Questions: Sinusoidal Response: Magnitude and Phase Angle

Question 1

A stable linear system with transfer function G(s) is driven by u(t) = 5 sin(3t). After all transients have died out, what is the form of the steady-state output?

• A. A sum of sinusoids at multiple frequencies, because the system's poles contribute natural frequencies to the output
• B. 5|G(3j)| sin(3t + ∠G(3j)) — a sinusoid at the same frequency, scaled in amplitude and shifted in phase
• C. A sinusoid at a frequency between 3 rad/s and the system's natural frequency
• D. 5 sin(3t) unchanged, because a stable system passes sinusoids without modification

Answer: (B)

For any stable linear time-invariant system, the steady-state response to a sinusoidal input is always a sinusoid at the same frequency. The system can only scale the amplitude by |G(jω)| and shift the phase by ∠G(jω) — it cannot generate new frequencies. The transient response (decaying exponentials from the poles) disappears as t → ∞. Option A describes transient behavior, not steady state. Option D is wrong because the system can attenuate or amplify the amplitude.

Question 2

For the transfer function G(s) = 10/(s + 10), what happens to the magnitude |G(jω)| as the input frequency ω increases from 0 to infinity?

• A. It stays constant at 1, because both numerator and denominator have magnitude 10 at DC
• B. It increases from 0 to 1, because higher frequencies excite the system more strongly
• C. It decreases from 1 toward 0, because the denominator magnitude grows with ω while the numerator stays constant at 10
• D. It oscillates, because the pole at s = −10 creates resonance near ω = 10

Answer: (C)

At ω = 0: |G(0)| = 10/10 = 1. As ω → ∞: |G(jω)| = 10/√(ω² + 100) → 0. This is a first-order low-pass filter that passes low-frequency sinusoids with gain ≈ 1 and attenuates high-frequency ones. The magnitude rolls off because the denominator magnitude |jω + 10| = √(ω² + 100) grows without bound while the numerator stays at 10. Any system with more poles than zeros exhibits this high-frequency attenuation.

Question 3

For a linear time-invariant system, the steady-state output in response to a sinusoidal input is always a sinusoid at the same frequency as the input.

• T. True
• F. False

Answer: True

Sinusoids are eigenfunctions of LTI systems: they pass through and emerge as sinusoids at the same frequency, only scaled in amplitude and shifted in phase. This eigenfunction property is why frequency-domain analysis is so powerful — you can fully characterize a system's behavior at any frequency with a single complex number G(jω). New frequencies can only be generated by nonlinear systems, where harmonic distortion occurs.

Question 4

To find the steady-state sinusoidal response of a system, you should solve the full differential equation, since evaluating G(jω) mainly gives an approximation.

• T. True
• F. False

Answer: False

Evaluating G(jω) is exact, not an approximation. The steady-state sinusoidal response is given precisely by A|G(jω)| sin(ωt + ∠G(jω)) — no differential equation solving is required. Solving the full differential equation would also yield transient terms (decaying exponentials), but for the steady-state sinusoidal response alone, evaluating G at s = jω is both sufficient and exact.

Question 5

Why is it sufficient to substitute s = jω into the transfer function G(s) to completely determine the steady-state response to any sinusoidal input at frequency ω?

Model answer: Complex exponentials e^(jωt) are eigenfunctions of linear time-invariant systems: when e^(jωt) is the input, the steady-state output is G(jω)·e^(jωt) — the same exponential scaled by the complex number G(jω). Since sin(ωt) = Im(e^(jωt)), the steady-state sinusoidal response inherits this scaling directly. The magnitude |G(jω)| scales the amplitude; the argument ∠G(jω) shifts the phase. The transfer function captures the complete input-output relationship in the Laplace domain, and the imaginary axis s = jω corresponds exactly to sinusoidal (eternal) excitation.

This eigenfunction property is why frequency-domain methods are so powerful in control design. Rather than solving a new differential equation for each input frequency, you evaluate a complex number. Sweeping ω from 0 to ∞ and plotting |G(jω)| in dB and ∠G(jω) in degrees traces out the Bode plot — a complete picture of how the system processes every sinusoidal frequency.