Modelling of dynamic systems
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| time constant/taɪm ˈkɒnstənt/ | hằng số thời gian |
| operating point | điểm làm việc |
| linearization/ˌlɪnɪəraɪˈzeɪʃn/ | tuyến tính hóa |
| natural frequency/ˈnætʃərəl ˈfriːkwənsi/ | tần số riêng |
| back-EMF | điện động lực ngược |
| state variable | biến trạng thái |
One skeleton: storage, flow, leakage
- A hot tank, a capacitor and a spinning rotor all write the same sentence: how fast the stored quantity changes = drive in − leakage out.
- The result is the first-order model $\tau\,\dfrac{dx}{dt} + x = K\,u(t)$: $\tau$ is the time constant 时间常数 (seconds to respond), $K$ the DC gain (final value per unit of input).
$R = 2$ kΩ, $C = 100$ µF: $\tau = RC = 0.2$ s. Step the input to 5 V and the capacitor reaches $5(1-e^{-1}) = 3.16$ V, which is 63.2%, after 0.2 s.
{exercise:uni-ee1041-tau}
An RC filter has R = 2 kΩ and C = 100 µF. After a 5 V step, the capacitor voltage at t = 0.2 s is about —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
tau = RC = 0.2 s, and one time constant brings the curve to 1 − e^(−1) = 63.2% of 5 V = 3.16 V. 36.8% is the fraction remaining, not reached.
Linearize about an operating point
- Real laws are curves. Pick the operating point 工作点 you work near and replace the curve by its slope: that is linearization 线性化. In words, linearization means the step of replacing a curved law by its straight-line slope model near the chosen point.
- The slope tells you how far the model is honest. A tank draining at $q = 2\sqrt{h}$ has slope $\frac{dq}{dh} = \frac{1}{\sqrt{h}}$: it is 0.5 near $h = 4$ m but 1/3 near $h = 9$ m.
A pendulum near rest: $\sin\theta \approx \theta$, so it behaves like a spring with natural frequency 固有频率 $\omega_n = \sqrt{g/l}$. For $l = 0.9$ m, $\omega_n = \sqrt{9.81/0.9} = 3.30$ rad/s, about 0.53 Hz.
{exercise:uni-ee1041-tank}
A tank drains at q = 2·sqrt(h). The linearised slope at h = 4 m is —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
dq/dh = 1/sqrt(h) = 1/2 at h = 4. The model is local: at h = 9 the slope is 1/3, so one line cannot serve both levels.
The DC motor: two laws, one loop
- Torque law $T = K\,i$ and Newton's law $J\dot\omega = T - B\omega$ couple the electrical and mechanical sides; the spinning armature pushes back a back-EMF 反电动势 $K\omega$, so KVL reads $V = Ri + K\omega$.
- With $J = 0.01$ kg·m², $B = 0.1$ N·m·s, $K = 0.5$, $R = 1$ Ω and inductance neglected, the two laws combine to
One pole at $s = -35$ rad/s: time constant $1/35 = 28.6$ ms, DC gain $50/35 = 1.43$ rad/s per volt. Sense check: $\dfrac{K}{BR+K^2} = \dfrac{0.5}{0.35} = 1.43$. The whole transient from three physical constants, and the back-EMF is a feedback loop the motor builds for free.
{exercise:uni-ee1041-motor}
With J = 0.01, B = 0.1, K = 0.5, R = 1 (SI units, L neglected), the motor transfer function is —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
K/[(Js+B)R + K²] = 0.5/(0.01s + 0.35); multiply top and bottom by 100 to get 50/(s+35). The pole is −35 rad/s and the DC gain is 1.43 rad/s per volt.
Keep the units and the point
- $J$ in kg·m², $B$ in N·m·s, $\omega$ in rad/s; the motor constant has the same value in N·m/A and V·s/rad.
- A linear model is honest only near its operating point. A state variable is a stored quantity whose value, with the input, fixes every future value of the system. The two states of the motor, $i$ and $\omega$, are the state variables 状态变量 that unit 10.8 formalises.
{exercise:uni-ee1041-point}
Dropping the back-EMF term K·ω from the motor KVL would —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Without the Kω feedback the loop that limits speed disappears: the denominator loses the K² term and speed has nothing to hold it down. The back-EMF is the motor's own negative feedback.