Time-domain response
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| overshoot/ˌəʊvəˈʃuːt/ | vượt quá |
| peak time | thời gian cực đại |
| settling time | thời gian ổn định |
| system type | loại hệ thống |
| steady-state error | sai số xác lập |
| velocity error constant | hằng số sai số vận tốc |
The second-order dictionary
- The prototype $\dfrac{\omega_n^{2}}{s^{2}+2\zeta\omega_n s+\omega_n^{2}}$ has poles at $-\zeta\omega_n \pm j\omega_n\sqrt{1-\zeta^{2}}$: distance from origin $\omega_n$, cosine of the angle $\zeta$.
- The dictionary: overshoot 超调量 $M_p = e^{-\pi\zeta/\sqrt{1-\zeta^{2}}}$, peak time 峰值时间 $t_p = \dfrac{\pi}{\omega_n\sqrt{1-\zeta^{2}}}$, settling time 调节时间 $t_s \approx \dfrac{4}{\zeta\omega_n}$ (2% band). In words, peak time is the clock reading at which the step response first reaches its highest point.
$\omega_n = 4$, $\zeta = 0.5$: poles $-2 \pm j3.46$, $M_p = 16.3\%$, $t_p = 0.907$ s, $t_s \approx 2.0$ s (2%) or 1.5 s (5%), rise 0.41 s.
{exercise:uni-ee1043-specs}
A system has ζ = 0.5 and ω_n = 4 rad/s. Its overshoot and 2% settling time are about —
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M_p = e^(−π·0.5/0.866) = 16.3%; t_s ≈ 4/(ζω_n) = 4/2 = 2.0 s. The same response peaks at t_p = 0.907 s.
Design the gain for a damping spec
- Match the closed denominator to $s^{2}+2\zeta\omega_n s+\omega_n^{2}$ and solve for the gain. For $G = \dfrac{K}{s(s+2)}$: $2\zeta\omega_n = 2$ and $\omega_n = \sqrt{K}$, so $\zeta = 1/\sqrt{K}$.
- Demand $\zeta = 0.45$: $K = 4.94$, $\omega_n = 2.22$ rad/s, predicted $M_p = 20.5\%$, $t_p = 1.58$ s, $t_s \approx 4.0$ s.
{exercise:uni-ee1043-design}
For G = K/[s(s+2)] in unity feedback, the gain K that gives ζ = 0.45 is —
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ζ = 1/√K, so K = 1/0.45² = 4.94 and ω_n = √K = 2.22 rad/s. Predicted: M_p 20.5%, t_s ≈ 4.0 s.
Error constants: read the loop, not the transient
- Count integrators in the open loop: that is the system type 系统类型. System type is the number of pure integrators in the open-loop forward path. Type 1 means zero steady-state error 稳态误差 to a step.
- The velocity error constant 速度误差常数 $K_v = \lim_{s\to0}\,sG(s)$ gives the ramp error $\dfrac{1}{K_v}$. The velocity error constant is the limiting value of the frequency-weighted open-loop gain at zero frequency, and smaller ramp errors need larger values of it.
The $K = 4.94$ design: $K_v = \dfrac{4.94}{2} = 2.47$, so a unit ramp leaves error $0.405$, while the step error is exactly zero. Verified by simulation: the step settles to 1.000 and peaks at 1.205.
{exercise:uni-ee1043-error}
That K = 4.94 design tracking a unit ramp settles to a steady-state error of —
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K_v = lim sG = K/2 = 2.47 and the ramp error is 1/K_v = 0.405. The step error is zero because one integrator makes it type 1; a ramp needs type 2 for zero error.
Two traps in the dictionary
- Peak time uses the damped frequency $\omega_d = \omega_n\sqrt{1-\zeta^{2}}$; the distance from the origin uses $\omega_n$. Mixing them moves every number.
- Quote the band with the settling time: the 2% and 5% estimates differ by a factor of 4/3.
{exercise:uni-ee1043-band}
For ζ = 0.5, ω_n = 4 rad/s, the 5% settling-time estimate is —
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The 5% band uses 3/(ζω_n) = 1.5 s against 2.0 s for 2%: a factor of 4/3. Always name the band with the number.