Convolution and laplace transform

Question

I was wondering why using convolution gives a different result:

Find the Laplace transformation of $e^t\sin t$.

Using convolution :

$1/s-1 *1/s^2-1$

using table: $1/(s-1)^2-1$

Answer 1

Using convolution, for $f(t) = e^{t},~ g(t) = \sin t$, we have

$$(f * g)(t) = \displaystyle \int_0^t e^{\tau} ~\sin(t - \tau)~d \tau = \dfrac{1}{2}(e^{t} + \sin t - \cos t)$$

Calculate the Laplace Transform of that convolution result.

Next, note that (compare to the previous result)

$$\mathcal{L}(f * g) = \mathcal{L}(f)~\mathcal{L}(g) = F(s)~G(s) = \frac{1}{s-1} ~ \frac{1}{s^2+1}$$

Update

Lastly

$$\mathcal{L} (e^t \sin t) = \dfrac{1}{(s-1)^2+1}$$

We derive this using entirely different methods than above, for example, using the definition of the Laplace Transform, we have

$$\mathcal{L}(e^t \sin t) = \displaystyle \int_0^{\infty} e^{-s t} f(t) dt = \int_0^{\infty} e^{-s t} (e^t \sin t)~dt = \dfrac{1}{s^2-2 s+2}$$

We can also use item $19.$ from the Laplace Table, we have

$$\mathcal{L} (e^{at} \sin (b t)) = \dfrac{b}{(s-a)^2 + b^2 } = \dfrac {1}{(s-1)^2 + 1^2 } = \dfrac {1 }{(s-1)^2 + 1 } = \dfrac{1}{s^2-2 s+2}$$

Do you now understand why you are having issues? In one case, we are finding the Laplace Transform of the convolution of two functions and in the other, we are finding the Laplace Transform of the product of two functions and these are different things.

Answer 2

On a hunch that may be OP was confused about $\displaystyle \mathcal{L}(e^t \sin t)\:$ and $\:\displaystyle \mathcal{L}(e^t * \sin t)$.

The convolution operation $\:\displaystyle e^t * \sin t \:$ is not same as multiplication operation $\displaystyle e^t \sin t\:$.

The convolution $\: f * g \:$ has a definition of its own, in the form of an integral:

$$f * g = \int_0^t f(t)g(t-\tau) d \tau$$ If $\:f = e^t \:$ and $\:g = \sin t \:$ then \begin{align} fg &= e^t \sin t \qquad \qquad \text{(Multiplication)}\\ f*g &= \frac{1}{2}(e^t + \sin t - \cos t) \qquad \text{(Convolution)} \end{align} Above two functions $fg$ and $f*g$ are generated from the same set, but they are entirely different functions now.

This answers your question as to why $\displaystyle \mathcal{L}(e^t \sin t)\:$ is different from $\:\displaystyle \mathcal{L}(e^t * \sin t)$.

A property of convolution is: $\quad\displaystyle \mathcal{L}(e^t * \sin t) = \mathcal{L}(e^t) \mathcal{L}(\sin t)=\displaystyle \frac{1}{s-1} \frac{1}{s^2-1}$.

RHS is ordinary multiplication and no "$*$". Note that same does not hold for multiplication, i.e. $\displaystyle \mathcal{L}(e^t \sin t) \neq \mathcal{L}(e^t) \mathcal{L}(\sin t)$. You have to use Integration or Translation theorem to calculate $\displaystyle \mathcal{L}(e^t \sin t) $