Laplace Transform Time Shift problem

[NewtonianAlch]

Homework Statement

Determine the Laplace transform:

g(t) = 2*e^{-4t}u(t-1)

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}}

 

[rude man]

Shifting is tricky.

You have used the formula L{f(t-a)U(t-a)} = e^-asL{f(t)}.

You need the formula for L{f(t)U(t-a)}.

 

[RoshanBBQ]

Homework Statement

Determine the Laplace transform:

g(t) = 2*e^{-4t}u(t-1)

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.