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John_henderson
Express the following in terms of Heavyside Function=> H(t-a) and find its laplace transform Where F(t) = t 0<t<1 0 1<t< 2 1 t>2
Hi guys, im new to this topic i would appreciate it if someone can show me in how to proceed forward with this problem.
It's been a while since I've done this, but here's what I'd work from: http://tutorial.math.lamar.edu/Classes/DE/StepFunctions.aspx
Thanks that site looks useful
The examples on the website are kinda difficult to understand, im still kinda confused....?
do you know the \[\mathcal L\{u_c(t)f(t-c)\}=e^{-cs}F(s)\]transform?
Yes i know the forumla but dont know how to apply it yet... Im i suppose to sub in the interval values in first.
For 0<t<1 we have f(t)=t, so let's start with that Now, since for 1<t<2 we have f(t)=0 we need to subtract the t we had at the beginning, and we need to initiate that using the step function with c=1. So, so far that gives us\[f(t)=t-tu_1(t)\]so far so good?
why is f(t)= 0 ? is that from the limits 0<t<1
your step function is\[f(t)=\left\{\begin{matrix}t&&0<t<1\\0&&1<t<2\\1&&t>2\end{matrix}\right.\]correct? if so, then for t between 1 and 2 your function should equal zero
yes thats correct, But why?
you asked why f(t)=0 for 1<t<2... that's why; it's stated in your step function (sorry for the slow replies, my connection is horrible right now) so do you see how\[f(t)=t-tu_1(t)\]satisfies the first two parts of the step function?
Oh right i see, so do i integrate now?
Well first off, I am thinking you want the laplace transforms, not regular integration. We need to do some more manipulation before we can do that. Secondly, we have not yet added the part that will make it such that f(t)=1 for t>2 any ideas how you might do that? (hint: it requires adding a single term, which is another step function)
but the all-important question is"what is a ?".
lol im so confused at the moment, im totally new to this!
no, you need to understand the heavyside function for H(t-1) what is the value when t=2 ?