Find the laplace transform of $u(-t+a)$, u is the step function

Question

I am not sure how to deal with the minus sign in front of t

But we can try:

$$U(s) = \int_0^\infty u(-(t-a)) e^{-st}dt$$

which leads to $$U(s) = \int_{0}^a e^{-st}dt$$

Is this correct?

Answer 1

Yes. We can write $\mathcal{U}(-(t-a))=\mathcal{U}(a-t)$ so the only part in the domain that isn't zero is when $t\in(0,a)$. Therefore, you obtain $$\int_0^ae^{-st}dt $$ as you have determined. Also, if you are a visual person, you can plot the step function to see what is occurring. In the plot below, I took $a = 3$ so we could take a look at the step function of $\mathcal{U}(3-t)$.