Difficulty: Easy
Correct Answer: Transportation lag
Explanation:
Introduction / Context:A transportation lag (dead time) element delays a signal by a fixed time L without changing its shape. In the frequency domain, dead time contributes phase lag that increases with frequency, but it does not attenuate amplitude in the ideal mathematical model.Given Data / Assumptions:
Concept / Approach:The frequency response of e^(−sL) evaluated at s = jω is e^(−jωL) whose magnitude is |e^(−jωL)| = 1 and whose phase is −ωL radians. Thus, the amplitude ratio (output amplitude / input amplitude) is unity for all frequencies, while the phase lag grows linearly with frequency.Step-by-Step Solution:
Start with G(jω) = e^(−jωL).Compute magnitude: |G(jω)| = 1.Compute phase: ∠G(jω) = −ωL.Conclude amplitude ratio is 1 for a pure dead time.Verification / Alternative check:Bode plots for dead time show a flat 0 dB magnitude line (AR = 1) and a phase that decreases linearly with frequency.
Why Other Options Are Wrong:
First/second order systems: their magnitudes vary with frequency (roll-off or resonance); not constant 1.None of these: incorrect because transportation lag matches the property.Common Pitfalls:Confusing practical dead time (with filters and measurement noise) with the ideal mathematical element; in practice, some attenuation can appear, but in theory AR = 1.
Final Answer:Transportation lag
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