A system has a transfer function of 1/(s + 2). What will be its output as a function of time when it is subject to a step input of 5V

The system's transfer function is given as H(s) = 1/(s + 2).

A step input can be represented in the Laplace domain as U(s) = 5/s.

The output in the Laplace domain is given by the product of the transfer function and the input, i.e., Y(s) = H(s) * U(s).

Substituting the given values, we get Y(s) = (1/(s + 2)) * (5/s) = 5/(s*(s + 2)).

To find the output as a function of time, we need to perform an inverse Laplace transform on Y(s).

The inverse Laplace transform of 5/(s*(s + 2)) can be found using partial fraction decomposition.

This gives us Y(s) = 5/s - 5/(s + 2).

Taking the inverse Laplace transform of each term separately, we get y(t) = 5 - 5e^(-2t) for t > 0.

So, the output of the system as a function of time when it is subject to a step input of 5V is y(t) = 5 - 5e^(-2t) for t > 0.