1. Not finding help here? Sign up for a free 30min 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!

Inverse laplace transform

  1. May 6, 2008 #1
    1. The problem statement, all variables and given/known data

    determine the inverse laplace transform f(t) of the function

    2. Relevant equations

    F(s) = 3/s - (4e^-s)/s^2 + (5e^-2s)/s^2


    3. The attempt at a solution

    i broke it up into the 3 part and got them seperatly which gave me

    f(t) = 3 - 4u(t-1)(t-1) + 5u(t-2)(t-2)

    Am i right sofar ?

    but i'm not sure how to graph the function?all tips welcome
     
  2. jcsd
  3. May 6, 2008 #2

    Defennder

    User Avatar
    Homework Helper

    Yes, that appears to be correct. I got the same answer. To graph it, think of the graph of f(t) as being split into 3 parts. 0<t<1, 1<t<2, t>2.

    For the first interval, notice that the unit step function has yet to be "activated", meaning to say that you consider only the part of f(t) which doesn't depend on the unit step function. That graph is f(t)=3. So for the first interval, it is just a straight horizontal line f(t)=3.

    For the 2nd interval, 2 things into play; both the f(t)=3 from the first part of the interval as well as the part of the function which is multiplied to u(t-1). The 3rd part of f(t) isn't represented in that interval because it hasnt' been "switched on" yet (only when t>2). means the graph of the function during that interval is made up of 3 -4(t-1).

    Finally for the final interval where t>2, all three parts of the function are active, so the final graph for t>2 is simply the graph of f(t)= 3- 4(t-1) + 5(t-2).

    Combine all of the above into 1 graph and there you have it. To check your answer, you can graph the function online here:
    http://fooplot.com/index.php?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Inverse laplace transform
Loading...