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For a transport lag element with frequency response G(jω) = e^(-jωT), what is the magnitude of the transfer function at all frequencies?

Difficulty: Easy

0
1
10
0.5
Depends on ωT

Correct Answer: 1

Explanation:

Introduction / Context:
Transport lag, also called dead-time, is common in process control systems where there is a pure time delay between input and output. Its frequency-domain representation is G(jω) = e^(-jωT). Understanding its magnitude and phase properties is critical in stability and frequency response analysis.

Given Data / Assumptions:

Transfer function G(jω) = e^(-jωT).
T = transport delay (positive constant).
ω = angular frequency.

Concept / Approach:
The exponential form e^(-jωT) has magnitude |e^(-jωT)| = 1 for all real values of ω and T, because the exponential is a purely imaginary phase shift (unit magnitude). Its phase is −ωT radians, while its amplitude remains constant.

Step-by-Step Solution:

Write G(jω) = cos(ωT) − j sin(ωT).Magnitude = √(cos²(ωT) + sin²(ωT)) = √1 = 1.Thus |G(jω)| = 1 independent of ω and T.

Verification / Alternative check:

Euler's identity confirms magnitude = 1. Simulation of Bode plot shows 0 dB flat magnitude line across all frequencies, with linearly decreasing phase.

Why Other Options Are Wrong:

0: implies no transmission, false. 10 or 0.5: arbitrary non-unity gain, false. 'Depends on ωT': wrong, only phase depends on ωT, magnitude does not.

Common Pitfalls:

Confusing magnitude with phase. Thinking dead-time attenuates magnitude—it only delays in time/phase.

For a transport lag element with frequency response G(jω) = e^(-jωT), what is the magnitude of the transfer function at all frequencies?

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