2006年杭电自动控制原理真题解析

r (ωc'' ) = 56.3o
相角裕度变大，稳定性强，截止频率减小 调节时间变大 超调量减小 系统稳态误差减小 八、解
sx ( s ) − x(0) = Ax( s ) + Bu ( s )
x( s ) = ( sI − A) −1 x(0) + ( sI − A) −1 Bu ( s)
x(t ) = Φ(t ) x(0) + ∫ Φ(t − τ ) Bu (τ )dτ
ωc = 5 = 2.24
相角裕度 (2) ω =
r (ωc ) = −90o − arctan 0.25ωc − arctan ωc = −5o
77 '' 1 时 L ( ω ) = 2 0 lg 20 7.7
1 77 = 20 lg 7.7 20
20 lg ω c'' − 20 lg
相角裕度
⇒ ω c'' = 0 .5
G1 ( s ) = 1 + 2 s
10 10 s ( s + 1) G2 ( s ) = = 10 1+ ks s ( s + 1) + 10ks s ( s + 1)
三、解
Φ( s) =
G1G2 10( s + 2) = 1 + G1G2 s [ s( s + 1) + 10ks ] + 10( s + 2)
0 a 1 − 4a A2 B −5 = 0 1 9 1 −3
S = B
AB
1 C 1 0 V = a −3 CA = −1 1 2 CA 1 −3 −4a +10
可控 R ( S ) = 3
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D( s ) = s 3 + (1 + 10k ) s 2 + 10s + 20
s3 s
2
1 1 + 10k −10 + 100k 10 + 10k 2
10 20 0
1 + 10k > 0 −10 + 100k > 0
ω = 0 时 G ( j 0) H ( j 0) = ∞
ω → ∞ 时 G ( j∞) H ( j∞) = 0
ω = 6 时 ∠G ( j 6) H ( j 6) = −180o ω ω g =1 2 3 k 与负实轴交点为 (− , j 0) 30
提示：
P = 0 要使系统稳定必须 R = 0
∴ 曲线不包围 (−1, j0) 点
可观测
S = −a − 1 ≠ 0 ⇒ a ≠ −1
V = −a + 3 ≠ 0
⇒ a ≠ −3
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5
4
eτ −t ∫0 Φ(t − τ ) Bu(τ )dτ = ∫0 0
t t
1 − e−t 0 dτ = 1 2τ − 2 t 2 − t (1 − e ) e 2
e x (t ) = 0
九、解
−t
1 − e−t 1 + e−t 0 2 + 1 1 −2 t = 1 5 −2 t −2 t e 3 − e + e 2 2 2 2
ω = 0.2 时 L(ω ) = 20 lg 5 + 40 lg 2 = 20 lg
k 0.2
⇒
k =4
2
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∴ G(s) =
六、解
4(2.5s + 1) s2 6 s (5s + 1)( + s + 1) 25 25 k jω ( jω + 2)( jω + 3)
k ∴ 30 <1
0 < k < 30
ω = 0-
ω = -∞ ω = +∞
ω = 0+
七、解
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3
20 lg G -20 -40 -20 -40 4
1 77
1 7.7
1
-40 -60 -60
ω
L
L''
(1) 40 lg ωc = 20 lg 5
杭州电子科技大学 2006 年攻读硕士学位研究生入学考试 861《自动控制原理》真题解析
2013 杭电自动化考研群(235627461)
一、解 控制对象：水箱
件：浮球 控制器：杠杆 执行器：气动阀门 被测量：液面高度 干扰量：流出水 测量元
二、解
G1
G2
H1G1
H2
G1
G2 1 + H1G1
H2
G2 C (s) 1 + H1G1 G1G2 = G1 g = GH R( s ) 1 + H1G1 + H 2G2 1+ 2 2 1 + H1G1
1 s 2ζ + s +1 2 ωn ωn
2
ωn = 5 ζ = 0.6
∴ 振荡环节为
25 s + 6s + 25
2
∴ G(s) =
k (2.5s + 1) s2 6 s (5s + 1)( + s + 1) 25 25
ω = 0.4 时 L(ω ) = 20 lg 5
ω = 2 时 L(ω ) = 0
s1 s0
⇒ k > 0.1
四、解
1.2 − 1 σ% = ×100% = 20% 1
−πζ
σ%= e
1−ζ 2
×100% ⇒ ζ = 0.46
β = arccos ζ = 1.08 rad
tr = 0.6
tr = π −β ωn 1 − ζ 2
⇒ ωn = 3.9
k1k2 T Φ( s) = k 1 s2 + s + 2 T T
0 t
0 s +1 [ sI − A] = s + 2 0
1 −1 [ sI − A] = s + 1
1 s + 2
e −t Φyright @tdshan. All Rights Reserved.
G ( jω ) H ( jω ) =
G ( jω ) H ( jω ) =
k ω ω2 + 4 ω2 + 9
ω ω − arctan 2 3
∠G ( j 0) H ( j 0) = −90o ∠G ( j∞) H ( j∞) = −270o G ( j 6) H ( j 6) = k 30
∠G ( jω ) H ( jω ) = −90o − arctan
c (∞ ) = lim sΦ ( s ) R ( s ) = k1 = 1
s →0
k2 = 15.2 T
五、解
1 = 3.744 T
T = 0.27
k2 = 4.06
低频段-20dB/dec 为积分环节
k 1 ， ω = 0.2 加入惯性环节 s 5s + 1
ω = 0.4 时加入微分环节 2.5s + 1 ， ω = 5 加入振荡环节