Laplce transform for 6t*(u(t)-u(t-1))

[rudyx61]

Im having trouble finding the laplace transfor for the title equation:

6t*(u(t)-u(t-1))

Im not sure how to go about, i was thinking of considering u(t) to be equal to 1 but that seemed a bit of a dodgy thing to do

Any one have any ideas or answers on how i should go about solving this problem

cheers
rudy

 

[HallsofIvy]

Well, by definition,
$$L(f(t))= \int_0^\infty e^{-st}f(t)dt$$
so
$$L(6t(u(t)- u(t-1))= \int_0^\infty e^{-st}6t(u(t)- u(t-1))dt$$
$$= 6\int_0^\infty e^{-st}tu(t)dt- 6\int_0^\infty e^{-st}tu(t-1)dt$$
The first is a standard integral. For the second, let r= t- 1 so that dt= dr, u(t-1)= u(r), and $e^{-st}= e^{-s(r+1)}= e^{-s}e^{-sr}$. The integral becomes
$$6e^{-s}\int_{-1}^\infty e^{-sr}(r+1)u(r)dr$$
and since the Laplace transform of u(t) only depends on the value of u for positive t, we can assume that u(t)= 0 for t< 0.

 

[rudyx61]

im not that good at integrating multiple variables so could you tell me how ud integrate the first term?

 

[rudyx61]

figured it out