2009年杭电自动控制原理真题解析

40(lg 2 − lg ω1 ) = 40
20 lg k = 40 0.2
⇒ ⇒
ω 1 = 0 .2
k = 20
G (s) =
十、解 (1)
20(0.5s + 1) s (5s + 1)(0.125s + 1)
ω = 0+
ω = +∞
−1
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1 s
x2
+
+
& x 1 = −2x 1 + 3x2 + u
& x 2 = x1 + u
y = x1
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& x −2 3 x1 1 1 = + u x 1 0 x2 1 2 &
a −1 a +1
B 的角度比 C 大
两端的电压值，使温度逐渐接近设定值，同时偏差减小直到温度达到设定值时，稳 定工作。 (2) 被控对象为电炉 被控量为炉温
设定电压的滑动变阻器起到设定炉温的作用。电压放大功率放大把小电压电流 信号放大后驱动执行器。 (3)
电压放大
功率放大
电机
传动装置
变压器
电阻丝
电炉
热电偶
四、解
1 sC1 1 sC 2
U o (s)
R1 U i (s)
R2
系统的频域图如上图(记住一定要画上图) R3 = 1 + R2 sC2 R4 = R1 R1sC1 + 1
U o (s) R3 s 2 R1 R2C1C2 + s ( R2C2 + R1C1 ) + 1 = = = 2 R1 1 U i ( s ) R3 + R4 s R1 R2C1C2 + s ( R2C2 + R1C1 + R1C2 ) + 1 + R2 + sC2 sR1C1 + 1
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0 s −1 ( sI − A + BK ) = 0 s −1 4 6 s + 4
x ( s ) = ( sI − A + BK ) −1 Bv
y(s) = Cx(s) = C (sI − A + BK )−1 Bv
x1 y = [10 0 0] x2 x3
百度文库
&= Ax + Bu y = Cx x &= Ax + B(v − Kx) = ( A − BK ) x + Bv x 设 K = [ k1 k2 k3 ]
0 0 s 0 0 0 1 ( sI − A + BK ) = 0 s 0 − 0 0 1 + 0 [ k1 0 0 s 0 −2 −3 1 k2 k3 ]
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六、解
要求 σ % = 9.5%
−πζ 1−ζ 2
t p = 0.8
σ% = e
tp = π
× 100% = 9.5% ⇒ ζ = 0.6
⇒ ωn = 5
ωn 1 − ζ 2
= 0.8
k1 k1 k1 s ( s + 1) Φ (s) = = = 2 k (1 + kt s ) s ( s + 1) + k1 (1 + kt s ) s + s (1 + k1kt s ) + k1 1+ 1 s ( s + 1)
二、单项选择题 1 2 3 4 5 6 7 8 9 10 三、解 (1) 当炉温偏离设定值时，产生一偏差电压经放大后带动点机运转，控制电阻丝
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D A D C D D B C B A
超前校正在 ω =1 处取得最大超前角 ϕ m = arcsin
det(sI − A+ BK) = s3 + (k3 + 3)s2 + (k2 + 2)s + k1
= ( s + 2)( s + 1 + j )( s + 1 − j ) = s 3 + 4s 2 + 6s + 4
∴
k1 = 4
k2 = 4
k3 = 1
K = [ 4 4 1]
0 1 0 &= ( A − BK ) x + Bv = 0 0 1 x + Bv x −4 −6 −4
1 1 S =[ B AB] = 1 1
R ( S ) = 1 不可控
x1 y = [1 0 ] x2
C 1 0 V = = CA −2 3
十二、解
R (V ) = 2 可观测
x1 0 1 0 x1 0 x = 0 0 1 x + 0 u 2 2 x3 0 −2 −3 x3 1
e(∞) = lim sE ( s ) =
s →0
b < ε0 K
∴ 应满足的条件为
b < ε0K T1 + T2 − T1T2 K > 0
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八、解
1 1.8 0.8 36 C ( s) = − + = s s + 4 s + 9 s( s + 4)( s + 9)
Φ( s) = C ( s) 36 = R ( s ) ( s + 4)( s + 9)
R( s) =
1 s
Φ ( jω ) =
Φ ( jω ) =
九、解
36 ( jω + 4)( jω + 9)
36 ω 2 + 42 ω 2 + 92
∠Φ ( jω ) = − arctan ω ω − arctan 4 9
2 k1 = ωn = 25
2ζωn = 1 + k1kt = 6
⇒ kt = 0.2
ts =
七、解
3.5 = 1.17 ζωn
首先要求系统稳定
D ( s ) = s (T1s + 1)(T2 s + 1) + K = T1T2 s 3 + (T1 + T2 ) s 2 + s + K
s3 s2 s1 s0
五、解 框图化简法：
1 + R2 sC2
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2
1 G1
H2
1 G4
G1
G2
G3
G4
1 G1
1 G2
H3
H1
G1
G2
G3
G4
H1 +
H3 H + 2 G1G2 G1G4
G1G2G3G4 C (s) = R ( s ) 1 + G1G 2 G3G4 H1 + G3G4 H 3 + G2G3 H 2
杭州电子科技大学 2009 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、判断题 1 2 3 4 5 6 7 8 9 10 × × × × √ √ × √ × √ PD 为相位超前校正方式 频域分析法 利用开环系统的频率特性 判断闭环系统的稳定性 反馈控制并不一定都是负反馈 只要存在非线性元件即为非线性系统 不能描述内部特性 典型二阶振荡环节 阻尼比越大 超调量越小
R(s) =
T1T2 T1 + T2 T1 + T2 − T1T2 K T1 + T2 K
a b as + b + = s s2 s2
1 K 0
T1 + T2 − T1T2 K > 0 K >0
E (s) =
1 R(s) = 1 + G(s) 1+
b K
1 as + b ⋅ 2 K s s (T1s + 1)(T2 s + 1)
5
N− = 1
N+ = 0
R = 2( N + − N − ) = −2
系统不稳定
P=0
Z = P−R=2
(2)
−1 ω = 0+
ω = +∞
N− = 0
N + = 0.5
R = 2( N + − N − ) = 1
P =1
Z = P−R=0
十一、解
系统稳定
u
+
x1
1 s
x1
x1 = y
2 3
x2
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8
信号流图法：
−H 2 1
G1
G2
G3
G4
1
−H3
− H1
∆ = 1 − [G1G2G3G4 (− H1 ) + G2G3 (− H 2 ) + G3G4 (− H 3 )
P 1 = G1G2 G3G4
Φ( s) =
∆1 = 1
G1G2G3G4 P 1∆1 = ∆ 1 + G1G 2 G3G4 H1 + G2G3 H 2 + G3G4 H 3
低频段为 −20dB / dec 所以为积分环节，转折频率 ω1 处变为 −40dB / dec 加入 1 1 ， ω = 2 处加入一阶微分环节 0.5s + 1 ， ω = 8 处加入 1 0.125s + 1 s +1 ω1
惯性环节
G(s) = s(
k (0.5s + 1) 1 s + 1)(0.125s + 1) ω1