Modelling of dynamic systems
導入済み| English | 意味 |
|---|---|
| state variable | state variable |
| time constant/taɪm ˈkɒnstənt/ | 時間定数 |
| linearization/ˌlɪnɪəraɪˈzeɪʃn/ | linearization |
| operating point | operating point |
| natural frequency/ˈnætʃərəl ˈfriːkwənsi/ | natural frequency |
| back-EMF | back-EMF |
Hold 100 km/h on a hill
- Cruise control holds a car at 100 km/h up a hill. A 3D printer places plastic to within 0.05 mm.
- Both start the same way: write what the physical system does as an equation you can compute with. This course calls that a model.
Store, flow, lose
- A hot tank stores heat, gains it from a heater and loses it to the room. A capacitor stores charge, gains it from a source and leaks it through a resistor.
- With one state variable 状态变量 $x$, both follow the first-order model $\tau\dfrac{dx}{dt} + x = K\,u$.
Match each part of a heated tank to its twin in an RC circuit.
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Both store, gain and lose, so both follow the same first-order model.
The time constant
- $\tau$ is the time constant 时间常数, in seconds; $K$ is the DC gain, the final $x$ per unit of input.
- After a step of size $U$, $x(t) = KU(1 - e^{-t/\tau})$. It reaches 63.2% of its final value at $t = \tau$.
A system has K = 2 and τ = 1 s. The input steps from 0 to 10. What is x after 1 s? Give 2 decimal places.
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The final value is $KU = 20$; after one time constant it has reached 63.2%: $20 \times 0.632 = 12.64$.
Linearization
- Real systems are curved. Linearization 线性化 picks an operating point 工作点 and uses the slope there.
- A pendulum near rest has $\sin\theta \approx \theta$, so it swings at the natural frequency 固有频率 $\omega_{n} = \sqrt{g/l}$: 3.30 rad/s for $l = 0.9$ m.
A pendulum is 2.45 m long. Take g = 9.81 m/s². What is its natural frequency, in rad/s? Give 2 decimal places.
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$\omega_{n} = \sqrt{9.81/2.45} = \sqrt{4.00} = 2.00$ rad/s.
A tank drains at q = 2√h. What is the slope dq/dh at h = 1 m?
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$dq/dh = 1/\sqrt{h} = 1$ at $h = 1$ m, twice the slope at 4 m.
Worked example: an RC circuit
- $R = 2$ kΩ, $C = 100$ µF: $RC\dfrac{dv}{dt} + v = v_{in}$, so $\tau = RC = 0.2$ s and $K = 1$.
- Step the input to 5 V: after 0.2 s the capacitor holds $5(1 - e^{-1}) = 3.16$ V, which is 63.2%.
R = 10 kΩ and C = 47 µF. What is the time constant, in s? Give 2 decimal places.
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$\tau = RC = 10\,000 \times 47 \times 10^{-6} = 0.47$ s.
The DC motor: two laws
- Current makes torque, $T = Ki$, and Newton's law gives $J\dot\omega = T - B\omega$.
- The spinning armature makes a back-EMF 反电动势 $K\omega$, so $V = Ri + K\omega$. The motor builds its own feedback loop.
Worked example: the motor's transfer function
- With $J = 0.01$, $B = 0.1$, $K = 0.5$, $R = 1$: $\dfrac{\Omega(s)}{V(s)} = \dfrac{K}{(Js + B)R + K^{2}} = \dfrac{50}{s + 35}$.
- One pole at $-35$: $\tau = 28.6$ ms and a DC gain of 1.43 rad/s per volt. This is the course's running example.
The motor's inertia doubles to J = 0.02 (B = 0.1, K = 0.5, R = 1). Where is the pole now? Give the magnitude, in rad/s.
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The denominator is $0.02s + 0.35$, so the pole is at $-0.35/0.02 = -17.5$. The DC gain stays 1.43: inertia changes the speed of response, not the final value.
Three common slips
- Dropping the back-EMF. The model becomes $50/(s + 10)$: 5 rad/s per volt and 100 ms, both wrong.
- One slope everywhere. The tank $q = 2\sqrt{h}$ has slope 0.5 at 4 m but 0.33 at 9 m.
- Mixed units. $J$ in kg·m², $B$ in N·m·s, $\omega$ in rad/s.
A linear model taken at one operating point is accurate at every operating point.
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The slope changes along the curve. The tank's slope is 0.5 at 4 m and 0.33 at 9 m, so each operating point needs its own model.
Recap
- Store, flow and lose give $\tau\dot{x} + x = Ku$; 63.2% at one time constant.
- Linearize at an operating point; the slope holds only near it.
- The motor model $50/(s + 35)$ runs through the course.