Laplace transform linearity problem

[Frankenstein19]

TL;DR

Don't understand why the Laplace transform for a u(t)*e^(-t/4) isn't (1/s)*(1/(s+1/4)). The book im reading says it's(1/(s+1/4))

I've included the problem statement and a bit about the function but my main issue is with the equation after "then" and the one with the red asterisk. I don't understand why the Laplace transform for a u(t)*e^(-t/4) isn't (1/s)*(1/(s+1/4)). The book I am reading says it's(1/(s+1/4)).

 

[Master1022]

Hi,

This follows from the definition of the Laplace Transform:
$$F(s) = \int_0^\infty f(t) e^{-st} dt$$

We know that $u(t) = 1$ for $t \geq 0$ so it just becomes a 1 in our integral:
$$F(s) = \int_0^\infty e^{-t/4} e^{-st} dt = \int_0^\infty e^{-t/4} e^{-st} dt = \int_0^\infty e^{-t(s + \frac{1}{4})} dt$$

and you can get to the result from there.

Hope that helps. What made you think it ought to have an extra $\frac{1}{s}$ term?

 

[pasmith]

It is not generally the case that the transform of a product is equal to the product of the transforms.

In this case from first principles:
$$\int_0^\infty u(t) e^{-at} e^{-st}\,dt = \int_0^\infty e^{-(s + a)t}\,dt = \frac{1}{s + a}$$ since $u(t) = 1$ for all $t \in (0, \infty)$.