Difficulty: Medium
Correct Answer: 2 A T
Explanation:
Introduction / Context:
A “dynamic error” question checks whether you can relate standard test inputs (step, ramp, impulse) to the transient/steady tracking error of common instrument models. A critically damped second-order instrument (two equal first-order lags) responding to a ramp input x(t) = A*t exhibits a constant lag proportional to the slope A and the sum of its time constants.
Given Data / Assumptions:
Concept / Approach:
For a first-order element with time constant τ, the steady ramp tracking error is A * τ. Two first-order elements in series (critically damped, equal τ = T) yield a steady ramp error equal to A * (T + T) = 2 A T. Intuitively, each lag adds its own share of delay proportional to the ramp slope; the errors add linearly for cascaded lags.
Step-by-Step Solution:
Represent the critically damped second-order system as G(s) = 1 / (1 + T s)^2.For input X(s) = A / s^2 (ramp), steady ramp error of one lag is A*T.Two identical lags in series → total steady ramp error = A*T + A*T = 2 A T.Therefore, dynamic error = 2 A T.
Verification / Alternative check:
Using final value theorem on the error transfer function E(s) = X(s) [1 − G(s)]; with G(s) = 1/(1+T s)^2 and X(s) = A/s^2, the limit t→∞ of e(t) evaluates to 2 A T, confirming the time-constant additivity rule for ramp inputs across cascaded first-order lags.
Why Other Options Are Wrong:
0.5 A T / A T / 1.5 A T: Underestimate the cumulative lag from two lags.A / T: Has wrong dimensions and ignores ramp-lag behavior.
Common Pitfalls:
Mixing up step error (which goes to zero for stable systems) with ramp error (nonzero for finite-order lags), or forgetting that for cascaded first-order lags the ramp errors add.
Final Answer:
2 A T
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