2020161自动控制原理(中英文)
自动控制原理（中英文）
自动控制原理（中英文）《自动控制原理》课程教学大纲课程编号：２020161课程类别：必修授课对象:本科三年级先修课程:复变函数，积分变换,信号与系统。
学分:4总学时:５6 课内学时:４８实验学时: ８一、课程性质、教学目得与任务课程性质:专业基础课，专业知识链条中得关键环节之一，自动控制原理就是仪器仪表类、测控类专业得重要基础课之一,这些专业主要学习信号传感(获取）、信号处理、控制及光机电系统等知识，而控制就是知识链条中得重要一环，随着科技发展，自动化、智能化已成为仪器、产品、系统等得重要功能，这就要求学生必须具备自动控制方面得知识。
教学目得与任务：培养学生自动控制原理得基础知识,学习掌握经典控制得基本理论、基本方法与控制系统得基本设计方法,重点学习分析与设计线性控制系统得基本理论、基本方法及控制系统设计方法.主要内容包括:控制系统得数学模型、控制系统得时域分析法、控制系统得根轨迹法、控制系统得频域分析法、控制系统得常用校正方法等。
二、教学基本要求学习经典控制得基本理论与基本方法,重点学习分析与设计线性控制系统得基本理论与基本方法。
主要内容包括：控制系统得数学模型、控制系统得时域分析法、控制系统得根轨迹法、控制系统得频域分析法、控制系统得常用校正方法等。
三、教学内容第一章控制系统得一般要概念 (4课时）自动控制得基本原理与方式,自动控制系统示例，自动控制系统得分类，对自动控制系统得基本要求1、基本概念;2、反馈系统基本组成；3、基本控制方式；4、控制系统分类:开环、闭环、复合控制;第二章控制系统得数学模型 (8课时）控制系统得时域数学模型,拉普拉斯变换，控制系统得复域数学模型，控制系统得状态空间模型,控制系统得结构图与信号流图２－１时域模型、微分方程表示方法;2－2 复域模型1、传递函数得定义与性质;2、传递函数得零、极点表示，开环增益、根轨迹增益等;3、典型环节得传递函数（比例、惯性、微分、积分、振荡）；2－3 控制系统得结构图与信号流图1、结构图得等效变换与化简2、信号流图组成与性质A．性质、术语（理解）B。
自动控制原理英文版课后全部_答案
Module3Problem 3.1(a) When the input variable is the force F. The input variable F and the output variable y are related by the equation obtained by equating the moment on the stick:2.233y dylF lk c l dt=+Taking Laplace transforms, assuming initial conditions to be zero,433k F Y csY =+leading to the transfer function31(4)Y k F c k s=+ where the time constant τ is given by4c kτ=(b) When F = 0The input variable is x, the displacement of the top point of the upper spring. The input variable x and the output variable y are related by the equation obtained by the moment on the stick:2().2333y y dy k x l kl c l dt-=+Taking Laplace transforms, assuming initial conditions to be zero,3(24)kX k cs Y =+leading to the transfer function321(2)Y X c k s=+ where the time constant τ is given by2c kτ=Problem 3.2 P 54Determine the output of the open-loop systemG(s) = 1asT+to the inputr(t) = tSketch both input and output as functions of time, and determine the steady-state error between the input and output. Compare the result with that given by Fig3.7 . Solution ：While the input r(t) = t , use Laplace transforms, Input r(s)=21sOutput c(s) = r(s) G(s) = 2(1)aTs s ⋅+ = 211T T a s s Ts ⎛⎫ ⎪-+ ⎪ ⎪+⎝⎭the time-domain response becomes c(t) = ()1t Tat aT e ---Problem 3.33.3 The massless bar shown in Fig.P3.3 has been displaced a distance 0x and is subjected to a unit impulse δ in the direction shown. Find the response of the system for t>0 and sketch the result as a function of time. Confirm the steady-state response using the final-value theorem. Solution ：The equation obtained by equating the force:00()kx cxt δ+=Taking Laplace transforms, assuming initial condition to be zero,K 0X +Cs 0X =1leading to the transfer function()XF s =1K Cs +=1C1K s C+The time-domain response becomesx(t)=1CC tK e -The steady-state response using the final-value theorem:lim ()t x t →∞=0lim s →s 1K Cs +1s =1K00000()()()1;11111()K t CK x x Cx t Kx X K Cs Kx Kx X C Cs K K s KKx x t eCδ-++=⇒++=--∴==⋅++-=⋅According to the final-value theorem:0001lim ()lim lim 01t s s Kx sx t s X C K s K→∞→→-=⋅=⋅=+ Problem 3.4 Solution:1.If the input is a unit step, then1()R s s=()()11R s C s sτ−−−→−−−→+ leading to,1()(1)C s s sτ=+taking the inverse Laplace transform gives,()1tc t e τ-=-as the steady-state output is said to have been achieved once it is within 1% of the final value, we can solute ―t‖ like this,()199%1tc t e τ-=-=⨯ (the final value is 1) hence,0.014.60546.05te t sττ-==⨯=(the time constant τ=10s)2.the numerical value of the numerator of the transfer function doesn’t affect the answer. See this equation, If ()()()1C s AG s R s sτ==+ then()(1)A C s s sτ=+giving the time-domain response()(1)tc t A e τ-=-as the final value is A, the steady-state output is achieved when,()(1)99%tc t A e A τ-=-=⨯solute the equation, t=4.605τ=46.05sthe result make no different from that above, so we said that the numerical value of the numerator of the transfer function doesn’t affect the answer.If a<1, as the time increase, the two lines won`t cross. In the steady state the output lags the input by a time by more than the time constant T. The steady error will be negative infinite.R(t)C(t)Fig 3.7 tR(t)C(t)tIf a=1, as the time increase, the two lines will be parallel. It is as same as Fig 3.7.R(t)C(t)tIf a>1, as the time increase, the two lines will cross. In the steady state the output lags the input by a time by less than the time constant T.The steady error will be positive infinite.Problem 3.5 Solution: R(s)=261s s+, Y(s)=26(51)s s s +⋅+=229614551s s s -+++ /5()62929t y t t e -∴=-+so the steady-state error is 29(-30). To conform the result:5lim ()lim(62929);tt t y t t -→∞→∞=-+=∞6lim ()lim ()lim ()lim(51)t s s s s y t y s Y s s s →∞→→→+====∞+.20lim ()lim ()lim [()()]161lim [()1]()lim (1)()5130ss t s s s s e e t S E S S Y S R S S G S R S S S S S→∞→→→→==⋅=⋅-=⋅-=⋅-⋅++=- Therefore, the solution is basically correct.Problem 3.623yy x += since input is of constant amplitude and variable frequency , it can be represented as:j tX eA ω=as we know ,the output should be a sinusoidal signal with the same frequency of the input ,it can also be represented as:R(t)C(t)t0j t y y e ω=hence23j tj tj tj yyeeeA ωωωω+=00132j y Aω=+ 0294Ayω=+ 2tan3w ϕ=- Its DC(w→0) value is 003Ay ω==Requirement 01122w yy==21123294AA ω=⨯+ →32w = while phase lag of the input:1tan 14πϕ-=-=-Problem 3.7One definition of the bandwidth of a system is the frequency range over which the amplitude of the output signal is greater than 70% of the input signal amplitude when a system is subjected to a harmonic input. Find a relationship between the bandwidth and the time constant of a first-order system. What is the phase angle at the bandwidth frequency ？ Solution ：From the equation 3.41000.71r A r ωτ22=≥+ (1)and ω≥0 (2) so 1.020ωτ≤≤so the bandwidth 1.02B ωτ=from the equation 3.43the phase angle 110tan tan 1.024c πωτ--∠=-=-=Problem 3.8 3.8 SolutionAccording to generalized transfer function of First-Order Feedback Systems11C KG K RKGHK sτ==+++the steady state of the output of this system is 2.5V .∴if s →0, 2.51104C R→=. From this ,we can get the value of K, that is 13K =.Since we know that the step input is 10V , taking Laplace transforms,the input is 10S.Then the output is followed1103()113C s S s τ=⨯++Taking reverse Laplace transforms,4/4332.5 2.5 2.5(1)t t C e eττ--=-=-From the figure, we can see that when the time reached 3s,the value of output is 86% of the steady state. So we can know34823(2)*4393τττ-=-⇒-=-⇒=, 4/3310.8642t t e ττ-=-=⇒=The transfer function is3128s +146s+Let 12+8s=0, we can get the pole, that is 1.5s =-2/3- Problem 3.9 Page 55 Solution:The transfer function can be represented,()()()()()()()o o m i m i v s v s v s G s v s v s v s ==⋅While,()1()111//()()11//o m m i v s v s sRCR v s sC sC v s R R sC sC =+⎛⎫+ ⎪⎝⎭=⎡⎤⎛⎫++ ⎪⎢⎥⎝⎭⎣⎦Leading to the final transfer function,21()13()G s sRC sRC =++ And the reason:the second simple lag compensation network can be regarded as the load of the first one, and according to Load Effect , the load affects the primary relationship; so the transfer function of the comb ination doesn’t equal the product of the two individual lag transfer functio nModule4Problem4.14.1The closed-loop transfer function is10(6)102(6)101610S S S S C RS s +++++==Comparing with the generalized second-order system,we getProblem4.34.3Considering the spring rise x and the mass rise y. Using Newton ’s second law of motion..()()d x y m y K x y c dt-=-+Taking Laplace transforms, assuming zero initial conditions2mYs KX KY csX csY =-+-resulting in the transfer funcition where2Y cs K X ms cs K +=++ And521.26*10cmkc ζ== Problem4.4 Solution:The closed-loop transfer function is210263101011n n d n W EW E W W E ====-=2121212K C K S S K R S S K S S ∙+==+++∙+Comparing the closed-loop transfer function with the generalized form,2222n n nCR s s ωξωω=++ it is seen that2n K ω= And that22n ξω= ; 1Kξ=The percentage overshoot is therefore21100PO eξπξ--=11100k keπ-∙-=Where 10%PO ≤When solved, gives 1.2K ≤（2.86）When K takes the value 1.2, the poles of the system are given by22 1.20s s ++=Which gives10.45s j =-±±s=-1 1.36jProblem4.5ReIm0.45-0.45-14.5 A unity-feedback control system has the forward-path transfer functionG (s) =10)S(s K+Find the closed-loop transfer function, and develop expressions for the damping ratio And damped natural frequency in term of K Plot the closed-loop poles on the complex Plane for K = 0,10,25,50,100.For each value of K calculate the corresponding damping ratio and damped natural frequency. What conclusions can you draw from the plot?Solution: Substitute G(s)=(10)K s s + into the feedback formula : Φ(s)=()1()G S HG S +.And in unitfeedback system H=1. Result in: Φ(s)=210Ks s K++ So the damped natural frequencyn ω=K ,damping ratio ζ=102k =5k.The characteristic equation is 2s +10S+K=0. When K ≤25,s=525K -±-; While K>25,s=525i K -±-; The value ofn ω and ζ corresponding to K are listed as follows.K 0 10 25 50 100 Pole 1 1S 0 515-+ -5 -5+5i 553i -+Pole 2 2S -10 515-- -5 -5-5i553i --n ω 010 5 52 10 ζ ∞2.51 0.5 0.5Plot the complex plane for each value of K:We can conclude from the plot.When k ≤25,poles distribute on the real axis. The smaller value of K is, the farther poles is away from point –5. The larger value of K is, the nearer poles is away from point –5.When k>25,poles distribute away from the real axis. The smaller value of K is, the further (nearer) poles is away from point –5. The larger value of K is, the nearer (farther) poles is away from point –5.And all the poles distribute on a line parallels imaginary axis, intersect real axis on the pole –5.Problem4.61tb b R L C b o v dv i i i i v dt C R L dt=++=++⎰Taking Laplace transforms, assuming zero initial conditions, reduces this equation to011b I Cs V R Ls ⎛⎫=++ ⎪⎝⎭20b V RLs I Ls R RLCs =++ Since the input is a constant current i 0, so01I s=then,()2b RLC s V Ls R RLCs==++ Applying the final-value theorem yields ()()0lim lim 0t s c t sC s →∞→==indicating that the steady-state voltage across the capacitor C eventually reaches the zero ,resulting in full error.Problem4.74.7 Prove that for an underdamped second-order system subject to a step input, thepercentage overshoot above the steady-state output is a function only of the damping ratio .Fig .4.7SolutionThe output can be given by222222()(2)21()(1)n n n n n n C s s s s s s s ωζωωζωζωωζ=+++=-++- (1)the damped natural frequencyd ω can be defined asd ω=21n ωζ- (2)substituting above results in22221()()()n n n d n d s C s s s s ζωζωζωωζωω+=--++++ (3) taking the inverse transform yields22()1sin()11tan n t d e c t t where ζωωφζζφζ-=-+--=(4)the maximum output is22()1sin()11n t p d p p d n e c t t t ζωωφζππωωζ-=-+-==-(5)so the maximum is2/1()1p c t eπζζ--=+the percentage overshoot is therefore2/1100PO eπζζ--=Problem4.8 Solution to 4.8:Considering the mass m displaced a distance x from its equilibrium position, the free-body diagram of the mass will be as shown as follows.aP cdx kxkxmUsing Newton ’s second law of motion,22p k x c x mx m x c x k x p--=++=Taking Laplace transforms, assuming zero initial conditions,2(2)X ms cs k P ++= results in the transfer function2/(1/)/((/)2/)X P m s c m s k m =++ 2(2/)(2/)((/)2/)k k m s c m s k m =++As we see2(2)X m s c s k P++= As P is constantSo X ∝212ms cs k ++ . When 56.25102cs m-=-=-⨯ ()25min210mscs k ++=4max5100.110X == This is a second-order transfer function where 22/n k m ω= and/2/22n c w m c k m ζ== The damped natural frequency is given by 2212/1/8d n k m c km ωωζ=-=-22/(/2)k m c m =- Using the given data,462510/2100.050.2236n ω=⨯⨯⨯== 462502.79501022100.05ζ-==⨯⨯⨯⨯ ()240.22361 2.7950100.2236d ω-=⨯-⨯= With these data we can draw a picture14.0501160004.673600p de s e T T πωτζωτ======222222112/1222()22,,,428sin (sin cos )0tan 7.030.02n n pp dd n dd n ntd d t t t n d p d d p ddd p p p nX k m c k P ms cs k k m s s s m m k c k c cm m m m km p x e tm p xe t t m t t x m ζωζωωωζωωωωζζωωωζωωωωωωωζω--===⋅=⋅++++++=-===∴==-+=∴=⇒=⇒= 其中Problem4.10 4.10 solution:The system is similar to the one in the book on PAGE 58 to PAGE 63. The difference is the connection of the spring. So the transfer function is2222l n d n n w s w s w θθζ=++222(),;p a m ld a m p m l m l l m mm l lk k k N RJs RCs R k k N k J N J J C N c c N N N θθωθωθ=+++=+=+===p a mn K K K w NJ R='damping ratio 2p a m c NRK K K J ζ='But the value of J is different, because there is a spring connected.122s m J J J J N N '=++Because of final-value theorem,2l nd w θθζ=Module5Problem5.45．4 The closed-loop transfer function of the system may be written as2221010(1)610101*********CR K K K S S K K S S K S S +++==+++++++ The closed-loop poles are the solutions of the characteristic equation6364(1010)3110210(1)n K S K JW K -±-+==-±+=+ 210(1)6310(1)E K E K +==+In order to study the stability of the system, the behavior of the closed-loop poles when the gain K increases from zero to infinte will be observed. So when12K = 3010E =321S J =-± 210K = 3110110E =3101S J =-± 320K = 21070E =3201S J =-±双击下面可以看到原图ReProblem5.5SolutionThe closed-loop transfer function is2222(1)1(1)KC K KsKR s K as s aKs Kass===+++++∙+Comparing the closed-loop transfer function with the generalized form, 2222nn nCR s sωξωω=++Leading to2nKa Kωξ==The percentage overshoot is therefore2110040%PO eξπξ--==Producing the result0.869ξ=（0.28）And the peak time241PnT sπωξ==-Leading to1.586nω=（0.82）Problem5.75.7 Prove that the rise time T r of a second-order system with a unit step input is given byT r = d ω1 tan -1n dζωω = d ω1 tan -1d ωζ21--Plot the rise against the damping ratio.Solution:According to (4.33):c(t)=1-2(cos sin )1n t d d e t t ζωζωωζ-+-. 4.33When t=r T ,c(t)=1.substitue c(t)= 1 into (4.33) Producing the resultr T =d ω1 tan -1n dζωω = d ω1 tan -112ζζ--Plot the rise time against the damping ratio:Problem5.9Solution to 5.9:As we know that the system is the open-loop transfer function of a unity-feedback control system.So ()()GH S G S = Given as()()()425KGH s s s =-+The close-loop transfer function of the system may be written as()()()()()41254G s C Ks R GH s s s K ==+-++ The characteristic equation is()()2254034100s s K s s K -++=⇒++-=According to the Routh ’s method, the Routh ’s array must be formed as follow20141030410s K s s K -- For there is no closed-loop poles to the right of the imaginary axis4100 2.5K K -≥⇒≥ Given that 0.5ζ=4103 4.752410n K K K ωζ=-=⇒=- When K=0, the root are s=+2,-5According to the characteristic equation, the solutions are349424s K =-±-while 3.0625K ≤, we have one or two solutions, all are integral number.Or we will have solutions with imaginary number. So we can drawK=102 -5 K=0K=3.0625K=2.5 K=10Open-loop polesClosed-loop polesProblem5.10 5.10 solution:0.62/n w rad sζ==according to()211sin()21n w t d e c w t ζφζ-=-+=- 1.2sin(1.6)0.4t e t φ-⋅+= 4t a n3φ= finally, t is delay time:1.23t s ≈(0.67)Module6Problem 6.3First we assume the disturbance D to be zero:e R C =-1011C K e s s =⋅⋅⋅+Hence:(1)10(1)e s s R K s s +=++ Then we set the input R to be zero:10()(1)C K e D e s s =⋅+⋅=-+ ⇒ 1010(1)e D K s s =-++Adding these two results together:(1)1010(1)10(1)s s e R D K s s K s s +=⋅-⋅++++21()R s s =; 1()D s s= ∴222110910(1)10(1)100(1)s s e Ks s s Ks s s s s s +-=-=++++++ the steady-state error:232200099lim lim lim 0.09100100ss s s s s s s e s e s s s s s →→→--=⋅===-++++Problem 6.4Determine the disturbance rejection ratio(DRR) for the system shown in Fig P.6.4+fig.P.6.4 solution ：from the diagram we can know :0.210.05mv K RK c === so we can get that()0.21115()0.05v m m OL n CL K K DRR cR ωω∆⨯==+=+=∆210.10.050.050.025s s =++, so c=0.025, DRR=9Problem 6.5 6.5 SolutionFor the purposes of determining the steady-state error of the system, we should get to know the effect of the input and the disturbance along when the other will be assumed to be zero.First to simplify the block diagram to the following patter:110s +2021Js Tddθoθ0.220.10.05s ++__+d T—Allowing the transfer function from the input to the output position to be written as01220220d Js s θθ=++ 012222020240*220220(220)dJs s Js s s Js s sθθ===++++++ According to the equation E=R-C:022*******(2)()lim[()()]lim[(1)]lim 0.2220220ssr d s s s Js e s s s s Js s Js s δδδθθ→→→+=-=-==++++问题;1. 系统型为2，对于阶跃输入，稳态误差为0.2. 终值定理写的不对。
《自动控制原理B(双语)》课程简介和教学大纲.doc
自动控制原理（B）课程简介课程编号：15J8047课程名称：（中文）自动控制原理（B）（英文）Introduction to Automatic Control学分/学时：42学时，实验6学时开课学期：秋季先修课程：高等数学、工程数学、大学物理内容简介：（500字以内）《自动控制原理》属于技术科学，是为专业课学习和参加控制工程实践而设置的，研究的对象是自动控制系统，研究的中心问题是控制系统在控制过程中的性能，学科的基本内容分为数学模型、工程分析计算方法和系统一般规律三个部分。
要求学生能够正确地理解和运用课程的基本概念及基本理论，逐步掌握“定性分析、定量估算和仿真实验"的研究问题的方法。
本课程内容主要为经典控制中的线性理论，课程前后内容联系紧密、系统性强，所附加的实验课，以学习控制系统的基本实验方法和培养测试、分析能力为主。
自动控制原理（B）课程教学大纲课程编号：15J8047课程名称：（中文）自动控制原理（B）（英文）Introduction to Automatic Control学分/学时：42学时，实验6学时先修课程：高等数学、工程数学、大学物理课程教学目标课程的性质：本科专业基本课，为必修课目的和任务：本课程是研究自动控制系统规律性的一门工程学科，是重要的专业基本课程。
通过本课程的教学工作，要求学生能够正确地理解和运用课程的基本概念和理论，逐步掌握“定性分析、定量估算和仿真实验”的研究问题的方法，具有从事与飞行器自动控制系统有■关的丁程实践与理论研究的基本知识，并为进一步学习现代控制理论打下基本。
教学内容及基本要求第一章自动控制的一般概念（2学时）1.1自动控制的任务1.2自动控制的基本方式1.3自动控制系统示例1.4对控制系统的性能要求第二章自动控制系统的数学模型（8学时）2.1控制系统微分方程的建立2.2非线性微分方程的线性化2.3拉普拉斯变换2.4传递函数2.5动态结构图2.6结构图的等效变换2.7典型反馈系统的几种传递函数第三章时域分析法（8学时）3.1典型响应及性能指标3.2 一、二阶系统分析3.3二阶系统分析3.4系统稳态误差分析及计算第四章根轨迹法（6学时）4.1根轨迹与根轨迹方程4.2根轨迹的绘制法则4.3系统闭环零、极点分布与阶跃响应的关系4.4系统阶跃响应的根轨迹分析第五章频率法（8学时）5.1频率特性5.2典型环节的频率特性5.3开环系统的频率特性5.4频率稳定判据5.5系统闭环频率特性与阶跃响应的关系5.5开环频率特性与系统阶跃响应的关系第六章控制系统的校正方法（4学时）6.1系统校正设计基础6.2串联校正6.3串联校正的理论设计方法6.4反馈校正6.5复合校正第七章状态空间分析方法（6学时）7.1状态空间方法基础7.2线性系统的可控性和可观测性7.3状态反馈与状态观测器7.4有界输入、有界输出稳定性7.5李雅普诺夫第二方法.教学安排及方式学时分配：讲课实验习题课第一章21第二章82第三章821第四章61第五章821第六章421第七章6课内外学时比：1:1.5实验安排：共三次实验，每次二小时。
《自动控制原理》试卷及答案(英文10套)
AUTOMATIC CONTROL THEOREM (1)⒈ Derive the transfer function and the differential equation of the electric network⒉ Consider the system shown in Fig.2. Obtain the closed-loop transfer function)()(S R S C , )()(S R S E . （12%） ⒊ The characteristic equation is given 010)6(5)(123=++++=+K S K S S S GH . Discuss the distribution of the closed-loop poles. (16%)① There are 3 roots on the LHP ② There are 2 roots on the LHP② There are 1 roots on the LHP ④ There are no roots on the LHP . K=?⒋ Consider a unity-feedback control system whose open-loop transfer function is )6.0(14.0)(++=S S S S G . Obtain the response to a unit-step input. What is the rise time for this system? What is the maximum overshoot? （10%）Fig.15. Sketch the root-locus plot for the system )1()(+=S S K S GH . ( The gain K is assumed to be positive.)① Determine the breakaway point and K value.② Determine the value of K at which root loci cross the imaginary axis.③ Discuss the stability. (12%)6. The system block diagram is shown Fig.3. Suppose )2(t r +=, 1=n . Determine the value of K to ensure 1≤e . (12%)Fig.37. Consider the system with the following open-loop transfer function:)1)(1()(21++=S T S T S K S GH . ① Draw Nyquist diagrams. ② Determine the stability of the system for two cases, ⑴ the gain K is small, ⑵ K is large. (12%)8. Sketch the Bode diagram of the system shown in Fig.4. (14%)⒈212121121212)()()(C C S C C R R C S C C R S V S V ++++=⒉ 2423241321121413211)()(H G H G G G G G G G H G G G G G G G S R S C ++++++=⒊ ① 0<K<6 ② K ≤0 ③ K ≥6 ④ no answer⒋⒌①the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2③⒍5.75.3≤≤K⒎ )154.82)(181.34)(1481.3)(1316.0()11.0(62.31)(+++++=S S S S S S GHAUTOMATIC CONTROL THEOREM (2)⒈Derive the transfer function and the differential equation of the electric network⒉ Consider the equation group shown in Equation.1. Draw block diagram and obtain the closed-loop transfer function )()(S R S C . （16% ） Equation.1 ⎪⎪⎩⎪⎪⎨⎧=-=-=--=)()()()()]()()([)()]()()()[()()()]()()[()()()(3435233612287111S X S G S C S G S G S C S X S X S X S G S X S G S X S C S G S G S G S R S G S X⒊ Use Routh ’s criterion to determine the number of roots in the right-half S plane for the equation 0400600226283)(12345=+++++=+S S S S S S GH . Analyze stability.（12% ）⒋ Determine the range of K value ,when )1(2t t r ++=, 5.0≤SS e . （12% ）Fig.1⒌Fig.3 shows a unity-feedback control system. By sketching the Nyquist diagram of the system, determine the maximum value of K consistent with stability, and check the result using Routh ’s criterion. Sketch the root-locus for the system （20%）（18% ）⒎ Determine the transfer function. Assume a minimum-phase transfer function.（10% ）⒈1)(1)()(2122112221112++++=S C R C R C R S C R C R S V S V⒉ )(1)()(8743215436324321G G G G G G G G G G G G G G G G S R S C -+++=⒊ There are 4 roots in the left-half S plane, 2 roots on the imaginary axes, 0 root in the RSP. The system is unstable.⒋ 208<≤K⒌ K=20⒍⒎ )154.82)(181.34)(1481.3)(1316.0()11.0(62.31)(+++++=S S S S S S GHAUTOMATIC CONTROL THEOREM (3)⒈List the major advantages and disadvantages of open-loop control systems. (12% )⒉Derive the transfer function and the differential equation of the electric network⒊ Consider the system shown in Fig.2. Obtain the closed-loop transfer function)()(S R S C , )()(S R S E , )()(S P S C . （12%）⒋ The characteristic equation is given 02023)(123=+++=+S S S S GH . Discuss the distribution of the closed-loop poles. (16%)5. Sketch the root-locus plot for the system )1()(+=S S K S GH . (The gain K is assumed to be positive.)④ Determine the breakaway point and K value.⑤ Determine the value of K at which root loci cross the imaginary axis. ⑥ Discuss the stability. (14%)6. The system block diagram is shown Fig.3. 21+=S K G , )3(42+=S S G . Suppose )2(t r +=, 1=n . Determine the value of K to ensure 1≤SS e . (15%)7. Consider the system with the following open-loop transfer function:)1)(1()(21++=S T S T S K S GH . ① Draw Nyquist diagrams. ② Determine the stability of the system for two cases, ⑴ the gain K is small, ⑵ K is large. (15%)⒈ Solution: The advantages of open-loop control systems are as follows: ① Simple construction and ease of maintenance② Less expensive than a corresponding closed-loop system③ There is no stability problem④ Convenient when output is hard to measure or economically not feasible. (For example, it would be quite expensive to provide a device to measure the quality of the output of a toaster.)The disadvantages of open-loop control systems are as follows:① Disturbances and changes in calibration cause errors, and the output may be different from what is desired.② To maintain the required quality in the output, recalibration is necessary from time to time.⒉ 1)(1)()()(2122112221122112221112+++++++=S C R C R C R S C R C R S C R C R S C R C R S U S U ⒊351343212321215143211)()(H G G H G G G G H G G H G G G G G G G G S R S C +++++= 35134321232121253121431)1()()(H G G H G G G G H G G H G G H G G H G G G G S P S C ++++-+=⒋ R=2, L=1⒌ S:①the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2⒍5.75.3≤≤KAUTOMATIC CONTROL THEOREM (4)⒈ Find the poles of the following )(s F :se s F --=11)( (12%)⒉Consider the system shown in Fig.1,where 6.0=ξ and 5=n ωrad/sec. Obtain the rise time r t , peak time p t , maximum overshoot P M , and settling time s t when the system is subjected to a unit-step input. （10%）⒊ Consider the system shown in Fig.2. Obtain the closed-loop transfer function)()(S R S C , )()(S R S E , )()(S P S C . （12%）⒋ The characteristic equation is given 02023)(123=+++=+S S S S GH . Discuss the distribution of the closed-loop poles. (16%)5. Sketch the root-locus plot for the system )1()(+=S S K S GH . (The gain K is assumed to be positive.)⑦ Determine the breakaway point and K value.⑧ Determine the value of K at which root loci cross the imaginary axis.⑨ Discuss the stability. (12%)6. The system block diagram is shown Fig.3. 21+=S K G , )3(42+=S S G . Suppose )2(t r +=, 1=n . Determine the value of K to ensure 1≤SS e . (12%)7. Consider the system with the following open-loop transfer function:)1)(1()(21++=S T S T S K S GH . ① Draw Nyquist diagrams. ② Determine the stability of the system for two cases, ⑴ the gain K is small, ⑵ K is large. (12%)8. Sketch the Bode diagram of the system shown in Fig.4. (14%)⒈ Solution: The poles are found from 1=-s e or 1)sin (cos )(=-=-+-ωωσωσj e e j From this it follows that πωσn 2,0±== ),2,1,0( =n . Thus, the poles are located at πn j s 2±=⒉Solution: rise time sec 55.0=r t , peak time sec 785.0=p t ,maximum overshoot 095.0=P M ,and settling time sec 33.1=s t for the %2 criterion, settling time sec 1=s t for the %5 criterion.⒊ 351343212321215143211)()(H G G H G G G G H G G H G G G G G G G G S R S C +++++= 35134321232121253121431)1()()(H G G H G G G G H G G H G G H G G H G G G G S P S C ++++-+=⒋R=2, L=15. S:①the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2⒍5.75.3≤≤KAUTOMATIC CONTROL THEOREM (5)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer function )()(S R S C , )()(S R S E . （18%）⒉ The characteristic equation is given 0483224123)(12345=+++++=+S S S S S S GH . Discuss the distribution of the closed-loop poles. (16%)⒊ Sketch the root-locus plot for the system )15.0)(1()(++=S S S K S GH . (The gain K is assumed to be positive.)① Determine the breakaway point and K value.② Determine the value of K at which root loci cross the imaginary axis. ③ Discuss the stability. (18%)⒋ The system block diagram is shown Fig.2. 1111+=S T K G , 1222+=S T K G . ①Suppose 0=r , 1=n . Determine the value of SS e . ②Suppose 1=r , 1=n . Determine the value of SS e . (14%)⒌ Sketch the Bode diagram for the following transfer function. )1()(Ts s K s GH +=, 7=K , 087.0=T . (10%)⒍ A system with the open-loop transfer function )1()(2+=TS s K S GH is inherently unstable. This system can be stabilized by adding derivative control. Sketch the polar plots for the open-loop transfer function with and without derivative control. (14%)⒎ Draw the block diagram and determine the transfer function. (10%)⒈∆=321)()(G G G S R S C ⒉R=0, L=3,I=2⒋①2121K K K e ss +-=②21211K K K e ss +-= ⒎11)()(12+=RCs s U s UAUTOMATIC CONTROL THEOREM (6)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer function )()(S R S C , )()(S R S E . （18%）⒉The characteristic equation is given 012012212010525)(12345=+++++=+S S S S S S GH . Discuss thedistribution of the closed-loop poles. (12%)⒊ Sketch the root-locus plot for the system )3()1()(-+=S S S K S GH . (The gain K is assumed to be positive.)① Determine the breakaway point and K value.② Determine the value of K at which root loci cross the imaginary axis. ③ Discuss the stability. (15%)⒋ The system block diagram is shown Fig.2. SG 11=, )125.0(102+=S S G . Suppose t r +=1, 1.0=n . Determine the value of SS e . (12%)⒌ Calculate the transfer function for the following Bode diagram of the minimum phase. (15%)⒍ For the system show as follows, )5(4)(+=s s s G ,1)(=s H , (16%) ① Determine the system output )(t c to a unit step, ramp input.② Determine the coefficient P K , V K and the steady state error to t t r 2)(=.⒎ Plot the Bode diagram of the system described by the open-loop transfer function elements )5.01()1(10)(s s s s G ++=, 1)(=s H . (12%)w⒈32221212321221122211)1()()(H H G H H G G H H G G H G H G H G G G S R S C +-++-+-+= ⒉R=0, L=5 ⒌)1611()14)(1)(110(05.0)(2s s s s s s G ++++= ⒍t t e e t c 431341)(--+-= t t e e t t c 41213445)(---+-= ∞=P K , 8.0=V K , 5.2=ss eAUTOMATIC CONTROL THEOREM (7)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer function)()(S R S C , )()(S R S E . （16%）⒉ The characteristic equation is given 01087444)(123456=+--+-+=+S S S S S S S GH . Discuss the distribution of the closed-loop poles. (10%)⒊ Sketch the root-locus plot for the system 3)1()(S S K S GH +=. (The gain K is assumed to be positive.)① Determine the breakaway point and K value.② Determine the value of K at which root loci cross the imaginary axis. ③ Discuss the stability. (15%)⒋ Show that the steady-state error in the response to ramp inputs can be made zero, if the closed-loop transfer function is given by:nn n n n n a s a s a s a s a s R s C +++++=---1111)()( ;1)(=s H （12%）⒌ Calculate the transfer function for the following Bode diagram of the minimum phase.（15%）w⒍ Sketch the Nyquist diagram (Polar plot) for the system described by the open-loop transfer function )12.0(11.0)(++=s s s S GH , and find the frequency and phase such that magnitude is unity. （16%）⒎ The stability of a closed-loop system with the following open-loop transfer function )1()1()(122++=s T s s T K S GH depends on the relative magnitudes of 1T and 2T . Draw Nyquist diagram and determine the stability of the system.（16%） ( 00021>>>T T K )⒈3213221132112)()(G G G G G G G G G G G G S R S C ++-++=⒉R=2, I=2,L=2 ⒌)1()1()(32122++=ωωωs s s s G⒍o s rad 5.95/986.0-=Φ=ωAUTOMATIC CONTROL THEOREM (8)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer function)()(S R S C , )()(S R S E . （16%）⒉ The characteristic equation is given 04)2(3)(123=++++=+S K KS S S GH . Discuss the condition of stability. (12%)⒊ Draw the root-locus plot for the system 22)4()1()(++=S S KS GH ;1)(=s H .Observe that values of K the system is overdamped and values of K it is underdamped. (16%)⒋ The system transfer function is )1)(21()5.01()(s s s s K s G +++=,1)(=s H . Determine thesteady-state error SS e when input is unit impulse )(t δ、unit step )(1t 、unit ramp t and unit parabolic function221t . (16%)⒌ ① Calculate the transfer function (minimum phase);② Draw the phase-angle versus ω (12%) w⒍ Draw the root locus for the system with open-loop transfer function.)3)(2()1()(+++=s s s s K s GH (14%)⒎ )1()(3+=Ts s Ks GH Draw the polar plot and determine the stability of system. (14%)⒈43214321432143211)()(G G G G G G G G G G G G G G G G S R S C -+--+= ⒉∞ K 528.0⒊S:0<K<0.0718 or K>14 overdamped ;0.0718<K<14 underdamped⒋S: )(t δ 0=ss e ; )(1t 0=ss e ; t K e ss 1=; 221t ∞=ss e⒌S:21ωω=K ; )1()1()(32121++=ωωωωs s ss GAUTOMATIC CONTROL THEOREM (9)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer function)(S C , )(S E . （12%）⒉ The characteristic equation is given0750075005.34)(123=+++=+K S S S S GH . Discuss the condition of stability. (16%)⒊ Sketch the root-locus plot for the system )1(4)()(2++=s s a s S GH . (The gain a isassumed to be positive.)① Determine the breakaway point and a value.② Determine the value of a at which root loci cross the imaginary axis. ③ Discuss the stability. (12%)⒋ Consider the system shown in Fig.2. 1)(1+=s K s G i , )1()(2+=Ts s Ks G . Assumethat the input is a ramp input, or at t r =)( where a is an arbitrary constant. Show that by properly adjusting the value of i K , the steady-state error SS e in the response to ramp inputs can be made zero. (15%)⒌ Consider the closed-loop system having the following open-loop transfer function:)1()(-=TS S KS GH . ① Sketch the polar plot ( Nyquist diagram). ② Determine thestability of the closed-loop system. (12%)⒍Sketch the root-locus plot. (18%)⒎Obtain the closed-loop transfer function )()(S R S C . （15%）⒈354211335421243212321313542143211)1()()(H G G G G H G H G G G G H G G G G H G G H G H G G G G G G G G G S R S C --++++-= 354211335421243212321335422341)()(H G G G G H G H G G G G H G G G G H G G H G H G G G H H G S N S E --+++--= ⒉45.30 K⒌S: N=1 P=1 Z=0; the closed-loop system is stable ⒎2423241321121413211)()(H G H G G G G G G G H G G G G G G G S R S C ++++++=AUTOMATIC CONTROL THEOREM (10)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer function)()(S R S C ,⒉ The characteristic equation is given01510520)(1234=++++=+S S KS S S GH . Discuss the condition of stability. (14%)⒊ Consider a unity-feedback control system whose open-loop transfer function is)6.0(14.0)(++=S S S S G . Obtain the response to a unit-step input. What is the rise time forthis system? What is the maximum overshoot? （10%）⒋ Sketch the root-locus plot for the system )25.01()5.01()(s S s K S GH +-=. (The gain K isassumed to be positive.)③ Determine the breakaway point and K value.④ Determine the value of K at which root loci cross the imaginary axis. Discuss the stability. (15%)⒌ The system transfer function is )5(4)(+=s s s G ,1)(=s H . ①Determine thesteady-state output )(t c when input is unit step )(1t 、unit ramp t . ②Determine theP K 、V K and a K , obtain the steady-state error SS e when input is t t r 2)(=. （12%）⒍ Consider the closed-loop system whose open-loop transfer function is given by:①TS K S GH +=1)(; ②TS K S GH -=1)(; ③1)(-=TS KS GH . Examine the stabilityof the system. （15%）⒎ Sketch the root-locus plot 。
自动控制原理(中英文对照李道根)习题2题解
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L1自动控制原理
自动控制系统
3
控制理论的发展
原始控制设备
没有理论指导 单输入(SI) 多输入输出（MIMO）
经典控制理论
现代控制理论
大系统理论/复杂系统理论 智能控制理论
4
控制理论的发展
经典控制理论
研究的主要对象是单输入、单输出——单变量系统。 如：调节电压改变电机的速度；调整方向盘改变汽 车的运动轨迹等。
d m r(t) d m-1r(t) dr(t) bm b b b 0 r(t) m-1 1 m m-1 dt dt dt
式中：r(t)——系统输入量； c(t)——系统输出量 主要特点是具有叠加性和齐次性。
14
非线性系统
特点：在构成系统的环节中有一个或一个以上的非 线性环节。 非线性的理论研究远不如线性系统那么完整，目前 尚无通用的方法可以解决各类非线性系统。 近似处理。
放大器
执行机构 （步进电机）
工作机床
切削刀具
图纸
指令
8
自动控制系统的分类Leabharlann 闭环控制系统（反馈控制系统）
特点：系统输出信号与测量元件之间存在反馈回 路。 “闭环”这个术语的含义，就是将输出信号 通过测量元件反馈到系统的输入端，通过比较、 控制来减小系统误差。
微型计算机 放大器 执行机构 工作机床 切削刀具 位移
15
自动控制系统的分类
其他分类方式
按系统数学模型参数特性分：定常系统和时变系统 按功能分：温度控制系统、速度控制系统、位置控制 系统等。 按元件组成分：机电系统、液压系统、生物系统等。
自动控制原理(中英文对照 李道根)习题5题解
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P5.5 Fig. P5.5 shows the polar plots of the open-loop transfer functions of some systems. Determine whether the closed-loop systems are stable. In each case, p is the number of the open-loop poles located in the right half s -plane, is the number of the integral factors in the open-loop transfer function.
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P5.3 Plot the asymptotic log-magnitude curves and phase curves for the following transfer functions (a) G ( s) H ( s) (c) G ( s) H ( s) (e) G ( s) H ( s)
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respectively. (a) In the case of r (t ) sin(t 30 ) , since 1 and 0 30 , we have
《自动控制原理》部分中英文词汇对照表(英文解释)
《自动控制原理》部分中英文词汇对照表AAcceleration 加速度Angle of departure分离角Asymptotic stability渐近稳定性Automation自动化Auxiliary equation辅助方程BBacklash间隙Bandwidth带宽Block diagram方框图Bode diagram波特图CCauchy’s theorem高斯定理Characteristic equation特征方程Closed-loop control system闭环控制系统Constant常数Control system控制系统Controllability可控性Critical damping临界阻尼DDamping constant阻尼常数Damping ratio阻尼比DC control system直流控制系统Dead zone死区Delay time延迟时间Derivative control 微分控制Differential equations微分方程Digital computer compensator数字补偿器Dominant poles主导极点Dynamic equations动态方程EError coefficients误差系数Error transfer function误差传递函数FFeedback反馈Feedback compensation反馈补偿Feedback control systems反馈控制系统Feedback signal反馈信号Final-value theorem终值定理Frequency-domain analysis频域分析Frequency-domain design频域设计Friction摩擦GGain增益Generalized error coefficients广义误差系数IImpulse response脉冲响应Initial state初始状态Initial-value theorem初值定理Input vector输入向量Integral control积分控制Inverse z-transformation反Z变换JJordan block约当块Jordan canonical form约当标准形LLag-lead controller滞后-超前控制器Lag-lead network 滞后-超前网络Laplace transform拉氏变换Lead-lag controller超前-滞后控制器Linearization线性化Linear systems线性系统MMass质量Mathematical models数学模型Matrix矩阵Mechanical systems机械系统NNatural undamped frequency自然无阻尼频率Negative feedback负反馈Nichols chart尼科尔斯图Nonlinear control systems非线性控制系统Nyquist criterion柰奎斯特判据OObservability可观性Observer观测器Open-loop control system开环控制系统Output equations输出方程Output vector输出向量PParabolic input抛物线输入Partial fraction expansion部分分式展开PD controller比例微分控制器Peak time峰值时间Phase-lag controller相位滞后控制器Phase-lead controller相位超前控制器Phase margin相角裕度PID controller比例、积分微分控制器Polar plot极坐标图Poles definition极点定义Positive feedback正反馈Prefilter 前置滤波器Principle of the argument幅角原理RRamp error constant斜坡误差常数Ramp input斜坡输入Relative stability相对稳定性Resonant frequency共振频率Rise time上升时间调节时间 accommodation timeRobust system鲁棒系统Root loci根轨迹Routh tabulation（array）劳斯表SSampling frequency采样频率Sampling period采样周期Second-order system二阶系统Sensitivity灵敏度Series compensation串联补偿Settling time调节时间Signal flow graphs信号流图Similarity transformation相似变换Singularity奇点Spring弹簧Stability稳定性State diagram状态图State equations状态方程State feedback状态反馈State space状态空间State transition equation状态转移方程State transition matrix状态转移矩阵State variables状态变量State vector状态向量Steady-state error稳态误差Steady-state response稳态响应Step error constant阶跃误差常数Step input阶跃输入TTime delay时间延迟Time-domain analysis时域分析Time-domain design时域设计Time-invariant systems时不变系统Time-varying systems时变系统Type number型数Torque constant扭矩常数Transfer function转换方程Transient response暂态响应Transition matrix转移矩阵UUnit step response单位阶跃响应VVandermonde matrix范德蒙矩阵Velocity control system速度控制系统Velocity error constant速度误差常数ZZero-order hold零阶保持z-transfer function Z变换函数z-transform Z变换。
自动控制原理 Chapter 1 All in English
1.1 System and System Analysis
A system is any collection of interacting elements for which there are cause-and-effect relationships among the variables.
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2. Friends after classes 3. No work, no gain. No hard work, no heavy gain. No zuo, no die. No big zuo, no bad die.
credit学分 hour学时 required course必修课 performance assessment成绩评定 assignment 作业 computer test 上机考试 final exam 期末考试 textbook教材 reference参考书
syllabus提纲 system系统 automatic Control Systems 自动控制系统 mechanical Engineering System机械工程系统 system analysis系统分析 principle原理
2018/10/22 12
Dynamic systems: variables are time-dependent.
(variables: variables & derivatives of one oபைடு நூலகம் more variables )
In the process of analyzing a system, two tasks must be performed: Modeling the system (1.2) Solving the model (1.3). The combination of the two steps (1.2&1.3) is referred to as system analysis.
自动控制原理（中英文对照李道根）习题3.题解
■SolutionsP3.1 The unit step response of a certain system is given by t t e e t c 21)(---+=, 0≥t (a) Determine the impulse response of the system.(b) Determine the transfer function )()(s R s C of the system.Solution:The impulse response is the differential of corresponding step response, i.e.t t e e t tt c t k 22)(d )(d )(--+-==δAs we know that the transfer function is the Laplace transform of corresponding impulse response, i.e.232422111]2)([)()(222++++=+++-=+-=--s s s s s s e e t L s R s C tt δP3.2Consider the system described by the block diagram shown in Fig. P3.2(a). Determinethe polarities of two feedbacks for each of the following step responses shown in Fig. P3.2(b), where “0” indicates that the feedback is open.Solution:In general we have(a) Block diagram.1.1.1.1.1(b) U nit-step resp onses(1)(2)(3)(4)(5)Figure P3.221020221)()(k k s k s k k s R s C ±±=Note that the characteristic polynomial is210202)(k k s k s s ±±=∆where the sign of s k 2is depended on the outer feedback and the sign of 21k k is depended on the inter feedback.Case (1).The response presents a sinusoidal. It means that the system has a pair of pure imaginary roots, i.e. the characteristic polynomial is in the form of 212)(k k s s +=∆. Obviously, the outlet feedback is “–”and the inner feedback is “0”.Case (2).The response presents a diverged oscillation.The system has a pair of complex conjugate roots with positive real parts, i.e. the characteristic polynomial is in the form of 2122)(k k s k s s +-=∆. Obviously, the outlet feedback is “+”and the inner feedback is “–”.Case (3).The response presents a converged oscillation. It means that the system has a pair of complex conjugate roots with negative real parts, i.e. the characteristic polynomial is in the form of 2122)(k k s k s s ++=∆. Obviously,both the outlet and inner feedbacks are “–”.Case (4).In fact this is a ramp response of a first-order system. Hence, the outlet feedback is “0”to produce a ramp signal and the inner feedback is “–”.Case (5).Considering that a parabolic function is the integral of a ramp function, both the outlet and inner feedbacks are “0”.P3.3Consider each of the following closed-loop transfer function. By considering the location of the poles on the complex plane, sketch the unit step response, explaining the results obtained.(a) 201220)(2++=s s s Φ,(b) 61166)(23+++=s s s s Φ(c) 224)(2++=s s s Φ,(d) )5)(52(5.12)(2+++=s s s s ΦSolution:(a) )10)(2(20201220)(2++=++=s s s s s ΦBy inspection, the characteristic roots are 2-, 10-. This is an overdamped second-order system. Therefore, considering that the closed-loop gain is 1=Φk , its unit step response can be sketched as shown.(b) )3)(2)(1(661166)(23+++=+++=s s s s s s s ΦBy inspection, the characteristic roots are 1-, 2-, 3-.Obviously, all three transient components are decayed exponential terms. Therefore, its unit step response, with a closed-loop gain 1=Φk , is sketched as shown..1.1(c) 1)1(4224)(22++=++=s s s s ΦThis is an underdamped second-order system, because its characteristic roots are j ±-1. Hence, transient component is a decayed sinusoid. Noting that the closed-loop gain is 2=Φk , the unit step response can be sketched as shown.(d) )5](21[(5.12)5)(52(5.12)(222++=+++=s s s s s s ）＋ΦBy inspection, the characteristic roots are 21j ±-, 5-. Since51.0-<<-, there is a pair of dominant poles,21j ±-, for this system. The unit step response, with a closed-loop gain 5.0=Φk , is sketched as shown.P3.4 The open-loop transfer function of a unity negative feedback system is)1(1)(+=s s s G Determine the rise time, peak time, percent overshoot and setting time (using a 5% setting criterion).Solution: Writing he closed-loop transfer function2222211)(nn n s s s s s ωςωωΦ++=++=we get 1=n ω, 5.0=ς. Since this is an underdamped second-order system with 5.0=ς, the system performance can be estimated as follows.Rising time .sec 42.25.0115.0arccos 1arccos 22≈-⋅-=--=πςωςπn r t Peak time .sec 62.35.011122≈-⋅=-=πςωπn p t Percent overshoot %3.16%100%100225.015.01≈⨯=⨯=--πςπςσe e p Setting time .sec 615.033=⨯=≈ns t ςω(using a 5% setting criterion)P3.5 A second-order system gives a unit step response shown in Fig. P3.5. Find the open-loop transfer function if the system is a unit negative-feedback system.Solution:By inspection we have%30%100113.1=⨯-=p σSolving the formula for calculating the overshoot,.1.1Figure P3.5.0.23.021==-ςπςσe p , we have 362.0ln ln 22≈+-=ppσπσςSince .sec 1=p t , solving the formula for calculating the peak time, 21ςωπ-=n p t , we getsec/7.33rad n =ωHence, the open-loop transfer function is)4.24(7.1135)2()(2+=+=s s s s s G n n ςωωP3.6A feedback system is shown in Fig. P3.6(a), and its unit step response curve is shown in Fig. P3.6(b). Determine the values of 1k , 2k ,and a .Solution:The transfer function between the input and output is given by2221)()(k as s k k s R s C ++=The system is stable and we have, from the response curve,21lim )(lim 122210==⋅++⋅=→∞→k sk as s k k s t c s t By inspection we have%9%10000.211.218.2=⨯-=p σSolving the formula for calculating the overshoot, 09.021==-ςπςσe p , we have608.0ln ln 22≈+-=ppσπσςSince .sec 8.0=p t , solving the formula for calculating the peak time, 21ςωπ-=n p t , we getsec/95.4rad n =ωThen, comparing the characteristic polynomial of the system with its standard form, we have.2.2(a)(b)Figure P3.622222n n s s k as s ωςω++=++5.2495.4222===n k ω02.695.4608.022=⨯⨯==n a ςωP3.7A unity negative feedback system has the open-loop transfer function)2()(k s s k s G +=(a) Determine the percent overshoot.(b) For what range of k the setting time less than 0.75 s (using a 5% setting criterion).Solution: (a)For the closed-loop transfer function we have222222)(nn n s s k s k s ks ωςωωΦ++=++=hence, by inspection,we getsec /rad k n =ω, 22=ςThe percent overshoot is%32.4%10021=⨯=-ςπςσe p (b) Since 9.022<=ς, letting.sec 75.025.033<⨯=≈kt ns ςω(using a 5% setting criterion)results in2275.06⎪⎪⎭⎫⎝⎛>k , i.e. 32>k P3.8For the servomechanism system shown in Fig. P3.8,determine the values of k and a that satisfy the following closed-loop system design requirements.(a) Maximum of 40% overshoot.(b) Peak time of 4s.Solution:For the closed-loop transfer function we have22222)(nn n s s k s k s ks ωςωωαΦ++=++=hence, by inspection, we getk n =2ω, αςωk n =2,and n n k ωςςωα22==Taking consideration of %40%10021=⨯=-ςπςσe p results in280.0=ς.In this case, to satisfy the requirement of peak time, 412=-=ςωπn p t , we haveFigure P3.8.sec /818.0rad n =ωHence, the values of k and a are determined as67.02==n k ω, 68.02==nωςαP3.9 The open-loop transfer function of a unity feedback system is)2()(+=s s k s G A step response is specified as:peak time s 1.1=p t , and percent overshoot %5=p σ.(a) Determine whether both specifications can be met simultaneously. (b) If the specifications cannot be met simultaneously, determine a compromise value for k so that the peak time and percent overshoot are relaxed the same percentage.Solution:Writing the closed-loop transfer function222222)(nn n s s k s s ks ωςωωΦ++=++=we get k n =ωand k 1=ς.(a) Assuming that the peak time is satisfiedsec1.1112=-=-=k t n p πςωπwe get 16.9=k . Then, we have 33.0=ςand%5%33%10021>=⨯=-ςπςσe p Obviously, these two specifications cannot be met simultaneously.(b) In order to reduce p σthe gain must be reduced. Choosing sec 2.221==p p t t results in04.31=k , 57.01=ς, %102%3.111=>=p p σσRechoosing sec 31.21.22==p p t t results in85.21=k , 59.01=ς, %10.51.2%0.101=<=p p σσLetting sec 255.205.23==p p t t results in941.23=k , 583.03=ς, %10.2505.2%5.103=≈=p p σσIn this way, a compromise value is obtained as941.2=k P3.10A control system is represented by the transfer function)13.04.0)(56.2(33.0)()(2+++=s s s s R s C Estimate the peak time, percent overshoot, and setting time (%5=∆), using the dominant pole method, if it is possible.Solution:Rewriting the transfer function as]3.0)2.0)[(56.2(33.0)()(22+++=s s s R s C we get the poles of the system: 3.02.021j s ±-=，, 56.23-=s . Then, 21，s can be considered as a pair of dominant poles, because )Re()Re(321s s <<，.Method 1. After reducing to a second-order system,the transfer function becomes13.04.013.0)()(2++=s s s R s C (Note: 1)()(lim 0==→s R s C k s Φ)which results in sec /36.0rad n =ωand 55.0=ς. The specifications can be determined assec 0.42112ςωπ-=n p t , %6.12%10021=⨯=-ςπςσe p sec 67.2011ln 12=⎪⎪⎪⎭⎫⎝⎛-=ς∆ςωn s t Method 2. Taking consideration of the effect of non-dominant pole on the transient components cause by the dominant poles, we havesec0.8411)(231=--∠-=ςωπn p s s t %6.13%10021313=⨯-=-ςπςσe s s s p sec 6.232ln 1313=⎪⎪⎭⎫⎝⎛-⋅=s s s t n s ∆ςωP3.11By means of the algebraic criteria, determine the stability of systems that have thefollowing characteristic equations.(a) 02092023=+++s s s (b) 025103234=++++s s s s (c) 021*******=+++++s s s s s Solution:(a) 02092023=+++s s s . All coefficients of the characteristic equation are positive. Using L-C criterion,1609120202>==D This system is stable.(b) 025103234=++++s s s s . All coefficients of the characteristic equation are positive. Using L-C criterion,15311002531103<-==D This system is unstable.(c) 021*******=+++++s s s s s . (It’s better to use Routh criterion for a higher-order system.)All coefficients of the characteristicequation are positive. Establish the Routh arrayas shown.There are two changes of sign in the first column, this system is unstable.P3.12The characteristic equations for certain systems are given below. In each case,determine the number of characteristic roots in the right-half s -plane and the number of pure imaginary roots.(a) 0233=+-s s (b) 0160161023=+++s s s (c) 04832241232345=+++++s s s s s (d) 0846322345=--+++s s s s s Solution:(a) 0233=+-s s . The Routh array shows that there are two changes of sign in the first column. So that there are two characteristic roots in the right-half s -plane.(b) 0160161023=+++s s s The 1s -row is an all-zero one and an auxiliary equation is made based on 2s -row162=+s Taking derivative with respect to s yields2=s The coefficient of this new equation is inserted in the1s row, and the Routh array is then completed. By inspection, there are no changes of sign in the firstcolumn, and the system has no characteristic roots in the right-half s -plane. The solution of the auxiliary are 4j s ±=, the system has a pair of pure imaginary roots.(c) 04832241232345=+++++s s s s s . The Routh array is established as follows.The 1s -row is an all-zero one and an auxiliary equation based on 2s -row is42=+s Taking derivative with respect to s yields2=s The coefficient of this new equation isinserted in the 1s row, and the Routh array is then completed. By inspection, there are no changes of sign in the first column, and the system has no characteristic roots in the right-half5s 1914s 21023s 402s 1021s -0.800s 23s 1-32s 0 0>⇒ε21s εε23--0s 23s 1162s 101⇒16016⇒1s 02⇒0s 165s 112324s 31⇒248⇒4861⇒3s 41⇒164⇒2s 41⇒164⇒1s 02⇒0s 4s -plane. The solution of the auxiliary are 2j s ±=, the system has a pair of pure imaginaryroots.(d) 0846322345=--+++s s s s s .The Routh array is established as follows.The 3s -row is an all-zero one and an auxiliary equationbased on 4s -row is04324=-+s s Taking derivative with respect to s yields643=+s s The coefficient of this new equation is inserted in the 3s row, and the Routh array is then completed. By inspection, the sign inthe first column is changed one time, and the system has one root in the right-half s -plane. The solution of the auxiliary are 121±=，s 243j s ±=，, the system has one pair of pure imaginary roots.P3.13The characteristic equations for certain systems are given below. In each case, determine the value of k so that the corresponding system is stable. It is assumed that k is positive number.(a) 02102234=++++k s s s s (b) 0504)5.0(23=++++ks s k s Solution: (a) 02102234=++++k s s s s .The system is stable if and only if⎪⎪⎩⎪⎪⎨⎧<⇒>=>9022*********k k D k i.e. the system is stable when 90<<k .(b) 0504)5.0(23=++++ks s k s . The system is stable if and only if⎪⎩⎪⎨⎧>-+⇒>-+⇒>+=>>+0)3.3)(8.34(05024041505.00,05.022k k k k k k D k k i.e. the system is stable when 3.3>k .P3.14The open-loop transfer function of a negative feedback system is given by)12.001.0()(2++=s s s Ks G ςDetermine the range of K and ςin which the closed-loop system is stable.Solution: The characteristic equation is2.001.023=+++K s s s ςThe system is stable if and only if5s 13-44s 21⇒63⇒-84-⇒3s 04⇒06⇒02s 3-8 1s 5000s -8⎪⎩⎪⎨⎧<⇒>-⇒>=>>ςςς200010200101.02.002.0,02K K .ς.K D k The required range is 020>>K ς.P3.15The open-loop transfer function of negative feedback system is given)12)(1()1()()(+++=s Ts s s K s H s G The parameters K and T may be represented in a plane with K as the horizontal axis and T as the vertical axis. Determine the region in which the closed-loop system is stable.Solution:The characteristic equation is)1()2(223=+++++K s K s T Ts Since all coefficients are positive, the system is stable if and only if)1)(2(01222>++⇒>++=K T K T KT D 022>++-T KT K 04)2()2(>+-+-T T K 4)1)(2(<--⇒K T The system is stable in the region 4)1)(2(<--K T , which is plotted as shown. (Letting 2-='T T and 1-='K K results in 4<''K T .)P3.16A unity negative feedback system has an open-loop transfer function)1)(1)(1()(2+++=Ts n nTs Ts Ks G where 10≤≤n , 0>K , T is a positive constant.(a) Determine the range of K and n so that the system is stable.(b) Determine the value of K required for stability for 1=n , 0.5, 0.1, 0.01, and 0.(c) Discuss the stability of the closed-loop system as a function of n for a constant K .Solution:The closed-loop characteristic equation is)1)(1)(1(2=+++K Ts n nTs Ts ＋i.e. 01)1()(22223333=+++++++K Ts n n s T n n n s T n ＋(a) The system is stable if and only if)1(1)1(233222>+++++=Tn n Tn K T n n n D i.e.)1(0)1()1(2223322>--++⇒>+-++K n n n n T K T n n n ⎪⎪⎭⎫⎝⎛-++⎪⎪⎭⎫ ⎝⎛+++<⇒-⎪⎪⎭⎫⎝⎛++<1111112222n n n n n n K n n n K ⎪⎭⎫ ⎝⎛++<⇒⎪⎪⎭⎫⎝⎛-+++++<2222211)1(11)1(n n K n n n n n n n K '21hence, the system is stable when ⎪⎭⎫ ⎝⎛++<<2211)1(0n n K .(b) The value of K required for stability for 1=n , 0.5, 0.1, 0.01, and 0are calculated as shown.80<<K for 1=n ,5.110<<K for 5.0=n ,21.1220<<K for 1.0=n ,102020<<K for 01.0=n ,∞<<K 0for 0=n .(c) For a constant K , the stability of the closed-loop system is related to the value of n , the larger the value of n ,the easier the system to be stable. (Stagger principle.)P3.17A unity negative feedback system has an open-loop transfer function)16)(13()(++=s s s Ks G Determine the range of k required so that there are no closed-loop poles to the right of the line 1-=s .Solution:The closed-loop characteristic equation is18)6)(3(0)16)(13(=+++⇒=+++K s s s K ss s i.e. 01818923=+++K s s s Letting 1~-=s s resulting in)1018(~3~6~018)5~)(2~)(1~(23=-+++⇒=+++-K s s s K s s s Using Lienard-Chipart criterion, all closed-loop poles locate in the right-half s ~-plane, i.e. to the right of the line 1-=s , if and only if⎪⎩⎪⎨⎧<⇒>-⇒>-=>⇒>-91408.1820311018695,010182K K K D K K The required range is 91495<<K , or 56.10.56<<K P3.18A system has the characteristic equation291023=+++k s s s Determine the value of k so that the real part of complex roots is 2-, using the algebraiccriterion.Solution:Substituting 2~-=s s into the characteristic equation yields02~292~102~23=+-+-+-k s s s ）（）（）（0)26(~~4~23=-+++k s s s The Routh array is established as shown.If there is a pair of complex roots with real part of 2-, then26=-k 3s 112s 426-k 1s 0si.e. 30=k . In the case of 30=k , we have the solution of the auxiliary equation j s ±=~, i.e. j s ±-=2.P3.19 An automatically guided vehicle is represented by the system in Fig. P3.19.(a) Determine the value of τrequired forstability.(b) Determine the value of τwhen one root of the characteristic equation is 5-=s , and the values of the remainingroots for the selected τ.(c) Find theresponse of the system to a step command for the τselected in (b).Solution:The closed-loop transfer function is10101010)()()(23+++==s s s s R s C s τΦ(a) The closed-loop characteristic equation is 010101023=+++s s s τSince all coefficients are positive, the system is stable if and only if1.0010110102>⇒>=ττD (b) Substituting 5~-=s s into the characteristic equation yields0105~105~105~23=+-+-+-）（）（）（s s s τ0)50135(~)2510(~5~23=-+-+-ττs s s In the case of 050135=-τ, i.e. 7.2=τ, we have 0~1=s , i.e. 51-=s . Solving the characteristic equation with 7.2=τ, i.e. 0~2~5~23=++-s s s results in 56.4~2=s and 44.0~3=s . Hence the remaining roots are 44.02-=s and 56.43-=s .(c) The closed-loop transfer function for 7.2=τis)5)(56.4)(44.0(10)(+++=s s s s ΦThe unit step response of the system is500.156.421.144.021.111)5)(56.4)(44.0(10)(+--+++-=⋅+++=s s s s s s s s s C tt t e e e t c 556.444.000.121.121.11)(----+-=Or, considering that there is a dominant pole for the system, we have127.2144.044.0)(+=+≈s s s Φte t c 44.01)(--≈P3.20A thermometer is described by the transfer function )11+Ts . It is known that, measuring the water temperature in a container, one minute is required to indicate 98% of the actual water temperature. Evaluate the steady-state indicating error of the thermometer if the container is heated and the water temperature is lineally increased at the rate of C/min 10 .travelFig.P3.19Solution:One minute required to indicate 98% of the actual water temperature means that the setting time is sec 604=≈T t s , i.e. the time constant of the thermometer issec15≈T The indicated error caused by the given ramp input, C/sec)(6010C/min)(10)( ==t t r , is222611611161)()()(sTs Ts s Ts s s C s R s E ⋅+=⋅+-=-=By inspection, a first-order system is always stable. Hence, the steady-state indicating error isC ss s s e s ss 5.26111515lim 20=⋅+⋅=→P3.21 Determine the steady-state error for a unit step input, a unit ramp input, and an acceleration input 22t for the following unit negative feedback systems. The open-loop transfer functions are given by(a) )12)(11.0(50)(++=s s s G ,(b) )5.0)(4(10)()(++=s s s s H s G (c) )11.0()15.0(8)(2++=s s s s G ,(d) )5)(1(10)(2++=s s s s G (e) )2004()(2++=s s s k s G Solution:(a) )12)(11.0(50)(++=s s s G . This is a second-order system and must be stable. Asa 0-type system,0=υ, the corresponding error constants are50=p K , 0=v K , 0=a K Consequently, the corresponding steady-state errors are0196.0501110.=+=+=p r ss K r ε, ∞==v v ss K v 0.ε, ∞==aa a ss K v .εrespectively.(b) )5.0)(4(10)()(++=s s s s H s G . The characteristic polynomial is40209)(23+++=s s s s τ∆Using L-C criterion,01402014092>==D the closed-loop system is stable. By inspection, system type 1=υand open-loop gain 5=K . Hence, the corresponding steady-state errors are0.=r ss ε, 2.01.==Kv ss ε, ∞=a ss .εrespectively.(c) )11.0()15.0(8)(2++=s s s s G . The characteristic polynomial is40209)(23+++=s s s s τ∆Using L-C criterion01402014092>==D the closed-loop system is stable. By inspection, system type 1=υand open-loop gain 5=K . Hence, the corresponding steady-state errors are0.=r ss ε, 2.01.==Kv ss ε, ∞=a ss .εrespectively.(d) )5)(1(10)(2++=s s s s G . The characteristic polynomial is1056)(234+++=s s s s ∆By inspection, this system is unstable (due to constructional instability).(e) )2004()(2++=s s s k s G . The characteristic polynomial isks s s s +++=2004)(23∆Using L-C criterionkkD -==800200142the closed-loop system is stable if and only if 8000<<k . This is a 1-type system with a open-loop gain 200k K =. In the case of 8000<<k , i.e. 40<<K ,the corresponding steady-state errors are0.=r ss ε, kK v ss 2001.==ε, ∞=a ss .εrespectively.P3.22 The open-loop transfer function of a unity negative feedback system is given by)1)(1()(21++=s T s T s Ks G Determine the values of K , 1T , and 2T so that the steady-state error for the input, bt a t r +=)(, is less than 0ε. It is assumed that K , 1T , and 2T are positive, a and b are constants.Solution:The characteristic polynomial is Ks s T T s T T s ++++=221321)()(∆Using L-C criterion, the system is stable if and only if2121212121212001T T T T K T KT T T T T K T T D +<⇒>-+⇒>+=Considering that this is a 1-type system with a open-loop gain K , in the case of 2121T T T T K +<, we have0..εεεεεbK Kbv ss r ss ss >⇒<=+=Hence, the required range for K is21210T T T T K b+<<εP3.23 The open-loop transfer function of a unity negative system is given by)1()(+=Ts s K s G Determine the values of K and T so that the following specifications are satisfied:(a) The steady-state error for the unit ramp input is less than 02.0.(b) The percent overshoot is less than %30and the setting time is less s 3.0.Solution:Assuming that both K and T are positive, the system must be stable. To meet the requirement on steady-state error, we have5002.010≥⇒≤==k KK v v ss εTo meet the second requirement, we have358.0%3021≥⇒≤=-ςσςπςe p and%)2(,10sec3.03=≥⇒≤≈∆ςωςωn ns t Considering that KT21=ςand TKn =ω, we get 95.1358.021≤⇒≥=KT KTς05.02010≤⇒=≥=T KT T K n ςωFinally, to met all specifications, the required ranges K and T are⎪⎩⎪⎨⎧≤≤≤T K T 95.15005.0P3.24 The block diagram of a control system is shown in Fig. P3.24, where)()()(s C s R s E -=. Select the values of τand b so that the steady-state error for a ramp input is zero.Figure P3.24Solution:Assuming that all parameters are positive, the system must be stable. Then, the error response is)()1)(1()(1)()()(21s R K s T s T b s K s C s R s E ⎥⎦⎤⎢⎣⎡++++-=-=τ)()1)(1()1()(2121221s R Ks T s T Kb s K T T s T T ⋅+++-+-++=τLetting the steady-state error for a ramp input to be zero, we get221212210.)1)(1()1()(lim )(lim sv Ks T s T Kb s K T T s T T s s sE s s r ss ⋅+++-+-++⋅==→→τεwhich results in⎩⎨⎧=-+=-00121τK T T Kb I.e. K T T 21+=τ, Kb 1=.P3.25 The block diagram of a compound system is shown in Fig. P3.25.Select the values ofa andb so that the steady-state error for a parabolic input is zero.Solution: The characteristic polynomial is 1012.0002.0)(23+++=s s s s ∆Using L-C criterion,1.01002.01012.02>==D the system is stable. The transfer function between error and input is given by10)102.0)(11.0()(10)102.0)(11.0()102.0)(11.0(101)102.0)(11.0()(101)()()(22++++-++=++++++-=-=s s s bs as s s s s s s s s s bs as s C s R s E 10)102.0)(11.0()101()101.0(002.023+++-+-+=s s s s b s a s Letting the steady-state error for a parabolic input to be zero yields010)102.0)(11.0()101()101.0(002.0lim 30230.=⋅+++-+-+⋅=→sa s s s sb s a s s s ass εwhich results inFigure P3.25⎩⎨⎧=-=-01012.00101a b i.e. 012.0=a , 1.0=b .P3.26 The block diagram of a system is shown in Fig. P3.26. In each case, determine the steady-state error for a unit step disturbance and a unit ramp disturbance, respectively.(a) 11)(K s G =, )1()(222+=s T s K s G (b) ss T K s G )1()(111+=, )1()(222+=s T s K s G , 21T T >Solution: (a) In this case the system is of second-order and must be stable. The transferfunction from disturbance to error is given by212212.)1(1)(K K Ts s K G G G s d e ++-=+-=ΦThe corresponding steady-state errors are12120.11)1(lim K s K K Ts s K s s p ss -=⋅++-⋅=→ε∞→⋅++-⋅=→22120.1)1(lim sK K Ts s K s s a ss ε(b) Now, the transfer function from disturbance to error is given by)1()1()(121222.+++-=s T K K s T s sK s d e Φand the characteristic polynomial is21121232)(K K s T K K s s T s +++=∆Using L-C criterion,)(121211212212>-==T T K K T K K T K K D the system is stable. The corresponding steady-state errors are01)1()1(lim 1212220.=⋅+++-⋅=→s s T K K s T s sK s s p ss ε121212220.11)1()1(lim K s s T K K s T s sK s s a ss -=⋅+++-⋅=→εFigure P3.26P3.27 The block diagram of a compound system is shown in Fig. P3.26, where1)(111+=s T K s G , )1()(222+=s T s K s G ,233)(K K s G =Determine the feedforward block transfer function )(s G d so that the steady-state error due tounit step disturbance is zero.Solution: the characteristic equation is 0121=+G G , i.e.21221321)(K K s s T T s T T ++++Using L-C criterion, the system is stable if and only if002121212121212>-+=+=T T K K T T T T K K T T D hence, the system is stable if212121T T T T K K +<The transfer function from disturbance to error is given by111)(1)1(1)(2211112322212123.+⋅++⎥⎦⎤⎢⎣⎡⋅+++-=+--=s T K s T K s G s T K K K s T s K G G G G G G G s dd de Φ21212113)1)(1()()1(K K s T s T s s G K K s T K d +++++-=When the system is stable, letting the steady-state error to be zero yields0)1)(1()()1(lim 0212121130=⋅⎥⎦⎤⎢⎣⎡+++++-⋅=→s d K K s T s T s s G K K s T K s d s ss ε[]0)()1(lim 21130=++→s G K K s T K d s i.e.213)(K K K s G d -=The feedforward block function is 213)(K K K s G d -=, where 212121T T TT K K +<.Figure P3.27。
自动控制原理专业词汇中英文对照.pdf
自动控制原理专业词汇中英文对照.pdf自动控制原理专业词汇中英文对照中文英文自动控制automatic control；cybernation 自动控制系统automatic control system自动控制理论 automatic control theory经典控制理论 classical control theory现代控制理论 modern control theory智能控制理论intelligent control theory 开环控制open-loop control闭环控制 closed-loop control输入量 input输出量 output给定环节 given unit/element比较环节 comparing unit/element放大环节 amplifying unit/element执行环节 actuating unit/element控制环节 controlling unit/element被控对象 (controlled) plant反馈环节 feedback unit/element控制器 controller扰动/干扰 perturbance/disturbance前向通道 forward channel反馈通道feedback channel 恒值控制系统constant control system随动控制系统servo/drive control system 程序控制系统programmed control system 连续控制系统continuous control system离散控制系统 discrete control system线性控制系统 linear control system非线性控制系统 nonlinear control system定常/时不变控制系统time-invariant control system 时变控制系统 time-variant control system 稳定性 stability快速性 rapidity准确性 accuracy数学模型 mathematical model微分方程 differential equation非线性特性 nonlinear characteristic线性化处理 linearization processing泰勒级数 Taylor series传递函数 transfer function比例环节 proportional element积分环节 integrating element一阶惯性环节 first order inertial element二阶惯性环节 second order inertial element二阶震荡环节second order oscillation element 微分环节differentiation element一阶微分环节 first order differentiation element二阶微分环节 second order differentiation element 延迟环节delay element动态结构图 dynamic structure block串联环节 serial unit并联环节 parallel unit信号流图 signal flow graph梅逊增益公式Mason’s gain formula时域分析法 time domain analysis method性能指标 performance index阶跃函数 step function斜坡函数 ramp function抛物线函数 parabolic function /acceleration function 冲击函数impulse function正弦函数 sinusoidal function动态/暂态响应 transient response静态/稳态响应 steady-state response 延迟时间 delay time上升时间 rise time峰值时间 peak time调节时间 settling time最大超调量 maximum overshoot稳态误差 steady-state error无阻尼 undamping欠阻尼 underdamping过阻尼 overdamping特征根 eigen root极点 pole零点 zero实轴 real axis虚轴 imaginary axis 稳态/静态分量 steady-state component 瞬态/暂态/动态分量transient component 运动模态motion mode衰减 attenuation系数 coefficient初相角 initial phase angle响应曲线 response curve主导极点 dominant pole 劳斯稳定判据 Routh stability criterion S平面 S plane胡尔维茨稳定判据Hurwitz stability criterion 测量误差measurement error扰动误差 agitation error结构性误差 structural error偏差 deviation根轨迹 root locus 常规根轨迹 routine root locus根轨迹方程 root locus equation 幅值 magnitude幅角 argument对称性 symmetry分离点 separation/break away point会合点 meeting/break-in point渐近线 asymptote出射角 emergence angle/angle of departure入射角incidence angle/angle of arrival 广义根轨迹generalized root locus零度根轨迹zero degree root locus 偶极子dipole/zero-pole pair 频域分析法frequency-domain analysis method 频率特性frequency characteristic极坐标系 polar coordinate system直角坐标系 rectangular coordinate system幅频特性 magnitude-frequency characteristic相频特性phase-frequency characteristic 幅相频率特性magnitude-phase frequency characteristic 最小相位系统minimum phase system非最小相位系统 nonminimum phase system奈奎斯特稳定判据Nyquist stability criterion 伯德定理Bode theorem稳定裕度 stability margin幅值裕度 magnitude margin 相位/相角裕度 phase margin对数幅频特性 log magnitude-frequency characteristic 无阻尼自然震荡角频率 undamped oscillation angular frequency 阻尼震荡角频率damped oscillation angular frequency 阻尼角damping angle带宽频率bandwidth frequency 穿越/截止频率crossover/cutoff frequency 谐振峰值 resonance peak系统校正 system compensation超前校正 lead compensation滞后校正 lag compensation自激震荡 self-excited oscillation死区特性 dead zone characteristic饱和特性 saturation characteristic间隙特性 backlash characteristic描述函数法 describing function method相平面法 phase plane method 采样控制系统 sampling control system数字控制系统 digital control system频谱 frequency spectrum 采样定理 sampling theorem信号重现 signal recurrence拉氏变换 Laplace transformZ变换 Z transform终值定理 final-value theorem差分方程 difference equation迭代法 iterative method 脉冲传递函数 pulse transfer function 零阶保持器 zero-order holder映射 mapping方框图 block diagram伯德图 Bode diagram特征方程 characteristic equation可控性 controllability临界阻尼 critical damping阻尼常数 damping constant阻尼比 damping ratio初始状态 initial state初值定理 initial-value theorem反Z变换 inverse Z-transformation负反馈 negative feedback正反馈 positive feedback 尼科尔斯图 Nichols chart部分分式展开partial fraction expansion 幅角原理argument principle相对稳定性 relative stability共振频率 resonant frequency劳斯表 Routh tabulation/array奇点 singularity渐进稳定性 asymptotic stability控制精度 control accuracy临界稳定性 critical stability耦合 coupling解耦 decoupling比例积分微分调节器proportional integral derivative regulator(PID) 串联校正 series/cascade compensation 单输入单输出 single input single output(SISO)多输入多输出 multi input multi output(MIMO)低通滤波器 low pass filter非线性系统 nonlinear system复合控制 compound control衰减振荡 damped oscillation主反馈 monitoring feedback 转折(交接)频率 break frequency 稳定焦点/节点 stable focus/node。
自动控制原理_英文论文
Automatic Control Applications In the social life班级：学号：：Programmable controller to control watersupply systemConstant pressure water supply system for a certain industry or a particular user is very important, for example in certain production processes, if the tap water supply or short-time shortages due to insufficient water, which may affect product quality, serious product scrap and damage to the equipment. When a fire occurs, if the water pressure is insufficient or no water supply, no rapid fire, can lead to significant economic losses and casualties. So some of the water area with constant voltage water supply system, has great economic and social significance.Mechanical technologyOld pressurized equipment, General starts or stops usingconstant-pressure water supply pressurizing station exit of pumps and regulating valve opening is to be achieved. Control system is the use of relay-contactor control circuits, this line complex, difficult to maintain, and operate trouble, workers to guards on duty 24 hours, labor intensive. It is necessary to reform, improve the level of automation.Electrical technologyPresented to a tap water pressure station quick starting of constant pressure water supply control system, the FP3 produced by Matsushita programmable logic controller (PLC) controls with Advantech industrial computer monitor, high degree of automation, the whole program to work automatically, clearly shows the real-time status of each device, and automatically adjust the water pressure. The system also has a wide range of protection, such as water pressure alarm fault alarm, water alarm, valves, pump motor current flow and processing alarm processing and so on.System structure and control requirements of mechanical and electrical engineering technology network，Constant pressure water supply system consists of the main loop, the alternate loop of water supply,Composed of 2 water tank and pump house, as shown in Figure 1.Pumping station equipped with a 1# ~ 6# a total of 6 sets of150kW pumps. There is more than one (V1 ~ V23) electric valvecontrols the water circuit and the flow of water.Requires the constant pressure water-supply system has the following basic operating functions.Electrical technical machinery technolog.When municipal water pressure is higher than the setting pressure 21.56x104Pa, directly by the municipal water supply in electrical technology. When city water is lower than the set pressure, but under the pressure of not less than 7.84x104Pa when using direct pumping pressurized water supply solutions. Of progressively starting 2 pumps to pipe network pressure. When city water higher than the set pressure is detected, then converted to city water supply directly. When the tap water pressure is lower than 2.94x104Pa, or when there is a negative pressure signal exactly, should immediately convert water pressure, but should ensure that the pool water level above the minimum water level conditions. Mechanical technology.When pumping or pumping water pressure water supply is used, should be able to automatically adjust the water pressure for a given value of its total exports, control deviation is less than or equal to 10%. Electrical technologyCAD/CAM technologyDesign of PLC control system CAD/CAM technologyConstant pressure water supply system for detection and control of more is a large control system. According to its characteristics, we have chosen Panasonic FP3 programmable controller is a controller.The controller with it can programming sequence controller comparedto, has some obviously of advantages, as FP3 used has module ofdesign, can according to actual needs flexible assembled, usingconvenient, I/O distribution used free programming way; capacity big, program volume only by scan cycle limit, and scan cycle can in mustrange within itself change; has A/D, and D/A, and pulse output, andlocation control, senior unit, can achieved "shared memory"; additionalso some special of function. Mechanical technologyConstant pressure water supply system of PLC system structure isshown in Figure 1-dashed border. Industrial computer to monitor theentire system, the display shows the total pressurized systemstructure, read the real-time status of each valve and pump, waterpressure and flow rate, the valve opening, the pool water level andother parameters, and real-time display alarm and fault record.Electrical technologyBoth analog input and switch input. Analog by a/d module input,total 27 channels. There are 96 I/O points.CAD/CAM technologyMechanical technologyPLC software designElectrical technologyAccording to constant pressure water supply system operational requirements, PLC control system to monitor City tap water, and waterhas to decide whether to start the water pump, or direct waterpumping programme or by pumping pressurized water full solution. Control system the procedure is more complex. Electrical/mechanical engineering technology networkIn the control process, water supply and pressure regulation isan important and one of the more distinctive design, focusing onsoftware design of automatic constant-pressure function.Due to the large water system piping length and diameter, the opening and closing of the valve, pipe network pressure is slow, so the system is a system with large time delay. And because it is based on the old equipment transformation, to make use of existing equipment, it does not use speed regulator, instead of using the various methods to adjust the water pressure. First used segment regulation method, put hydraulic deviation is divided into four segment, that 10%, and 20%, and 30%, and 40%, dang detection to deviation smaller Shi, output of control volume (butterfly valve of incremental) smaller, and operation cycle also larger; In addition, when the deviation is less than or equal to ± 10%, coupled with the fuzzy control, according to d EK=EK-EK-1 to determine whether regulating butterfly valve opening, to further reduce the errors to ensure its error less than or equal to ± 10% requirements. Pressure is regula ted by multiple methods combination of outlet pressure can be satisfied with the effect. Electrical technical machinery technologyConclusionThe design of constant pressure water supply control system with PLC have successfully applied to an industrial zone, results showed that the system satisfies its design requirements, with convenient operation, high reliability, data integrity and monitor timely advantage and significantly reduce the labor intensity of workers, shorten the operating time, operators, maintainers, managers at home. The successful design of the monitoring system, as well as similar systems of old equipment modification to provide a good experience.。
The Principle of Automatic Control 自动控制原理
The Principle of Automatic Control 自动控制原理 sensors actuators communications computing architectures and interfacing algorithms
A Control System is an interconnection of components forming a system configuration that will provide a desired system response. The basis for system analysis and design is linear system theory. Process or plant to be controlled can be represented by cause-effect relationship.
The Principle of Automatic Control 自动控制原理
Feedback concept is the foundation for control system analysis and design Feedback control system is to control the process by using the difference between the output and reference input Negative feedback (refer to P7) Positive feedback
自动控制-中英翻译
automatic controlAutomatic control is relatively artificial control terms ，which refers to correlate with no people directly involved in conditions, the additional equipments or devices to make the machine, equipment or the production process of a certain job of state or parameters automatically set to run the rule.Automatic control technology research will benefit mankind from complicated, risky, tedious work environment free of control and greatly improve the efficiency. Automatic control is a branch of engineering science. It involves using a feedback of dynamic system of the principle of automatic influence, in order to make the output value close with that we set value. From the theory of methods, the mathematical theory is a foundation of Automatic control. As we known today, as automatic control is in the middle of the twentieth century from the control of a branch. The conclusion is based by Norbert wiener, Rudolf kalman proposes.1.The first generation process Control System(PCS)The first generation process Control System is based on 5-13 psi Pneumaticrebefo signal standard ( Pneumatic Control System) PCS before 150 years ,which includes simple on-site operation mode, control theory preliminary form. There does not has been the concept of the control room.2.The second generation process Control System(ACS)The second generation process Control System (ACS or Analog Control System) Analog to produce stats are based on 0-10 mA or 4-20 mA's current Analog signals, the obvious progress of the Control System is that to rule the whole firmly automatic Control field in the whole 25 years. It represents the arrival of the era of electrical automatic control. Control theory has the significant development, the establishment of the three big cybernetics laid the foundation of modern control; The establishment of control function, control room separate model has been used today.3.The third generation of process Control System(CCS)The third generation of process Control System, Computer Control System (CCS) began in t he 70’s ,the application of digital Computer has a great technical advantage, people in measurement, simulation and logic Control field, whichpioneered the use of the third generation process Control System, Computer Control System (CCS). This is called the third generation process control system is automatic control a revolution in the field, it give full play to the computer specialty, so it is widely acknowledged that computer can do all things, naturally produced is called "the centralized control" of the central control computer system, it should be pointed out that the signal transmission system is still the most used with 4-20 mA analog signals, but soon after that, as people focus and reliability of the control aspects of the problem, the out of control risk also focused on, a little wrong will make the whole system to paralysis. So it was quickly developed into the distributed control system (DCS).4.The fourth generation process Control System(DCS)With the rapid development of semiconductor manufacturing technology, the microprocessor to the widespread use of computer technology greatly increased, the reliability of the currently used is the fourth generation process Control System (DCS, or Distributed digital Control System), it is the main features of the whole Control System that there is no longer only a computer, it is a Control System by a computer and some intelligent instruments and intelligent components comprise. So the way of distributed control became the most important characteristic. Another important development exception is the signal transmission among of them are not based on 4-20 mA analog signals, and gradually digital signal to replace analog signals.5.The fifth generation process Control System (FCS)The development of FCS from the DCS, like DCS from CCS over the development as, there is a qualitative leap, that is "Distributed control" developed to " focused control". The way of data transmission is "bus" way. The so-called field-bus is intelligent measurement and control equipment conect all the digital, two-way transmission, with many node of the structure of the branch communications link. Say simply traditional control is a loop, and FCS technology is modules such as controller, actuators, detector etc on a bus to hang up realize communication, of course, transmission is the digital signal. The main bus has Profibus, LonWorks, etc. But the real with DCS FCS difference is to have a more FCS wide development space. Because the traditional DCS technical level while continuously improved, but themost low-end only communication network to the control station level, the control station and the field measurement instrument, the contact between the actuators are still used one-to-one transmission of 4-20 mA analog signals, high cost, low efficiency, maintenance difficulties, can't play the field instruments, to realize the potential of the intelligent field device the work state of comprehensive monitoring and deep management..In the modern science and technology in many fields, automatic control technology is playing a more and more important role. Automatic control theory is the study of the automatic control of common laws technology science. Its initial development stage, is based on the theory of the feedback of automatic adjustment principle, mainly for industrial control, during world war two, in order to design and manufacture the plane and Marine autopilot, gun positioning system, radar tracking system based on feedback and other military equipment, the principle of further promote and perfect the development of the theory of automatic control. After the war, in order to form the full automatic control theory system, which is the transfer function is the foundation of classical control theory, it mainly studies single input and single output, the linear system analysis and set constant design problem.Automatic control of the development, from the start of the happen to form a control theory, the whole that process. Speak Automatic control is refers to such feedback control system, this is a controller object with a control of the control object, the output signal get it back, after come back for measuring with the signal are compared. According to the error tell controller, which is within the machine work. Let the controller to complete the control effect, make the deviation eliminate or make the controlled objects output tracking what I need to be requirements of the signal. The controlled objects output in general is a physical quantities, for example say me to control a machine speed, is need to come out, to measure the speed control.Saying to hear I have to mention is the "engineering cybernetics qian xuesen". Qian xuesen, written in 1954, when his book in the United States, we wrote the book here also can't get in English, but the former Soviet union very seriously, the former Soviet union immediately translate it into Russian. We see of the former Soviet union was in 1956 put his Russian, translation come out, we see at that time is the Russianversion of, this is in about twenty s the formation process of the main process, the experience is concluded.The robust leaves (Lurie) in the former Soviet union in 1944 about him out to everybody now make nonlinear may know, out to a robust leaves in question, this problem has been not solve, he later to write a book, is to keep the problem. The problem, start to when? I said something about this here two style, British and American of just everybody has to come out, make the person is engineering in make control, the former Soviet union is the application with mechanics in make home mathematician control, so two played the role is not the same. His 1951 book at that time is very hard to understand, hard to read. This work, he brought out the robust, Lurie), (until 1960, someone out to solve a solution, we may know, is that the absolute Popov) Popov (stability. Then out of the stability and is robust to solve this problem, Lurie (in). That is the book was 20 s, twenty century in the late '40 s, 50 s, some of the work has been influence of the twentieth century, but also affects the 60 s to present some nonlinear theory, is his work for the foundation. The twentieth century is in the 40 s, front of the development of the technology is a process, slowly forms a theory. I'm used to in such a schedule to said 50 s, most of the time, call classical control theory; 60 s call the state space method, is actually state space method, but you know, at that time, the name of the called it the modern control theory; Then in the 70 s is the modern the frequency domain method, such a process.The next now is to explain the state space method, finished that modern control theory. The state space method who first brought out? The third book is just of the former Soviet union, these scholars. They make applied mathematics and mechanics, they never get the is used is the state space method. In 1960, it introduced to kalman English world, so the world it is spending in English, that means the country want English country, because it was all don't know, kalman is a Slavic name. In 1960, he, he put the state space method introduced to the United States. But add people to this hype, modern control theory to hype seem very much as gods, was also some people expected is relatively large. This is the development of the ten years later, he found too, like the expected only so! Some problem you didn't also solved. So the time, and some people back to the frequency domain method, is the earliest 50 s is thefrequency domain method, 60 s, 70 s state space and back to the frequency domain method.Of course, this is the spiral, this time the frequency domain method and then add a name, so that the modern the frequency domain method. Think frequency domain to consider the design problems or more appropriate, consider some of the design requirements, I just have this frequency domain method. Just in the modern the frequency domain method development on this momentum, 1981, and wrote you this no said robust, we now everyone make control theory know to robust design. Say you this modern the frequency domain method not robustness, when people do not believe, after the 80 s, the arguments that dispute slowly forms. By 1991, is now of course is that you may be someone the term does not necessarily unification, someone called it the modern control theory, Postmodern after keep control. We now have to go back to see, why do you say that no robustness? One will say, we from the many variable system, it is actually much more variable system into the system. Many variables don't appropriate, there is a lot of input, output a. Many avariable a problem, called the coupling, is between input and output coupling each other. Control, intuitive request was for decoupling control, decoupling control later! Is this the 1 to 1 can form output feedback system, the 2! with the 2 of output can form feedback systems, and the design will be easier.自动控制自动控制（automatic control）是相对人工控制概念而言的，指的是在没有人直接参与的情况下，利用外加的设备或装置，使机器、设备或生产过程的某个工作状态或参数自动地按照预定的规律运行。
自动控制原理中英文对照
自动控制原理中英文对照Automatic Control Principles 自动控制原理Introduction 简介Automatic control principles refer to the principles and theories that govern the design, development, and implementation of automated control systems. These systems are used in a variety of fields, including manufacturing, transportation, aerospace, and more. The goal of automatic control principles is to create systems that can operate independently and make decisions based on the input they receive.自动控制原理是指掌握设计、开发和实现自动控制系统的原理和理论。
这些系统应用于各种领域，包括制造业、交通运输、航空航天等。
自动控制原理的目标是创建能够独立运作并根据所接收的输入做出决策的系统。
Types of Control Systems 控制系统的类型There are two main types of control systems: open-loop and closed-loop. Open-loop systems are those that operate without any feedback, meaningthat they do not adjust their output based on the input they receive. Closed-loop systems, on the other hand, use feedback to adjust their output based on the input they receive.控制系统主要有两种类型：开环和闭环。
自动控制原理(上)-第1章-自动控制系统的一般概念
1.2.1 人工控制与自动控制
在控制被控对象抵消外界干扰的过程中，若控制本身与人 工操作有关，便称为人工控制，而若是没有人类的直接操作， 即纯粹依靠自动装置来完成控制过程中的调节，则称该控 制 为自动控制。
图1-1 人工控制的电热水壶
图1-2 自动控制的电热水壶
1.2.2 自动控制系统的表示方法
1.3.3 自动控制系统的基本组成
图1-7 典型的自动控制系统的基本组成
1.4 控制系统的基本要求
控制系统能够正常工作的最基本条件便是稳定。 稳定性通常指系统在受到干扰后能够恢复到平衡状态的能力。 如果系统能够恢复到平衡状态，则称该系统是稳定的，否则称该 系统不稳定。系统从受到干扰到恢复稳定状态，其被控量呈衰减 震荡或非周期过程。
1.3.2 闭环控制系统
图1-5 直流电机闭环转速控制系统
1.3 闭环控制系统
图1-6 直流电机闭环转速控制系统方框图
1.3 闭环控制系统
控制器与被控对象之间不仅存在正向控制作用，而且 还存在被控对象到控制器的反向联系。把这种控制过程称 为闭环控制按闭环控制方式组成的系统称为闭环控制系统。
由于闭环控制系统是根据偏差进行控制的，只要被控 量偏离给定值，系统就会自动纠偏，所以说闭环控制系统 具有很强的纠偏功能，对于干扰具有良好的适应性。
本章主要介绍了自动控制和自动控制系统的基本概念，使 读者对自动控制系统的组成结构、基本原理、主要性能指标、 类别以及控制理论的发展有个初步了解，为以后章节的制”是一个较为常见的词汇，可以将其理解为，一个 对象为了某个特定的目的，在另一个对象上施加的作用，这些 特定目的可能是将电压、电流、水位、温度、位移、转速等物 理量尽可能维持在某一范围，进而使得生产过程、生产设备或 是生产工具能够以正常的工作条件运行，而这些生产过程、生 产设备便是施加作用的对象，这些作用可以是属于物理、化学、 生物学等方面的作用。
自动控制原理专业词汇中英文对照
自动控制原理专业词汇中英文对照中文 英文自动控制 automatic control；cybernation 自动控制系统 automatic control system自动控制理论 automatic control theory经典控制理论 classical control theory现代控制理论 modern control theory智能控制理论 intelligent control theory 开环控制 open-loop control闭环控制 closed-loop control输入量 input输出量 output给定环节 given unit/element比较环节 comparing unit/element放大环节 amplifying unit/element执行环节 actuating unit/element控制环节 controlling unit/element被控对象 (controlled) plant反馈环节 feedback unit/element控制器 controller扰动/干扰 perturbance/disturbance前向通道 forward channel反馈通道 feedback channel 恒值控制系统 constant control system随动控制系统 servo/drive control system 程序控制系统 programmed control system 连续控制系统 continuous control system离散控制系统 discrete control system线性控制系统 linear control system非线性控制系统 nonlinear control system定常/时不变控制系统 time-invariant control system 时变控制系统 time-variant control system 稳定性 stability快速性 rapidity准确性 accuracy数学模型 mathematical model微分方程 differential equation非线性特性 nonlinear characteristic线性化处理 linearization processing泰勒级数 Taylor series传递函数 transfer function比例环节 proportional element积分环节 integrating element一阶惯性环节 first order inertial element二阶惯性环节 second order inertial element二阶震荡环节 second order oscillation element 微分环节 differentiation element一阶微分环节 first order differentiation element二阶微分环节 second order differentiation element 延迟环节 delay element动态结构图 dynamic structure block串联环节 serial unit并联环节 parallel unit信号流图 signal flow graph梅逊增益公式 Mason’s gain formula时域分析法 time domain analysis method性能指标 performance index阶跃函数 step function斜坡函数 ramp function抛物线函数 parabolic function /acceleration function 冲击函数 impulse function正弦函数 sinusoidal function动态/暂态响应 transient response静态/稳态响应 steady-state response 延迟时间 delay time上升时间 rise time峰值时间 peak time调节时间 settling time最大超调量 maximum overshoot稳态误差 steady-state error无阻尼 undamping欠阻尼 underdamping过阻尼 overdamping特征根 eigen root极点 pole零点 zero实轴 real axis虚轴 imaginary axis 稳态/静态分量 steady-state component瞬态/暂态/动态分量 transient component 运动模态 motion mode衰减 attenuation系数 coefficient初相角 initial phase angle响应曲线 response curve主导极点 dominant pole 劳斯稳定判据 Routh stability criterion S平面 S plane胡尔维茨稳定判据 Hurwitz stability criterion 测量误差 measurement error扰动误差 agitation error结构性误差 structural error偏差 deviation根轨迹 root locus 常规根轨迹 routine root locus根轨迹方程 root locus equation 幅值 magnitude幅角 argument对称性 symmetry分离点 separation/break away point会合点 meeting/break-in point渐近线 asymptote出射角 emergence angle/angle of departure入射角 incidence angle/angle of arrival 广义根轨迹 generalized root locus零度根轨迹 zero degree root locus 偶极子 dipole/zero-pole pair 频域分析法 frequency-domain analysis method 频率特性 frequency characteristic极坐标系 polar coordinate system直角坐标系 rectangular coordinate system幅频特性 magnitude-frequency characteristic相频特性 phase-frequency characteristic 幅相频率特性 magnitude-phase frequency characteristic 最小相位系统 minimum phase system非最小相位系统 nonminimum phase system奈奎斯特稳定判据 Nyquist stability criterion 伯德定理 Bode theorem稳定裕度 stability margin幅值裕度 magnitude margin 相位/相角裕度 phase margin对数幅频特性 log magnitude-frequency characteristic 无阻尼自然震荡角频率 undamped oscillation angular frequency 阻尼震荡角频率 damped oscillation angular frequency 阻尼角 damping angle带宽频率 bandwidth frequency 穿越/截止频率 crossover/cutoff frequency 谐振峰值 resonance peak系统校正 system compensation超前校正 lead compensation滞后校正 lag compensation自激震荡 self-excited oscillation死区特性 dead zone characteristic饱和特性 saturation characteristic间隙特性 backlash characteristic描述函数法 describing function method相平面法 phase plane method 采样控制系统 sampling control system数字控制系统 digital control system频谱 frequency spectrum 采样定理 sampling theorem信号重现 signal recurrence拉氏变换 Laplace transformZ变换 Z transform终值定理 final-value theorem差分方程 difference equation迭代法 iterative method 脉冲传递函数 pulse transfer function零阶保持器 zero-order holder映射 mapping方框图 block diagram伯德图 Bode diagram特征方程 characteristic equation可控性 controllability临界阻尼 critical damping阻尼常数 damping constant阻尼比 damping ratio初始状态 initial state初值定理 initial-value theorem反Z变换 inverse Z-transformation负反馈 negative feedback正反馈 positive feedback 尼科尔斯图 Nichols chart部分分式展开 partial fraction expansion 幅角原理 argument principle相对稳定性 relative stability共振频率 resonant frequency劳斯表 Routh tabulation/array奇点 singularity渐进稳定性 asymptotic stability控制精度 control accuracy临界稳定性 critical stability耦合 coupling解耦 decoupling比例积分微分调节器 proportional integral derivative regulator(PID) 串联校正 series/cascade compensation 单输入单输出 single input single output(SISO)多输入多输出 multi input multi output(MIMO)低通滤波器 low pass filter非线性系统 nonlinear system复合控制 compound control衰减振荡 damped oscillation主反馈 monitoring feedback 转折(交接)频率 break frequency稳定焦点/节点 stable focus/node。
自动控制原理 英文版
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1) Identify the output and input variables of the control systems.
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自动控制原理 ()
2020/3/3
第一章 自动控制系统的基本概念
1.1 开环控制系统和闭环控制系统
2.闭环控制系统
例1-2 温度闭环控制系统 （1）人工闭环控制
人的功能用一些 物理装置来实现
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第一章 自动控制系统的基本概念
1.1 开环控制系统和闭环控制系统
控制目标：要求炉子的温度恒定在期望的数 值上。
Classical Control Theory 第二阶段：现代控制理论（20世纪60～70年代）
Modern Control Theory 第三阶段：智能控制理论（20世纪70年代～ 今）
2020/3/3
Intelligent Control Theory
第一章 自动控制系统的基本概念
古典控制理论（20世纪40～6S0i年ng代le）Input Single
控制目标：要求炉子的温度恒定在期望的数值上。
系统 组成：1-自耦变压器 2-电阻炉（被控对象）
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第一章 自动控制系统的基本概念
1.1 开环控制系统和闭环控制系统
专业术语
被控量（输出量）output——系统要控制的物理量。
T
给定量或控制量 （输入量）input——决定被控量的物理量。 u c
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第一章 自动控制系统的基本概念
第一章 自动控制系统本结构和特点及 其工作原理；
❖ 了解闭环控制系统的组成和基本环节；
❖ 掌握反馈控制系统的基本要求及反馈控 制系统的作用；
❖ 学会分析自动控制系统的类型及本质特 征。
2020/3/3
第一章 自动控制系统的基本概念
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《自动控制原理》课程教学大纲课程编号：2020161课程类别：必修授课对象：本科三年级先修课程：复变函数，积分变换，信号与系统。
学分：4总学时：56 课内学时：48 实验学时： 8一、课程性质、教学目的与任务课程性质：专业基础课，专业知识链条中的关键环节之一，自动控制原理是仪器仪表类、测控类专业的重要基础课之一，这些专业主要学习信号传感（获取）、信号处理、控制及光机电系统等知识，而控制是知识链条中的重要一环，随着科技发展，自动化、智能化已成为仪器、产品、系统等的重要功能，这就要求学生必须具备自动控制方面的知识。
教学目的与任务：培养学生自动控制原理的基础知识，学习掌握经典控制的基本理论、基本方法和控制系统的基本设计方法，重点学习分析和设计线性控制系统的基本理论、基本方法及控制系统设计方法。
主要内容包括：控制系统的数学模型、控制系统的时域分析法、控制系统的根轨迹法、控制系统的频域分析法、控制系统的常用校正方法等。
二、教学基本要求学习经典控制的基本理论和基本方法，重点学习分析和设计线性控制系统的基本理论和基本方法。
主要内容包括：控制系统的数学模型、控制系统的时域分析法、控制系统的根轨迹法、控制系统的频域分析法、控制系统的常用校正方法等。
三、教学内容第一章控制系统的一般要概念（4课时）自动控制的基本原理与方式，自动控制系统示例，自动控制系统的分类，对自动控制系统的基本要求1、基本概念；2、 反馈系统基本组成；3、 基本控制方式；4、 控制系统分类：开环、闭环、复合控制；第二章 控制系统的数学模型 （8课时）控制系统的时域数学模型，拉普拉斯变换，控制系统的复域数学模型，控制系统的状态空间模型，控制系统的结构图与信号流图2-1 时域模型、微分方程表示方法；2-2 复域模型1、 传递函数的定义与性质；2、 传递函数的零、极点表示，开环增益、根轨迹增益等；3、 典型环节的传递函数（比例、惯性、微分、积分、振荡）；2-3 控制系统的结构图与信号流图1、 结构图的等效变换与化简2、 信号流图组成与性质A ．性质、术语（理解）B ．由结构图转化为信号流图方法C ．梅逊公式第三章 线性系统的时域分析法 （10课时）线性系统时间响应的性能指标，一阶系统的时域分析，二阶系统的时域分析，高阶系统的时域分析，线性系统的稳定性分析，线性系统的稳态误差计算。
3-1 线性系统时间响应的性能标r t ，p t ，s t ，%σ3-2 一阶系统的单位阶跃响应，3-3 二阶系统的时域响应1、二阶系统的标准数学模型 闭环传递函数形式，表示为单位反馈系统形式2、二阶系统单位阶跃响应（重点：欠阻尼情形）r t ，p t ，s t ，%σ3、二阶系统性能改善A、比例一微分控制掌握原理、特点B、测速反馈控制掌握原理、特点及性能参数计算3-4高阶系统时域分析，主导闭环极点概念3-5线性系统稳定性分析劳斯判据及其应用3-6线性系统的稳态误差1、误差传递函数计算2、利用终值定理求稳态误差3、系统类型（型别）4、典型参考信号输入下的稳态误差；误差系数5、减小稳态误差方法：提高型别、提高开环增益、采用复合控制6、扰动误差的传函、减小扰动误差方法A、增加扰动作用点之前积分环节数目B、增加扰动作用点之前环节的增益C、采用复合控制技术第四章根轨迹（6课时）根轨迹方程，根轨迹绘制的基本法则，广义根轨迹，系统性能的分析与估算，基于根轨迹的控制系统校正方法，MATLAB语言根轨迹分析法。
4-1 根轨迹方程（相角、幅值条件）4-2 根轨迹绘制的基本法则，绘制根轨迹第五章线性系统的频域分析（10课时）频率特性，典型环节和开环系统频率特性，奈奎斯特稳定判据，稳定裕度，闭环频率特性，系统时域指标估算，传递函数的实验确定法，MATLAB语言频域分析法。
5-2 频率特性1、基本概念系统频率特性与传函间关系2、频率特性表示方法幅相曲线；幅频特性；对数幅频特性曲线；相频特性曲线；对数相频特性曲线5-3 典型环节和开环系统频率特性1、典型环节的幅频、相频特性2、开环对数幅频、幅相曲线绘制对数幅频特性采用渐近线法3、由幅频曲线或相频曲线确定最小相位系统传递函数的方法5-4 奈奎斯特稳定性判据1、奈代判据，实际使用判据，幅相曲线对称性2、对数频率特性稳定性判据 I型以上辅助线作法5-5稳定裕度概念：截止频率；相角交界频率；相角裕度；幅值裕度5-6闭环频率特性（了解）5-7系统时域指标与频域指标关系了解它们之间的联系；第六章线性系统的校正方法（10课时）系统的设计与校正问题，常用校正装置及其特性，串联校正，反馈校正，复合校正，PID调节器，基于MATLAB语言的校正分析法。
6-1 校正方法串联，反馈，前馈，复合，原理，特性6-2常用校正装置及其特性（传函，特性，重要公式）1、无源超前2、无源迟后3、无源迟后—超前6-3串联校正1、串联超前方法，实用范围2、串联迟后方法，实用范围3、串联迟后—超前方法，实用范围6-4反馈校正（原理，优点）6-5复合校正A．按扰动补偿（原理，扰动误差传函推导，误差全补偿条件）B．按输入补偿（原理，误差传函推导，误差全补偿条件）6-6 PID控制原理（P，PD，I，PI，PID的特点）实验安排：实验二直流电机速度控制实验（8课时）1、开环控制、闭环控制系统设计2、控制系统的校正设计3、PID调节器开放性、综合设计实验，要求学生独自完成实验前准备工作（实验目的、原理分析，实验设计、实验步骤、参数计算等），实际动手实验，实验结果分析等。
四、学时分配总学时 56学时讲课48学时实验8学时课外学时 64学时五、教材与参考资料胡寿松等，《自动控制原理》，国防工业出版社，第4版，2002[1] 自动控制原理，清华大学出版社，吴麒主编；[2] 自动控制原理，中国科学技术大学出版社，庞国仲编；[3] 自动控制原理试题精选与答题技巧，哈尔滨工业大学出版社，王彤主编；[4] 自动控制原理常见题型解析及模拟题，西北工业大学出版社，史忠科，卢京潮编著；[5] MATLAB 语言工具箱—TOOLBOX实用指南，西北工业大学出版社，施阳李俊等编著；[6] MATLAB 语言—演算纸式的科学工程计算语言，中国科学技术大学出版社，张培强主编；[7] 系统分析与仿真—MATLAB语言及应用，国防科技大学出版社，黄文梅杨勇等编著。
六、成绩评定期末成绩80%，平时成绩（作业、平时测验及课堂情况）及综合实验能力20%。
大纲撰写人：段发阶、吴斌大纲批准人：段发阶制定日期：2011年6月TJU Syllabus for “Automatic control principle”Code:Category: compulsoryFor: juniorPrerequisite:complex variables functions, integral transformation, Signals and SystemsCredits: 4Semester Hour: 56 Lecture： 48 Computer Lab：81.Course nature, teaching goal and missionCourse nature: As professional basic course, one of the key links of professional knowledge chain, automatic control theory is one of most important basic courses in majors of instruments and measure & control. These majors mainly study the knowledge of signal sensing (for), signal processing, control and optomechatronics system and so on, and control is an important link in the chain knowledge. Along with the development of science and technology, automation and intelligent have become important function in instrument, production, system, etc. This requires the students must have knowledge of automatic control aspects. Teaching goal and mission: Cultivate students the basic knowledge of automatic control principle. Learn to master the basic theory and basic method of classical control, and the basic design method of the control system. Learn mainly analysis and design of linear control system of basic theory, the basic method and control system design method. The main contents include: the mathematical model of control system, the time domain analysis method of the control system, the root locusmethod of the control system, the method of frequency domain of the control system, the commonly used correction methods of the control system, etc.2.Teaching basic requirementsThe basic requirements are learning the basic theories and methods of classical control, mainly learning the basic theory and basic methods of analysis and design of linear control system. The main contents include: the mathematical model of control system, the time domain analysis method of the control system, the root locus method of the control system, the frequency domain method of the control system, the commonly used correction methods of the control system, etc.3. Content of courses(4 hours)Chapter One, the general concepts of Control systemThe basic principle and means of automatic control, automatic control system examples, the classification of the automatic control system, the basic requirements of automatic control system1.Basic concept2.The basic structure of feedback system3.The basic control mode4. Control system classification: open loop, closed loop, composite control*Chapter Two, The mathematical model of the control system (8 hours) The mathematical model of time domain of the control system, Laplace transform, the complex domain mathematical model of the control system, the state space model of the control system, the structure and signal flow graph of the control system 2-1 The time domain model, differential equation expression method2-2 Complex domain model1.the definition and properties of the transfer function;2.zero and poles expression, the open loop gain, root locus gain, etc.of the transfer function3.the transfer function’s typical link (proportion, inertia,differential and integral, oscillation);2-3 The structure and signal flow diagram of the control system1. The equivalent transformation and simplification of the structurechart2. Signal flow chart composition and properties3. Properties, term (understand)a. The method of structure chart translate into signal flow graphb. Mason formulaChapter Three, time domain analysis method of the linear system (10 hours)The performance index of the linear system time response, analysisof time domain of the first-order system, analysis of time domain ofthe second-order systems, analysis of time domain of the high ordersystems, linear system stability analysis, the steady-state errorcalculation of linear system3-1 The performance standard of time response of the linear systemr t ，p t ，s t ，%σ3-2 The first-order system ’s unit step response ]3-3 Time domain response of the second order system1. The closed-loop transfer function form of the second order system ’s standard mathematical model, expresses as the form of unit feedbacksystem2. Unit step response of the second order system (Key: owe damping case)r t ，p t ，s t ，%σ3. System performance improved of the second order systemA. Proportion-differential control, master principle andcharacteristicsB. Velocity measurement feedback control, master the calculation ofprinciple, characteristics and performance parameter3-4 Analysis of time domain of the high order system, the concept of the leading close-loop poles3-5Linear system stability analysisLaws criterion and its application3-6 Steady-state error of linear system1.The calculation of error transfer functioning final-value theorem for steady-state error3.System type(pattern)4.The steady-state error of typical reference signal input; Errorcoefficient5.Reduce the steady-state error method: improve pattern, improve the openloop gain, and use composite control6. The transfer function of agitation error, the method of reduce agitation error(1). Increase integral element number before the agitationapplication point(2). Increase integral element gain before the agitationapplication pointAdopted complex control technologyChapter Four, Root locus (6 hours)Root locus equation, basic law of root locus rendering, generalized root locus, system performance analysis and estimate, the control system correction methods based on root trajectory, MATLAB language root locus analysis method.4-1 Root locus equation (Angle, amplitude condition)4-2 Basic law of root locus rendering, draw root locus*Chapter Five, frequency domain analysis of the linear systemFrequency characteristics, typical links and open loop system frequency characteristics, Nyquist stability criterion, the margin of the stable, closed-loop frequency characteristics, the system time domain indexes estimation, the experiment confirm method of the transfer function, the frequency domain method of MATLAB language5-2 Frequency characteristics1.The basic concept, the relationship between frequency characteristics andtransfer function of the system2.The expression method of frequency characteristics, phase curve; Amplitudefrequency characteristics; Logarithm amplitude frequency characteristics curve; Phase frequency characteristic curve; Logarithm phase frequency characteristic curve5-3Typical links and open loop system frequency characteristics1.Typical link’s amplitude frequency, phase frequency characteristics3.The open loop logarithm amplitude and frequency, amplitude and phase curvedrawing, logarithm amplitude frequency characteristics uses theasymptote method2.The minimum phase system transfer function method determined by amplitudefrequency curve or phase frequency curve5-4Nyquist stability criterion1.Nyquist criterion, actual use criterion, amplitude and phase curvesymmetry2.The stability criterion of log frequency characteristics auxiliary lineused in type I and above5-5 The concept of stability margin: cut-off frequency; Phase angle border frequency; the margin of the phase angle; amplitude margin5-6 Closed-loop frequency characteristics (know)5-7 the relationship between system time domain index and system frequency domain indexUnderstand the relationship between them*Chapter Six, Linear system calibration method (10 hours)The system design and correct problem, commonly used correction device and its characteristics, series correction, feedback correction, compound adjustment, PID regulator, the correction analysis method based on MATLAB language.6-1 Correction method series, feedback, feedforward, composite,principle, characteristics6-2 commonly used calibration device and its characteristics (transfer function, characteristics, important formula)1.Passive ahead2.Passive lag3.Passive lag-ahead6-3 Series correction1. Serial advanced method, practical scope2. Serial lag method, practical scope3. Serial lag-advanced method, practical scope6-4 Feedback correction (principle, advantages)6-5 Compound correctionA. According to the disturbance compensation (principle, the transferfunction derivation of disturbance error, error full compensationconditions)B. According to the input compensation (principle, the transferfunction derivation of error, error full compensation conditions) 6-6 PID control principle (the characteristics of P, PD, I, PI, PID) Experimental arrangement:Lab 2, DC motor speed control experiment (8 hours)1. Design of open loop control, closed loop control system2. Correction design of control system3. PID regulatorOpen, integrated design experiments, students are required to completethe preparation work alone before (experiment aim, principle analysis,experiment design, experiment steps, parameter calculation, etc.),practical experiments, the analysis of experimental results, etc.4. Semester Hour Structure5. Text-Book & Additional ReadingText-Book:“Automatic control principle”, Shousong Hu, etc. Defense industry press, version 4, 2002Additional Reading:[1] “Automatic control principle”, Qi Wu, etc. Qinghua Press[2] “Automatic control principle”, Guozhong Pang, etc. China scienceand technology university press[3] “Automatic control principle questions selected and examinationskills”, Tong Wang, etc. Harbin industrial university press[4] “Automatic control principle common question analytical andsimulation topic”, Zhongke Shi, Jingchao Lu, etc. northwestern university press[5] “MATLAB language TOOLBOX-TOOLBOX practical guide”, ShiYang northrop,etc. Northwestern university press[6] “MATLAB language-the operations paper type science and engineeringcalculation language”, PeiJiang Zhang, China science and technologyuniversity press;[7] “System analysis and simulation---MATLAB language and application”,WenMei Huang, Yong Yang, etc. National defense science and technology university press.6. GradingThe final examination 80%, Ordinary times (assignments, exams and class performance) and comprehensive experiment ability 20%.Data: 2011.6。