实验二北京科技大学自控实验(3)解析
【自我实践4-1】某单位负反馈系统的开环传递函数()(1)(2)
k
G s s s s =
++,求(1) 当k=4时,
计算系统的增益裕度,相位裕度,在Bode 图上标注低频段斜率,高频段斜率及低频段、高频段的渐近相位角。(2) 如果希望增益裕度为16dB ,求出响应的k 值,并验证。 (1)当K=4时>> num=[4]; den=[1,3,2,0]; G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G) bode(num,den) grid
title(′Bode Diagram of G(s)=4/[s(s+1)(s+2)] ′) G =
4
----------------- s^3 + 3 s^2 + 2 s
Continuous -time transfer function.
Gm =1.5000,Pm =11.4304,Wcg =1.4142,Wcp =1.1431 title(′Bode Diagram of G(s)=4/[s(s+1)(s+2)] ′)
低频段斜率为-20dB/dec ,高频段斜率为-60dB/dec ,低频段渐近相位角为-90度,高频段的渐近相位角为-270度。增益裕度GM=1.5000dB/dec ,相位裕度Pm=11.4304度 (2)当增益裕度为16dB 时,算得K=0.951,对应的伯德图为:>> num=[0.951]; den=[1,3,2,0]; G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G) bode(num,den) grid
title(′Bode Diagram of G(s)=4/[s(s+1)(s+2)] ′) G = 0.951 ----------------- s^3 + 3 s^2 + 2 s
Continuous -time transfer function.
Gm =6.3091,Pm =54.7839,Wcg =1.4142,Wcp =0.4276 title(′Bode Diagram ′)
【自我实践4-2】系统开环传递函数()(0.51)(0.11)
k
G s s s s =
++,试分析系统的稳定性。
计算可得当K=12时系统的增益裕度,相位裕度为0.对应的程序为: >> num=[12];
den=[0.05,0.6,1,0]; G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G) bode(num,den) grid G =
12
---------------------- 0.05 s^3 + 0.6 s^2 + s
Continuous -time transfer function.
Gm =1Pm =9.5374e -06Wcg =4.4721Wcp =4.4721
当K=10时即当k<12时的特例,对应的程序为:>> num=[10]; den=[0.05,0.6,1,0];
G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G)
bode(num,den)
grid
G =
10
----------------------
0.05 s^3 + 0.6 s^2 + s
Continuous-time transfer function.
Gm =1.2000Pm =3.9431Wcg =4.4721Wcp =4.0776
系统产生衰减震荡,增益裕度和相角裕度都大于0,系统稳定。
当K=20时即当k>12时的特例,对应的程序为:>> num=[20]; den=[0.05,0.6,1,0];
G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G)
bode(num,den)
grid
G =
20
----------------------
0.05 s^3 + 0.6 s^2 + s
Continuous-time transfer function.
Gm =0.6000,Pm =-10.5320,Wcg =4.4721,Wcp =5.7247
此时的增益裕度和相角裕度都小于0,系统不稳定。
【自我实践4-3】某单位负反馈系统的开环传递函数
31.6
()
(0.011)(0.11)
G s
s s s
=
++
,求(1)绘制
Bode图,在幅频特性曲线上标出低频段斜率、高频段斜率、开环截止频率和中频段穿越频率;在幅频特性曲线标出:低频段渐近相位角、高频段渐近相位角和-180?线的穿越频率。(2)计算系统的相位裕度γ和幅值裕度h,并确定系统的稳定性。
程序为:>> num=[31.6];
den=[0.001,0.11,1,0];
G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G)
bode(num,den)
grid
title(′Bode Diagram ′) G =
31.6
------------------------
0.001 s^3 + 0.11 s^2 + s
Continuous -time transfer function.
Gm =3.4810,Pm =22.2599,Wcg =31.6228,Wcp =16.3053
低频段斜率为-20dB/dec ,高频段斜率为-60dB/dec ,低频段渐近相位角为-90度,高频段的渐近相位角为-270度。增益裕度GM=3.4810dB/dec ,相位裕度Pm=22.2599度,-180度线穿越频率为Wcg =31.6228。由伯德图可知系统的相位裕度和幅值裕度都大于零,故系统稳定。
【自我实践4-4】某单位负反馈系统的开环传递函数2(1)
()(0.11)k s G s s s +=+,令k=1作bode 图,
应用频域稳定判据确定系统的稳定性,并确定使系统获得最大相位裕度的增益k 值。 当K=1时的伯德图程序为: >> num=[1,1]; den=[0.1,1,0,0]; G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G) bode(num,den) grid G =s + 1 ------------- 0.1 s^3 + s^2
Continuous -time transfer function.
Gm =0,Pm =44.4594,Wcg =0,Wcp =1.2647
由图可知系统的相位裕度和幅值裕度都大于零,故系统稳定。由计算得当k=3.16系统获得最大相位裕度.
【综合实践】试观察下列典型环节BODE图形状,分析参数变化时对BODE图的影响,填写下表。
(1)比例环节:K(K=10、K=30)
K=10 num=[10];
den=[0,1];
G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G)
bode(num,den)
grid
title(Bode Diagram)
K=30
num=[30]; den=[0,1]; G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G) bode(num,den) grid
title(Bode Diagram)
(2) 惯性环节:1
Ts K
(K=1、K=10、T=0.1、1) K=1,T=0.1 cp]=margin(G) bode(num,den) grid
title(′Bode Diagram ′) G =
1 --------- 0.1 s + 1
Continuous -time transfer function
K=1,T=1
>> num=[1];
den=[1,1];
G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G) bode(num,den)
grid
title(′Bode Diagram ′)
G =1
-----
s + 1
Continuous-time transfer function. Gm =InfPm = -180Wcg =NaN
K=10,T=0.1
>> num=[10];
den=[0.1,1];
G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G)
bode(num,den)
grid
title(′Bode Diagram ′)
G =10
---------
0.1 s + 1
Continuous-time transfer function.
Gm = InfPm =95.7406Wcg =NaNWcp =99.4731
K=10,T=1
>> num=[10];
den=[1,1];
G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G)
bode(num,den)
grid
title(′Bode Diagram ′)
G =10
-----
s + 1
Continuous-time transfer function.
Gm =InfPm =95.7406Wcg =NaNWcp =9.9473
(3) 积分环节:
s
K
(K=1、K=10) K=1
>> num=[1]; den=[1,0]; G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G) bode(num,den) grid
title(′Bode Diagram ′) G =1 - s
Continuous -time transfer function. Gm = InfPm =90Wcg = NaNWcp =1
K=10
>> num=[10];
den=[1,0];
G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G)
bode(num,den)
grid
title(′Bode Diagram ′)
G =10
--
s
Continuous-time transfer function. Gm =InfPm =90Wcg =NaNWcp =10.0000
(4)微分环节:Ks(K=1、K=10)K=1
>> num=[1,0];
den=[0,1];
G=tf(num,den)
[Gm,Pm,Wcg,Wcp]=margin(G)
bode(num,den)
grid
title(′Bode Diagram ′)
G =s
Continuous-time transfer function.
Gm =InfPm =-90Wcg =NaNWcp =1