传递函数与方框图
引入| English | 释义 | 拼音 |
|---|---|---|
| block diagram | 方框图 | fāng kuāng tú |
| transfer function | 传递函数 | chuán dì hán shù |
| feedback/ˈfiːdbæk/ | 反馈 | fǎn kuì |
| characteristic equation/ˌkærɪktəˈrɪstɪk ɪˈkweɪʒn/ | 特征方程 | tè zhēng fāng chéng |
| damping ratio | 阻尼比 | zǔ ní bǐ |
Systems as boxes and arrows
- A poster of a car's cruise control is a block diagram 方框图, not a page of equations. Engineers talk about systems this way.
- Each box holds a transfer function 传递函数: output over input in the s-domain. Three rules turn any diagram into one box.
Three rules
- Series: multiply, $G = G_{1}G_{2}$. Parallel: add, $G = G_{1} + G_{2}$.
- Feedback 反馈: divide by the loop, $T = \dfrac{G}{1 + GH}$.
G₁ = 2/(s + 1) and G₂ = 3/(s + 2) are in series. What is the DC gain (s = 0) of the pair?
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Series multiplies: $G_{1}(0)G_{2}(0) = 2 \times 1.5 = 3$.
Where the feedback rule comes from
- The error is $e = r - Hy$ and the output is $y = Ge$. So $y = G(r - Hy)$, which gives $y(1 + GH) = Gr$.
- $1 + GH = 0$ is the characteristic equation 特征方程. Its roots are the closed-loop poles, so it holds the whole stability story.
G = 10/(s + 1) runs in unity feedback. What is the closed loop's DC gain? Give 3 decimal places.
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$T(0) = G(0)/(1 + G(0)) = 10/11 = 0.909$. The output settles 9% short of the input.
Reading a second-order loop
- Match the denominator to $s^{2} + 2\zeta\omega_{n}s + \omega_{n}^{2}$. $\omega_{n}$ says how fast the loop is.
- The damping ratio 阻尼比 $\zeta$ says how much it rings. Designs usually aim for $0.4 \le \zeta \le 0.7$.
A closed loop is 9/(s² + 3s + 9). What is its damping ratio? 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
$\omega_{n} = 3$ and $2\zeta\omega_{n} = 3$, so $\zeta = 0.5$.
Worked example: one loop
- $\dfrac{1}{s+1}$ and $\dfrac{1}{s+2}$ in series with gain 6, unity feedback: $T(s) = \dfrac{6}{(s+1)(s+2) + 6} = \dfrac{6}{s^{2} + 3s + 8}$.
- $\omega_{n} = \sqrt{8} = 2.83$ rad/s and $\zeta = \dfrac{3}{2 \times 2.83} = 0.53$: inside the design region, about 14% overshoot.
In the one-loop example the gain rises from 6 to 14. What happens to ζ?
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The denominator becomes $s^{2} + 3s + 16$: $\omega_{n} = 4$ and $\zeta = 3/8 = 0.375$. More gain, less damping.
Worked example: nested loops
- Inner loop first: $\dfrac{2/s}{1 + (2/s)(0.5)} = \dfrac{2}{s + 1}$. In series with $\dfrac{1}{s+1}$ it gives $\dfrac{2}{(s+1)^{2}}$.
- Close the outer loop: $T = \dfrac{2}{s^{2} + 2s + 3}$, so $\omega_{n} = 1.73$ rad/s and $\zeta = 0.577$.
An inner loop is 4/s with feedback gain 2. After collapsing it to a/(s + b), what is b?
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$\dfrac{4/s}{1 + 8/s} = \dfrac{4}{s + 8}$, so $b = 8$.
Three common slips
- Closing with $G/(1 + G)$ when $H \ne 1$. The loop carries $GH$.
- A lost minus sign. Adding at the summer gives $1 - GH = 0$, and every stability claim flips.
- Outer loop first. It still works, but the algebra explodes. Inner loops first.
The summer adds the feedback instead of subtracting it. The characteristic equation becomes
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With $e = r + Hy$ you get $y(1 - GH) = Gr$. Positive feedback flips the stability story.
Match each connection to its rule.
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
Every block diagram reduces with these three rules, inner loops first.
Recap
- Series multiplies, parallel adds, feedback divides by $1 + GH$.
- $1 + GH = 0$ gives the closed-loop poles.
- Match $s^{2} + 2\zeta\omega_{n}s + \omega_{n}^{2}$ to read speed and damping.