Question

The value of the steady state error for first order system, \(\frac{1}{sT+1}\) with Unit Ramp Function will be

• A. \(\frac{1}{T}\)
• B. T
• C. \(\frac{1}{T} e^{\frac{-t}{T}}\)
• D. \(T(1 - e^{\frac{-t}{T}})\)

Hint:

For a Type 0 system (like this one) in a unity feedback loop, the steady-state error for a ramp input is infinite. If you get such a question with finite answers, it likely refers to the limiting difference between the input and the open-loop response.

Answer 1

The Correct Option is B

Note: This question is slightly ambiguous. Standard analysis of a unity feedback system with this transfer function yields an infinite steady-state error. However, the options provided suggest the question is asking for the limiting difference between the input ramp and the open-loop response of the system. Step 1: Define the input and the system's open-loop response.

Input signal: Unit Ramp, \(r(t) = t\), so \(R(s) = \frac{1}{s^2}\).
System Transfer Function: \(G(s) = \frac{1}{sT+1}\).
The system's output (open-loop response) is \(Y(s) = G(s)R(s)\).

Step 2: Calculate the output in the Laplace domain. \[ Y(s) = \frac{1}{sT+1} \cdot \frac{1}{s^2} = \frac{1}{s^2(sT+1)} \]
Step 3: Find the time-domain output \(y(t)\) using inverse Laplace transform (via partial fraction expansion). \[ \frac{1}{s^2(sT+1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{sT+1} \] Solving for the coefficients gives: \(A = -T\), \(B = 1\), \(C = T^2\). \[ Y(s) = \frac{-T}{s} + \frac{1}{s^2} + \frac{T^2}{sT+1} = \frac{-T}{s} + \frac{1}{s^2} + \frac{T}{s+1/T} \] Taking the inverse Laplace transform: \[ y(t) = -T + t + T e^{-t/T} \]
Step 4: Calculate the error function and its steady-state value. The error is the difference between the input and 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 \(t \to \infty\). \[ e_{ss} = \lim_{t \to \infty} (T - T e^{-t/T}) = T - T(0) = T \]