Final answer:
The steady-state output of a system given the transfer function G(s) = 20/(s+5) with a unit step input can be found using the final value theorem, and the steady-state gain is calculated as G(0) = 4.
Step-by-step explanation:
To find the steady-state output of a system excited by a unit step input, we look at the final value theorem in Laplace transforms, which states that the steady-state value of the function as time approaches infinity (i.e., steady-state value) is the limit of s multiplied by the Laplace transform of the function as s approaches zero.
The given transfer function is G(s) = 20/(s+5). Applying a unit step input to the system, U(s), which is 1/s in Laplace form, the output Y(s) becomes G(s) multiplied by U(s), or Y(s) = G(s)*U(s) = (20/s) * (1/(s+5)).
To find the steady-state gain, we set s to 0 in G(s), which yields G(0) = 20/5 = 4. Thus, the steady-state gain of the system is 4.