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

[zachdr1]

Homework Statement

$$F(s) = \frac{s^2e^{-4s}}{s^2+9}$$

Homework Equations

$$\mathcal{L} ({f(t-a)\mathcal{H}(t-a)}) = e^{-as}F(s)$$

The Attempt at a Solution

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

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

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

 

[mighty2000]

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.