I don't understand how to laplace transform heaviside functions

[1MileCrash]

Homework Statement

(6-t)heaviside(t-2)

This is just one term of the real problem I'm working, but it will serve to help me figure this out.

Homework Equations

The Attempt at a Solution

http://www.wolframalpha.com/input/?i=laplace+transform+{(6-t)heaviside(t-2)}

I really, REALLY don't understand how to do this, at all.

My textbook makes the following claims:

heaviside(t-c) laplace transform: e^(-c)/s
heaviside(t-c)f(t-c) laplace transform: e^(-cs)F(s) where F denotes the transform of f

Ok, fine. Apparently I am not very math literate because that makes me do the following:

(6-t)heaviside(t-2) = (4-(t-2))heaviside(t-2) = 4heaviside(t-2) - (t-2)heaviside(t-2)

By the rules above, the laplace transform I want to say is:

4e^(-2s)/s - e^(-2s)(1/s^2 - 2/s)


Which is wrong. I have NO idea what to do with this, I don't understand what these rules are telling me to do.


Thanks for any advice.

 

[1MileCrash]

*sigh* nevermind, I see that I am transforming the shifted function and I shouldn't be.

This is annoying.

 

[LCKurtz]

Homework Statement

(6-t)heaviside(t-2)

This is just one term of the real problem I'm working, but it will serve to help me figure this out.

Homework Equations

The Attempt at a Solution

http://www.wolframalpha.com/input/?i=laplace+transform+{(6-t)heaviside(t-2)}

I really, REALLY don't understand how to do this, at all.

My textbook makes the following claims:

heaviside(t-c) laplace transform: e^(-c)/s
heaviside(t-c)f(t-c) laplace transform: e^(-cs)F(s) where F denotes the transform of f

Ok, fine. Apparently I am not very math literate because that makes me do the following:

(6-t)heaviside(t-2) = (4-(t-2))heaviside(t-2) = 4heaviside(t-2) - (t-2)heaviside(t-2)

By the rules above, the laplace transform I want to say is:

4e^(-2s)/s - e^(-2s)(1/s^2 - 2/s)Which is wrong. I have NO idea what to do with this, I don't understand what these rules are telling me to do.

Thanks for any advice.

I will use $u(t)$ for heaviside(t). You have the formula$$
\mathcal L(f(t-c)u(t-c) = e^{-sc}\mathcal L(f(t))=e^{-sc}F(s)$$The problem with using this for taking transforms is that usually you are asked to take the transform of $f(t)u(t-c)$. In your example where $f(t) = 6 -t$ that is why they jump through hoops to write $6-t =4-(t-2)$, so it is written as a function of $(t-2)$ to match the $u(t-2)$.

Let's look at what you would get taking the transform of $f(t)u(t-c)$:$$
\mathcal L(f(t)u(t-c) = \int_0^\infty e^{-st}f(t)u(t-c)\, dt =\int_c^\infty e^{-st}f(t)\, dt$$Now let $v = t-c$ so we get$$
\int_0^\infty e^{-s(v+c)}f(v+c)\, dv =e^{-sc}\int_0^\infty e^{-sv}f(v+c)\, dv =
e^{-sc}\int_0^\infty e^{-st}f(t+c)\, dt=e^{-sc}\mathcal L(f(t+c))$$This says that if you want to transform $f(t)u(t-c)$ you can do the following steps:
1. Replace $t$ by $t+c$ in the formula for $f(t)$.
2. Transform that.
3. Multiply the result by $e^{-sc}$.
Try it and see if you like it.

[Edit] I see you solved it while I was typing. Try my suggestion anyway, you might like it better.