1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I don't understand how to laplace transform heaviside functions

  1. May 4, 2013 #1
    1. The problem statement, all variables and given/known data

    (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.

    2. Relevant equations



    3. 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.
     
    Last edited: May 4, 2013
  2. jcsd
  3. May 4, 2013 #2
    *sigh* nevermind, I see that I am transforming the shifted function and I shouldn't be.

    This is annoying.
     
  4. May 4, 2013 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: I don't understand how to laplace transform heaviside functions
Loading...