Just realized I had a typo in the original post. Lines 5 and 6 I typed "sin" instead of "cos." It's all changed now and my question still stands.

5. Sep 16, 2011

### irtsrider

When you write (2s-1)/(s-1)^2+1, it wont give you cos because you should have nA/A^2+1 then you will have a cos so you should have n(s-1)/(s-1)^2+1 which n=2 here. Since you will have 2(s-1)=2s-2 and to make it equal to 2s in line 2 you should write like 2s-2+2 the separate to 2s-2 and 2 then you will have the answer

6. Sep 16, 2011

### cdotter

Oh, I think you must be doing it differently? My table has:

$$e^{at} sin(bt) = \frac{b}{(s-a)^2+b^2}, s>a$$
$$e^{at} cos(bt) = \frac{s-a}{(s-a)^2+b^2}, s>a$$

I'm matching it to that. Should I be doing it a "longer" way? I still don't see how my original method is wrong. 2s+1-1=2s, just as 2s+2-2=2s and 2(s+1)-2=2s.

Last edited by a moderator: Sep 16, 2011
7. Sep 16, 2011

### glebovg

(2s + 1)/(s^2 - 2s + 2) = (2s + 1)/[(s - 1)^2 + 1] = [2(s - 1) + 3]/[(s - 1)^2 + 1]

Which equals 2*(s - 1)/[(s - 1)^2 + 1] + 3*1/[(s - 1)^2 + 1].

Use the appropriate formulas from the table of Laplace transforms.

8. Sep 16, 2011

### cdotter

I still don't see it but thank you for your time.

9. Sep 16, 2011

### irtsrider

Thats it for cos you should have s-a in numerator and and s-a is s-1 when you write it like 2s-1 it is not s-a it is 2s-a and for solving it you should have 2(s-a) not 2s-a. Right?

10. Sep 16, 2011

### cdotter

I understand why your method works.

Why won't the method I used in the first post work? Is it not algebraically sound and does it not fit the form given in the Inverse Laplace Transform tables? What am I doing wrong?

11. Sep 16, 2011

### glebovg

Essentially all you have to do is make the denominator and the numerator look similar. The easiest way to do it is to complete the square.

You need (s - 1) in the denominator and in the numerator to match the formulas in the table of Laplace transforms.

12. Sep 16, 2011

### cdotter

I must be missing something major because I still don't get it. I think all of my steps match the formulas in my table.

13. Sep 16, 2011

### glebovg

Have a look at the 4th line. (2s - 1)/[(s - 1)^2 + 1] ≠ 2*[(s - 1 )/(s - 1)^2 + 1].

L^-1{2* (s - 1)/[(s - 1)^2 + 1]} = 2e^t*cos(t).

14. Sep 16, 2011

### cdotter

Thank you, that was exactly what I needed to see -- where I made the algebra mistake.[URL]http://smiliesftw.com/x/bowdown.gif[/URL]

Last edited by a moderator: Apr 26, 2017
15. Sep 16, 2011

### glebovg

Next time use the method I described above it is much easier.

16. Sep 16, 2011

### cdotter

Ok, another quick question. Say I have this:

$$L^{-1} \{ \frac{1}{(s+1)^2+4} \}$$

This is very close to

$$e^{at} sin(bt) = \frac{b}{(s-a)^2+b^2}, s>a$$

So to get the b equal I would multiply the transform by 2?

$$2L^{-1}= \{ \frac{1}{(s+1)^2+4} \} \rightarrow f(t) = 2e^{-t} sin(2t)$$

But my book says the answer is $$f(t) = \frac{1}{2}e^{-t} sin(2t)$$

Am I making another dumb mistake? Why is the book multiplying by 1/2 and not 2?

edit: Nevermind, I see it now.

Thank you everyone for your help.

Last edited: Sep 16, 2011