Difficulty: Easy
Correct Answer: Gets attenuated (magnitude decreases)
Explanation:
Introduction / Context: The frequency response of linear time-invariant (LTI) systems is foundational for control and signal processing. A first-order process behaves like a low-pass filter. Understanding what such a system does to a sinusoidal input—without solving differential equations every time—helps predict attenuation and phase lag across frequencies.
Given Data / Assumptions:
Concept / Approach: For LTI systems, a sinusoidal input at frequency ω produces a sinusoidal output at the same frequency, scaled in magnitude by |G(jω)| and shifted in phase by ∠G(jω). For a first-order low-pass, |G(jω)| = K / sqrt(1 + (ωτ)^2), which is ≤ K, showing attenuation that increases with ω. The frequency does not change; only amplitude is reduced and phase lags.
Step-by-Step Solution:
Represent the input as sin(ωt) and evaluate the steady-state output using G(jω).Compute magnitude ratio: |G(jω)| = K / sqrt(1 + (ωτ)^2) → less than or equal to K.Conclude amplitude decreases with increasing ω; frequency remains ω (unchanged).Verification / Alternative check: Bode plots of a first-order system show a 0 dB plateau at low ω and a −20 dB/decade roll-off past the corner at 1/τ, confirming attenuation at higher frequencies.
Why Other Options Are Wrong:
Frequency increases/decreases — LTI systems do not change input frequency.Gets amplified — a passive first-order low-pass does not amplify; it attenuates above the corner.Frequency doubles while amplitude unchanged — violates linear frequency invariance.Common Pitfalls: Confusing transient behaviour with steady-state sinusoidal response; only amplitude and phase change at the same frequency in steady state.
Final Answer: Gets attenuated (magnitude decreases)
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