Determine the Laplace transform ofg(t)=2e12−2tu6(t)

The Laplace transform of a function g(t) is defined as:

L{g(t)} = ∫_0^∞ e^(-st)g(t) dt

Given the function g(t) = 2e^(12-2t)u6(t), where u6(t) is the unit step function, we can break this down into two parts: the exponential function and the unit step function.

The Laplace transform of the exponential function 2e^(12-2t) is given by:

L{2e^(12-2t)} = 2 ∫_0^∞ e^(-st)e^(12-2t) dt = 2 ∫_0^∞ e^(12-(s+2)t) dt = 2/(s+2) * e^(12-(s+2)t) |_0^∞ = 2/(s+2) * [0 - e^12] = -2e^12/(s+2)

The unit step function u6(t) delays the function by 6 units, so the Laplace transform of u6(t) is given by:

L{u6(t)} = e^(-6s)

Therefore, the Laplace transform of the given function g(t) = 2e^(12-2t)u6(t) is given by the product of the Laplace transforms of the two parts:

L{g(t)} = L{2e^(12-2t)} * L{u6(t)} = -2e^12/(s+2) * e^(-6s) = -2e^(12-6s)/(s+2)

So, the Laplace transform of the given function is -2e^(12-6s)/(s+2).