Inverse laplace tables, where am I going wrong?

In summary, my book has an answer of 2e^t*cos(t). Am I making a mistake in the algebra?Instead of 2s+1-1 put 2s+2-2 then you will habe 2(s-1)+2 and the you will have the answerInstead of 2s+1-1 put 2s+2-2 then you will habe 2(s-1)+2 and the you will have the answer
  • #1
cdotter
305
0

Homework Statement


Find the inverse Laplace transform of [tex]\frac{2s+1}{s^2-2s+2}[/tex]

Homework Equations


The Attempt at a Solution


[tex]
\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*}
[/tex]

My book has an answer of [itex]2e^t cos(t) + 3e^t sin(t)[/itex]. Am I making a mistake in the algebra?
 
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  • #2
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
irtsrider said:
Instead of 2s+1-1 put 2s+2-2 then you will habe 2(s-1)+2 and the you will have the answer

Ok I see how that works, but what about my answer is wrong? What algebra mistake am I making?
 
  • #4
Just realized I had a typo in the original post. :cry: Lines 5 and 6 I typed "sin" instead of "cos." It's all changed now and my question still stands.
 
  • #5
When you write (2s-1)/(s-1)^2+1, it won't 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
irtsrider said:
When you write (2s-1)/(s-1)^2+1, it won't 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

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

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

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.
 
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  • #7
(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
I still don't see it but thank you for your time. :redface:
 
  • #9
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
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
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
glebovg said:
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.

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
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
glebovg said:
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).

Thank you, that was exactly what I needed to see -- where I made the algebra mistake.[URL]http://smiliesftw.com/x/bowdown.gif[/URL]
 
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  • #15
Next time use the method I described above it is much easier.
 
  • #16
Ok, another quick question. Say I have this:

[tex]L^{-1} \{ \frac{1}{(s+1)^2+4} \}[/tex]

This is very close to

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

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

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

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

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.
 
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1. What is an inverse Laplace table?

The inverse Laplace table is a mathematical tool used to find the original function from its Laplace transform. It contains a list of commonly used Laplace transforms and their corresponding inverse functions.

2. How do I use an inverse Laplace table?

To use an inverse Laplace table, you first need to have the Laplace transform of a function. Then, you can look for the Laplace transform in the table and find its inverse function. This will give you the original function.

3. Can an inverse Laplace table be used for all functions?

No, an inverse Laplace table can only be used for functions that have a Laplace transform. This means that the function must be continuous and have a finite number of discontinuities.

4. What are the common mistakes when using an inverse Laplace table?

The most common mistake is using the wrong Laplace transform from the table. It is important to double check that the Laplace transform being used is for the correct function. Another mistake is not considering the region of convergence, which can lead to incorrect results.

5. Are there any tips for using an inverse Laplace table effectively?

Yes, it is important to practice and become familiar with the table. Also, always double check the Laplace transform being used and consider the region of convergence. It can also be helpful to use multiple tables or online resources for a wider range of Laplace transforms.

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