Control systems (steady-state error): for a type 1 system (one integrator in the forward path), which statements are correct about steady-state error for standard inputs?

Introduction / Context:
System type (the number of pure integrators) determines tracking performance for standard test inputs. A type 1 system contains one integrator and exhibits characteristic steady-state errors for step, ramp, and acceleration inputs. This question checks conceptual mastery of error constants without requiring algebraic derivation.

Given Data / Assumptions:

• Type 1 means exactly one pole at the origin in the open loop.
• Standard inputs: step (position), ramp (velocity), and parabolic/acceleration.
• Closed-loop is stable so steady-state values are meaningful.

Concept / Approach:
For type 1: position error constant Kp is finite, so step error = 0; velocity error constant Kv is finite, giving finite, non-zero ramp error; acceleration error constant Ka = 0, so error to acceleration grows without bound (infinite). Thus statements about zero step error and non-zero/infinite acceleration error are correct; zero ramp error is incorrect.

Step-by-Step Solution:
Identify system type N = 1 (one integrator).Recall results: step error = 0; ramp error = finite non-zero; acceleration error = infinite.Match options: (a) true, (b) false, (c) true.Choose combined option (a) and (c).

Verification / Alternative check:
Using error constant relations: e_ss(step) = 1/(1+Kp) → 0 for type 1; e_ss(ramp) = 1/Kv → finite; e_ss(acceleration) = ∞ since Ka = 0. This aligns with the qualitative reasoning above.

Why Other Options Are Wrong:
(b) Ramp error is not zero for type 1. (e) Invalid because (a) and (c) hold.

Common Pitfalls:
Mixing up type 0/1/2 behaviors; assuming that additional gain eliminates ramp error without integral action of sufficient order.

Final Answer:
(a) and (c)