Solved Examples

Today, we're discussing the Convolution theorem, which is crucial in finding inverse Laplace transforms.

What exactly do we mean by convolution in this context?

Great question! Convolution combines two functions into a new function by integrating the product of one function and a time-reversed version of the other.

Can you give us a formula for convolution?

Absolutely! The convolution of functions $f(t)$ and $g(t)$ is defined as $(f*g)(t) = \int_0^t f(\tau) g(t - \tau) d\tau$.

What kind of problems can we solve using this theorem?

Convolution is particularly useful when handling products of Laplace transforms in differential equations and circuit analysis.

Now, let’s have a look at the first solved example to see how we can apply these concepts.

For our first example, we want to find the inverse Laplace transform of $\frac{1}{s(s+1)}$.

How do we start this problem?

We’ll use known transforms. We know that $\mathcal{L}^{-1}\{\frac{1}{s}\} = 1$ and $\mathcal{L}^{-1}\{\frac{1}{s+1}\} = e^{-t}$.

So we can use convolution here?

Exactly! We express $(f * g)(t)$ as $\int_0^t 1 \cdot e^{-(t - \tau)} d\tau$.

What do we get after integrating?

We find that $(f*g)(t) = 1 - e^{-t}$. That's our final result.

To summarize, we utilized the convolution theorem to find the inverse Laplace transform, simplifying the initial product.

Now let's tackle our second example, where we find the inverse Laplace transform of $\frac{1}{s^2(s+2)}$.

I noticed that it’s a bit more complicated compared to the first one.

You're correct! We'll break it down. Let $F(s) = \frac{1}{s^2}$ and $G(s) = \frac{1}{s+2}$.

What are the known inverses?

We have $\mathcal{L}^{-1}\{\frac{1}{s^2}\} = t$ and $\mathcal{L}^{-1}\{\frac{1}{s+2}\} = e^{-2t}$.

Does this mean we'll use convolution again?

Exactly! We’ll set up the integral $\int_0^t \tau e^{-2(t - \tau)} d\tau$.

So how do we evaluate this integral?

We can apply integration by parts. This will ultimately give us the final expression.

To recap, we demonstrated how to apply the convolution theorem for more complex cases.