Final Answer:
The steady-state error of the closed-loop system to a unit ramp input is -1/2.
Step-by-step explanation:
Step 1: Closed-loop transfer function
The closed-loop transfer function is:
T(s) = G(s) / (1 + G(s)H(s))
where G(s) is the open-loop transfer function and H(s) is the feedback function. In a unity negative feedback system, H(s) = -1.
Therefore, the closed-loop transfer function simplifies to:
T(s) = G(s) / (1 - G(s))
Substituting the given loop transfer function L(s) for G(s):
T(s) = L(s) / (1 - L(s))
= s²(s + 15) / (s²(s + 15) + s + 10)
Step 2: Response to a unit ramp input
The Laplace transform of a unit ramp input is:
R(s) = 1/s²
The output of the closed-loop system, Y(s), is the product of the input R(s) and the closed-loop transfer function T(s):
Y(s) = T(s) * R(s)
= (s²(s + 15) / (s²(s + 15) + s + 10)) * (1/s²)
= s²(s + 15) / (s⁴ + 15s³ + 10s²)
Step 3: Steady-state error
The steady-state error is the difference between the final value of the input and the final value of the output. For a unit ramp input, the final value is the slope of the ramp, which is 1.
To calculate the steady-state error, we need to find the final value of the output. We can do this by finding the limit of Y(s) as s approaches 0.
lim_(s→0) sY(s) = lim_(s→0) s * (s²(s + 15) / (s⁴ + 15s³ + 10s²))
= 15/10
= 3/2
Therefore, the final value of the output is 3/2. Since the final value of the input is 1, the steady-state error is:
Steady-state error = 1 - 3/2 = -1/2
The steady-state error is -1/2. This means that the system will always be 0.5 units below the desired output level in the steady state.