时域响应
引入| English | 释义 | 拼音 |
|---|---|---|
| overshoot/ˌəʊvəˈʃuːt/ | 超调量 | chāo diào liàng |
| step response/step rɪˈspɒns/ | 阶跃响应 | jiē yuè xiǎng yìng |
| peak time | 峰值时间 | fēng zhí shí jiān |
| settling time | 调节时间 | tiáo jié shí jiān |
| steady-state error | 稳态误差 | wěn tài wù chā |
| system type | 系统类型 | xì tǒng lèi xíng |
| velocity error constant | 速度误差常数 | sù dù wù chā cháng shù |
| rise time | 上升时间 | shàng shēng shí jiān |
"Settles in half a second"
- A drone maker promises: settles within half a second, overshoots less than 10%. Those are time-domain specifications.
- This lesson is the dictionary between pole positions and those numbers.
The second-order prototype
- $G(s) = \dfrac{\omega_{n}^{2}}{s^{2} + 2\zeta\omega_{n}s + \omega_{n}^{2}}$ has poles at $-\zeta\omega_{n} \pm j\omega_{n}\sqrt{1 - \zeta^{2}}$.
- Their distance from the origin is $\omega_{n}$, and $\zeta$ is the cosine of their angle. A ray from the origin is a line of constant damping.
A pole pair has ω_n = 6 rad/s and ζ = 0.5. What is the real part of the poles?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
The real part is $-\zeta\omega_{n} = -0.5 \times 6 = -3$.
Four specifications
- After a unit step response 阶跃响应: overshoot 超调量 $M_{p} = e^{-\pi\zeta/\sqrt{1-\zeta^{2}}}$ and peak time 峰值时间 $t_{p} = \dfrac{\pi}{\omega_{d}}$, with $\omega_{d} = \omega_{n}\sqrt{1 - \zeta^{2}}$.
- Settling time 调节时间 $t_{s} \approx \dfrac{4}{\zeta\omega_{n}}$ for a 2% band, or $\dfrac{3}{\zeta\omega_{n}}$ for 5%.
What overshoot, in %, does ζ = 0.7 give? Give 1 decimal place.
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
$M_{p} = e^{-\pi \times 0.7/0.714} = e^{-3.08} = 0.046$, so 4.6%.
A loop has ζ = 0.5 and ω_n = 8 rad/s. Estimate its 2% settling time, in s.
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
$t_{s} \approx 4/(\zeta\omega_{n}) = 4/4 = 1$ s. The 5% estimate would be 0.75 s.
Steady-state error
- Count the integrators in the open loop: that is the system type 系统类型. A type-1 loop has zero steady-state error 稳态误差 to a step.
- Its velocity error constant 速度误差常数 $K_{v} = \lim_{s \to 0} sG(s)$ sets the ramp error $1/K_{v}$.
The open loop is 10/[s(s + 5)]. What steady-state error does a unit ramp leave?
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$K_{v} = \lim sG = 10/5 = 2$, so the ramp error is $1/2 = 0.5$. The step error is zero: the loop is type 1.
Worked example: four numbers from two
- $\omega_{n} = 4$, $\zeta = 0.5$: poles $-2 \pm j3.46$. $M_{p} = e^{-1.814} = 16.3\%$ and $t_{p} = \pi/3.46 = 0.907$ s.
- $t_{s} \approx 4/(0.5 \times 4) = 2.0$ s for 2%, or 1.5 s for 5%. The 10–90% rise time 上升时间 is 0.41 s.
For ω_n = 4 rad/s and ζ = 0.5, which frequency gives the peak time?
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The response rings at the damped frequency, and the first peak comes half a cycle in, at $\pi/\omega_{d}$.
Worked example: choose the gain
- $G = \dfrac{K}{s(s+2)}$ in unity feedback gives $s^{2} + 2s + K$, so $\zeta = 1/\sqrt{K}$. For $\zeta = 0.45$, $K = 4.94$ and $\omega_{n} = 2.22$ rad/s.
- Then $M_{p} = 20.5\%$, $t_{p} = 1.58$ s and $t_{s} \approx 4.0$ s. $K_{v} = 4.94/2 = 2.47$, so a ramp leaves an error of 0.405.
For G = K/[s(s + 2)] in unity feedback, what K gives ζ = 0.5?
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
$\zeta = 1/\sqrt{K} = 0.5$ gives $K = 4$, with $\omega_{n} = 2$ rad/s and 16.3% overshoot.
Three common slips
- $\omega_{n}$ for $\omega_{d}$. Peak time uses the damped frequency $\omega_{d}$.
- $t_{s}$ without its band. 2% and 5% differ by a factor of 4/3.
- Formulas without a dominant pair. Extra poles or zeros near the pair change the response.
Raising K in the loop K/[s(s + 2)] makes the step response ring more.
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$\zeta = 1/\sqrt{K}$ falls as $K$ rises, so the overshoot grows. The ramp error $2/K$ falls: the classic trade-off.
Recap
- Poles in, specifications out: $M_{p}$, $t_{p}$, $t_{s}$.
- A ray from the origin is constant $\zeta$, so constant overshoot.
- Count integrators for the type; $1/K_{v}$ is the ramp error.