# Laplace transform of Heaviside function

1. Nov 13, 2013

### gravenewworld

1. The problem statement, all variables and given/known data

What is the laplace transform of H(-t-17)

2. Relevant equations

Shifting theorem:

L(H(t-a)) = (e^-as)/s

3. The attempt at a solution

This is the only part of the problem that I can not get (this part is from a larger differential equation I'm trying to solve). I'm can't seem to figure out how the Laplace transform changes if there is a negative sign in front of the t. I'm not sure if there are some properties of the Heaviside function that will allow me to solve this that I simply don't know and that we didn't go over in class.

Last edited: Nov 13, 2013
2. Nov 13, 2013

### Staff: Mentor

It's "Heaviside" which is now fixed.

Your relevant equation is incorrect and has two errors. It should be L(H(t - a)) = (e-as)/s.

H(t) is the unit step function. H(t - a) is the translation of the unit step function by a units. H(-t - 17) = H(-(t + 17)). This involves a reflection across the vertical axis, followed by a translation to the left by 17 units.

3. Nov 13, 2013

### pasmith

If $t > 0$ then $-t -17 < 0$ so $H(-t-17) = 0$ for all $t > 0$. Since the Laplace transform only cares about the values of a function in $t > 0$ the Laplace transform of $H(-t-17)$ is equal to the Laplace transform of the zero function.

4. Nov 13, 2013

### gravenewworld

Thank you both.

I was actually able to pull up a nice identity online for H(x), where H(x) = (x + |x|)/2x. It helped a lot as well.