Control systems: for a type 2 system (with two integrators in the open-loop path), which statements about steady-state error are correct?

Introduction / Context:
System “type” in classical control refers to the number of pure integrators in the open-loop transfer function. This directly determines steady-state error characteristics for standard inputs (step, ramp, and parabolic/acceleration). Type 2 systems contain two integrators and thus excel at tracking low-order inputs with zero error.

Given Data / Assumptions:

• Type 2 implies two integrators in the forward path.
• Steady-state error depends on error constants (position, velocity, acceleration).
• Standard inputs considered: step (position), ramp (velocity), parabolic (acceleration).

Concept / Approach:
For a type N system: step error is zero for N ≥ 1, ramp error is zero for N ≥ 2, and parabolic error is zero for N ≥ 3 (otherwise finite). Consequently, a type 2 system achieves zero steady-state error for step and ramp inputs but exhibits finite non-zero error for parabolic inputs. This is a fundamental result derived from final value considerations and error constant definitions (Kp, Kv, Ka).

Step-by-Step Solution:
Relate system type (number of integrators) to input order (step = 0, ramp = 1, parabolic = 2).Apply the rule: if system type ≥ input order + 1, steady-state error = 0; otherwise finite.For type 2: step (order 0) → zero error; ramp (order 1) → zero error; parabolic (order 2) → finite non-zero error.Thus, all listed statements are correct.

Verification / Alternative check:
Using final value reasoning with appropriate error constants confirms the same results for canonical test inputs.

Why Other Options Are Wrong:
Each individual statement (A–C) is true; claiming otherwise or selecting “none” would contradict standard control theory.

Common Pitfalls:
Confusing system type with closed-loop order; forgetting that more integrators improve low-order tracking but may challenge stability margins.

Final Answer:
All of the above.