Bode and Nyquist analysis
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| phase margin | bù pha |
| gain crossover | tần số cắt biên độ |
| gain margin | biên lợi ích |
| encirclement | vòng bao |
| Nyquist criterion | tiêu chuẩn Nyquist |
Bode: two lines, two margins
- Each pole adds $-20$ dB/decade and up to $-90°$; each zero the opposite. Phase margin 相位裕度 PM is read at the gain crossover 增益穿越 ($|L| = 1$); gain margin 幅值裕度 GM at the phase crossover ($\angle L = -180°$). Gain crossover is the frequency where the open-loop magnitude has fallen to exactly one.
- The working rule between domains: $\zeta \approx PM/100$. A 50° margin behaves like $\zeta = 0.5$, an overshoot near 16%.
![Straight-line Bode plot of 100/[s(s+10)] with margins annotated](/handout-media/university_electrical_engineering/assets/10-bode-margins.png?v=1793069544)
$L = \dfrac{100}{s(s+10)}$: gain crossover solves $\omega\sqrt{\omega^{2}+100} = 100$, giving $\omega_c = 7.86$ rad/s; phase $-128.2°$ there, so $PM = 51.8°$ and GM is infinite (the phase never reaches $-180°$). The exact closed loop is $\zeta = 0.5$: the rule's 0.52 is a working estimate.
{exercise:uni-ee1046-pm}
A loop has phase margin 50 degrees. The working rule predicts —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
ζ ≈ PM/100 = 0.5, and the second-order overshoot at ζ = 0.5 is e^(−π/√3) = 16.3%. The two domains must tell one story.
Nyquist counts RHP poles exactly
- Plot $L(j\omega)$, count clockwise encirclements 包围 $N$ of the point $-1$, and $Z = N + P$: closed-loop RHP poles = encirclements plus open-loop RHP poles. This is the Nyquist criterion 奈奎斯特准则. An encirclement is a full clockwise loop of the minus-one point by the Nyquist curve as the frequency rises.
- Gain scales the whole curve uniformly, so one shape slides through $-1$ as $K$ grows.
![Nyquist plot of K/[s(s+2)(s+4)] for three gains](/handout-media/university_electrical_engineering/assets/10-nyquist-criterion.png?v=1793069544)
The running plant at $\omega = \sqrt{8} = 2.83$ rad/s has $|L| = K/48$. At $K = 20$ the curve passes right of $-1$: stable. At $K = 48$ it passes through $-1$: marginal. At $K = 96$ it wraps twice, so $Z = 2 + 0 = 2$: two closed-loop RHP poles, agreeing with Routh.
{exercise:uni-ee1046-nyquist}
The loop K/[s(s+2)(s+4)] at K = 96 shows the Nyquist curve wrapping twice clockwise around −1. The closed loop has —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Z = N + P with N = 2 clockwise encirclements and P = 0 (no open-loop RHP poles) gives two closed-loop RHP poles. Routh on s³+6s²+8s+96 agrees.
Margins belong to the loop
- Read PM and GM off the open-loop $L$, never off the closed-loop response.
- Tie the story together: for $L = \dfrac{20}{s(s+2)(s+4)}$, GM $= 48/20 = 2.4$ (7.6 dB) and $PM = 25.7°$ at $\omega_c = 1.73$ rad/s; the rule predicts $\zeta \approx 0.26$ and simulation gives 48% overshoot. Three methods, one answer.
{exercise:uni-ee1046-gm}
For L = 20/[s(s+2)(s+4)], the phase crosses −180° at ω = √8 where |L| = 20/48. The gain margin is —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
GM is the factor you can multiply the gain by before instability: 48/20 = 2.4, and 20·log₁₀2.4 = 7.6 dB. At the same gain PM = 25.7°, predicting ζ ≈ 0.26 and a 48% overshoot in simulation.