In a linear time-invariant (LTI) system, if the input is x(t) = A sin(ωt), what is the general expression for the steady-state output y(t) in terms of the frequency response G(jω)?

Difficulty: Easy

Correct Answer: |G(jω)| A sin[ωt + ∠G(jω)]

Explanation:


Introduction / Context:
Frequency response analysis in control and signal processing allows prediction of the output of an LTI system given a sinusoidal input. The output retains the same frequency as the input, but its amplitude and phase are modified according to the system's transfer function evaluated at that frequency.


Given Data / Assumptions:

  • Input x(t) = A sin(ωt).
  • System transfer function frequency response: G(jω).
  • Assume linear, time-invariant, stable system with well-defined steady state.


Concept / Approach:
The output steady-state sinusoid is scaled by the magnitude of G(jω) and phase-shifted by its angle. Therefore, y(t) = |G(jω)| A sin(ωt + ∠G(jω)). Frequency does not change, only amplitude and phase.


Step-by-Step Solution:

Represent input as phasor: X(jω) = A ∠0°.Output phasor: Y(jω) = G(jω) × X(jω).Magnitude: |Y| = |G(jω)| A; Phase: ∠Y = ∠G(jω).Convert back to time domain: y(t) = |G(jω)| A sin(ωt + ∠G(jω)).


Verification / Alternative check:

Simulation of sinusoidal response in MATLAB/Scilab yields amplitude gain |G(jω)| and phase shift ∠G(jω), matching the formula.


Why Other Options Are Wrong:

Option (a): φ = tan⁻¹|G(jω)| is meaningless; phase is not tan⁻¹ of magnitude.Option (c): Doubles frequency (2ω), which does not occur in LTI systems.Option (d): A G(jω) sin[…] incorrectly multiplies sinusoid by a complex number directly.Option (e): Cosine form is valid but requires phase redefinition; the standard representation is sine with phase shift.


Common Pitfalls:

Forgetting that LTI systems preserve frequency; only amplitude and phase change.


Final Answer:

|G(jω)| A sin[ωt + ∠G(jω)]

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