Note: This question presents a slight ambiguity. A standard unity feedback analysis of the provided transfer function results in an infinite steady-state error. However, the answer choices suggest the question seeks the limiting difference between the input ramp and the system's open-loop response.Step 1: Define the input signal and the open-loop system response.
Input signal (Unit Ramp): \(r(t) = t\), thus \(R(s) = \frac{1}{s^2}\).
System Transfer Function: \(G(s) = \frac{1}{sT+1}\).
Open-loop output: \(Y(s) = G(s)R(s)\).
Step 2: Determine the output in the s-domain.\[ Y(s) = \frac{1}{sT+1} \cdot \frac{1}{s^2} = \frac{1}{s^2(sT+1)} \]
Step 3: Obtain the time-domain output \(y(t)\) using the inverse Laplace transform (via partial fraction decomposition).\[ \frac{1}{s^2(sT+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{sT+1} \]Solving for the coefficients yields: \(A = -T\), \(B = 1\), \(C = T^2\).\[ Y(s) = \frac{-T}{s} + \frac{1}{s^2} + \frac{T^2}{sT+1} = \frac{-T}{s} + \frac{1}{s^2} + \frac{T}{s+1/T} \]Applying the inverse Laplace transform:\[ y(t) = -T + t + T e^{-t/T} \]
Step 4: Determine the error function and its steady-state value.The error is defined as the input minus the output:\[ e(t) = r(t) - y(t) = t - (-T + t + T e^{-t/T}) = T - T e^{-t/T} \]The steady-state error is the limit of \(e(t)\) as \(t\) approaches infinity.\[ e_{ss} = \lim_{t \to \infty} (T - T e^{-t/T}) = T - T(0) = T \]