Inverse Laplace Help: F(s)=e^-4s(s^2/(s^2+9))

In summary, the correct solution is f(t) = \delta(t-4) - 3\sin(3t-12)\mathcal{H}(t-4), and your mistake was in assuming that the inverse Laplace transform of \frac{s^2}{s^2+9} is \cos(3t) instead of \cos(3t-12).
  • #1
zachdr1
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Homework Statement


[tex] F(s) = \frac{s^2e^{-4s}}{s^2+9}[/tex]

Homework Equations


[tex] \mathcal{L} ({f(t-a)\mathcal{H}(t-a)}) = e^{-as}F(s) [/tex]

The Attempt at a Solution



[tex] F(s) = e^{-4s}(\frac{s^2}{s^2+9}) [/tex]
using [tex] \mathcal{L} {f(t-a)\mathcal{H}(t-a)} = e^{-as}F(s) [/tex]
[tex] f(t) = cos(3t-12)\mathcal{H}(t-4) [/tex]

I do not understand why this is incorrect. My professor got [tex]f(t) = \delta(t - 4) - 3sin(3t - 12)\mathcal{H}(t-4) [/tex]

EDIT: Wow I get it now, it's because [tex] \mathcal{L}(cos(kt)) = \frac{s}{s^2+k} [/tex] not [tex]\mathcal{L}(cos(kt)) = \frac{s^2}{s^2+k} [/tex] stupid mistake lol
 
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  • #2


It appears that your professor's solution is correct. Let's break down the steps to see why your attempt is incorrect.

First, we have the given function F(s) = \frac{s^2e^{-4s}}{s^2+9} which can be rewritten as F(s) = e^{-4s}(\frac{s^2}{s^2+9}).

Next, we use the Laplace transform property \mathcal{L} {f(t-a)\mathcal{H}(t-a)} = e^{-as}F(s) to get f(t) = \frac{s^2}{s^2+9}.

Now, we need to find the inverse Laplace transform of \frac{s^2}{s^2+9}. Using the table of Laplace transforms, we can see that the inverse Laplace transform of \frac{s^2}{s^2+9} is \cos(3t).

However, we need to take into account the time shift of t-4. This means that our final solution should be f(t) = \cos(3(t-4)) = \cos(3t-12).

Therefore, the correct solution is f(t) = \delta(t-4) - 3\sin(3t-12)\mathcal{H}(t-4). The first term accounts for the time shift of t-4, and the second term is the inverse Laplace transform of \frac{s}{s^2+9}, which is \sin(3t).

So, your mistake was in assuming that the inverse Laplace transform of \frac{s^2}{s^2+9} is \cos(3t) instead of \cos(3t-12). This is because the inverse Laplace transform of \frac{s^2}{s^2+9} is \cos(3t) when there is no time shift, but in this case, we have a time shift of -12.
 

Related to Inverse Laplace Help: F(s)=e^-4s(s^2/(s^2+9))

1. What is the inverse Laplace transform of the given function?

The inverse Laplace transform of F(s)=e^-4s(s^2/(s^2+9)) is f(t)=3e^(2t)sin(3t).

2. How do I solve for the inverse Laplace transform?

To solve for the inverse Laplace transform, you can use partial fraction decomposition and the properties of the Laplace transform to simplify the function into a form that can be easily inverted. Then, use a table of Laplace transforms to find the inverse function.

3. What is the significance of the parameter "s" in the function?

The parameter "s" in the function represents the variable in the Laplace domain. It is a complex variable that allows us to transform a function from the time domain to the Laplace domain, making it easier to solve.

4. Can the inverse Laplace transform be used to solve differential equations?

Yes, the inverse Laplace transform is a useful tool in solving differential equations. By transforming a differential equation into the Laplace domain, it can be solved algebraically, and then the inverse Laplace transform can be used to find the solution in the time domain.

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

Yes, there are other methods for finding the inverse Laplace transform, such as using contour integration, the convolution theorem, or the Bromwich integral. These methods may be more complex, but they can be useful for more complicated functions that cannot be easily solved using partial fraction decomposition.

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