Question

The steady-state error due to unit acceleration input for a type 2 system is:

• A. Zero
• B. Infinity
• C. \( \frac{1}{K_a} \)
• D. \( \frac{1}{K_v} \)

Hint:

For a Type 2 system, the steady-state error for an acceleration input is finite and given by \( \frac{1}{K_a} \), whereas for a Type 1 system, it would be infinite.

Answer 1

The Correct Option is C

Step 1: The steady-state error (\( e_{ss} \)) for a given system is determined using the Final Value Theorem: \[ e_{ss} = \lim_{s \to 0} s E(s) \] 
Step 2: For a Type 2 system, the open-loop transfer function contains two poles at the origin, meaning it has two integrators. 
Step 3: The steady-state error for different inputs is determined using error constants: 
- Position error constant \( K_p \) (for step input) 
- Velocity error constant \( K_v \) (for ramp input) 
- Acceleration error constant \( K_a \) (for parabolic input) 
Step 4: For a unit acceleration input (\( R(s) = \frac{1}{s^3} \)), the steady-state error is given by: \[ e_{ss} = \frac{1}{K_a} \] where \( K_a \) is the acceleration error constant. 
Step 5: Since a Type 2 system has two integrators, it can track acceleration inputs with finite error, given by \( \frac{1}{K_a} \).