# Laplace transform Heaviside step function

1. Feb 28, 2009

### Pietair

1. The problem statement, all variables and given/known data
What is the Laplace transform of f(t) = 0 for 0 < t < 2 and f(t) = (4-t) for 2 < t < 3 and f(t) = 1 for 3 < t < 4 and f(t) = (5-t) for 4 < t < 5 and f(t) = 0 for t > 5?

2. Relevant equations

3. The attempt at a solution
f(t) = H(t-2)(4-t) - H(t-3)(4-t) + H(t-3) - H(t-4) + H(t-4)(5-t) - H(t-5)(5-t)
f(t) = H(t-2)(4-t) + H(t-3)(2t-7) - H(t-4)(t-4) + H(t-5)(t-5)

How can I rewrite H(t-2)(4-t) + H(t-3)(2t-7) to a form suitable for Laplace transform?

2. Feb 28, 2009

### Tom Mattson

Staff Emeritus
For $H(t-2)(4-t)$, you'd like to see a $t-2$ in the second factor right? So just put one there, and balance it out elsewhere in the expression. Like this:

$H(t-2)(4-(t-2)-2)$

See how I added and subtracted 2? This simplifies to:

$H(t-2)(2-(t-2))$

Do the same for $H(t-3)(2t-7)$.

3. Feb 28, 2009

### Pietair

Now I am getting:
$H(t-2)(2-(t-2)) + H(t-3)(2(t-3)-1) - H(t-4)(t-4) + H(t-5)(t-5)$

And the Laplace transform:
$((2/s)-(1/s^2))e^(-2s) + ((2/s^2)-(1/s))e^(-3s) + ((-1/s^2)e^(-4s)) + ((1/s^2)e^(-5s))$

Is this correct?

4. Feb 28, 2009

### Tom Mattson

Staff Emeritus
Looks good to me.

5. Feb 28, 2009

### Pietair

Thanks a lot!