Question

Steady-state error of the following second-order system for a unit ramp input is \[ \frac{C(s)}{R(s)} = \frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}. \] The steady-state error is:

• A. \(\frac{\zeta}{2\omega_n}\)
• B. \(\frac{2\zeta}{\omega_n}\)
• C. \(\frac{\omega_n}{2\zeta}\)
• D. \(\frac{2\omega_n}{\zeta}\)

Hint:

For a standard Type-1 system, the velocity error coefficient is \(K_v = \frac{\omega_n}{2\zeta}\). Since steady-state error for a unit ramp input is given by \(e_{ss} = \frac{1}{K_v}\), directly reciprocating \(K_v\) gives \(\frac{2\zeta}{\omega_n}\).

Answer 1

The Correct Option is B

Concept: The steady-state error \(e_{ss}\) is computed using the Final Value Theorem of Laplace transforms: \[ e_{ss} = \lim_{s \to 0} s \cdot E(s) \] where \(E(s)\) represents the error signal, defined as the difference between input reference and feedback output. For a unity feedback loop configuration: \[ E(s) = R(s) - C(s) = R(s)\left[1 - \frac{C(s)}{R(s)}\right] \]

Step 1: Substitute the given system transfer function into the error equation. We are given: \[ \frac{C(s)}{R(s)} = \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} \] Thus, the error expression becomes: \[ E(s) = R(s) \left[ 1 - \frac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} \right] \] Taking a common denominator inside the brackets: \[ E(s) = R(s) \left[ \frac{(s^2 + 2\zeta\omega_n s + \omega_n^2) - \omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2} \right] = R(s) \left[ \frac{s^2 + 2\zeta\omega_n s}{s^2 + 2\zeta\omega_n s + \omega_n^2} \right] \]

Step 2: Evaluate for a unit ramp input. For a unit ramp input, the Laplace transform is given by \(R(s) = \frac{1}{s^2}\). Substituting this value: \[ E(s) = \frac{1}{s^2} \cdot \frac{s(s + 2\zeta\omega_n)}{s^2 + 2\zeta\omega_n s + \omega_n^2} = \frac{s + 2\zeta\omega_n}{s(s^2 + 2\zeta\omega_n s + \omega_n^2)} \]

Step 3: Apply the Final Value Theorem. Now, let's determine the value of the steady-state error: \[ e_{ss} = \lim_{s \to 0} s \cdot E(s) = \lim_{s \to 0} s \cdot \left[ \frac{s + 2\zeta\omega_n}{s(s^2 + 2\zeta\omega_n s + \omega_n^2)} \right] \] Canceling out the variable \(s\) from the numerator and denominator: \[ e_{ss} = \lim_{s \to 0} \frac{s + 2\zeta\omega_n}{s^2 + 2\zeta\omega_n s + \omega_n^2} \] Evaluating the limit by substituting \(s = 0\): \[ e_{ss} = \frac{0 + 2\zeta\omega_n}{0 + 0 + \omega_n^2} = \frac{2\zeta\omega_n}{\omega_n^2} = \frac{2\zeta}{\omega_n} \] This expression matches Option (B).