# 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?

• 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.

• Thank you. However, my fundamental doubt remains-- why is it that it that the time domain function can be multiplied by a unit step instead of convolved by it when making the system causal? Why is that operation in time domain a multiplication and not a convolution? Is it because the unit step does not alter the function in any way and only shifts it (or truncates it) to be non zero at t>0? Apr 11 '16 at 10:48
• You must understand (and perhaps is what has confused) that convolution is an operation (which becomes a multiplication in the frequency domain). While the unit step function alters the function (truncates it) to be zero at t<0, to force mathematically a system to be causal. However, in essence, causality and the convolution are two totally different concepts.
– Noir
Apr 11 '16 at 14:45
• Thanks for hanging in there with me! Let me try to frame my misunderstanding in a different way. If u(t) (hypothetically) were not a unit step , but were instead, say, u(t)=3t^2+t+1, we would not be be able to multiply h(t) by u(t) (in the time domain), because u(t) actually convolves with h(t) in this example with u(t)=3t^2+t+1. Assuming this is correct, my fundamental question is why then can we then multiply h(t) by u(t) (in the time domain)when u(t) is the unit step? Under what conditions do we convolve vs multiply when combining functions in the time domain? Apr 12 '16 at 2:51
• I got it. Let me explain: You are mistaking the input signal system with the unit step function. The unit step function only serves to force (mathematically) to the system function $h(t)$ to behave as a causal system. In these terms, the right thing would be to define: $$\mbox{input signal }\rightarrow x(t)=3t^2+t+1$$ $$\mbox{system function }\rightarrow h(t)=\cdots$$ $$\mbox{system function (forced to be causal) }\rightarrow h(t)u(t) = h_2(t)$$ And only now you can make the convolution between $x(t)$ and $h(t)$ (or $x(t)$ and $h_2(t)$, if your original $h(t)$ is a causal system).
– Noir
Apr 12 '16 at 3:48
• Ok ,got it. I know the definition of the unit step, my intent was to try to determine if the reason we could multiply by it had something to do with the fact that it is a constant, 1, for t>0- hence the change in its definition for the sake of asking the question. I now understand that that is not the reason. The reason is that the u(t) is not the input signal, but a mathematical mechanism to force the system to be causal.--Thank you! Apr 12 '16 at 11:00