Reply to Ackbeet @ 13

Sorry, I missed a 2 out of my expression.

I should have put 1 - H(t-1) + 2H(t-2).

The 1 I admit is lazy, H(t) would be better, but since the Laplace transform is defined for zero to infinity they are in effect the same aren't they ? (They have the same LT).

Can you confirm that x(t) is defined as being equal to 1 for 0<=t<=1, zero for 1<t<2 and 2 for t>=2 ?