Final answer:
The phase margin for the transfer function KG(s) is independent of the natural frequency ωₙ and only depends on the damping ratio ζ. By setting the magnitude of the transfer function equal to 1, the phase margin can be calculated and shown to be independent of ωₙ.
Step-by-step explanation:
To show that for the transfer function KG(s) = w²ₙ / (s² + 2ζωₙs), the phase margin is independent of ωₙ and is given by PM = tan⁻¹(2ζ / √(√(4ζ⁴ + 1) - 2ζ²)), we need to analyze the frequency response of the system.
The phase margin (PM) is the difference in phase between the response and -180 degrees at the frequency where the gain is one (or 0 dB). To get this, we set the magnitude of the open-loop transfer function equal to 1 and solve for phase at that frequency.
To find the frequency for which the magnitude is equal to 1, we solve:
- |KG(jω)| = √((w²ₙ - ω²)^2 + (2ζωₙω)^2) = 1
We then substitute back to determine the phase at this frequency and find the PM.
However, notice that ωₙ is not present in the final PM formula, indicating that the PM is independent of the natural frequency (ωₙ) and is only a function of the damping ratio (ζ).