View Single Post
P: 280
 Quote by NewtonianAlch 1. The problem statement, all variables and given/known data Determine the Laplace transform: g(t) = 2*e$^{-4t}$u(t-1) 3. The attempt at a solution Essentially we're told for a time shift we multiply the Laplace transform pair of the function (without the delay) by e$^{-as}$ So here a = 1 (for the delay) The Laplace transform for e$^{-4t}$ is $\frac{1}{s + 4}$ Multiplying we should get e$^{-s}$($\frac{2}{s + 4}$) However the answer is e$^{-s}$*($\frac{2}{s + 4}$) * $\frac{1}{e^{4}}$
It's because you must be in the form
$$e^{-4(t-a)}u(t-a)$$
Take a look at your exponential. It isn't time-shifted by a to the right. What you can do is use the fact that exponentials multiplied add their exponents and the fact that e^b/e^b = 1, so you can multiply by it without changing your values. So you choose e^b so that you can add its exponents and arrive to the needed shift. The e^b in the denominator is factored outside of the linear inverse Laplace operator and you go from there.