Frequency response of a first-order linear system: the steady-state phase shift between sinusoidal input and output varies between which lower and upper limits (in radians) as excitation frequency goes from very low to very high?

Introduction / Context:First-order systems (e.g., thermal elements, RC low-pass circuits, stirred-tank dynamics) exhibit a characteristic frequency response that includes a magnitude roll-off and a phase lag. Knowing the phase limits is crucial for loop stability assessment and Bode plot interpretation in control engineering.

Given Data / Assumptions:

• Standard first-order transfer function: G(s) = K / (1 + τ s).
• Input is sinusoidal; we consider steady-state sinusoidal response.
• We analyze the phase shift limit as ω → 0 and ω → ∞.

Concept / Approach:The frequency response G(jω) = K / (1 + j ω τ). The phase angle φ(ω) = - arctan(ω τ). As excitation frequency increases, arctan(ω τ) grows from 0 to π/2. Hence, φ(ω) decreases from 0 to -π/2. Therefore the phase shift spans the interval [-π/2, 0], with 0 at low frequency and approaching -π/2 at high frequency.

Step-by-Step Solution:Write G(jω) = K / (1 + jωτ).Compute phase: φ = - arctan(ωτ).Evaluate limits: ω→0 ⇒ arctan(0)=0 ⇒ φ→0; ω→∞ ⇒ arctan(∞)=π/2 ⇒ φ→-π/2.Conclude bounds are -π/2 to 0 radians.

Verification / Alternative check:Bode plots for first-order low-pass elements universally show -45° at ω=1/τ, 0° at low frequency, and -90° at high frequency, confirming the derived limits.

Why Other Options Are Wrong:-∞, π/2: Nonsensical for bounded first-order phase.-π/2, π/2 and 0, π/2: Imply leading phase or positive upper bound not present in lag elements.-π, 0: Too large a lag for first-order; -π is typical of additional dynamics.

Common Pitfalls:Mixing up sign conventions or confusing phase of lead/lag compensators. Always check the transfer function form and remember arctan behavior.

Final Answer:-π/2, 0