The Second Translation Theorem for Laplace Transforms

[@essayprometheus]

This is the best video I have found that explains and uses the 2nd TT with a practical example. And its not too easy or insanely difficult, so you really get the idea of how to use it. Thank you so much man, u rock

[@engi101g_nevarez3]

honestly you have some of the best videos i wish i could have you as my actual proffesor. what ever they pay they need to pay you more

[@diegogallego9370]

You explained it very clear. Now i understand better my book. Thank you!

[@arathi2501]

Thank you!!! Finally understood how to solve this

[@sarahhassan4953]

love this!

[@peilin2428]

thank you sir

[@matthewel-sirafy7266]

hi, could you help in explaining to me how to solve for the laplace of: 3cos(t)U(t-pi)

[@carultch]

Given: 3*cos(t)*u(t - pi) When multiplying a function g(t) by the unit step function shifted by a distance of c to the right, u(t - c), the corresponding Laplace transform is: G(s)*e^(-c*s) This means the given function's Laplace transform will be: L{3*cos(t)} * e^(-pi*s) Now we just need to find L{3*cos(t)}. Since the Laplace transform is a linear operator, we can pull the 3 out in front, and get: 3*L{cos(t)}. The Laplace of cos(t) we an look up directly, which is s/(s^2 + 1). Construct all of the above to get the solution: [3*s/(s^2 + 1)]* e^(-pi*s)