武汉科技大学2020年《827自动控制原理》考研专业课真题试卷【答案】

0
ess essr 0 Kt
1
R(s)
N (s) 0,
s2
1 KKt s
E(s)
s 2 (s 25)
R(s) 1 KKt s K (0.02s 1) s 2 (s 25) s 2 (s 25)
s(s 2 25s KKt ) s 3 25s 2 K (K t 0.02)s K
ess
essr
lim sE(s) lim s
n1 2
12
0.14s 1 0.42
0.4 T 0.16s
5 3
(1)
%
% e / 1 2 e 0.4 / 1 0.42 25.4%
4 4T 4 0.16
ts
n
1.6s 0.4
T
0.16
tp
d
n1 2
12
1 0.42
0.55s
T
10
T
%
(2)
T
ts
2
5
tp
(3)
T
ts
tp ( 25 )
1
N (s)
2 KK
3
K0 K (K t 0.02) 0 25K (K t 0.02) K
K0 K t 0.02
GK (s)
K (0.02s 1) s 2 (s 25) 1 KKt
s(s 25)
1 (0.02s 1)
Kt
s( 1 s2
25 s 1)
KKt
KKt
v 1, K K
1 Kt
1
kp
, kv
KK
, Kt
ka
2020
B)
827
(6
5
1. B 2. C 3. 4.A 5.A 6.C
(4
5
1
30 ) 20 )
2
3
4
c(t) 1 2e 2t e t
Gs
3s 2
(s 1)(s 2)
(30 )
C(s)
1
(s)
R(s) T 2s2 2 Ts 1
1 0.2 T 0.08s
0.8 T 0.08s
%
ts 10
2
0.4 T 0.04s
R(s) E(s) 0.02s 1
K
C(s)
s2 (s 25)
Kts
1
15 2 10
K Kt r(t) t1(t) n(t) 0
3 10
r(t) 0 n(t) 1(t)
1
1 KKt s
K(0.02s 1) 0
s 2 (s 25) s 2 (s 25)
1
e ss e ss
s 3 25s 2 K(Kt 0.02)s K 0
5
12
100
100(s 1)
G(s)
s2
2
c 100
( c) ( c ) 1800 arctan c 1800 89.40 90.60
3
( c)
( c ) 1800 ( c ) 89.40
4
5
( 20 )
6
1 10
C(Z) / R(Z) E(Z) / R(Z)
2 10
KT
r(t) e(t)
T
K
1 e Ts s
1
c(t)
(s 2)
6
G(z)
C(z) R(z)
0.5K (1 e 2T ) ;
z e 2T 0.5K (1 e 2T ) ;
z e 2T 0.5K (1 e 2T )
e 2T 0.5K (1 e 2T ) 1
E(z) R(z)
z e 2T z e 2T 0.5K (1 e 2T )
5
5
K s 3 25s 2 K (K t 0.02)s K
E(s)
K
1
ess
essn
lim s
N(s)
s 0 N(s)
lim s[ s 0 s3
25s 2
K(Kt
0.02)s
] Ks
1
25
7
db
40dB / dec 40
20dB / dec
20
0
1
10
100
7
1
G(s) 10
2
c
( c ) 10
3
( c)
% e / 1 2 e 0.8 / 1 0.82 1.5%
4 4T 4 0.08
ts
n
0.4s 0.8
5
T
0.08
tp
d
n1 2
12
0.42s 1 0.82
0.4
2
T 0.04s
% e / 1 2 e 0.4 / 1 0.42 25.4%
4 4T 4 0.04
ts
n
0.4s 0.4
5
T
0.04
tp
d
s(s 2 25s KKt )
1
s0
s 0 s 3 25s 2 K (K t 0.02)s K s 2
Kt
3
5
n(t) 1(t)
G1 s 0.02s 1
0 , K 1,
s1
s1
essn
lim
N(s)
s0K
lim s 01 s
1
K
E(s)
s 2 (s 25)
N (s) 1 KKt Biblioteka Baidu K (0.02s 1) s 2 (s 25) s 2 (s 25)
0.4 T 0.16s
%
ts 10
3
T
10
(s)
T 2s2
1 2 Ts 1
s2
2 n
2 ns
2 n
1
5
1 nT
0.2
1
T 0.08s
% e / 1 2 e 0.2 / 1 0.22 52.7%
4 4T 4 0.08
ts
n
1.6s 0.2
5
T
0.08
tp
d
n1 2
12
0.26s 1 0.22
0.8 T 0.08s