线性控制课件 module 21 08-4-27-english_new

合集下载

线性控制课件 module 13 08-4-27-english

线性控制课件 module 13 08-4-27-english

Application to Stability

Suppose F(s) equals to the characteristic equation F(s) = 1+GH(s) The roots of above equation are the closed loop poles. Unstable system=>roots are in the right half of s plane. Apply Cauchy’s theorem to determine the stability of system. (1) define a clockwise path in s plane that encircles the whole of the right half of s plane. (2) observing the number of clockwise encirclements of the F(s) plane.
(a) GH(s)-> ∞ when θ varying. When s locate at C, GH(s) move to C’. (b) As s move form C to D, θ : 0->90; in the GH(s) plane, this corresponds to an infinite radius quadrant from C’ to D’ . (c) As s move form C to B, θ : 0->-90; in the GH(s) plane, this corresponds to an infinite radius quadrant from C’ to B’ . - Note: In here, a counterclockwise movement around the semicircle in the s plane produce a clockwise movement in the GH(s) plane. - The open loop frequency response does no enclose (-1,0), the system is stable.

现代控制理论线性控制系统的能控性与能观性基础知识资料PPT课件

现代控制理论线性控制系统的能控性与能观性基础知识资料PPT课件

u(t)
x(t0 )
x2
x0 x(t f ) 0
所有非零状态
x0 在t0 时刻能控 系统在t0 时刻完全能控
所有时刻
系统一致能控
x1
x(t1)
t0
x(t2 )
t1
线性定常 系统的能 控性与 t0 无关
t
t2
第11页/共45页
x(t0 ) 0 x(t1) 0 x(t0 ) 0 x(t1) 0
第1页/共45页
能控性和能观测性的基本概念:
20世纪60年代初,由卡尔曼提出, 与状态空间描述相对应。
卡尔曼
能控性:反映了控制输入对系统状态的制约能力。 输入能否控制状态(控制问题)
能观测性:反映了输出对系统状态的判断能力。 状态能否由输出反映(估计问题)
第2页/共45页
由于系统需用状态方程和输出方程两个方程来描述输入-输出 关系,状态作为被控量,输出量仅是状态的线性组合,于是有 “能否找到使任意初态转移到任意终态的控制量”的问题,即能 控性问题。并非所有状态都受输入量的控制,或只存在使任意初 态转移到确定终态而不是任意终态的控制。还有“能否由测量到 的输出量来确定出各状态分量”的问题,即能观测性问题。
a2
1 a2 a1 a22
rankM 3 n 故系统的状态完全能控!
此形式的状态方程为能控标准型
第35页/共45页
[例] 判别如下系统的能控性
x1 1 2 2 x1 2
x 2
0
1
1
x2
0
u
x3 1 0 1 x3 1
[解]:
2 4 0
M b Ab A2b 0 1
0 0 2
3
4 1 0

线性控制课件 module 8 08-4-11-english_new

线性控制课件 module 8 08-4-11-english_new

1 . 1 1 3 G s s lim s 2 . G s 1 0
s 0
Setting
lim s2 . G s acceleration ka
s 0
error constant
Then
e ss
1 ka
(加速度误差系数)
G s
K s
n
G 0 s
C. Steady-state Errors
ess lim s E s lim s
s 0 s 0
1 1 G s
Rs
Impulse input
R s 1
ess lim s . E s lim s.
s 0 s 0
1 .1 1 G s


- There is a distinction between the type number and the order number of a system.
is a type 2 system, but it is a third order system. • The steady state error is achieved by final value theorem
G s
K
s
n
G0 s
Type 0: Type 1: Type 2:
kp lim G s lim K K ,
s 0 s 0
1 ess 1 kp 1 0 ess 1
K , kp lim G s lim s 0 s 0 s
When
Rs
A s
K p lim G s
s 0

现代控制理论精品ppt课件(英文版)Chapter

现代控制理论精品ppt课件(英文版)Chapter
5) Controllable canonical form
2) Observable canonical form
SISO system
Then the system is called observable canonical form
Theorem: if system (A, B, C) is completely observable, then
Observabiltiy Criterion:
1) For any LTI continuous system with m dimension output
The necessary and sufficient condition of system being completely observable is
1) Controllability structure decomposition
Theorem: if the n-dimension LTI system (A, B, C) is not completely controllable
then there exists a nonsingular linear transform system to be
1. Calculate matrix
2. Calculate invert of 3. Set 4. Calculate 5. Observable canonical form
• 5. System structure decomposition
If the LTI system is not completely controllable or observable.
2) If the system has distinct eigenvalue

线性控制系统分析与设计 中英文课件第1章

线性控制系统分析与设计  中英文课件第1章

前向单元(系统动态环节). 前向单元(系统动态环 节) 在激励信号作用下产生期望输出的单元。该单元 起控制输出的作用,因此它可能是一个功率放大器。
输出(被控变量) Output (controlled variable). 该输 出量必须维持在所规定的值上,即跟随指令输入
开环控制系统(Open-loop control system). 输出信 号对输入信号不产生影响的系统
School of Mechanical Engineering
湖南工业大学机械工程学院
Fundamentals of Control Theory
Chap 1 INTRODUCTION
闭环控制系统 : 输出信号对输入信号有直接影响
的系统
单输入单输出系统(SISO)的功能方框图 :
特点: 输出信号被反馈 到输入端并与输入信号进行比较
School of Mechanical Engineering
湖南工业大学机械工程学院
Fundamentals of Control Theory
Chap 1 INTRODUCTION
References
1. 杨叔子,《机械工程控制基础》,第四版, 华中理工大学出版社
2. 绪方盛彦, 《现代控制理论》, 科学出版社 3. 李友善, 《自动控制原理(上册) 》, 国防工业出版社 4. 张伯鹏, 《控制工程基础》, 机械工业出版社 5. 阳含和, 《机械控制工程(上册) 》, 机械工业出版社 6. 姚伯威, 《控制工程基础》, 电子科技大学出版社 7. 薛定宇, 《控制系统计算机辅助设计—MATLAB语言及
应用》, 清华大学出版社
School of Mechanical Engineering

线性控制课件 module 20 08-4-27-english_new

线性控制课件 module 20 08-4-27-english_new

(a)
Amplitude
0.8 0.6
compensated
0.4 0.2 0
0
1
2
3
4
5
6
7
8
9
10
Time (s)
(b)
Fig.20.6 Closed-loop step response of compensated system
Comment on Phase Lag compensation (P433)
- PM=30º requirement (4) can’t be meet.
- Introduce a phase lead compensator to increase PM in order to meet requirement (4). - Because the lead-lag filter does not modify the slope of uncompensated magnitude at high frequency. Using requirement (2), let the gain pass through – 30db at ω=100.
20 log 10 40
100
( rad / s )
60
1
10
100
M db
40
20 0
40db
0
gc

-90 -180
PM 50

Fig.20.4 Determining the new gain crossover frequency.
Determine τ : the two break frequencies are at a lower frequency than ωgc’=3. This is determine by: (1) The higher break frequency 1/ τ should at a frequency that the residual phase is no more than 10% for gain crossover frequency. Usually 1/ τ = ωgc’/10. (2) The lower break frequency should not be too small to keep the bandwidth.

线性控制课件 module 16 08-4-15-english_new

线性控制课件 module 16 08-4-15-english_new
20 log 10 NK 20 log 10 K 20 log 10 N 2o log 10 NK K db N db
M db
lo g 1 0
K3
Kc

K2
K1 -90 -180
-270
Im
-1 Im Kc K2 K3 K3 K1

Re
Kc K1 K2
Re
Fig.16.2 Bode,Nyquist,and root locus of system with variable K.
M db
0db GM
lo g 1 0

180
0
PM
lo g 1 0
Fig.16.3 Gain and phase margins
System Type and Steady State Error from Bode Diagrams
A general
GH s
open loop transfer
p1 1 s / p2 1 s / pk
K b 1 s / z1 1 s / z 2 1 s / z m s
n
1 s /
When ω approach 0
GH s Kb s
n
Type
0 system
n 0 G H (s) K b
Kdb is about –18db at -180° , if system is unstable
0 18 N db K c NK 0.794
N db 20 log 10 N 18 N 7.94
The result is close to the previous result K=0.832. The error is due to the straight line approximation to the phase angle plot in the Bode diagram.

线性控制课件 module 11 08-4-27-english_new

线性控制课件 module 11 08-4-27-english_new
Requirement: 1. Overshoot less than 10%=> K<2.89 2. Ramp input steady state error less than 10%=> K>20. 3. Dominant constant time <0.1s => poles lie to the left of s=-10.
R K2 s2
-
s 1
K1
c
Inner loop:
Gin s K1 s 1 K1
- Assume K1=2, the open-loop transfer function of the complete system.
GH s
s 2
K2

2 s3
If K2=5,
4
2
Re
-6 -4 -2 -2 2 4
-4
Fig.SP 11.2.5
Assume that b=5, a=1 The intersecting point on the real axis:
a
1 2
5 0.5 0.5 1 1.5
Im
4 2.64
The system is stable when K>8
Note: the root locus only yield the closed loop poles. The zeros influence the system performance.
Performance Requirements as Complex-plane Constraints
Suggest suitable values of a and b that stabilize the system , and for those values, find the range of K for which the system is stable.

线性控制课件 module 10 08-4-15-english_new

线性控制课件 module 10 08-4-15-english_new

asymptotes
360 nm

- Example: Angle between asymptotes = 360º /(n-m)=180º and , the Asymptote are parallel to imaginary axis. Negative real axis is not asymptote.
Rule #5:
The angle of emergence:
pc 180 pi zi
i 1 i 1 n 1 m
The angle of entry:
zc 180 pi zi
i 1 i 1 n m 1
Rule #6 : Imaginary-axis crossing points
Rule #4: The intersection point at the
real axis
pi zi a nm
Pi the real part of open loop poles; Zi the real part of open loop zeros; -Example:
- Next step: Determine the angle of emergence from
Linear Control Systems Engineering 线性控制系统工程
Module 10
Rules for Plotting the Root Locus (绘制根轨迹的规则)
根轨迹示例
j
j j
j
j
0
j
0 0
0
0
0
同学们,头昏了吧?
j
j
j j
0

线性控制课件 module 17_18 08-4-21-english_new

线性控制课件 module 17_18 08-4-21-english_new

-
- To find the frequency when the magnitude is unity
- Substitute it to phase expression
- M=1
- For 0<ξ<0.6, the relationship between the phase margin and the damping ratio: - The relationship between the gain crossover frequency ωgc and the undamped natural frequency ωn.
The approximate damping ratio

PM 100 0.4
The frequency ratio (from Fig.17.6)
r
gc n
0.85
n 15.2 rad/s
1.4 1.2 1 Amplitude 0.8 0.6 0.4 0.2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Time(s) Fig.SP17.2.2 1
From Fig 175: For Fig 176:
PO PM

gc n

4 1 2
4
2
r
When very small gc n
t (3/ ) gc s n
The Response of Higher Order Systems

20 s1 s / 10
The break frequencies are ω =1 and 10, the diagram will be plotted in the frequency range

线性控制课件 module 14 08-4-27-english

线性控制课件 module 14 08-4-27-english

3. Determine the frequency that makes the phase angle equals -180 °
4. Use the above frequency to determine the corresponding magnitude.
5. If the magnitude is less than unity and the critical point passes to the left, the system is stable; otherwise it is not.
Linear Control Systems Engineering 线性控制系统工程
Module 14
Nyquist Analysis and Relative Stability (奈奎斯特分析和相对稳定性)
Module 14 Nyquist analysis and stability
Conditional stability
The
procedure to calculate the pahse margin
1. From the transfer function, derive expressions for the magnitude and phase as functions of frequency.
2. Determine the frequency ωu that makes the magnitude unity.
3. Use the above frequency to determine the corresponding phase Φ. 4. Calculate the phase margin as

现代控制理论基础线性定常系统的综合PPT课件

现代控制理论基础线性定常系统的综合PPT课件
任意配置后零极点对消可能导致能控性发生变化10原受控系统ducxbuax二反馈至输入矩阵二反馈至输入矩阵bb前端的系统前端的系统将系统的输出量乘以相应的反馈系数馈送到输入端与参考输人相加其和作为受控系统的控制输入
5.1 线性反馈控制系统的基本结构
• 带输出反馈结构的控制系统 • 带状态反馈结构的控制系统 • 带状态观测器结构的控制系统 • 解耦控制系统
• 状态观测器: • 状态观测器基于可直接量测的输出变量y和控制变量u来估计状态变量, 是一个物理可实现的模拟动力学系统。
20
第20页/共47页
状态重构: 不是所有的系统状态物理上都能够直接测量得到。需要从系统的 可量测参量,如输入u和输出y来估计系统状态 。
状态观测器: 状态观测器基于可直接量测的输出变量y和控制变量u来估计状态 变量,是一个物理可实现的模拟动力学系统。
(4)确定K阵
由 f *( ) f ( ) 得:6 k 14, 5 k 60, 1 k 200
3
2
1
求得:k1 199, k2 55, k3 8
所以状态反馈矩阵K为: K [199 55 8]
17
第17页/共47页
三、状态反馈下闭环系统的镇定问题
镇定的概念:一个控制系统,如果通过反馈使系统实现渐近稳
5.2 带输出反馈系统的综合
一、反馈至输入矩阵B后端的系统
将系统的输出量乘以相应的负反馈系数,馈送到状态微分处。
v
x
B u
x C
y
A
H
原受控系统
0
( A,
B, C )

x y
Ax Cx
Bu
输出反馈控制规律:u Bv Hy
输出反馈系统状态空间描述为:
  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
K p s Ki C 3 2 R s s K p s K i
Kp=0 s will disappear from the denominator system is unstable (Routh’s criterion) Conclusion: PI control is able to eliminate steady state error but has little effect to transient response.


o
R
+ -
K p Kd s
+ +
1 Js 2 cs
C
Fig.21.11 Secondorder PD control system.
The transfer function of the controller:
o K Kd s i
p
For second order system ( damping ratio is meaningless for first order system)
Ki Kp 1
If the pole close to imaginary axis, the damping ratio will be small, and it cause an oscillatory response. Note: I control is not used alone.

For second order system:
K p s Ki C 3 2 R s s K p s K i C s 3 2 T s s K p s K i i
R is unit step input: T is unit step disturbance: Draw the root locus using:
o Ki Kp Kd s i s
o 1 K p (1 Td s) i Ti s
Trial-and-Error Method
(1)Select Kp firstly to achieve the transient response desired while setting Ti and Td to zero (2)Then Ti is adjusted to satisfy steady state error requirements. This will degrade the transient response due to the inclusion of a closed loop zero. (3)Select a suitable Td to restored the transient response.
For second order system.
C K 2 R s s K C 1 2 T s s K
R is unit step input Css t 1 T is unit step disturbance. 1
Css t K
If K is made large, the disturbance approach 0. But the large K should cause an oscillatory response. Conclusion: P control is of limited success in trying to obtain good performance in terms of steady state error, disturbance rejection and transient response.
Css t 1
Compare it to standard second order system.
2 s 2 2n s n Js 2 c K d s K p
c Kd 2n J 2 n K p / J
The above obtained Kp can satisfy damping ratio. But a zero is exist, and it must be taken into account in determining the transient response.
Proportional-Plus-Derivative Control (PD)

PD control can influence the transient response by changing the damping ratio.
i
Kp
Kd s
Fig.21.10 PD controller T
The transfer function of the controller:
Ki o K p i s o K p s Ki i s
K p s Ki C 2 R s s 1 K p K i C s 2 T s s 1 K p K i
Controller e K×e
K
Fig.21.2 Proportional controller
T R + +
+ -
-
K
1 1 s
C
Fig.21.3 First-order type 1 system with disturbance.
For first order system.
R is unit step input: T is unit step disturbance:
The
output is proportional to the input plus the integral of the input.
i
Kp
+ +
o
Ki s
Fig.21.6 PI controller T R
+ -
K P s Ki s
+ +
1 1 s
Fig.21.7 PI control of first- order sysrem.
C
Kp
+ +
1 Js c
1 s
Kr
Fig. 21.12 Derivative control using a tachometer
Kd
Ktach
Fig.21.13 Implementation of tachometer gain K r

The derivative control is never used alone but is always with a proportional component. The reason is to avoid noise. The differentiator can magnify noise.
Linear Control Systems Engineering 线性控制系统工程
Module 21
Multimode Controllers (多模控制器)
21 Multimode Controllers
• Multimode control design:
change the type of the system to meet desired performance requirements.
Css t 1 Css t 0
GH s K p s Ki s 2 1 s
To make the system stable, the zero must closer to the imaginary axis than system pole. (otherwise the asymptote may be at the right of imaginary axis).
C K / 1 K R 1 s / 1 K C 1 / 1 K T 1 s / 1 K
K Css t 1 K 1 Css t 1 K
If K is made large, the output for input approach 1 and for disturbance approach 0. But the K is finite.
For first order system:
R is unit step input: T is unit step disturbance:
Css t 1 Css t 0
The syom 0 to 1, due to the controller.
Fig.21.16 Quarter-decay response.

Ziegler and Nichols first method The method only applies if the open loop step response of the plat has no overshoot. To obtain maximum slope P and the time L
• Mode control – Design the controller in the loop.
R + - Controller Plant C
Feedback transducer
Fif.21.1 Definition of controller
Proportional Control

The output is proportional to the input.
T R
+ -
K p s Ki s
+ +
1 s (1 s )
C
Fig.21.8 PI control of secend-order sistem. Im
×
﹥ ﹤
Re
1
Ki K p
Fig.21.9 Root locus of PI-controlled second-order system
相关文档
最新文档