Laplace and unit step- multiplication vs convolution Please be gentle if the question is stupid.
When using the laplace transform, you often multiply the function of interest by a shifted unit step function to operate on the positive portion of the function since the Laplace transform is defined from time=0 to infinity.
Why can we do this multiplication?  Why is it not a convolution?
 A: *

*Your questions are not stupid questions. In fact, they are the most important questions, which you should not forget when making an analysis of signals and systems. This is because if you apply erroneously the unit step function to a particular signal, the analysis will surely be wrong (in engineering, can make drop a bridge, almost without exaggeration ...).

*If you are not completely internalized or do not understand the answers here, check my answer to the following question that explains a little more detailed the matter: Unilateral Laplace Transform vs Bilateral Fourier Transform.

1. Why can we do this multiplication?
This multiplication is done to make a system acquires the behavior of a physical system, or more specifically, the behavior of a causal system, since:
$$
h(t) \mbox{ is a casual system} \quad\Leftrightarrow\quad \forall t\in\mathbb{R},\, t < 0:\quad h(t) = 0
$$
$$
\therefore\quad h(t)u(t) \mbox{ is always a casual system}
$$
Note that you should not always multiply a system for the unit step function. In fact, if a system is not causal, then you never should multiply the system function $h(t)$ by the unit step function. In summary:


*

*Causal System (Non-anticipative System): always multiply by the unit step.

*No-Causal System (Anticipative System): never multiply by the unit step.


The reason for this is explained in my answer in the link above.

2. Why is it not a convolution?
Because the objective of the Laplace transform is just avoid convolution. Convolution is difficult to calculate and needs a lot of computing power, while a transformed simplifies the process of convolution to a simple multiplication.
$$
y(t) = h(t) \ast x(t) \quad\xrightarrow{\mathcal{L}}\quad Y(s) = H(s) X(s)
$$
Again, the reason for this is explained in my answer in the link above.
