自控原理总复习.ppt

1、自控原理总复习,控制理论主要研究内容:,1 稳定性 2 动态特性 3 稳态特性,一 控制系统的分析,二 控制系统的设计,Generalized feedback control system,Differencing junction,Output,Controller,Input,Plant,Transducer,Error,Control signal,+,Forward path,Feedback path,Fig 1.3Generalized feedback control system,Generalized terminology,Plant: 被控对象 the major co

2、mponent being controlled，its transfer function is fixed. Controller: 控制器 the control device designed by the engineer. Feedback path: 反馈通道 the output variable fed back to be compared with desired value. Forward path: 前向通道 Transducer: 变送器 Differencing junction: 差分连接点，减法器,Module 2 Transfer Functions an

3、d Block Diagram Algebra （传递函数和方框图运算）, Block Diagram Algebra,A. The simplification of a complex feedback control systems a. Serial Elements,b. Paralel Elements,x,y,x,y,x,y,+,+,x,y,x,y,+,+,x,y,c. General feedback case,x,z,y,+,+,x,y,z,+,+,x,x,y,x,y,y,x,z,y,+,+,x,z,y,+,+,x,y,y,x,y,x,e. Multiple inputs,S

4、teps of Block Diagram Simplification,Step4: Move summing junctions to the left and pick off points to the right (rule 4-7),Step1: Combin all serial blocks (rule 1),Step2: Combin all parallel blocks (rule 2),Step3: Close all inner loops (rule 3),Sample Problem 2.3,Simplify the block diagram and deter

5、mine the closed loop transfer function C/R.,R,s1,s2,s3,C,+,_,+,+,+,_,_,Module 3 First-order Systems (一阶系统),The Generalized First-Order Systems:,Transfer function Block diagram,Fig.3.4 First-order system transfer function, The Response of First-Order System,Some Common Inputs:,a. Impulse（脉冲） Signal,b

6、. Step （阶跃） Signal,c. Ramp （斜坡）Signal,d. Parabole（抛物线 ）Signal,e. Harmonic （谐波）Signal,First-Order Feedback Systems,a. Open loop control,the response of a unit step input,k k , ,Fig.3.10 First-order feedback control system,b. Closed loop control,Let =1 and K=1,Let =1 and K=10,Fig.3.11 Step response of

7、 feedback system,K=10,K=1,0.5,c(t),t,r(t),open-loop response,1,0.909,Important observations: (P45.),1. The feedback system has a steady-state error between the input and output. 2. The feedback system response is faster than the open loop system.,If K is increased to 10, output is close to 1 Steady-

8、state error is less than 10%,and 10 times the response speed.,Module 4 Second-order Systems,(二阶系统),Standard form of the second-order system:, the damping ratio (阻尼比）,There are two characteristic parameters here:,n the undamped natural frequency (无阻尼自然频率),Open-loop transfer function,Closed-loop trans

9、fer function, Step Response,When the second-order system is subjected to a step input,The characteristic equation and the roots:,The step response for the four cases of :,When = 0 the undamped case,1,When = 1 the critically damped case,When 1 the overdamped case,When 0 1 the underdamped case,表3-2 二阶

10、系统的阻尼系数与单位阶跃响应关系,阻尼系数,特征方程根,根在复平面上位置,单位阶跃响应,= 0,(无阻尼),1,0,欠阻尼,0,S1,s2,=1,1,(临界阻尼),(过阻尼),0,0,0,S1,S1,S1,s2,s2,s2,y (t),y (t),y (t),y (t),0,0,0,0,1,1,1,1,t,t,t,t,Module 5 Second-Order System Time -Domain Response （二阶系统时域响应）,Relationship Between System Poles and Transient Response,Characteristic equati

11、on:,Fig.5.4 Relationship between poles and time-domain parameters,Considering the step time-domain response,t, Time Domain Performance Specifications,Consider a underdamped system is subjected to a step input, the performance items include: 1 Rise time（上升时间）Tr 2 Peak time（峰值时间） Tp 3 Settling time(调节

12、时间）Ts 4 Percentage Overshoot(超调量) PO%,1. Rise time（上升时间）Tr,The time taken for the system to first achieve the final value of the output.,t,Tr Rise time,Tp Peak time,Ts(2%) Settling time,Fig.5.6 Representative second-order step response,Overshoot,Steady-state output,+2%,c(t),1.0,2. Peak time（峰值时间） Tp

13、,The time to the maximum of the first overshoot,t,Tr Rise time,Tp Peak time,Ts(2%) Settling time,Fig.5.6 Representative second-order step response,Overshoot,Steady-state output,+2%,c(t),1.0,3. Equivalent time constant(等效时间常数）,4. Settling time (调节时间）Ts,The elapsed time before the output becomes bound

14、ed by two equispaced limits on either side of the steady-state output.,t,Tr Rise time,Tp Peak time,Ts(2%) Settling time,Fig.5.6 Representative second-order step response,Overshoot,Steady-state output,+2%,c(t),1.0,5. Percentage Overshoot(超调量) PO%,t,Tr Rise time,Tp Peak time,Ts(2%) Settling time,Fig.5

15、.6 Representative second-order step response,Steady-state output,+2%,c(t),1.0,Module 7 Higher-Order Systems （高阶系统）,Module 7 Higher-order Systems, Reduction to Lower Order Systems,Example: A Higher-order System,There are two forms:,Evans form ,Bode form ,1. For the Evans form, if,2. For the bode form

16、, if,then,then,Conclusion: the Bode form must be used in order to correctly obtain the transfer function of the reduced-order system.,Linear Control Systems Engineering 线性控制系统工程,Module 8 System Type: Steady-State Errors （系统的型：稳态误差）,Module 8 System Type: Steady-State Error,Considering a generalized f

17、eedback control system:,A. Definition,1 E = R - C,2 Ea = R - C,When H(s) = 1, E = Ea,Then,B. System Type,The generalized open-loop transfer function:,When n = 0, type 0 n = 1, type 1 n = 2, type 2,So, the system type is defined by the value of n,C. Steady-state Errors, Impulse input,(,),s,See P132.

18、Table 8.1 and 1,2,3 points,Acceleration,Ramp,Step,Impulse,System type,Input,TABLE 8.1 Steady-state error as a function of system type and input,Linear Control Systems Engineering 线性控制系统工程,Module 9 Rouths Method,(劳斯判据） Root Locus: Magnitude and Phase Equation （根轨迹：幅值和相角方程）, Stability of a system:,The

19、 system become unstable as soon as one closed-loop pole is located in the right-hand half of the complex plane,Rouths Stability Criterion,Write the characteristic equation in polynomial form:,Step 1 : All the coefficient are positive; (necessary condition),Step 2 : There are no sign changes in the f

20、irst column of the Routh array. (sufficient condition),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）,Rule #1: Start and finish points,Rule #2: Root locus on the real axis,Rule #3: The angle between adjacent asymptotes,Rule #4: The intersection poi

21、nt at the read axis,Rule #5: The angle of emergence:,The angle of entry:,Rule #6 : Imaginary-axis crossing points,Setting 1+GH(s) = 0 substituting s = j Equating the real and imaginary parts to zero:,Get and K,Rule #7 : The points of emergence or entry on the real axis.,Method 1 graphic representati

22、on,Method 2,From 1+GH(s) = 0,Get K = f(s),K is differentiated and equated to zero:,Yieds the values: s = .,Rule #8 : The angle between the direction of emergence (or entry) of q coincident poles(or zeros) on the real axis.,Linear Control Systems Engineering 线性控制系统工程,Module 12 Frequecy response and n

23、yquist diagrams （频率响应和 nyquist 图）,Transfer function,Magnitude M,Phase,Asymptotes,TABLE 12.1.Frequency response of transfer function elements,中文版教材图521,3 start point,4 final point,n m,Linear Control Systems Engineering 线性控制系统工程,Module 13 Nyquist Stability Criterion （Nyquist 稳定判据）,Conformal mapping: C

24、auchys theorem (保角映射：柯西定理）,Consider a F(s):,Example,S-plane,F(s)-plane,For s1,E(-1.5,j0) ,E(-0.667,j0),G(0.5 , j0) ,G(0.48 , j0),Re,Im,Re,Im,-3,-2,-1,1/2,A,B,C,D,F,D,C,F,B,A,S plane,F(s) plane,Fig.13.4 Mapping the contour to the contour,For s2,For s3,C,E,D,A,B,F,Im,Re,A,B,C,D,-3,-2,-1,1/2,-1/2,E,-5/

25、2,F,-3/2,s,Im,Re,s plane,F(s) plane,Fig.13.5 Trajectory encircling a pole of F(s),1,1/2,-1,-2,-3,Re,Im,s,A,B,C,D,D,B,A,C,s plane,F(s) plane,Im,Re,Fig.13.6 Trajectory encircling no singularities of F(s),Application to stability Suppose F(s) = 1+GH(s) F(s)的极点等于开环传函的极点，极点数用P表示； F(s)的零点等于闭环传函的极点，零点数用Z表示

26、； 结论：系统稳定的条件转化为F（s）的零点数Z在s右边平面为0，即Z=0时闭环稳定。,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,Consider:,Finding Kc,P.274. 15,Re,Im,-1,K=0.832,K=0.3,K=0.1,K=0.05,Fig.14.1 Nyquist

27、diagrams for different values of K,Gain margin:,Gain and phase margins. (performance specifications),Example:,Phase margin:,The procedure to calculate the phase margin 1. From the transfer function, derive expressions for the magnitude and phase as functions of frequency. 2. Determine the frequency

28、u that makes the magnitude unity. 3. Use the above frequency to determine the corresponding phase . 4. Calculate the phase margin as 5. If PM is positive, system is stable; otherwise it is not.,Linear Control Systems Engineering 线性控制系统工程,Module 15 Bode diagrams,伯德图,Elemental bode diagram,1.,2.,Probl

29、em: Plot the Bode diagram of the system described by the open loop transfer function elements Solution: Elements: a gain of 10, a zero, a pole and an integrator Step 1: Calculate the break frequency,Sample Problem 15.1,Step 2: Determine the frequency rage to be plotted All break frequencies in the decade 1-10, So the Bode plot will be draw for Step 3: Plot the straight line magnitude approximations of each element. Step 4: Graphically add all element magnitude. Step 5: Plot deviations between true and appr