线性控制课件 module 4 08-4-11-english_new
合集下载
线性控制课件 module 21 08-4-27-english_new
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:
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:
4.1线性控制系统课件
2 1.5
1
0.5
0
-0.5
-1
-1.5
-2
0
1
2
3
4
5
6
7
8
9
10
6.白噪声函数
u (t ) ∽N(0,b)
对于任意 t 大于等于0, u (t )是高斯随机过程， 该函数常常用于刻画系统噪声和测量噪声。
40
30
20
10
0
-10
-20
-30
-40
0
1
2
3
4
5
6
7
8
9
10
二. 动态性能与稳态性能
0,t 0 (t ) , ,t 0
-
(t)dt 1
通过脉冲响应的分析可以确定系统的模型。
5. 正弦函数 u(t ) A sin(t )
其中A为幅值，ω为角频率，φ为初始相角。 正弦函数在研究系统的频率响应时将是重 要的输入函数。
例4.1 设单位反馈控制系统的开环传递函数为 G(s) 1 / Ts，求当
r (t ) sin wt 时控制系统的稳态误差。
解： 因为 E (s)
1 R( s ) 1 G( s)
例4.1 设单位反馈控制系统的开环传递函数为 G(s) 1 / Ts，求当
r (t ) sin wt 时控制系统的稳态误差。 1 ws 解： 因为 E (s) R( s ) 1 G( s) ( s 1/ T )( s 2 w2 )
可得 E(s) R(s) /(1 G(s) H (s)) 由Laplace变换的性质，易知
ess lim e(t ) lim L-1 [E( s)] lim L-1 ( R( s) /(1 G( s) H ( s)))
1
0.5
0
-0.5
-1
-1.5
-2
0
1
2
3
4
5
6
7
8
9
10
6.白噪声函数
u (t ) ∽N(0,b)
对于任意 t 大于等于0, u (t )是高斯随机过程， 该函数常常用于刻画系统噪声和测量噪声。
40
30
20
10
0
-10
-20
-30
-40
0
1
2
3
4
5
6
7
8
9
10
二. 动态性能与稳态性能
0,t 0 (t ) , ,t 0
-
(t)dt 1
通过脉冲响应的分析可以确定系统的模型。
5. 正弦函数 u(t ) A sin(t )
其中A为幅值，ω为角频率，φ为初始相角。 正弦函数在研究系统的频率响应时将是重 要的输入函数。
例4.1 设单位反馈控制系统的开环传递函数为 G(s) 1 / Ts，求当
r (t ) sin wt 时控制系统的稳态误差。
解： 因为 E (s)
1 R( s ) 1 G( s)
例4.1 设单位反馈控制系统的开环传递函数为 G(s) 1 / Ts，求当
r (t ) sin wt 时控制系统的稳态误差。 1 ws 解： 因为 E (s) R( s ) 1 G( s) ( s 1/ T )( s 2 w2 )
可得 E(s) R(s) /(1 G(s) H (s)) 由Laplace变换的性质，易知
ess lim e(t ) lim L-1 [E( s)] lim L-1 ( R( s) /(1 G( s) H ( s)))
线性控制系统理论与方法动态系统模型ppt课件
17
定义 3.8、(传递函数矩阵)对于 MIMO 线性定常系统，令输入
变量组为{u1,u2, ,u p} 输出变量组为{y1, y2, , yq} ，且假设
系统的初始条件为零，gij (s) 表示系统的第 j 个输入端到第 i 个输
出端的传递函数，其中 j 1, 2, , p, i 1, 2, , q ，则
（3.6）
18
(3.6)的向量形式为 y(s)
g11 (s
)
g1p(s)u(s) G(s)u(s)
g q1 (s) g qp (s)
称由上式定义的 G(s) 为系统的传递函数矩阵，当 G(s) 至少有一
个传递函数中分子分母次数相同时，称 G(s) 为真有理分式阵，否
13
当n为无穷大时，相应的系统称
为无穷维系统。所有的集中参数 系统都是有穷维系统，而所有分 布参数系统都是无穷维系统。
状态变量组的选取是不惟 一的，系统的任意两个状态变量
组间是线性变换的关系。
14
三、两类动态模型的定义：
定义 3.7、单变量标量（SISO）线性定常系统的输入／输出模 型为
y (n) an1 y (n1) a1 y a0 y bmu (m) bm1u (m1) b1u b
则称 G(s) 为严格真有理分式阵。显然 G(s) 为 q p 的一个有理分
式矩阵，对于因果系统，当且仅当 G(s) 为真的或严格真的时，它
才是物理上可以实现的。
19
当且仅当 limG(s) 0 或 limG(s) D(D 0) 时，相应的
s
s
传递函数阵G(s) 为严格真的或真的。
图3.1 系统的方框图表示
定义 3.8、(传递函数矩阵)对于 MIMO 线性定常系统，令输入
变量组为{u1,u2, ,u p} 输出变量组为{y1, y2, , yq} ，且假设
系统的初始条件为零，gij (s) 表示系统的第 j 个输入端到第 i 个输
出端的传递函数，其中 j 1, 2, , p, i 1, 2, , q ，则
（3.6）
18
(3.6)的向量形式为 y(s)
g11 (s
)
g1p(s)u(s) G(s)u(s)
g q1 (s) g qp (s)
称由上式定义的 G(s) 为系统的传递函数矩阵，当 G(s) 至少有一
个传递函数中分子分母次数相同时，称 G(s) 为真有理分式阵，否
13
当n为无穷大时，相应的系统称
为无穷维系统。所有的集中参数 系统都是有穷维系统，而所有分 布参数系统都是无穷维系统。
状态变量组的选取是不惟 一的，系统的任意两个状态变量
组间是线性变换的关系。
14
三、两类动态模型的定义：
定义 3.7、单变量标量（SISO）线性定常系统的输入／输出模 型为
y (n) an1 y (n1) a1 y a0 y bmu (m) bm1u (m1) b1u b
则称 G(s) 为严格真有理分式阵。显然 G(s) 为 q p 的一个有理分
式矩阵，对于因果系统，当且仅当 G(s) 为真的或严格真的时，它
才是物理上可以实现的。
19
当且仅当 limG(s) 0 或 limG(s) D(D 0) 时，相应的
s
s
传递函数阵G(s) 为严格真的或真的。
图3.1 系统的方框图表示
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
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.
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
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.
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
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
-
- 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 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
【VIP专享】线性控制系统课件Module4(免费)
Rs
n2
Cs
s2 2ns
DC motor position control system The block diagram of this system
Fig.4.13 position control system block diagram
1. Motor field winding electrical circuit
m
K pKaKm NRJ
c
NR
2 KpKaKmJ
Step Response
When the second-order system is subjected to a step input
Cs
s
s2
n2 2n s
n2
1 s
s
s n
n
2
d2
s
n
n
2
d2
ct 1
e nt
1 2
sin dt
Second-order electrical system example:
Ui
Uo
The transfer function is:
The UO(s) is written as:
Uo s Ui s
R Ls
1
1 Cs
Cs
1
Uo s Ui s
LCs2
1 RCs
1
s2
LC Rs
1
L LC
Standard form of the second-order system:
s1,2 jn
ct 1 cosnt
When ζ=1 —the critically damped case
s1,2 n
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
相关主题
- 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、"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、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)。
1, 2
= — ζ ωn ±ωn
ζ
2
－ 1
When 0 < < 1 the underdamped case
s
1, 2
= — ζ ωn ±ωn
e
n t 2
ζ
2
－ 1
c t 1
1
sin d t
C(t)
1
t
1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 =0.2 =0.4 =0.6 =0.8 =1.2 =2 =4 =1.0
Apply above equation to the mechanical system
Referring all variables to motor shaft
The final equation
Note: In here the gearbox reduce the load inertia and load shaft damping (阻尼) by a factor of N2. Usually N is several thousand, so that load inertia and damping may be neglected. The transfer function of torque τm to speed ωm .
0 , =1 , 1 , 0 1
s
— 1, 2
n n
2
－ 1
When = 0 the undamped case
1
c(t ) = 1 cos ωnt －
t
s
— 1, 2
n n
2
－ 1
When = 1 the critically damped case
Motor field winding electrical circuit
Take Laplace transform
- Electrical -> mechanical: the field current I establishes a magnetic flux(磁通量）that interacts with the armature field to produce a mechanical torque.
Linear Control Systems Engineering
线性控制系统工程
Module 4 Second-order Systems
(二阶系统)
Module 4
Second-Order Systems (二阶系统）
Second-order system: a system is represented by a second order differential equation.
The time constant of the electrical system is much smaller than that of the mechanical system.
In physically, the current in the field is established much more quickly than the motor achieve a steady state speed. Based on above observation, ignore the inductance of the field circuit.
The behavior of second-order system is different from that of first-order system, and it may cause oscillation response (振荡响应） and overshoot(系统超调).
• Terminology: described in a qualitative manner 定性描述 bipolar voltage 双极性电压
• Problem: a spring-mass damper system subject to a external force p. - Show the transfer function between the input p and output distance x; - Develop expressions for the damping ratio and damped natural frequency. - Determine the damped natural frequency, the undamped natural frequency, the damping ratio and the location of poles. Data: k=10N/m, c=4N-s/m, m=1kg
2 n
Second-Order Systems Example
• DC motor position control system
• System analysis (1)Value set and feedback component , summing junction Set and feedback : potentiometer
C ( s) =
ω
S ( s + 2ζ
2
2 n
ω
n
s + ωn)
2
=
1 S
－
S + 2ζ
2
ω
n 2 n
(S + ζ ωn) + ω
n t 2
(1 ζ ) －
2
c t L
1
C s 1
e
1
sin d t 1-
2
d n 1
(2) Amplifier
(3)Motor( includes electrical and mechanical portion) – Electrical analysis The motor is field controlled. stator : 定子 rotor : 转子 Stationary magnetic field: 固定磁场
the exponentn
3. The overshoot depends on the value of
60
•The relationship between overshoot and damping ratio when system is underdamped.
50
40
30
20
the undamped natural frequency
The damping ratio
The damped natural frequency
If the input is a unit step
Based on final-value theorem
Sample Problem 4.2
2 n s
There are two characteristic parameters here:
ζ — the damping ratio (阻尼比）
ωn — the undamped natural frequency (无阻尼自然频率)
For the R-L-C circuit, Compare with the standard form:
1
2
3
4
5
6
7
8
9
10
Time (s)
Fig.4.16 Second-order step response
We can recognize from here (P65.)
1. The system oscillates at frequency d 2. The response decays due to
10
0
0.2
0.4
0.6
0.8
1.0
Damping ratio
Fig.4.17 Relationship between overshoot and damping ratio
Sample Problem 4.1
• Problem: a closed loop transfer function
Second order electrical system Example:
R L C
ui
uo
capacitor: 电容 impedance: 阻抗
The Uo(s) is written as:
R L C
U o s U i s R Ls 1 Cs 1 Cs
ui
Determine the undamped natural frequency, the damping ratio, and damped natural frequency. What is the steady-state output for a unit step input. • Solution:
(4)The tachometer
(5)Position measuring
• The complete system block diagram.
Transfer function
◆ Step Response
When
the second-order system is subjected to a step input
S1, 2 n
C(t)
c t 1 e －
nt
1 nt
1
t
sHale Waihona Puke 1, 2= — ζ ωn ±ωn