Difficulty: Easy
Correct Answer: e^{-τ s}
Explanation:
Introduction / Context:
Frequency-domain fingerprints help identify elementary dynamic elements without time-domain tests. A pure time delay (dead time) has a distinctive Bode plot: flat magnitude and linearly increasing phase lag with frequency. This question asks you to map that signature to the correct transfer function.
Given Data / Assumptions:
Concept / Approach:
A pure time delay of amount τ has transfer function G(s) = e^{−τ s}. In the frequency domain, G(jω) has magnitude |G| = |e^{−j ω τ}| = 1 for all ω and phase ∠G = −ω τ (radians), a straight line through the origin with slope −τ. By contrast, first-order lag 1/(τ s + 1) shows decreasing magnitude and asymptotic −90° phase; second-order terms add curvature and possible resonance; an integrator K/s has magnitude decreasing 20 dB/decade and −90° phase, not constant magnitude.
Step-by-Step Solution:
Verification / Alternative check:
Plot Bode: delay line shows 0 dB across, with a straight-line phase lag increasing with ω; others do not.
Why Other Options Are Wrong:
Common Pitfalls:
Approximating delay by extra poles; only e^{−τ s} preserves unit magnitude.
Final Answer:
e^{-τ s}
Discussion & Comments