Control systems (steady-state error): for a type 0 system (no pure integrator), which statement correctly describes steady-state error to standard inputs?

Introduction / Context:
Knowing how system type affects tracking error is crucial for selecting appropriate controllers. A type 0 system contains no integrator in the open loop and therefore exhibits limited tracking performance compared to higher-type systems.

Given Data / Assumptions:

• Type 0: open-loop transfer has no pole at the origin.
• Standard test inputs: step, ramp, acceleration (parabolic).
• Closed-loop stability is assumed.

Concept / Approach:
For type 0, the position error constant Kp is finite, so step error is finite and non-zero (not zero). The velocity error constant Kv = 0, thus ramp error diverges (infinite). Acceleration input is even more demanding; error is unbounded/infinite. Hence among the options, only the statement asserting non-zero (infinite) error to acceleration is generally true.

Step-by-Step Solution:
Recall: type 0 → step error finite non-zero; ramp error infinite; acceleration error infinite.Match options: (a) false, (b) false, (c) true.Select (c).

Verification / Alternative check:
Using error constants: e_ss(step) = 1/(1+Kp) → finite; e_ss(ramp) = ∞ because Kv = 0; e_ss(acceleration) = ∞ because Ka = 0. This confirms the classification above.

Why Other Options Are Wrong:
(a) Step error is not zero for type 0. (b) Ramp error is not zero; it diverges. (d) Combining (a) with anything cannot be right. (e) Invalid because (c) is correct.

Common Pitfalls:
Assuming high proportional gain alone eliminates step error entirely; without integral action, offset typically remains.

Final Answer:
Error for an acceleration input is always non-zero (in practice unbounded/infinite)