二阶系统动态性能分析

δ1%= e
ts =
−
π 1 ζ 2 1−ζ1
×100%=16%
4
ωn1ζ1
=
4 =1 s) ( 0.5×8
Monday, October 17, 2011
11
How to improve the performance of second order system
a. Derivative feedback of output
e − ζω n t 1−ζ
2
sin( ω d t + tg −1
1− ζ
2
ζ
) ≤ ∆%
for convenience
e − ζω n t s 1−ζ 2
= ∆%
2
ts = −
Monday, October 17, 2011
ln(
1−ζ
× ∆ %)
ζω
n
6
衰减振荡瞬态过程的性能指标 Transient response specification
cos β = ζ
β is damped angle
td =
Monday, October 17, 2011
1+ 0.7ξ
ωn
2
Transient response specification Peak time ：t p ,for c′(t p ) = 0 when t = t p
c (t ) = 1 − e −ζω n t 1−ζ 2 sin(ω d t + β ) , t ≥ 0
Qts = 4
ωnζ
(or
3
ωnζ
),∴ωn is increased, ts . For a certain
ζ
Usually, letting ζ =0.4~0.8, then the overshoot is 25%~1.5%.
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8
Example
Consider a system shown in fig.
Obtain ① ζ and ωn ; ② δ % and ts
K =16,T = 0.25
R (s )
③for δ %=16%,K=?, When T is same. Solution：① ω n = ②
K = T
−
K s(Ts + 1)
C (s )
16 1 1 = 8, ζ = = = 0 . 25 0 . 25 2 KT 2 16 × 0 . 25
R (s )
2 ωn s ( s + 2ζω n )
C (s )
-
-
τs
将输出信号的导数反馈到 系统输入端并与误差信号 比较。 比较。保持无阻尼振荡频 率不变，增大系统阻尼， 率不变，增大系统阻尼， 减小超调， 减小超调，同时减小调节 时间。 时间。
b. Proportional plus derivative of error
1− ζ 2 ωd = = tgβ thus： tg (ωd t p + β ) = ζωn ζ
故峰值时间满足的条件是 tg (ωd t p + β ) = tgβ
ωd t p = nπ (n = 0,1,2,...)
π t p is first time of peak，so n=1，we have： t p = = ωn 1−ζ 2 ωd
−
δ %×100%= e
ζ π 1− 2 ζ
×100% = 44%
4 4 = = 2s, (for∆ = 2) ω ζ 8×0.25 n ts = 3 3 = =1.5s, (for∆ = 5) ωnζ 8×0.25
③
−
ζπ
1− ζ
2
Q δ % = 16 %, ∴ 0 . 16 = e
4
衰减振荡瞬态过程的性能指标 Transient response specification
100 90 80 70 60 50 40 30 20 10 0
δ%
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Monday, October 17, 2011
ζ
5
衰减振荡瞬态过程的性能指标 Transient response specification
Settling time t s： according definition ,when t≥ts, |c(t)-c(∞)|≤ c(∞) ×∆%。 If the final value is unity, the error between the steady state and the response is equal to
Second order system and transient response specification Following shall obtain the rising time \ peak time \ maximum overshoot and settling time
c ( t ) = 1 − e − ζω n t (cos ω d t + sin ω d t ) , 1−ζ 2
−1
ζ
t≥0
c(t ) = 1 −
e −ζωnt 1− ζ
2
sin( 1 − ζ ωn t + tg
2
1− ζ 2
ζ
), t ≥ 0
rising time t r : according definition, for c ( t r ) = 1 when t = t r
c ( t ) = 1 − e − ζω n t r (cos ω d t r +
R (s )
2 ωn s ( s + 2ζω n )
Wn2 (1 + τs ) G ( s) = 2 s + (2ζWn + Wn2τ ) s + Wn2 C (s )
-
τs
+
相当于给系统增加了一个 闭环零点， 闭环零点，同时保持系统 自然频率不变， 自然频率不变，增大系统 阻尼比。 阻尼比。 Wτ ζd =ζ + n 2
Monday, October 17, ห้องสมุดไป่ตู้011
12
1
1− e −ζω n t 1−ζ 2
0
t
1−
1 1−ζ 2
ts t's
Response curve within a pair of envelope curve of
1±
Monday, October 17, 2011
e − ζω n t 1−ζ 2
7
[In Summary] Damped ratio ζ is very important parameter of second order system. It determines the performance of the system. For ζ > 1 the curve is monotonically and no oscillation and no overshoot. For ζ = 0 ,curve is oscillation for ever. For (0 < ζ < 1) ,when ζ is small, large oscillation and settling time and bad performance.
−
= 1− e
ζπ 1−ζ 2
(cos π +
ζ
1−ζ 2
sin π ) = 1 + e
−
ζπ 1−ζ 2
δ% =
c(t p ) − c(∞) c (∞ )
×100% = (c(t p ) − 1) ×100%
−
ζπ
1−ζ 2
thus： %= e δ
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×100%
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π
3
Transient response specification Maximum overshoot δ % ：
π Substitute t p = into c(t ), hence c(t p ) = cmax ωd ζ −ζω n t p cmax = c(t p ) = 1 − e (cos ω d t p + sin ω d t p ) 2 1−ζ
1− ζ 2 where β = tg −1
ζ
c′(t ) = −
− ζωne
−ζω n t p
1−ζ 2
sin(ω d t p + β ) −
e
−ζω n t p
1−ζ 2
ωd ⋅ cos(ωd t p + β ) = 0
ζωn sin(ωd t p + β ) −ωd ⋅ cos( d t p + β ) = 0 ω
ζ
cos ω d t r +
sin ω d t r = 0 1−ζ 2
ζ
1−ζ
2
sin ω d t r ) = 1
tg ω d t r = −
Monday, October 17, 2011
1−ζ
2
ζ
thus：t r =
1
ω
tg
−1
(−
1−ζ 2
d
ζ
)
1
衰减振荡瞬态过程的性能指标 Second order system and transient response specification