Convolution Theorem

The convolution theorem states that the Laplace transform of the convolution of two functions $f(t)$ and $g(t)$ is the product of their individual Laplace transforms. That is:

$\mathcal{L}\{f(t) * g(t)\} = F(s)G(s),$ where: $f(t) * g(t) = \int_0^t f(\tau)g(t-\tau) \, d\tau.$

The inverse Laplace transform then is given by: $\mathcal{L}^{-1}\{F(s)G(s)\} = f(t) * g(t).$

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Problem: Inverse Laplace Transform

We need to compute: $\mathcal{L}^{-1}\left(\frac{6s}{(s^2 - 16)^2}\right).$

Step 1: Rewrite the given expression

The denominator $(s^2 - 16)^2$ suggests repeated roots. We rewrite $s^2 - 16$ as $(s - 4)(s + 4)$, so: $\frac{6s}{(s^2 - 16)^2} = \frac{6s}{((s - 4)(s + 4))^2}.$

Step 2: Factorize and relate to convolution theorem

This is already the Laplace domain representation of the product of two functions. To simplify: $\frac{6s}{(s^2 - 16)^2} = \frac{3}{(s - 4)^2} + \frac{3}{(s + 4)^2}.$

Step 3: Compute individual inverse Laplace transforms

1. For $\frac{1}{(s - a)^2}$, the inverse Laplace is: $\mathcal{L}^{-1}\left(\frac{1}{(s - a)^2}\right) = t e^{at}.$

2. Using this:

   • For $\frac{3}{(s - 4)^2}$, we get $3t e^{4t}$.
   • For $\frac{3}{(s + 4)^2}$, we get $3t e^{-4t}$.

Step 4: Combine results

The final inverse Laplace transform is: $\mathcal{L}^{-1}\left(\frac{6s}{(s^2 - 16)^2}\right) = 3t e^{4t} + 3t e^{-4t}.$

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Final Answer

$\mathcal{L}^{-1}\left(\frac{6s}{(s^2 - 16)^2}\right) = 3t \left(e^{4t} + e^{-4t}\right).$

Would you like to see more detailed intermediate steps or have additional examples?

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Related Questions

1. How does the convolution theorem apply to partial fraction decomposition?
2. What are the steps to find $\mathcal{L}^{-1}$ of more complex rational functions?
3. How is convolution used in systems involving impulse response?
4. What is the significance of exponential terms in Laplace transforms?
5. How would the answer change if the numerator were $s^2$ instead of $6s$?

Tip

When working with repeated roots in the Laplace domain, always check for patterns that simplify into standard inverse Laplace pairs.