Question

Laplace of unit step function

Answer 1

Typing out question now...

Answer 2

Given a graph, I see that the function u(t)=0 when \[0 \le t \le 1\] and u(t)=1 when \[1<t \le 2\] and u(t)=0 when \[t >2\] I am supposed to find the Laplace of it but I can't even write out the equation first. I believe it should be of the form u(t)f(t-a) but the more I try to read about unit step functions, the more confused I am becoming. How do I write out an equation that I can take the Laplace of given the information above?

Answer 3

I would start from:
\[L(f(t))=\int\limits_0^{\infty} e^{-st}f(t)dt\]
But I don't know much :/

Answer 4

So you would have something like:
\[\int\limits_0^1 e^{-st}(0)dt+\int\limits_1^2 (1)(e^{-st})dt+\int\limits (0)dt\]
So the way I see it you have the middle integral only which is:
\[-e^{-st}|_1^2=-(e^{-2s}-e^{-s})=e^{-s}-e^{-2s}\]
Shrug/ I've never done laplacian stuff I'm just going off wikipedia lol :P

Answer 5

Ah, it didn't even occur to me that 0, 1, and 0 were my f(t)'s! :) Aside from a little error (the integral of e^-st is (-1/s)e^-st, not just -e^-st), that method produces the correct answer I need. Thank you so much, I really appreciate it! :)

Answer 6

Oh yeah, you're right, I forgot my constant :P Sorry haha. Glad I could help :P