Help with inverse Laplace transforms

In summary, Inverse laplace transforms are not being solved and the student is stuck on finding the answer for equation 2.27 for s=-1.
  • #1
rusty009
69
0

Homework Statement



I have three inverse laplace transforms I can't solve, they are,

i)
(s-1)/(s^2 + 8s + 17)

ii)
(s+3)/(s^2 + 4s)

iii)
2/[(s+1)*(s^2 + 1)]

Homework Equations



The laplace transform table,

http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms

The Attempt at a Solution

i) I completed the square and got,

(s-1)/(s^2 + 8s + 17)

-->(s-1)/[(s+4)^2 -1]

then split up into, s/[(s+4)^2 -1] and -1/[(s+4)^2 -1]

my answer was e^-4t *cos(t) - e^-4t *sin(t)
but the answer is e^-4t *cos(t) - 5e^-4t *sin(t) , I do not know where the 5 has come from ?ii) I'm not sure where to start on this one

(s+3)/(s^2 + 4s) , I have taken out a factor of 1/s but I am not sure were to go from there.iii) For this one, I'm not having problem with the actual laplace but rather partial fractions, what I have done,

2/[(s+1)*(s^2 + 1)] = A/(s+1) + B/(s^2 + 1)

A(s^2 + 1) + B(s+1) = 2

for s=-1 I found A= 1

but I'm stuck at finding B, how do I make (s^2 + 1) = 0

Thanks a lot in advance !
 
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  • #2
Part 3:

It should be A/(s+1) + (Bs + c)/(s2 + 1)

Hence

2 = A.(s2 + 1) + (Bs +c).(s + 1)

You can then find all the values using s = -1, 0 and 1.
 
  • #3
i) I completed the square and got,

(s-1)/(s^2 + 8s + 17)

-->(s-1)/[(s+4)^2 -1]

No, (s+4)^2 - 1 = s^2 + 8s + 16 - 1 = s^2 + 8s + 15, you have 17

ii) I'm not sure where to start on this one

(s+3)/(s^2 + 4s) , I have taken out a factor of 1/s but I am not sure were to go from there

Note that s^2 + 4s = (s+2)^2 - 4. Does that help?

2/[(s+1)*(s^2 + 1)]

Your setup is wrong, s^2 + 1 has no real roots therefore the proper way to try to split it up is to rewrite it as A/(s+1) + (Bs + C)/(s^2 + 1). Work with that.
 
  • #4
(s-1)/(s^2 + 8s + 17)

-->(s-1)/[(s+4)^2 -1]

then split up into, s/[(s+4)^2 -1] and -1/[(s+4)^2 -1]

1) (s+4)^2 -1 = s^2+8s+16-1 .. but that's not 17?
2) Is s/[(s+4)^2 -1] = sin .. ? You should look at your numerator

NoMoreExams beat me so only look at 2 ..
 
  • #5
rusty009 said:
iii) For this one, I'm not having problem with the actual laplace but rather partial fractions, what I have done,

2/[(s+1)*(s^2 + 1)] = A/(s+1) + B/(s^2 + 1)

A(s^2 + 1) + B(s+1) = 2

for s=-1 I found A= 1

but I'm stuck at finding B, how do I make (s^2 + 1) = 0

Thanks a lot in advance !

First it is just wrong as NotStine mentioned. Second, you might want to review some partial fractions. You not always make things equal to 0 .. Use linear algebra
 
  • #6
Also if you ARE allowed to use a table, you can use this: http://www.vibrationdata.com/Laplace.htm

i) (s-1)/(s^2 + 8s + 17) = (s-1)/((s + 4)^2 + 1). By equation 2.27, lambda = -1, alpha = 4, beta = 1. So plugging that in we get e^(-4t)[cos(t) - 5sin(t)]

etc.
 

What is an inverse Laplace transform?

An inverse Laplace transform is a mathematical operation that takes a function in the Laplace domain and converts it back into the time domain. It is the inverse of the Laplace transform, which is used to convert a function from the time domain to the Laplace domain.

Why are inverse Laplace transforms useful?

Inverse Laplace transforms are useful because they allow us to solve differential equations in the time domain by converting them into algebraic equations in the Laplace domain. This makes it easier to solve complex equations and analyze the behavior of systems in the time domain.

How do you perform an inverse Laplace transform?

To perform an inverse Laplace transform, you first need to have a function in the Laplace domain. Then, you can use a table of Laplace transforms or algebraic manipulation to find the inverse Laplace transform. Some functions may require the use of partial fraction decomposition or other techniques to find the inverse Laplace transform.

What are some common techniques for finding inverse Laplace transforms?

Some common techniques for finding inverse Laplace transforms include using a table of Laplace transforms, using partial fraction decomposition, using the convolution integral, and using the residue theorem. The technique used depends on the complexity of the function in the Laplace domain.

Are there any limitations to inverse Laplace transforms?

Yes, there are some limitations to inverse Laplace transforms. Some functions may not have an inverse Laplace transform or may require advanced techniques to find the inverse. Additionally, the inverse Laplace transform may not exist if the function has a singularity or pole at the origin in the Laplace domain.

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