Bode and Nyquist analysis
導入済み| English | 意味 |
|---|---|
| phase margin | 位相余裕 |
| gain crossover | ゲインクロスポイント |
| gain margin | ゲイン余裕 |
| phase crossover | phase crossover |
| Nyquist criterion | ナイキスト判別法 |
| encirclements | encirclements |
Measure, don't guess
- In industry you rarely know where the poles are. You measure a frequency response instead.
- Margins say how close a loop is to instability, in decibels and degrees. Every datasheet for a control loop quotes them.
The Bode plot
- A Bode plot draws the loop gain $L(j\omega)$ as magnitude in dB and phase in degrees, against log frequency.
- Each pole adds a $-20$ dB/decade slope and up to $-90°$ of phase; each zero does the opposite.
Above both of its pole frequencies, a loop with two poles and no zeros falls at what slope, in dB per decade?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Each pole adds −20 dB per decade, so two poles give −40.
Two margins
- Phase margin 相位裕度: at the gain crossover 增益穿越 ($|L| = 1$), how far the phase is above $-180°$.
- Gain margin 幅值裕度: at the phase crossover 相位穿越 ($\angle L = -180°$), how far $|L|$ is below 1. The working rule is $\zeta \approx PM/100$.
A loop has a phase margin of 60°. Estimate the closed loop's damping ratio. Give 1 decimal place.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
$\zeta \approx PM/100 = 0.6$. It is a working rule, good to a few hundredths for typical loops.
The Nyquist criterion
- The Nyquist criterion 奈奎斯特准则 plots $L(j\omega)$ and counts clockwise encirclements 包围 $N$ of the point $-1$.
- Then $Z = N + P$ closed-loop poles lie in the right half plane, where $P$ counts the open-loop RHP poles.
The open loop has one RHP pole (P = 1), and the Nyquist plot circles −1 once clockwise (N = 1). How many RHP poles does the closed loop have?
Difficulty 2/5 · 140 coins + a small bonus on first completion today
$Z = N + P = 1 + 1 = 2$. To stabilise this plant, the plot must circle −1 once anticlockwise (N = −1).
Worked example: margins by arithmetic
- $L = \dfrac{100}{s(s+10)}$: $|L| = 1$ at $\omega_{c} = 7.86$ rad/s. The phase there is $-128.2°$, so $PM = 51.8°$; the gain margin is infinite.
- The closed loop $\dfrac{100}{s^{2} + 10s + 100}$ has $\zeta = 0.5$ exactly; the rule predicted 0.52.
![Straight-line Bode plot of 100/[s(s+10)] with the phase margin marked.](/handout-media/university_electrical_engineering/assets/10-bode-margins.png?v=1793069544)
For L = 4/[s(s + 2)], the gain crossover is at 1.572 rad/s. What is the phase margin, in degrees? Give 1 decimal place.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
The phase is $-90° - \arctan(1.572/2) = -128.2°$, so $PM = 51.8°$. Like the worked example, this loop has $\zeta = 0.5$.
Worked example: one plant, three ways
- $L = \dfrac{20}{s(s+2)(s+4)}$: the phase is $-180°$ at 2.83 rad/s, where $|L| = 20/48$. So $GM = 2.4$ (7.6 dB), and $PM = 25.7°$: 48% overshoot.
- The Nyquist curve crosses the real axis at $-K/48$: right of $-1$ at $K = 20$, through it at 48, and twice round it at 96, giving 2 RHP poles. Routh agrees.
![Nyquist plots 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 K/[s(s + 2)(s + 4)] runs at K = 12. What is its gain margin, in dB? Give the nearest whole number.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
$GM = 48/12 = 4$, and $20\log_{10}4 = 12.0$ dB.
Three common slips
- Margins from the closed loop. They belong to the loop gain $L$, before the loop closes.
- Mixed-up crossovers. PM goes with the gain crossover; GM with the phase crossover.
- Forgetting $P$. If the plant has RHP poles, $N$ alone is not the answer.
Phase and gain margins are read from the closed-loop frequency response.
Difficulty 1/5 · 100 coins + a small bonus on first completion today
They belong to the loop gain L(jω), before the loop closes. That is what makes them a prediction.
Match each quantity to where you read it.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
Two crossovers give the two margins; Nyquist's count gives the exact answer.
Recap
- PM at $|L| = 1$; GM at $\angle L = -180°$.
- $\zeta \approx PM/100$ carries margins into the time domain.
- Nyquist: $Z = N + P$, counted around $-1$.