Frequency response of a first-order system (τ = 1 min) A first-order process with time constant τ = 1 minute is tested at input frequency ω = 1 rad/min. What is the steady-state phase shift (in degrees) between output and sinusoidal input?

Difficulty: Easy

Correct Answer: -45°

Explanation:


Introduction / Context:
First-order dynamics are ubiquitous in process units (tanks, heat exchangers). Their frequency response is simple yet powerful: knowing the phase shift and attenuation at a given frequency helps in controller tuning and estimating closed-loop behaviour. This question focuses on computing the phase at a specific frequency using the well-known formula for a first-order lag.


Given Data / Assumptions:

  • Transfer function form: G(jω) = K / (1 + jωτ).
  • Time constant τ = 1 min; test frequency ω = 1 rad/min.
  • We assume standard steady-state sinusoidal response.


Concept / Approach:
For G(jω) = K / (1 + jωτ), the phase is φ(ω) = −tan^−1(ωτ). This is independent of K. Substituting τ = 1 and ω = 1 gives φ = −tan^−1(1) = −45°. The negative sign indicates the output lags the input, characteristic of a causal lag element.


Step-by-Step Solution:

Write the phase formula: φ = −tan^−1(ωτ).Insert values: ωτ = 1 * 1 = 1.Compute: tan^−1(1) = 45°.Attach sign: output lags → φ = −45°.


Verification / Alternative check:
The magnitude at this frequency is |G(j1)| = K / √(1 + 1^2) = K/√2, confirming a −3 dB point when ω = 1/τ. The corner frequency 1/τ = 1 rad/min also typically corresponds to a −45° phase, consistent with the result.


Why Other Options Are Wrong:

45° — sign error; the lag is negative.−90° — would be the asymptotic phase at very high frequency for a first-order lag, not at ω = 1/τ.−180° — never reached by a single first-order lag.0° — only at ω → 0.


Common Pitfalls:
Forgetting the negative sign (lag), or confusing the break frequency with the high-frequency asymptote.


Final Answer:
−45°

More Questions from Process Control and Instrumentation

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion