Finding the inverse laplace transform of (2/(s+2)^4) using Convolution theorem.

In summary, the conversation discusses finding the inverse laplace transform of (2/(s+2)^4) using a given table of identities. The individual first attempts to use a convolution theorem, but the other person advises them to discard the flawed table and use the correct identity, which involves an integral with both functions depending on the integration variable. The first person thanks the other and asks if there is an easier way, to which the other person suggests deriving the identity and having a longer table with fewer errors.
  • #1
theBEAST
364
0

Homework Statement


Find the inverse laplace transform of (2/(s+2)^4) using the given table of identities:

Homework Equations


Here are the given identities:
FWlY5.png


The Attempt at a Solution


Alright, I realize that there is a simple identity that I can use with a factorial symbol, but this identity is not on our formula sheet, so I decided to try it with convolution theorem.

xiuk2.jpg


What did I do wrong and also is there an easier way given the table of identities? Thanks!
 
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  • #2
Where did you get that flawed table? Discard it.

[tex]G(s)H(s) \rightarrow \int_0^t g(t-\tau)h(\tau) \mathop{d\tau}[/tex]

Clearly both functions should depend on the integration variable.
 
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  • #3
lurflurf said:
Where did you get that flawed table? Discard it.

[tex]G(s)H(s) \rightarrow \int_0^t g(t-\tau)h(\tau) d\tau[/tex]

Clearly both functions should depend on the integration variable.

Ahh, thanks!

This is the table that we get for our midterms and final lol

Is there an easier way to solve this with another identity? The integral is pretty messy.
 
  • #4
That integral is not so bad, and you wanted to avoid using

[tex]\frac{1}{(s+a)^{n+1}} \rightarrow \frac{t^n}{n!} e^{-a \mathop{t}}[/tex]

you could derive it. It helps to have a longer table with fewer errors.
 
Last edited:

Related to Finding the inverse laplace transform of (2/(s+2)^4) using Convolution theorem.

1. What is the Convolution theorem and how does it relate to finding the inverse Laplace transform?

The Convolution theorem is a mathematical concept that states that the inverse Laplace transform of a product of two functions is equal to the convolution of the individual inverse Laplace transforms of the functions. In other words, it allows us to find the inverse Laplace transform of a complicated function by breaking it down into simpler functions.

2. How do we apply the Convolution theorem to finding the inverse Laplace transform of (2/(s+2)^4)?

In order to apply the Convolution theorem, we first need to express the given function in terms of simpler functions using partial fraction decomposition. In this case, we can break down the function into (1/(s+2)^4) and 2, which have known inverse Laplace transforms. We then use the Convolution theorem to find the inverse Laplace transform of (1/(s+2)^4) and then multiply it by the inverse Laplace transform of 2 to get the final result.

3. What is the inverse Laplace transform of (1/(s+2)^4)?

The inverse Laplace transform of (1/(s+2)^4) is (1/6)t^3e^(-2t).

4. Can we use the Convolution theorem for any function to find its inverse Laplace transform?

No, the Convolution theorem can only be applied to functions that can be decomposed into simpler functions using partial fraction decomposition. If a function cannot be decomposed in this way, then the Convolution theorem cannot be used to find its inverse Laplace transform.

5. Are there any other methods for finding the inverse Laplace transform besides the Convolution theorem?

Yes, there are other methods such as using tables of known Laplace transforms, using the definition of the Laplace transform, and using the Laplace transform properties. However, the Convolution theorem is a useful tool for finding the inverse Laplace transform of more complicated functions.

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