自动控制原理答案(第六章)

Thus, the conditions for stability are: K > 2 and K < is unstable for all values of K .
−2.9055.
(f)
s
4
+ 12 . 5 s + s + 5 s + K = 0
3
Routh Tabulation:
s s s
K
2
=
K
−1
K
s
1
−9 K − 1
K
−1
−9 K −1 > 0
s
0
10
2 2
The conditions for stability are: K > 0, K > 1, and − 9 K − 1 > 0 . Since K is always positive, the last condition cannot be met by any real value of K . Thus, the system is unstable for all values of K .
≅−
10
ε
Two sign changes in first column. Two roots in RHP.
s
0
5
3
6-3 (a)
s
4
+ 25 s + 15
s
2
+ 20 s + K = 0
Routh Tabulation:
72
s s s s
4
1 25 375
15 20
K
3
2
− 20
25
= 14.2
s
3
− 33 −396 + 24 −33 − 541 .1 + 528
11 .27
− 48 = 11 .27 = −1.16
16
s
2
s
1
0
s
0
0
Four sign changes in the first column. Four roots in RHP.
(f)
s
4
+ 2 s + 10
3 4
s
2
+ 20 s + 5 = 0
4
1 12 . 5 12 . 5
1 5
K
3
2
−5
= 0 .6 = 5 − 20 .83
K
12 . 5 s
1
3
− 12 . 5 K
0 .6
K
5
− 20 . 83 K > 0
or K
< 0 .24
s K K >0 The condition for stability is 0 < K < 0.24. When K = 0.24 the system is marginally stable. The auxiliary
0
equation is A ( s ) = 0 . 6 s oscillation is 0.632 rad/sec.
2
+ 0 .24 = 0 .
2
The solution of the auxiliary equation is s
2
= − 0 .4.
The frequency of
6-4
The characteristic equation is Ts
2
= 20
s
2
+ 100 = 0 .
The solution of the auxiliary
= −5 .
2
The frequency of oscillation is 2.236 rad/sec. K
(e)
s
4
+ Ks + 5 s + 10 s + 10
3
=0
10 K K
Routh Tabulation:
3
+ ( 2T +1) s + ( 2 + K ) s + 5 K = 0
Routh Tabulation:
74
s s
3
T 2T ( 2T
K
+2
5K
T T
>0 > −1 /
2
2
+1
s
1
+ 1 )( K + 2 ) − 5 KT
2T
+1
K (1 − 3 T ) K
+ 4T + 2 > 0
s
0
5K
>0
Roots: −0 .29 , − 1. 788 , 0 . 039 + j 3 .105 , 0 . 039 − j 3 .105
Routh Tabulation:
s s s 1 2 20 10 20 5
3
2
− 20
2
=0
5
s
2
ε
20
5
Replac e 0 in last row by
ε
s
1
ε − 10 ε
5
+ 600
s
4
+ 50000
s
3
+ Ks
2
+ 24
Ks
+ 80
K
=0
Routh Tabulation:
s s
5
1 600 3
50000 K
7
24 K 80 K
4
s
3
× 10
−K
00 K
7
14320 K 600
K
< 3 × 10
7
600 s
2
214080 3
−K
2
× 10 − K
(b) Poles are at s = − 5, − j 2 , j 2 (c) Poles are at s = − 0 .8688 , 0 .4344 + j 2 . 3593 , 0 .4344 − j 2 . 3593 (d) Poles are at s = − 5, − 1 + j , − 1 − j (e) Poles are at s = − 1.3387 , 1. 6634 + j 2 . 164, 1. 6634 − j 2 .164 (f) Poles are at s = − 22 . 8487 ± j 22 . 6376 , 21 . 3487 ± j 22 . 6023 6-2 (a)
2
+ 11 . 36 = 0 .
The solution of A(s) = 0 is s
2
= − 0 .8 .
The
(b)
s
4
+ Ks + 2 s + ( K + 1) s + 10 = 0
3 2
Routh Tabulation:
s s s
4
1 K 2K K
2
10 K K
3
+1
10
>0 >1
2
2
− K −1
2
marginally stable. The auxiliary equation is A ( s ) The frequency of oscillation is 1.7026 rad/sec.
= 3.4495
s
2
+ 10 = 0 .
The solution is s
2
= −2 .899
.
(d)
0
s
0
450 s
2
(b)
s
3
+ 25
3
+ 10 s + 50 = 0
Roots: − 24. 6769 , − 0 .1616 + j 1.4142 , − 0 .1616 − j 1.4142
Routh Tabulation:
s s s 1 25 250 10 50
2
1
− 50
25
=8
No sign changes in the first column. No roots in RHP.
Chapter 6
6-1 (a)
Poles are at s
STABILITY OF LINEAR CONTROL SYSTEMS
= 0 , − 1. 5 +
j 1. 6583 ,
− 1. 5 −
j 1. 6583
One poles at s = 0. Marginally stable . Two poles on j ω axis. Marginally stabl e . Two poles in RHP. Unstable . All poles in the LHP. Stable . Two poles in RHP. Unstable . Two poles in RHP. Unstable .
.6 0
25 s
0
10
(d)
2s
4
+ 10
s
3
+ 5 . 5 s + 5 . 5 s + 10 = 0
2
Roots: − 4.466 , − 1.116 , 0 .2888 + j 0 . 9611 , 0 .2888 − j 0 . 9611
Routh Tabulation:
71
s s s s
4
(c)
s
3
+ ( K + 2 ) s + 2 Ks + 10 = 0
2 3
Routh Tabulation:
s s 1 K 2K 2K 10 K
2
+2
2
> −2
2
s
1
+ 4 K − 10
K
+2
K
+ 2 K −5 > 0
s
0
10
The conditions for stability are: K > −2 and K + 2 K − 5 > 0 or (K +3.4495)(K − 1.4495) > 0, or K > 1.4495. Thus, the condition for stability is K > 1.4495. When K = 1.4495 the system is
K
K K
1
284
− 25
= 20 − 1. 76
20
− 1. 76 >0
K
>0
or K
< 11 . 36
14.2 s
0
K
K
Thus, the system is stable for 0 < K < 11.36. When K = 11.36, the system is marginally stable. The auxiliary equation equation is A ( s ) = 14.2 s frequency of oscillation is 0.894 rad/sec.
s
2
+ 16 s + 16 = 0
Roots: − 1.222 ± j 0 .8169 , 0 . 0447 ± j 1.153 , 0 .1776 ± j 2 .352
Routh Tabulation:
s s s 1 2 16 8 15 20 16 16
5
4
− 15
2
= 0 .5
40
− 16
2
= 12
s
3
+ 20
s
2
+ 5 s + 10
K
=0
Routh Tabulation:
73
s s s
3
1 20 100
5 10 K K
2
1
− 10
20
= 5 − 0 .5 K
5
− 0 .5 K > 0 >0
or K
< 10
s
0
10 K
K
The conditions for stability are: K > 0 and K < 10. Thus, 0 < K < 10. When K = 10, the system is marginally stable. The auxiliary equation is A ( s ) equation is s
s s s
4
1 K 5K
5 10 10 K
3
>0 − 10 > 0
or K
2
− 10
K
5K
>2
50 K s
1
− 100
K 5K
− 10
K
2
− 10
K
=
50 K
− 100 − 10
5K
K
3
− 10
5K
− 10 − K > 0
3
s
0
10 K
K
>0
The last condition is written as
2 10 55
5 .5 5 .5
10
3
2
− 11
= 4.4 = − 75 . 8
10
1
10 24.2
− 100
4.4 s
0
10
Two sign changes in the first column. Two roots in RHP.
(e)
s
6
+ 2 s + 8 s + 15
5 4 6
s
3
+ 20
0
s
0
50 s
2
(c)
s
3
+ 25
3
+ 250
s
+ 10 = 0
Roots: − 0 . 0402 , − 12 .48 + j 9 . 6566 , − j 9 . 6566
Routh Tabulation:
s s s 1 25 6250 250 10
2
1
− 10
= 249
No sign changes in the first column. No roots in RHP.
s
3
+ 25
3
s
2
+ 10 s + 450 = 0
Roots: − 25 . 31 , 0 .1537 + j 4.214, 0 .1537 − 4.214
Routh Tabulation:
s s s 1 25 250 10 450
2
1
− 450
25
= −8
Two sign changes in the first column. Two roots in RHP.
The conditions for stability are:
T > 0, K > 0, and K
<
4T 3T
+2 −1
. The regions of stability in the
T -versus-K parameter plane is shown below.
6-5 (a)
Characteristic equation: s
The conditions for stability are: K > 0, K > 2, and 5 K
ቤተ መጻሕፍቲ ባይዱ
β
K
+ 2 . 9055
γ ε
2
K
2
− 2 . 9055
K
+ 3 .4419
ϕ
− 10 − K > 0 .
3
<0.
The second-order term is positive for all values of K . Since these are contradictory, the system