Question

The steady state error for a type 1 system with input of \[ \frac{At^2}{2}u(t) \] is

• A. 0
• B. \(\frac{1}{2}\)
• C. 1
• D. \(\infty\)

Hint:

Steady-state error for unity feedback systems: \[ \begin{array}{|c|c|c|c|} \hline \text{System Type} & \text{Step} & \text{Ramp} & \text{Parabola} \hline 0 & \text{Finite} & \infty & \infty 1 & 0 & \text{Finite} & \infty 2 & 0 & 0 & \text{Finite} \hline \end{array} \] A Type-1 system always has infinite steady-state error for a parabolic input.

Answer 1

The Correct Option is D

Concept: The input \[ r(t)=\frac{At^2}{2}u(t) \] is a parabolic input. The steady-state error of a unity feedback system depends on the system type and the corresponding static error constant.

Step 1: Identify the input type.
Given, \[ r(t)=\frac{At^2}{2}u(t) \] which is a parabolic input. Its Laplace transform is \[ R(s)=\frac{A}{s^3}. \]

Step 2: Recall steady-state error for parabolic input.
For a unity feedback system, \[ e_{ss}=\frac{A}{K_a} \] where \[ K_a=\lim_{s\to0}s^2G(s). \]

Step 3: Determine \(K_a\) for a Type-1 system.
A Type-1 system contains one pole at the origin. Therefore, \[ K_a=0. \]

Step 4: Compute the steady-state error.
Substituting, \[ e_{ss}=\frac{A}{0} \] which becomes unbounded. Hence, \[ e_{ss}=\infty. \]

Step 5: Final Answer.
A Type-1 system cannot track a parabolic input with finite steady-state error. Therefore, \[ \boxed{e_{ss}=\infty} \] and \[ \boxed{\text{Correct Option (D)}} \]