I wanna know how can I get Laplace transform by differentiation of this unit step function
$$f(t)= u(t)-u(t-k)+2(u(t-k)-u(t-2k))+3(u(t-2k)-u(t-3k))+\ldots$$
I found that $$F(s) = \frac1s(1+e^{-ks}+(e^{-ks})^2+\ldots)=\frac1{s(1-e^{-ks})}$$
But by differentiation will be equal to zero :/
Any help and is there a specific name for this function