1、Automatic Control Theory,School of Information Science and Engineering,CSU Sunday, September 8, 2019,Chapter3 Time Domain Analysis of Linear System,Introduction 3.1 Performance indexes of system time response 3.2 First-order system 一阶系统 3.3 Second-order system 二阶系统 3.4 Higher-order system 高阶系统 3.5 S
2、tability analysis of linear system 3.6 Steady-state error calculating of linear system 3.7 Time domain design of control system,Introduction,Three problem of control theory:1、modeling;2、analysis;3、synthesis and design;The content of this chapter is system analysis in time domain.,Request of control
3、system:1. Stability The fundamental requirement of a control system. A system is stable if all the poles of the closed-loop transfer function have negative real parts.2. Transient-state performance damping degree and response speed.3. Steady-state performance *4. Robustness,Return,3.1.1 Typical inpu
4、t signals 典型输入信号 3.1.2 Transient-state and steady-state process动态过程与稳态过程 3.1.3 Transient-state and steady-state performance 动态性能与稳态性能,Return,3.1 Performance indexes of system time response,3.1.1 Typical input signals,Acceleration,Return,3.1.2 Transient-state and steady-state process,1. Transient-sta
5、te process 动态过程Systems output transits from initial state to steady state effect on some typical input signal. It can provide information such as stability, response speed, damping degree, and so on.Real control system must be stable.,Return,2. Steady-state process 稳态过程 Steady-state process is the b
6、ehavior mode of systems output under typical input signal when time tend towards infinite. It can also be called steady-state response, and it represents the degree that the systems output reflects its input.,3.1.3 Transient-state and steady-state performance,1. Dynamic performance 动态性能 Condition: s
7、ystem must be a stable system. Dynamic performance specifications 动态性能指标 reflect the change status of dynamic process to unit step signal with zero initial condition.,The typical response curve,Dynamic performance specification: Rise time 上升时间 tr：the shortest time to achieve the final or steady-stat
8、e values 10% to 90%. Or the first time to achieve the steady-state value. Delay time 延迟时间 td：the time for the system output to achieve half of the final value for the first time. Peak time 峰值时间 tp：the time for the system output to achieve the max value.,Overshoot 超调量 %：percentage overshoot of first
9、peak.,*Oscillation times 振荡次数 N：the times of curve oscillating before ts.,Settling time 调节时间 ts ：the time for the system output to settle down to within a tolerance band of the final value. Normally between , i.e.,B,The transient response of the system may be described in terms of two factors:1. The
10、 swiftness of response, as represented by the rise time and the peak time.2. The damping degree of the response, as represented by the overshoot.3. Settling time is a synthesis specification of swiftness and damping degree.There are contradictory requirements and a compromise must be obtained.,Retur
11、n,2. Steady state performance 稳态性能The steady-state error ess is a performance index to describe systems steady-state performance. Usually it is calculated when the input signal is step, ramp or acceleration. Steady-state error is the error when the time is large and the transient response has decaye
12、d, leaving the continuous response. (Only a stable system have steady-state error),3.2 First-order system,Return,3.2.1 Mathematic model 3.2.2 Unit step response 单位阶跃响应 3.2.3 Unit impulse response 单位脉冲响应 3.2.4 Unit ramp response 单位斜坡响应 3.2.5 Unit acceleration response,3.2.1 Mathematic model,First ord
13、er systems (Inertial惯性或惰性): systems described by a single first order differential equation. Find the time response to impulse, step, ramp and acceleration input.,T is the time-constant.,Differential equation:,Transfer function (zero-initial condition):,first order systems,Return,3.2.2 Unit step res
14、ponse,Concept: The response to unit step signal with zero initial condition, marked as h(t).,ts=3T (5%) ts=4T (2%),num=1;den1=1 1; t=0:0.02:50; figurestep(num,den1,t);grid%阶跃响应 hold on den2=2 1; den3=10 1; step(num,den2,t); step(num,den3,t);,Return,T =1s,T =2s,T =10s,3.2.3 Unit impulse response,Conc
15、ept: The response of a linear constant system to unit impulse signal with zero initial condition, marked as g(t).1. The relationship between g(t) and G(s),2. First-order system,Conclusion: Transfer function of a plant is the Laplace transform of its impulse response.,num=1;den1=1 1; t=0:0.02:10; fig
16、ureimpulse(num,den1,t);grid%脉冲响应,Return,3.2.4 Unit ramp response,Concept: The response of a linear constant system to unit ramp signal with zero initial condition.,Steady-state part,Transient-state part,Ramp response,input,output,Steady state error ! The value is T.,The steady state error ess in the
17、 ramp response of first-order system is:,Return,3.2.5 Unit acceleration response,Concept: The response of a linear constant system to unit acceleration signal with zero initial condition.,Acceleration Response,A first-order system can follow a impulse signal, a step signal, and a ramp signal (steady
18、 state error exists in the ramp response), but can not follow an acceleration signal !,The relationship between input signals, or output signals of a linear control system,And we can find,It suits every linear time-invariant system.,Return,3.3 Second-order system,Return,3.3.1 Mathematic model 3.3.2
19、Unit step response 3.3.3 Dynamic process analysis (under-damped) *3.3.4 Dynamic process analysis (over-damped) *3.3.5 Unit ramp response 3.3.6 Performance improvement *3.3.7 Response process of non-zero initial condition,欠阻尼,过阻尼,3.3.1 Mathematic model,Differential equation,- Damping ratio 阻尼比 wn- Na
20、tural frequency 无阻尼自然振荡角频率,Block diagram,Characteristic equation：,poles:,Return,3.3.2 Unit step response,1. The influence to roots distributing 根的分布 and stability 稳定性 of 1 over-damped 过阻尼 two unequal negative real roots stable 两个不相等的负实根 稳定=1 critically damped临界阻尼 two equal negative real roots stable
21、 两个相等的负实根 稳定,01 under-damped 欠阻尼 complex conjugate pairs of negative real stable 一对具有负实部的共轭复根 稳定=0 un-damped 无阻尼 two pure imaginary roots critically stable 两个纯虚根 临界稳定0 negative-damped 负阻尼 roots of positive real unstable 具有正实部的根 不稳定,unstable,stable,Step Response,-10,-1,01,2. Unit step response of ove
22、r-damped (1),num1=4;den1=1 6 4; t=0:0.02:10; figurestep(num1,den1,t);grid hold on num2=1; den2=0.191 1; num3=1; den3=1.3091 1; step(num2,den2,t); step(num3,den3,t);,3. Unit step response of critically damped (=1),4. Unit step response of under-damped (01),阻尼振荡频率,5. Unit step response of un-damped (=
23、0),zita0=0;zita1=0.1;zita2=0.2;zita3=0.3;zita4=0.4; zita5=0.5;zita6=0.6;zita7=0.7; zita8=0.8;zita9=1.0; zita10=2.0; T=1; t=0:0.02:12; num=1;den0=T*T 2*zita0*T 1; den1=T*T 2*zita1*T 1; den2=T*T 2*zita2*T 1; den3=T*T 2*zita3*T 1; den4=T*T 2*zita4*T 1; den5=T*T 2*zita5*T 1; den6=T*T 2*zita6*T 1; den7=T
24、*T 2*zita7*T 1; den8=T*T 2*zita8*T 1; den9=T*T 2*zita9*T 1; den10=T*T 2*zita10*T 1; figurestep(num,den0,t);grid hold on step(num,den1,t); step(num,den2,t); step(num,den3,t); step(num,den4,t); step(num,den5,t); step(num,den6,t); step(num,den7,t); step(num,den8,t); step(num,den9,t); step(num,den10,t);
25、,Return,3.3.3 Dynamic process analysis (under-damped),Instructional objectives:Calculate time response of second-order systems (under-damped).Apply the formulas for the overshoot, rise time, settling time, peak time and steady state error.,01,1、Dynamic performance indexes 1) delay time td,2) rise ti
26、me tr,3) peak time tp,4) overshoot %,5) settling time ts,The effects of and n,If n is fixed ，% ts (0 0.707) ts (0.707 1) , % ts (0 0.707) ts (0.707 1) If is fixed n , % invariable ts 、tr 、tp ,Optimum damping ratio: =0.707,(1) if the input is unit step, when KA =200，try to determine the value of ts 、
27、 tp 、 % (2)when KA=1500 or KA=13.5，what is the effect of the each vale of gain?,Example: Consider the open-loop transfer function of a system (unit feedback),Solution,When KA=200,Conclusion,KA,n, %, ts invariable; KA, , n, over-damped, without peak-time and overshoot, ts.,Return,*3.3.4 Dynamic proce
28、ss analysis (over-damped),Return,*3.3.5 Unit ramp response,Steady-state part:,Transient-state part:,1. Under-damped systems unit ramp response,Steady-state error:,The maximal error:,The maximal departure value of error response:,Settling time:,Error response curve:,Steady-state error:,2. Critical-da
29、mped systems unit ramp response,Settling time:,3. Over-damped systems unit ramp response,Steady-state error:,Return,3.3.6 Performance improvement,1. Adjust gain K,2. PD control of error,Td is differential time constant.,Closed-loop TF:,So systems performance can be improved.,3. Speed feedback of out
30、put,Closed-loop TF:,Exp.3-5 (a) is a proportional control system, (b) is a speed feedback system. Determine Kt to make,Calculate every performance indexes.,Solution:,Closed-loop TF of (a):,Closed-loop TF of (b):,Return,3.4 High-order system,*1. Unit step response of third-order system,Exp.3-6,num=5
31、25 30;den=1 6 10 8; sys=tf(num,den);%高阶系统建模,Transfer function:5 s2 + 25 s + 30 - s3 + 6 s2 + 10 s + 8,den1=1 6 10 8 0; z,p,k=tf2zp(num,den) %对传递函数因式分解,z =-3.0000-2.0000 p =-4.0000 -1.0000 + 1.0000i-1.0000 - 1.0000i k =5,sys1=zpk(z,p,k),Zero/pole/gain:5 (s+3) (s+2) - (s+4) (s2 + 2s + 2),r,p,k=residue
32、(num,den1) %部分分式展开,r =-0.2500 -1.7500 - 0.2500i-1.7500 + 0.2500i3.7500 p =-4.0000 -1.0000 + 1.0000i-1.0000 - 1.0000i0 k =,step(sys),2. Unit step response of high-order system,If all closed-loop poles lie in the left half s-plane, the relative magnitudes of the residues determine the relative importa
33、nce of the components in the expanded form of c(t).,Analysis：1) If there is a closed-loop zero close to a pole, then the residue at this pole is small and the coefficient of the transient response term corresponding to this pole becomes small. So a pair of closely located pole and zero will effectiv
34、ely cancel each other.,2) If a pole is located very far from the origin, the residue at this pole may be small.,The transients corresponding to such a remote pole are small and last a short time.So, terms in the expanded form of c(t) having very small residue contribute little to the transient respo
35、nse, and these terms may be neglected, if this is done, the higher-order system may be approximated by a lower-order one.,3. Dominant closed-loop poles 闭环主导极点If the ratios of the real parts of other poles exceeds 5 times to the pole(s), and there are no zeros nearby the pole(s), then the closed-loop
36、 poles nearest the imaginary axis will dominate in the transient response. Quite often the dominant poles occur in the form of complex conjugate pair, so the higher-order system may be approximated by a second-order system with a pair dominant poles.,other poles,Dominant poles,zeros,Instructional ob
37、jectives:Identify the dominant poles of a high-order system. Determine the damping ratio and natural frequency of a second order system.apply the formulas.,Example: Step response,Exp.3-7,Zero/pole/gain:8 (s+2.1) - (s+8) (s+2) (s2 + s + 1),sys=zpk(-2.1,-8 -2 -0.5+0.866*j -0.5-0.866*j ,8) %原四阶系统建模,sys
38、1=tf(1.05,1 1 1) %近似的二阶系统,Transfer function:1.05 - s2 + s + 1,step(sys,b-,sys1,r:),Return,3.5 Stability analysis of linear system,3.5.1 Basic concept of stability 3.5.2 The sufficient and necessary condition of stability of linear system 3.5.3 Routh-Hurwitz criterion(1877/1895) 3.5.4 Special instanc
39、e of Routh criterion 3.5.5 Application of Routh criterion,Return,3.5.1 Basic concept of stability,1. The stability of motion,Stable: If the circular cone is resting on its base and is tipped slightly, it returns to its original equilibrium 平衡 position. Neutral中立: If the cone rests on its side and is
40、 displaced slightly, it rolls with no tendency to leave the position on its side. Unstable: If the cone is placed on its tip and release, it falls onto its side.,Unstable balance point,Stable balance point,If a differential equation of x have solution x(t)，due to initial condition x(t0)=x0, and as t
41、o any positive number 0 there always exists a positive number (). As long as | | and the corresponding solution duing to satisfied the expression | |=t0) with initial condition x0 varing to , we call the state vector x(t) is stable, otherwise, it is unstable.,2. The definition of stability (Lyapunov
42、,1892),If = , it is said to be large bound stability.大范围稳定,Then we called x(t) have asymptotic stability. 渐近稳定 As to engineering practice, the concept of stability is the asymptotic stability, all other conditions belong to instability.,If there exists or,As to a linear system, if it is stable with
43、initial condition x0, then it is stable with any other initial condition, so it can be said to a stable system, namely, all motion to linear system is stable if one of its motions is stable.,3. Definition of stabilityBounded-input bounded-output stability (BIBO)有界输入有界输出稳定性: any bounded input results
44、 in bounded output. (outside)*Asymptotic stability 渐近稳定性 (inside),Return,3.5.2 The sufficient and necessary condition of stability of linear system,1. The sufficient and necessary condition for a feedback system to be stable is that all poles of the system (roots of the characteristic equation) need
45、 to have negative real parts, i.e. be on the left half of the complex plane.,Eigenvalue criterion (1) all poles have negative real parts stable (2) one pole have positive real parts unstable (3) on the imaginary(jw) axis with all other roots in the left half-plane critical stable,System is stable!,E
46、xample 1: Servo-system:,Characteristic equation:,characteristic roots:,All poles have negative real parts,n=0.025 0.55 1.5 1 1; p=roots(n) %求特征根,p =-18.9373 -2.6060 -0.2284 + 0.8709i-0.2284 - 0.8709i,System is unstable!,Example 2: Servo-system:,Characteristic equation:,characteristic roots:,Two pole
47、s have positive real parts.,2. The first method of Lyapunov,The first theorem of Lyapunov The actual system is stable if the linearize systems roots are located in the left half of the s-plane. The second theorem of LyapunovThe actual system is unstable if one or more roots of the linearize system a
48、re found in the right half of the s-plane.,Nonlinear system can be linearized with small range theory.,If one root of the linearize system locate on imaginary axis of the s-plane, in order to determine the stability, we must use the second method of Lyapunov. Because on that condition, the stability
49、 is relative to the high-order terms which are neglected.,Nonlinear system can not be linearizedCriterion: the second method of Lyapunov (chapter 9)The stability of the system is determined by checking on the properties of the Lyapunov function of the system. It can also be applied to linear systems.,
