【自动控制原理胡寿松第四版课后答案】 自控开环闭环
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1-3
解:系统的工作原理为:当流出增加时,液位降低,浮球降落,控制器通过移动气动阀门的 开度,流入量增加,液位开始上。当流入量和流出量相等时达到平衡。当流出量减小时,系 统的变化过程则相反。
流出量
希望液位
图一
1-4 (1) (2) (3) (4) (5) (6)
非线性系统 非线性时变系统 线性定常系统 线性定常系统 线性时变系统 线性定常系统
2-1 解:
显然,弹簧力为 kx(t ) ,根据牛顿第二运动定律有:
d 2 x(t )
F (t ) − kx(t) = mdt 2
移项整理,得机械系统的微分方程为:
2 d x(t )
2+ kx(t ) = F (t )
dt
对上述方程中各项求拉氏变换得:
ms 2 X (s) + kX (s) = F (s)
所以,机械系统的传递函数为:
G(s) =
X (s) 1
=
F (s) ms 2 + k
2-2 解一:
由图易得:
i1 (t )R1 = u1 (t ) − u2 (t ) uc (t ) + i1 (t )R2 = u2 (t )
duc (t )
i1 (t ) = Cdt
由上述方程组可得无源网络的运动方程为:
du1 (t ) du2 (t )
+ + C ( R1 + R u ) = CR2 u1 (t ) 2 )2 (t dtdt
对上述方程中各项求拉氏变换得:
C (R1 + R2 )sU 2 (s) + U 2 (s) = CR2 sU1 (s) + U1 (s)
所以,无源网络的传递函数为:
U (s) 1 + sCR2
G(s) = 2 =
U1 (s) 1 2
解二(运算阻抗法或复阻抗法):
1+ R2
U (s ) 1 + R Cs 2 2 1 U 1 + ( R + R 1 (s) R 1 2 )Cs+ R1 2
Cs
2-5 解:按照上述方程的顺序,从输出量开始绘制系统的结构图,其绘制结果如下图所示:
依次消掉上述方程中的中间变量 X 1 , X 2 , X 3 , 可得系统传递函数为:
G1 (s)G2 (s)G3 (s)G4 (s) C(s) =
R(s) 1 + G2 (s)G3 (s)G6 (s) + G3 (s)G4 (s)G5 (s) + G1 (s)G2 (s)G3 (s)G4 (s)[G7 (s) − G8 (s)]
2-6 解:
① 将 G1 (s) 与 G1 (s) 组成的并联环节和 G1 (s) 与 G1 (s) 组成的并联环节简化,它们的 等效传递函数和简化结构图为:
G12 (s) = G1 (s) + G2 (s)
G34 (s) = G3 (s) − G4 (s)
② 将 G12 (s), G34 (s) 组成的反馈回路简化便求得系统的闭环传递函数为:
G12 (s) G1 (s) + G2 (s)C(s) =R(s) 1 + G12 (s)G34 (s) 1 + [G1 (s) + G2 (s)][G3 (s) − G4 (s)]
2-7 解:
由上图可列方程组:
[E (s)G1 (s) − C (s)H 2 (s)]G2 (s) = C (s) C (s)
= E (s) R(s) − H1 (sG2 (s)
联列上述两个方程,消掉 E (s) ,得传递函数为:
G1 (s)G2 (s) C(s)
=
R(s) 1 + H1 (s)G1 (s) + H 2 (s)G2 (s)
联列上述两个方程,消掉 C (s) ,得传递函数为:
1 + H 2 (s)G2 (s) E(s)
=
R(s) 1 + H1 (s)G1 (s) + H 2 (s)G2 (s)
2-8 解:
将①反馈回路简化,其等效传递函数和简化图为:
0.4 2s + 1 = 1 G (s) = 1
0.4 * 0.5 5s + 3 1 +
2s + 1
将②反馈回路简化,其等效传递函数和简化图为:
1 2
5s + 3 s + 0.3s + 1 G (s) = =2 3 2
0.4 5s + 4.5s + 5.9s + 3.4 1 + 2
(s + 0.3s + 1)(5s + 3)
将③反馈回路简化便求得系统的闭环传递函数为:
0.7 * (5s + 3) 3 2 Θ o (s) 3.5s + 2.1 = =
Θi (s) 0.7 * Ks(5s + 3) 5s 3 + (4.5 + 3.5K )s 2 + (5.9 + 2.1K )s + 3.4
1 +
5s
3-3
解:该二阶系统的最大超调量:
σ p = e
−ζ−ζ
2
π /
*100%
当σ = 5% 时,可解上述方程得:
p
ζ =
0.69
当σ = 5% 时,该二阶系统的过渡时间为:
p
t s ≈ 3
ζ
wn
所以,该二阶系统的无阻尼自振角频率 wn ≈3= 3-4 解:
ζ
t s
3= 2.17
0.69 * 2
由上图可得系统的传递函数:
⑴ 若
10 * (1 + Ks)
C (s) s(s + 2) 10 * (Ks + 1)== 2 =
s + 2 * (1 + 5K )s + 10 R(s) 1 +s(s + 2)
所以 wn =,ζwn = 1 + 5K
K ≈ 0.116 = 0.5 时,
ζ
所以 K ≈ 0.116 时,= 0.5
ζ
⑵ 系统单位阶跃响应的超调量和过渡过程时间分别为:
σ p =
e
−ζ
π /
2
*100% = e
−0.5*3.14 /
*100% ≈ 16.3%
ts = 3 =ζ
≈ 1.9 0.5 *
3
wn
⑶ 加入 (1 + Ks ) 相当于加入了一个比例微分环节,将使系统的阻尼比增大,可以有效
地减小原系统的阶跃响应的超调量;同时由于微分的作用,使系统阶跃响应的速度(即变
化率)提高了,从而缩短了过渡时间:总之,加入 (1 + Ks ) 后,系统响应性能得到改善。
3-5 解:
由上图可得该控制系统的传递函数:
C(s)
10K1 R(s) =τ + 1)s +
s 2 + (101
二阶系统的标准形式为:
10K
C (s)
R(s)= w 2n
s 2 + 2w
ζw s + 2n
n
所以
wn 2 = 10K1 2ζwn = 10τ
+ 1
由
−
σ p=−ζπ / e
ζ 2
*100%
t π
p =wn − ζ 2
σ p =
9.5%
t 0.5
p = 可得
ζ =
0.6
wn = 7.85
w2 n = 10K1
ζ =
由
0.6
2ζw和
n = 10τ
wn = 7.85
可得:
+ 1
K1 = 6.16
τ = 0.84
t s ≈ 3
ζ
wn
= 0.64
3-6 解:⑴ 列出劳斯表为:
因为劳斯表首列系数符号变号 2 次,所以系统不稳定。 ⑵ 列出劳斯表为:
因为劳斯表首列系数全大于零,所以系统稳定。 ⑶ 列出劳斯表为:
因为劳斯表首列系数符号变号 2 次,所以系统不稳定。
3-7 解:系统的闭环系统传递函数:
K (s +1)
C (s) s(2s +1)(Ts +1) K (s +1)
= =
R(s) 1 + K (s +1)
s(2s +1)(Ts +1)
K (s +1)
= 3
2Ts+ (T + 2)s 2 + (K +1)s + K
列出劳斯表为:
s3 2T K +1
s2 T + 2 K
(K +1)(T + 2) − 2KT s1
sK
(K + 1)(T + 2) − 2KT > 0 , K > 0 T > 0 ,T + 2 > 0 ,T + 2
T > 0 K > 0 , (K + 1)(T + 2) − 2KT > 0
(K +1)(T + 2) − 2KT = (T + 2) + KT + 2K − 2KT
= (T + 2) − KT + 2K = (T + 2) − K (T − 2) > 0 K (T − 2)
3-9 解:
由上图可得闭环系统传递函数:
C (s) KK2 K3
=R(s) 2 3 2 3 2 3
K
代入已知数据,得二阶系统特征方程:
(1 + 0.1K )s2 − 0.1Ks − K =
列出劳斯表为:
s2 1 + 0.1K s1 − 0.1K s0
− K
− K
可见,只要放大器 −10
3-12 解:系统的稳态误差为:
sess = lim e(t ) = lim sE (s) = lim R(s) s →0 1 t →∞ s→0 + G0 (s)
⑴ G0 (s) =
10
s(0.1s + 1)(0.5s + 1)
系统的静态位置误差系数:
K p = lim G 0(s) = lim
s →0
10
= ∞
s →0 s(0.1s + 1)(0.5s + 1)
系统的静态速度误差系数:
K v = lim sG0(s) = lim
s →0
10s
= 10
s →0 s(0.1s + 1)(0.5s + 1)
系统的静态加速度误差系数:
K s G a = lim
2 s→0
0(s) = lim
10s 2
= 0
s→0 s(0.1s + 1)(0.5s + 1)
当 r (t ) = 1(t ) 时, R(s) =
1
s
s1
ess = *= 0
s→0 10 s1 +s(0.1s + 1)(0.5s + 1)
当 r (t ) = 4t 时, R(s) =
4
s 2
e ss = lim
s 4
2= 0.4
s →0 1 +s(0.1s + 1)(0.5s + 1) s 3
当 r (t ) = t 时, R(s) = 2
2
s 2
ess = lim3= ∞
s →0 10 s
1 +
s(0.1s + 1)(0.5s + 1)
当 r(t) = 1(t) + 4t + t 时, R(s) = 2
1 4 2s s 2 s 3
ess = 0 + 0.4 + ∞ = ∞
3-14 解:
由于单位斜坡输入下系统稳态误差为常值=2,所以系统为 I 型系统
设开环传递函数 G(s) =
K
s(s2 + as + b)
⇒
K
= 0.5 b
闭环传递函数 φ(s) =G(s) =
1 + G(s) s3 + as2 + bs + K
Q s = −1 ± j 是系统闭环极点,因此
s3 + as2 + bs + K = (s + c)(s2 + 2s + 2) = s3 + (2 + c)s2 + (2c + 2)s + 2c
⎧K = 0.5b ⎪K = 2c ⎪ ⎨
b = 2c + 2 ⎪ ⎪⎩a = 2 + c
所以 G(s) =
⇒
⎧K = 2
⎪a = 3 ⎪ ⎨
b = 4 ⎪ ⎪1 ⎩c =
2
。 2
s(s+ 3s + 4)
4-1
(a)
(b)
(c)
(d)
4-2
p p 1 = 0,2 = 0, p3 = −1
1. 实轴上的根轨迹
(−∞, −1) (0, 0)
1
2. n − m = 3
3 条根轨迹趋向无穷远处的渐近线相角为
180°(2q + 1)
(q = 0,1)
ϕ =±= ±
60a °
,180° 3
渐近线与实轴的交点为
∑n
m
p
i
− ∑ zi
σ =i =1
a
j =1
=0 − 0 −1 = − 1
m
3 3
3. 系统的特征方程为
1+G(s) = 1 +
K
s2
(s +1)
= 0
即
K = − s2
(s +1) = −s3 − s2
dK = − 3s2 − 2s =
s(3s + 2) = 0
0ds
根 s1 = 0 (舍去)
s2 = −
4. 令 s = jω
代入特征方程 1+G(s) = 1 +
K
s2
(s +1)
= 0 s2 (s +1) + K =0 ( jω )2 ( jω +1) + K =0
−ω 2
( jω +1) + K =0
K − ω 2 − jω =0 ⎧⎨K − ω 2 =0 ⎩ω
= 0
ω=0
(舍去)
与虚轴没有交点,即只有根轨迹上的起点,也即开环极点
p 1,2 = 0
在虚轴上。
2
5-1
G(s) =
5 0.25s +1 5 G( jω ) =
0.25 jω +1
A(ω ) ϕ(ω) = − arctan(0.25ω)
ω=4
输入 r(t) = 5 cos(4t − 30°) = 5 sin(4t +
60°) A(4) =系统的稳态输出为
= ϕ(4) = − arctan(0.25 * 4) = −45°
c(t ) = A(4) * 5 cos[4t − 30° + ϕ(4)]
= 5 cos(4t − 30° − 45°)
= 17.68 cos(4t − 75°) = 17.68 sin(4t +15°)
sin α = cos(90° −α ) = cos(α − 90°) = cos(α + 270°)
c(t ) = A(4) * 5 sin[4t + 60° + ϕ(4)]
= 5 sin(4t + 60° − 45°)
= 17.68 sin(4t +15°)
1
G( jω ) =(1 + jω )(1 + j 2ω )
或者,
5-3
1
(2) G(s) =(1 + s)(1 + 2s)
A(ω ) ϕ(ω) = − arctan ω − arctan 2ω
2 )
ϕ(ω) = − arctan ω − arctan 2ω = −90° arctan ω + arctan 2ω = 90°
ω = 1/(2ω)
2
ω = 1/ 2
A(ω ) == 0.47 3
与虚轴的交点为(0,-j0.47)
(ω)
1
(3) G(s) =
1
s(1 + s)(1 + 2s) 1 G( jω ) =
jω (1 + jω )(1 + j2ω )
1 ϕ(ω) = −90° − arctan ω − arctan 2ω A(ω ) 2 )
ϕ(ω) = −90° − arctan ω − arctan 2ω = −180° ω = 1/(2ω)
2 ω = 1/ 2
arctan ω + arctan 2ω = 90°
A(ω ) 2
= = 0.67
3
与实轴的交点为(-0.67,-j0)
)
ω (4) G(s) =
1 1 G( jω ) = s2 (1 + s)(1 + 2s) ( jω )2 (1 + jω )(1 + j 2ω )
ϕ(ω ) = −180° − arctan ω − arctan 2ω A(ω ) 2 )
ϕ(ω) = −180° − arctan ω − arctan 2ω = −270° ω = 1/(2ω)
2 ω = 1/ 2
arctan ω + arctan 2ω = 90°
A(ω ) = 0.94
与虚轴的交点为(0,j0.94)
ω
)
2
5-4
(2)ω1 = 0.5 ,ω2 = 1 , k = 1 ,υ = 0
(3)ω1 = 0.5 ,ω2 = 1 , k = 1 ,υ = 1
(4)ω1 = 0.5 ,ω2 = 1 , k = 1 ,υ = 2
5-6
G(s) = 1
是一个非最小相位系统
s −1
3
o 1 G( jω ) ==−1 − jω ) j ( −180+arctgω )
jω −1 1 G(s) = 是一个最小相位系统 s +1
1 G( jω ) ==− jω ) − jarctgω
jω +1
5-8(a)
ω = 0
−
ω = ∞
-1
X (ω )
ω = 0
+
系统开环传递函数有一极点在 s 因此乃氏回线中半径为无穷小量ε 的半圆 平面的原点处,弧 对应的映射曲线是一个半径为无穷大的圆弧:
ω :0− → 0+ ;θ :-90°→ 0°→ +90°; ϕ(ω) :+90°→ 0°→ -90° N=P-Z, Z=P-N=0-(-2)=2 闭环系统有 2 个极点在右半平面,所以闭环系统不稳定 (b)
jY (ω )
ω = 0−
ω = ∞
ω = 0+
4
系统开环传递函数有 2 个极点在 s 平面的原点处,因此乃氏回线中半径为无穷小量ε 的半圆
弧对应的映射曲线是一个半径为无穷大的圆弧:
ω :0− → 0+ ;θ :-90°→ 0°→ +90°; ϕ(ω) :+180°→ 0°→ -180° N=P-Z, Z=P-N=0-0=0 闭环系统有 0 个极点在右半平面,所以闭环系统稳定 5-10
(1) G(s)H (s) =K Ts +1= K
2.28K s +1 =s + 2.28
2.28
0°
(2) G s ( ) H s ( ) K1 K 1
=2.28K
= =
s s +1
2.28
−90°
1
(3) G(s)H (s) = K τ s +1
K s +1 = 0.5=
4K (s + 0.5) 2
2
2
s +1
2
b
5
11120 = a −20 lg K + 20 = 40 0.5 0.5 0.5 1
−20 lg K = 20 lg
0.5
20 lg(K )−1 = 20 lg 2 G(s)H (s) =
K = 1/ 2 = 0.5
4K (s + 0.5) 2(s + 0.5)2
s2 (s + 2) s(s + 2)
−
90°
5-11
jY (ω)
ω = 0
G(s)H (s) =
K K ⇒ G( jω )H ( jω ) s(s +1)(3s +1) = jω ( jω +1)(3 jω +1)
arctan ω + arctan 3ω = 90°
ϕ(ω ) = −90° − arctan ω − arctan 3ω = −180° ω = 1/(3ω)
2
ω = 1/ 3
A(ω ) 3
= K = 1
4
Kc = 4/3 = 1.33
6
6-2 (1)
2
6 ω G(s) =2 =2
2
s(s+ 4s + 6) s(s+ 2ξωωn s + n )
2
ω = ω 2.45, 6 n ξ =n
2ξω =4
n
4 2ωn
= 0.816 6
K = 1
所以,ωc = 1 20lgK = 0
⎛ 2ξω / ω ⎞ ⎛ 2 * 0.816 *1/ 2.45 ⎞
ϕ (ω ) = −90° − c n= −90° − arctg
⎜ c ⎜ 1 −1/ 2.452 ⎟ 2 2 ⎟ 1 − ω / ω
⎝ ⎠ ⎝ c n ⎠
⎛ 2 * 0.816 *1 / 2.45 ⎞ ⎛ 0.666 ⎞
= −90° − arctg = −90° − arctg = −90° − arctg 0.7995⎜1 −1 / 2.452⎟ ⎜0.833 ⎟
⎝ ⎠ ⎝ ⎠ = −90° − 38.64° = −128.64°
γ = 180° + ϕ (ωc ) = 180° −128.64° = 51.36°
[1**********]-10--30-40
(2)
ω1 = 1, ω2 =1/0.2=5
⎛ 2ξω / ω ⎞ ⎛ ω ⎞ ⎛ ω ⎞
c
− arctg c ϕ (ω ) = −90° − c n+ arctg
⎜ ⎜ ⎟ ⎜ ⎟ c 2 2 ⎟ 1 − ω / ω ω ω ⎝ c n ⎠ ⎝ 1 ⎠ ⎝ 2 ⎠
1 ⎞ ⎛ ⎛ 1 ⎞
= −128.64° + − ⎜⎟ ⎜⎟ = −128.64° + 45° −11.31° = −94.95°
1 ⎠ 5 ⎠ ⎝ ⎝
γ = 180° + ϕ (ωc ) = 180° − 94.95° = 85.05°
1
课后答案网
L(ω) (dB )
50 403020100-10-20--40
G(s) =
ω = 1,
10
s(0.5s +1)(0.1s +1)
20 lg K =20lg10=20dB
ω1 = 1/ 0.5 = 2, ω2 = 1 / 0.1 = 10
ω1 = 2 时, L(ω1 ) = 20 − 20(lg 2 − lg1) = 20lg10 − 20 lg 2 = 20lg5 = 14dB ω2 = 10 时, L(ω2 ) = 14 − 40(lg10 − lg 2) = −13.96dB
所以,ω1
L(ω1 ) = 40(lg ωc − lg 2) = 40(lg ωc / 2) = 14dB
ωc = 4.48
ϕ (ωc ) = −90° − arctg 0.5ωc − arctg 0.1ωc = −90° − arctg 2.24 − arctg 0.448
= −90°− 65.94°− 24.13° = −180.07°
γ = 180° + ϕ (ωc ) = 180° −180.07° = −0.07°
L (ω )(dB)
[1**********]-10 -20 -30 -40
(2)
G(s)Gc (s) =
ω = 1,
10(0.33s +1)
20 lg K =20lg10=20dB
ω1 = 1 / 0.5 = 2, ω2 = 1/ 0.33 = 3, ω3 = 1 / 0.1 = 10, ω4 = 1/ 0.033 = 30
ω2 = 3 时, L(ω1 ) − L(ω2 ) = 40(lg ω2 − lg ω1 ) 14 − L(ω2 ) = 40(lg 4.35 − lg 2) L(ω2 ) = 7dB
L(ω3 = 10) − L(ω2 = 3) = −20(lg ω3 − lg ω2 ) = −3.37dB
所以ω2
L(ω2 ) = 20(lg ωc 2 − lg ω2 ) = 20(lg ωc 2 / 3) = 7dB
ωc 2 = 6.72
ϕ (ωc ) = −90° − arctg 0.5ωc 2 − arctg 0.1ωc 2 + arctg 0.33ωc 2 − arctg 0.033ωc
2
= −90° − arctg 3.36 − arctg 0.672 + arctg 2.22 − arctg 0.222
= −90°− 73.43°− 33.90°+ 65.75°−12.52° = −144.1°
γ 2 = 180° + ϕ (ωc 2 ) = 180° −144.1° = 35.9°
L(ω )(dB)
50 40 20100--20-30-40
-40dB /dec
20dB /dec
3
解:系统的工作原理为:当流出增加时,液位降低,浮球降落,控制器通过移动气动阀门的 开度,流入量增加,液位开始上。当流入量和流出量相等时达到平衡。当流出量减小时,系 统的变化过程则相反。
流出量
希望液位
图一
1-4 (1) (2) (3) (4) (5) (6)
非线性系统 非线性时变系统 线性定常系统 线性定常系统 线性时变系统 线性定常系统
2-1 解:
显然,弹簧力为 kx(t ) ,根据牛顿第二运动定律有:
d 2 x(t )
F (t ) − kx(t) = mdt 2
移项整理,得机械系统的微分方程为:
2 d x(t )
2+ kx(t ) = F (t )
dt
对上述方程中各项求拉氏变换得:
ms 2 X (s) + kX (s) = F (s)
所以,机械系统的传递函数为:
G(s) =
X (s) 1
=
F (s) ms 2 + k
2-2 解一:
由图易得:
i1 (t )R1 = u1 (t ) − u2 (t ) uc (t ) + i1 (t )R2 = u2 (t )
duc (t )
i1 (t ) = Cdt
由上述方程组可得无源网络的运动方程为:
du1 (t ) du2 (t )
+ + C ( R1 + R u ) = CR2 u1 (t ) 2 )2 (t dtdt
对上述方程中各项求拉氏变换得:
C (R1 + R2 )sU 2 (s) + U 2 (s) = CR2 sU1 (s) + U1 (s)
所以,无源网络的传递函数为:
U (s) 1 + sCR2
G(s) = 2 =
U1 (s) 1 2
解二(运算阻抗法或复阻抗法):
1+ R2
U (s ) 1 + R Cs 2 2 1 U 1 + ( R + R 1 (s) R 1 2 )Cs+ R1 2
Cs
2-5 解:按照上述方程的顺序,从输出量开始绘制系统的结构图,其绘制结果如下图所示:
依次消掉上述方程中的中间变量 X 1 , X 2 , X 3 , 可得系统传递函数为:
G1 (s)G2 (s)G3 (s)G4 (s) C(s) =
R(s) 1 + G2 (s)G3 (s)G6 (s) + G3 (s)G4 (s)G5 (s) + G1 (s)G2 (s)G3 (s)G4 (s)[G7 (s) − G8 (s)]
2-6 解:
① 将 G1 (s) 与 G1 (s) 组成的并联环节和 G1 (s) 与 G1 (s) 组成的并联环节简化,它们的 等效传递函数和简化结构图为:
G12 (s) = G1 (s) + G2 (s)
G34 (s) = G3 (s) − G4 (s)
② 将 G12 (s), G34 (s) 组成的反馈回路简化便求得系统的闭环传递函数为:
G12 (s) G1 (s) + G2 (s)C(s) =R(s) 1 + G12 (s)G34 (s) 1 + [G1 (s) + G2 (s)][G3 (s) − G4 (s)]
2-7 解:
由上图可列方程组:
[E (s)G1 (s) − C (s)H 2 (s)]G2 (s) = C (s) C (s)
= E (s) R(s) − H1 (sG2 (s)
联列上述两个方程,消掉 E (s) ,得传递函数为:
G1 (s)G2 (s) C(s)
=
R(s) 1 + H1 (s)G1 (s) + H 2 (s)G2 (s)
联列上述两个方程,消掉 C (s) ,得传递函数为:
1 + H 2 (s)G2 (s) E(s)
=
R(s) 1 + H1 (s)G1 (s) + H 2 (s)G2 (s)
2-8 解:
将①反馈回路简化,其等效传递函数和简化图为:
0.4 2s + 1 = 1 G (s) = 1
0.4 * 0.5 5s + 3 1 +
2s + 1
将②反馈回路简化,其等效传递函数和简化图为:
1 2
5s + 3 s + 0.3s + 1 G (s) = =2 3 2
0.4 5s + 4.5s + 5.9s + 3.4 1 + 2
(s + 0.3s + 1)(5s + 3)
将③反馈回路简化便求得系统的闭环传递函数为:
0.7 * (5s + 3) 3 2 Θ o (s) 3.5s + 2.1 = =
Θi (s) 0.7 * Ks(5s + 3) 5s 3 + (4.5 + 3.5K )s 2 + (5.9 + 2.1K )s + 3.4
1 +
5s
3-3
解:该二阶系统的最大超调量:
σ p = e
−ζ−ζ
2
π /
*100%
当σ = 5% 时,可解上述方程得:
p
ζ =
0.69
当σ = 5% 时,该二阶系统的过渡时间为:
p
t s ≈ 3
ζ
wn
所以,该二阶系统的无阻尼自振角频率 wn ≈3= 3-4 解:
ζ
t s
3= 2.17
0.69 * 2
由上图可得系统的传递函数:
⑴ 若
10 * (1 + Ks)
C (s) s(s + 2) 10 * (Ks + 1)== 2 =
s + 2 * (1 + 5K )s + 10 R(s) 1 +s(s + 2)
所以 wn =,ζwn = 1 + 5K
K ≈ 0.116 = 0.5 时,
ζ
所以 K ≈ 0.116 时,= 0.5
ζ
⑵ 系统单位阶跃响应的超调量和过渡过程时间分别为:
σ p =
e
−ζ
π /
2
*100% = e
−0.5*3.14 /
*100% ≈ 16.3%
ts = 3 =ζ
≈ 1.9 0.5 *
3
wn
⑶ 加入 (1 + Ks ) 相当于加入了一个比例微分环节,将使系统的阻尼比增大,可以有效
地减小原系统的阶跃响应的超调量;同时由于微分的作用,使系统阶跃响应的速度(即变
化率)提高了,从而缩短了过渡时间:总之,加入 (1 + Ks ) 后,系统响应性能得到改善。
3-5 解:
由上图可得该控制系统的传递函数:
C(s)
10K1 R(s) =τ + 1)s +
s 2 + (101
二阶系统的标准形式为:
10K
C (s)
R(s)= w 2n
s 2 + 2w
ζw s + 2n
n
所以
wn 2 = 10K1 2ζwn = 10τ
+ 1
由
−
σ p=−ζπ / e
ζ 2
*100%
t π
p =wn − ζ 2
σ p =
9.5%
t 0.5
p = 可得
ζ =
0.6
wn = 7.85
w2 n = 10K1
ζ =
由
0.6
2ζw和
n = 10τ
wn = 7.85
可得:
+ 1
K1 = 6.16
τ = 0.84
t s ≈ 3
ζ
wn
= 0.64
3-6 解:⑴ 列出劳斯表为:
因为劳斯表首列系数符号变号 2 次,所以系统不稳定。 ⑵ 列出劳斯表为:
因为劳斯表首列系数全大于零,所以系统稳定。 ⑶ 列出劳斯表为:
因为劳斯表首列系数符号变号 2 次,所以系统不稳定。
3-7 解:系统的闭环系统传递函数:
K (s +1)
C (s) s(2s +1)(Ts +1) K (s +1)
= =
R(s) 1 + K (s +1)
s(2s +1)(Ts +1)
K (s +1)
= 3
2Ts+ (T + 2)s 2 + (K +1)s + K
列出劳斯表为:
s3 2T K +1
s2 T + 2 K
(K +1)(T + 2) − 2KT s1
sK
(K + 1)(T + 2) − 2KT > 0 , K > 0 T > 0 ,T + 2 > 0 ,T + 2
T > 0 K > 0 , (K + 1)(T + 2) − 2KT > 0
(K +1)(T + 2) − 2KT = (T + 2) + KT + 2K − 2KT
= (T + 2) − KT + 2K = (T + 2) − K (T − 2) > 0 K (T − 2)
3-9 解:
由上图可得闭环系统传递函数:
C (s) KK2 K3
=R(s) 2 3 2 3 2 3
K
代入已知数据,得二阶系统特征方程:
(1 + 0.1K )s2 − 0.1Ks − K =
列出劳斯表为:
s2 1 + 0.1K s1 − 0.1K s0
− K
− K
可见,只要放大器 −10
3-12 解:系统的稳态误差为:
sess = lim e(t ) = lim sE (s) = lim R(s) s →0 1 t →∞ s→0 + G0 (s)
⑴ G0 (s) =
10
s(0.1s + 1)(0.5s + 1)
系统的静态位置误差系数:
K p = lim G 0(s) = lim
s →0
10
= ∞
s →0 s(0.1s + 1)(0.5s + 1)
系统的静态速度误差系数:
K v = lim sG0(s) = lim
s →0
10s
= 10
s →0 s(0.1s + 1)(0.5s + 1)
系统的静态加速度误差系数:
K s G a = lim
2 s→0
0(s) = lim
10s 2
= 0
s→0 s(0.1s + 1)(0.5s + 1)
当 r (t ) = 1(t ) 时, R(s) =
1
s
s1
ess = *= 0
s→0 10 s1 +s(0.1s + 1)(0.5s + 1)
当 r (t ) = 4t 时, R(s) =
4
s 2
e ss = lim
s 4
2= 0.4
s →0 1 +s(0.1s + 1)(0.5s + 1) s 3
当 r (t ) = t 时, R(s) = 2
2
s 2
ess = lim3= ∞
s →0 10 s
1 +
s(0.1s + 1)(0.5s + 1)
当 r(t) = 1(t) + 4t + t 时, R(s) = 2
1 4 2s s 2 s 3
ess = 0 + 0.4 + ∞ = ∞
3-14 解:
由于单位斜坡输入下系统稳态误差为常值=2,所以系统为 I 型系统
设开环传递函数 G(s) =
K
s(s2 + as + b)
⇒
K
= 0.5 b
闭环传递函数 φ(s) =G(s) =
1 + G(s) s3 + as2 + bs + K
Q s = −1 ± j 是系统闭环极点,因此
s3 + as2 + bs + K = (s + c)(s2 + 2s + 2) = s3 + (2 + c)s2 + (2c + 2)s + 2c
⎧K = 0.5b ⎪K = 2c ⎪ ⎨
b = 2c + 2 ⎪ ⎪⎩a = 2 + c
所以 G(s) =
⇒
⎧K = 2
⎪a = 3 ⎪ ⎨
b = 4 ⎪ ⎪1 ⎩c =
2
。 2
s(s+ 3s + 4)
4-1
(a)
(b)
(c)
(d)
4-2
p p 1 = 0,2 = 0, p3 = −1
1. 实轴上的根轨迹
(−∞, −1) (0, 0)
1
2. n − m = 3
3 条根轨迹趋向无穷远处的渐近线相角为
180°(2q + 1)
(q = 0,1)
ϕ =±= ±
60a °
,180° 3
渐近线与实轴的交点为
∑n
m
p
i
− ∑ zi
σ =i =1
a
j =1
=0 − 0 −1 = − 1
m
3 3
3. 系统的特征方程为
1+G(s) = 1 +
K
s2
(s +1)
= 0
即
K = − s2
(s +1) = −s3 − s2
dK = − 3s2 − 2s =
s(3s + 2) = 0
0ds
根 s1 = 0 (舍去)
s2 = −
4. 令 s = jω
代入特征方程 1+G(s) = 1 +
K
s2
(s +1)
= 0 s2 (s +1) + K =0 ( jω )2 ( jω +1) + K =0
−ω 2
( jω +1) + K =0
K − ω 2 − jω =0 ⎧⎨K − ω 2 =0 ⎩ω
= 0
ω=0
(舍去)
与虚轴没有交点,即只有根轨迹上的起点,也即开环极点
p 1,2 = 0
在虚轴上。
2
5-1
G(s) =
5 0.25s +1 5 G( jω ) =
0.25 jω +1
A(ω ) ϕ(ω) = − arctan(0.25ω)
ω=4
输入 r(t) = 5 cos(4t − 30°) = 5 sin(4t +
60°) A(4) =系统的稳态输出为
= ϕ(4) = − arctan(0.25 * 4) = −45°
c(t ) = A(4) * 5 cos[4t − 30° + ϕ(4)]
= 5 cos(4t − 30° − 45°)
= 17.68 cos(4t − 75°) = 17.68 sin(4t +15°)
sin α = cos(90° −α ) = cos(α − 90°) = cos(α + 270°)
c(t ) = A(4) * 5 sin[4t + 60° + ϕ(4)]
= 5 sin(4t + 60° − 45°)
= 17.68 sin(4t +15°)
1
G( jω ) =(1 + jω )(1 + j 2ω )
或者,
5-3
1
(2) G(s) =(1 + s)(1 + 2s)
A(ω ) ϕ(ω) = − arctan ω − arctan 2ω
2 )
ϕ(ω) = − arctan ω − arctan 2ω = −90° arctan ω + arctan 2ω = 90°
ω = 1/(2ω)
2
ω = 1/ 2
A(ω ) == 0.47 3
与虚轴的交点为(0,-j0.47)
(ω)
1
(3) G(s) =
1
s(1 + s)(1 + 2s) 1 G( jω ) =
jω (1 + jω )(1 + j2ω )
1 ϕ(ω) = −90° − arctan ω − arctan 2ω A(ω ) 2 )
ϕ(ω) = −90° − arctan ω − arctan 2ω = −180° ω = 1/(2ω)
2 ω = 1/ 2
arctan ω + arctan 2ω = 90°
A(ω ) 2
= = 0.67
3
与实轴的交点为(-0.67,-j0)
)
ω (4) G(s) =
1 1 G( jω ) = s2 (1 + s)(1 + 2s) ( jω )2 (1 + jω )(1 + j 2ω )
ϕ(ω ) = −180° − arctan ω − arctan 2ω A(ω ) 2 )
ϕ(ω) = −180° − arctan ω − arctan 2ω = −270° ω = 1/(2ω)
2 ω = 1/ 2
arctan ω + arctan 2ω = 90°
A(ω ) = 0.94
与虚轴的交点为(0,j0.94)
ω
)
2
5-4
(2)ω1 = 0.5 ,ω2 = 1 , k = 1 ,υ = 0
(3)ω1 = 0.5 ,ω2 = 1 , k = 1 ,υ = 1
(4)ω1 = 0.5 ,ω2 = 1 , k = 1 ,υ = 2
5-6
G(s) = 1
是一个非最小相位系统
s −1
3
o 1 G( jω ) ==−1 − jω ) j ( −180+arctgω )
jω −1 1 G(s) = 是一个最小相位系统 s +1
1 G( jω ) ==− jω ) − jarctgω
jω +1
5-8(a)
ω = 0
−
ω = ∞
-1
X (ω )
ω = 0
+
系统开环传递函数有一极点在 s 因此乃氏回线中半径为无穷小量ε 的半圆 平面的原点处,弧 对应的映射曲线是一个半径为无穷大的圆弧:
ω :0− → 0+ ;θ :-90°→ 0°→ +90°; ϕ(ω) :+90°→ 0°→ -90° N=P-Z, Z=P-N=0-(-2)=2 闭环系统有 2 个极点在右半平面,所以闭环系统不稳定 (b)
jY (ω )
ω = 0−
ω = ∞
ω = 0+
4
系统开环传递函数有 2 个极点在 s 平面的原点处,因此乃氏回线中半径为无穷小量ε 的半圆
弧对应的映射曲线是一个半径为无穷大的圆弧:
ω :0− → 0+ ;θ :-90°→ 0°→ +90°; ϕ(ω) :+180°→ 0°→ -180° N=P-Z, Z=P-N=0-0=0 闭环系统有 0 个极点在右半平面,所以闭环系统稳定 5-10
(1) G(s)H (s) =K Ts +1= K
2.28K s +1 =s + 2.28
2.28
0°
(2) G s ( ) H s ( ) K1 K 1
=2.28K
= =
s s +1
2.28
−90°
1
(3) G(s)H (s) = K τ s +1
K s +1 = 0.5=
4K (s + 0.5) 2
2
2
s +1
2
b
5
11120 = a −20 lg K + 20 = 40 0.5 0.5 0.5 1
−20 lg K = 20 lg
0.5
20 lg(K )−1 = 20 lg 2 G(s)H (s) =
K = 1/ 2 = 0.5
4K (s + 0.5) 2(s + 0.5)2
s2 (s + 2) s(s + 2)
−
90°
5-11
jY (ω)
ω = 0
G(s)H (s) =
K K ⇒ G( jω )H ( jω ) s(s +1)(3s +1) = jω ( jω +1)(3 jω +1)
arctan ω + arctan 3ω = 90°
ϕ(ω ) = −90° − arctan ω − arctan 3ω = −180° ω = 1/(3ω)
2
ω = 1/ 3
A(ω ) 3
= K = 1
4
Kc = 4/3 = 1.33
6
6-2 (1)
2
6 ω G(s) =2 =2
2
s(s+ 4s + 6) s(s+ 2ξωωn s + n )
2
ω = ω 2.45, 6 n ξ =n
2ξω =4
n
4 2ωn
= 0.816 6
K = 1
所以,ωc = 1 20lgK = 0
⎛ 2ξω / ω ⎞ ⎛ 2 * 0.816 *1/ 2.45 ⎞
ϕ (ω ) = −90° − c n= −90° − arctg
⎜ c ⎜ 1 −1/ 2.452 ⎟ 2 2 ⎟ 1 − ω / ω
⎝ ⎠ ⎝ c n ⎠
⎛ 2 * 0.816 *1 / 2.45 ⎞ ⎛ 0.666 ⎞
= −90° − arctg = −90° − arctg = −90° − arctg 0.7995⎜1 −1 / 2.452⎟ ⎜0.833 ⎟
⎝ ⎠ ⎝ ⎠ = −90° − 38.64° = −128.64°
γ = 180° + ϕ (ωc ) = 180° −128.64° = 51.36°
[1**********]-10--30-40
(2)
ω1 = 1, ω2 =1/0.2=5
⎛ 2ξω / ω ⎞ ⎛ ω ⎞ ⎛ ω ⎞
c
− arctg c ϕ (ω ) = −90° − c n+ arctg
⎜ ⎜ ⎟ ⎜ ⎟ c 2 2 ⎟ 1 − ω / ω ω ω ⎝ c n ⎠ ⎝ 1 ⎠ ⎝ 2 ⎠
1 ⎞ ⎛ ⎛ 1 ⎞
= −128.64° + − ⎜⎟ ⎜⎟ = −128.64° + 45° −11.31° = −94.95°
1 ⎠ 5 ⎠ ⎝ ⎝
γ = 180° + ϕ (ωc ) = 180° − 94.95° = 85.05°
1
课后答案网
L(ω) (dB )
50 403020100-10-20--40
G(s) =
ω = 1,
10
s(0.5s +1)(0.1s +1)
20 lg K =20lg10=20dB
ω1 = 1/ 0.5 = 2, ω2 = 1 / 0.1 = 10
ω1 = 2 时, L(ω1 ) = 20 − 20(lg 2 − lg1) = 20lg10 − 20 lg 2 = 20lg5 = 14dB ω2 = 10 时, L(ω2 ) = 14 − 40(lg10 − lg 2) = −13.96dB
所以,ω1
L(ω1 ) = 40(lg ωc − lg 2) = 40(lg ωc / 2) = 14dB
ωc = 4.48
ϕ (ωc ) = −90° − arctg 0.5ωc − arctg 0.1ωc = −90° − arctg 2.24 − arctg 0.448
= −90°− 65.94°− 24.13° = −180.07°
γ = 180° + ϕ (ωc ) = 180° −180.07° = −0.07°
L (ω )(dB)
[1**********]-10 -20 -30 -40
(2)
G(s)Gc (s) =
ω = 1,
10(0.33s +1)
20 lg K =20lg10=20dB
ω1 = 1 / 0.5 = 2, ω2 = 1/ 0.33 = 3, ω3 = 1 / 0.1 = 10, ω4 = 1/ 0.033 = 30
ω2 = 3 时, L(ω1 ) − L(ω2 ) = 40(lg ω2 − lg ω1 ) 14 − L(ω2 ) = 40(lg 4.35 − lg 2) L(ω2 ) = 7dB
L(ω3 = 10) − L(ω2 = 3) = −20(lg ω3 − lg ω2 ) = −3.37dB
所以ω2
L(ω2 ) = 20(lg ωc 2 − lg ω2 ) = 20(lg ωc 2 / 3) = 7dB
ωc 2 = 6.72
ϕ (ωc ) = −90° − arctg 0.5ωc 2 − arctg 0.1ωc 2 + arctg 0.33ωc 2 − arctg 0.033ωc
2
= −90° − arctg 3.36 − arctg 0.672 + arctg 2.22 − arctg 0.222
= −90°− 73.43°− 33.90°+ 65.75°−12.52° = −144.1°
γ 2 = 180° + ϕ (ωc 2 ) = 180° −144.1° = 35.9°
L(ω )(dB)
50 40 20100--20-30-40
-40dB /dec
20dB /dec
3
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