自动控制原理英文版
《自动控制原理》绪论、第1、2章(新)
1、线性系统— 系统中各组成部分或元件特性可以用线性微分方程式 来描述的系统。 特点： （1）满足叠加原理。（对线性系统，初始条件为零时，几 个输入信号同时作用在系统上所产生的输出信号，等 于各输入信号单独作用时产生的输出信号的和。） （2）系统的运动方程式可以用线性微分方程式来描述，暂 态特性与初始条件无关。 （3）系统为线性定常系统。 2、非线性系统 —当系统中存在非线性元件或具有非线性特性，其运 动方程用非线性微分方程式来描述。 特点： 不满足叠加原理；暂态特性与初始条件有关。
第四章
4.1 4.2 4.3
根轨迹法的基本概念 根轨迹的绘制法则 用根轨迹法分析系统的暂态特性
( 2) (4) (2)
第五章 5.1 5.2 5.3 5.4 5.5 5.6
频率法 12学时 频率特性的基本概念 频率特性的基本方法 典型环节的频率特性 系统开环频率特性的绘制 用频率法分析控制系统的稳定性 系统暂态特性和开环闭环特性的关系 6学时
3、典型的非线性环节特性
4、两者的关系（参考教材Page6)
二、 连续数据系统和离散数据系统
1 、连续数据系统—— 信号为模拟的连续函数。 2、离散数据系统 —— 系统中一处或多处，信号以序列 或数码形式传递。 3、两者研究方法比较 连续：微分方程 — 拉氏变换 — 传递函数和频率特性 分析 离散：差分方程 — Z变换 —— 脉冲传递函数和频率 特性分析
自动控制原理教学大纲
自动控制原理Automatic Control Principle一、课程基本信息二、课程简介自动控制原理是一门专业基础必修课，属于经典控制理论，主要处理单输入单输出定常反馈控制系统。
通过对本课程的学习，使学生掌握系统数学模型的建立方法，学会经典控制理论的三种分析方法，即时域法，根轨迹法和频域法，围绕三个性能指标，对控制系统进行分析，并在此基础上，学会控制系统的设计与综合，继而培养学生在实际中分析问题和解决问题的能力。
该课程为现代控制理论及智能控制理论等后继课程打下了必要的理论基础。
English Course IntroductionAutomatic Control Principle is a compulsory course in basic professional studies, which belongs to classical control theory. It mainly deals with the single input and single output steady feedback control system. Through the study of this course, the students can master the method of establishing the mathematical model of the system and learn the three analysis methods of the classical control theory, namely, the time domain method, the root locus method and the frequency domain method, and analyze the control system around three performance indexes, on this basis, learn the design and synthesis of the control system, and then train students in the actual analysis of problems and problem-solving ability. This course lays a necessary theoretical foundation for the following courses such as Modern Control Principle and Intelligent Control Principle.三、教学目的通过本课程的学习，使学生了解和掌握自动控制理论的基本概念、主要原理和分析方法，了解自动控制技术发展的概况，为学习后继课程以及从事与本专业有关的自动控制技术工作打下一定的基础。
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。
自动控制原理重点英文单词+汉语翻译
Closed-loop control systems 闭环控制系统Open-loop control systems开环控制系统Process 过程Linear system线性系统Nonlinear system非线性系统Continuous system 连续系统Discrete system离散系统Stability 稳定性Steady-state performance 稳态性能Transient performance 暂态特性mathematical model 数学模型differential equations 微分方程Transfer function传递函数zeros and poles of transfer function传递函数的零极点Inverse proportion part 比例环节Inertia part惯性环节Integral part 积分环节Derivative part 微分环节Vibrate part震荡环节Delay part 滞后环节Block diagram (Equivalent transformation)方框图(Unit) negative (positive) feedback loop 负（正）反馈回路Mason formula 梅逊公式disturbance 干扰Step signal 阶跃信号Ramp signal(speed function signal) 斜波信号Parabola signal(acceleration signal) 加速度信号pulse signal脉冲信号Sinusoidal signal 正弦信号Delay time延迟时间Rise time 上升时间Peak time峰值时间Settling time 稳定时间Percent overshoot超调量Steady-state error稳态误差position error coefficient Kp speed error coefficient Kv acceleration error coefficient Kafirst-order system一阶系统S econd-order system 二阶系统high-order system 高阶系统Dominant pole主导极点Underdamped欠阻尼Critically Damped临界阻尼Overdamped 过阻尼Undamped无阻尼的Routh-Hurwitz stability criterion劳斯稳定性判据R outh array 劳斯表Character equation 特征方程root locus 根轨迹open-loop zeros and poles 开环零极点Magnitude and angle requirements of root locus幅值与相角frequency character 频率特征(inverse) Laplace transformation 拉普拉斯（反）变换Nyquist plot奈奎斯特图Bode diagram波德图Logarithmic magnitude frequency character对数幅值频率特性Logarithmic phase frequency character对数相频特征Nyquist stability criterion奈奎斯特稳定判据cutoff frequency 剪切频率Phase margin 相位裕量Gain margin 增益裕量Cutoff frequencyCascade phase-lead compensation串联超前矫正Cascade phase-lag compensation 串联滞后校正Cascade phase-lag and -lead compensation串联滞后-超前矫正sample control system 采样控制系统digital control system 数控系统discrete control system离散控制系统Shannon sampling theorem 香农采样定理Zero-order hold 零阶保持sampling period 采样周期Sampling frequency 采样频率Z-transform z变换Z-inverse transform z逆变换pulse transfer function脉冲传递函数bilinear transform双线性变换。
《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。
自动控制原理第八版英文版教学设计 (2)
Teaching Design for Automatic Control Principles 8thEdition (English Version)1. IntroductionAutomatic control principles are the foundation of modern control engineering. With the rising demand for control engineering talents, it is essential to design a suitable teaching plan for students who are interested in this field. This teaching design is med at providing an understanding of the basic concepts of automatic control principles and their applications.2. ObjectivesThe primary objectives of the teaching design are:•To equip students with knowledge, concepts, and analytical skills required in the automatic control principles field.•To facilitate students with the application of control theory methods to the design and analysis of control systems.•To establish a foundation for further study in advanced control techniques.3. Teaching MethodologyThe teaching of automatic control principles will use the flipped classroom approach. This approach will enable students to engage in a more interactive and practical learning experience where they can applythe theoretical principles they have learned in class to real-world applications.3.1 Pre-class LearningBefore class, students are expected to study the related learning materials, which includes studying the assigned chapters from the textbook – Automatic Control Principles, 8th edition, written by the renowned author, G. Franklin, J. Da Powell, and A. Emami-Naeini.3.2 In-class LearningClassroom lectures will not be used to cover the fundamental concepts of automatic control principles. Rather, the students will participate in interactive sessions, where they will have a chance to engage in problem-based learning activities.3.3 After-class LearningAfter class, students will have access to online resources such as recorded classroom sessions, tutorials, and online assessments. The materials will be made avlable to students via the school’s online learning management system.4. Course OutlineThe course outline will cover the following key topics:•Introduction to automatic control principles and its applications.•Mathematical modeling of dynamic systems.•Time-domn analysis of control systems.•Root-locus analysis of control systems.•Frequency-domn analysis of control systems.•Design of classical control systems.•Introduction to modern control principles.•Digital control systems.5. Teaching Plan•Week 1: Introduction to Automatic Control Principles•Week 2-3: Mathematical Modeling of Dynamic Systems•Week 4-5: Time-domn Analysis of Control Systems•Week 6-7: Root-locus Analysis of Control Systems•Week 8-9: Frequency-domn Analysis of Control Systems•Week 10-11: Design of Classical Control Systems•Week 12-13: Introduction to Modern Control Principles•Week 14-15: Digital Control Systems6. Course AssessmentCourse assessment will be divided into formative and summative assessments. The formative assessments will include online quizzes, homework assignments, and problem-solving in the classroom activities, while the summative assessment will include a final examination.7. ConclusionIn conclusion, this teaching design is med at equipping studentswith theoretical concepts and analytical skills required in the automatic control principles field. The teaching will be organized using the flipped classroom approach, which will be more interactive and practical for students to participate in problem-based learningactivities. With this approach, students will have an opportunity to apply the theoretical principles they have learned in class to real-life problems. The course ms to establish a strong foundation for further study in advanced control techniques.。
《自动控制原理》
《自动控制原理》教学大纲一、课程基本信息：1、课程英文名称：Principles of Automatic Control2、课程类别：技术基础课程3、课程学时：总学时64，实验学时84、学分：45、先修课程：《电路原理》、《信号与系统》、《复变函数与积分变换》等6、适用专业：测控技术与仪器7、大纲执笔：自动化教研室罗敏8、大纲审批：电子信息工程学院学术委员会9、制定(修订)时间：2007年10月二、课程的目的与任务：随着生产和科学技术的发展，自动控制技术在国民经济和国防建设中所起的作用越来越大。
自动控制技术的应用不仅使生产过程实现了自动化，极大的提高了劳动生产率和产品质量，改善了劳动条件，并且在人类探索新能源，发展空间技术和改善人民物质生活都起着极为重要的作用。
自动控制原理是电子信息类专业的技术基础课（专业基础平台课），是必修课，是以原理为主的理论性课程；主要讲述自动控制原理与控制系统设计、实验等内容。
根据自动控制技术发展的不同阶段，自动控制原来可分为古典控制理论和现代控制理论两大部分。
本课程主要介绍古典控制理论，其主要内容是以传递函数为基础，研究单输入单输出自动控制系统的分析和设计问题。
这些理论研究较早，现在已经比较成熟，并且在工程实践中得到了广泛的应用。
主要目的是培养学生掌握经典控制论中线性定常连续、单输入单输出闭环控制系统的工作原理、分析和综合，掌握反馈控制原理的应用以及分析和设计的一般规律，使其具有分析和设计自动控制系统的初步能力，使学生对系统的认识上升到更高的层次。
三、课程的基本要求：本课程是电子信息类专业重要的技术基础课。
要求在理解有关自动控制系统的基本概念、建立控制系统数学模型的基础上，掌握并灵活运用时域法、根轨迹法和频率法进行系统分析的思路和方法，基本明确三种方法各自的特点及其内在联系。
基本掌握运用频率法进行串联校正设计的能力。
通过对离散系统的学习，掌握离散控制系统的特点，理解脉冲传递函数和系统稳定性分析等知识。
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2021双语cha幻灯片PPT
控制系统简介
2
The Principle of Automatic Control 2008
Course Description
• This course is concerned with the analysis and design of control system
Manipulated variable 〔操纵量〕:按照指定规律变化使得被控 量到达指定目标的变量(varied by controller so as to affect the controlled variable)
e.g. Temperature control system (空调〕
Speed control system 〔汽车〕
转速0~1000转 扰动
P
电位器 ur
电压 uk 放大器
可控硅 ua 直流 n
功放
电动机
13
The Principle of Automatic Control 2008
Comments on open-loop control system
• The output is not measured and compared with the input, no feedback
comparing and using the difference (error) as a means of control –feedback control system
15
The Principle of Automatic Control 2008
Block diagram
r
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
y
y 0 y
y0
df dx
x0
f ( x)
y k x
A
x
x 0 x0 x
Example 2.5 : Elasticity equation
F ( x ) kx
suppose : k 12.65; 1.1;
F ( x ) kx
'
operating point x0 0.25
1) Identify the output and input variables of the control systems.
2) Write the differential equations of each system’s component in terms of the physical laws of the components. * necessary assumption and neglect. * proper approximation. 3) Dispel the intermediate(across) variables to get the input-output description which only contains the output and input variables. 4) Formalize the input-output equation to be the “standard” form: Input variable —— on the right of the input-output equation . Output variable —— on the left of the input-output equation. Writing the equation—according to the falling-power order.
自动控制系统英文介绍
自动控制系统英文介绍全文共四篇示例，供读者参考第一篇示例：Automatic Control System IntroductionIn conclusion, automatic control systems play a crucial role in regulating the operation of machines and processes in various applications. They help improve efficiency, accuracy, and safety, making them essential in modern industrial and technological systems. The continuous advancement in automation technology will further enhance the capabilities of automatic control systems, leading to more efficient and intelligent systems in the future.第二篇示例：Automatic Control System IntroductionAn automatic control system is a system designed to control and regulate processes or systems without human intervention. It is an essential technology in various fields, including manufacturing, aerospace, automotive, and other industries.第三篇示例：There are several key components to an automatic control system. The first is the control algorithm, which is the set of rules that determine how the system should respond to changes in the input variables. This algorithm is often implemented using a computer program that runs on a microcontroller or a PLC (programmable logic controller).第四篇示例：Automatic control system is a technology that enables the control of machines, processes, and systems without human intervention. It relies on the use of sensors, actuators, controllers, and communication devices to monitor and adjust the operation of a system in real time. The goal of an automatic control system is to ensure the system operates efficiently, safely, and reliably without the need for constant human intervention.。
完整word版,《自动控制原理》试卷及答案(英文10套)
AUTOMATIC CONTROL THEOREM(1)⒈Derive the transfer function and the differential equation of the electric network shown in Fig.1. (12% )R1C1V1(S)C2V2(S) R2⒉Consider the system shown in Fig.2. Obtain the closed-loop transferfunction C (S),E( S).（12%）R(S) R(S)G4R ECG1G2G3H2H1⒊ The characteristic equation is given 1 GH (S) S35S2(6 K )S 10K 0. Discuss the distribution of the closed-loop poles. (16%)①There are 3 roots on the LHP ② There are 2 roots on the LHP②There are 1 roots on the LHP ④ There are no roots on the LHP . K=?⒋Consider a unity-feedback control system whose open-loop transfer function is1G(S). Obtain the response to a unit-step input. What is the rise time for S( S0.6)this system? What is the maximum overshoot?（ 10%）5. Sketch the root-locus plot for the systemK. ( The gain K is GH (S)S(S1)assumed to be positive.)①Determine the breakaway point and K value.②Determine 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. Supposer( 2 t),n. Determine1the value of K to ensure e SS1 .(12%)NR E4CKS 2S(S 3)7. Consider the system with the following open-loop transfer function:GH (S)K① Draw Nyquist diagrams. ② Determine the.S(T1S1)(T2 S 1)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%)R(S)S 2S3C(S) (S2)(S 5)(S10)⒈V2(S)R2C1C2 S C1V1( S) ( R1R2 )C1C2 S C1 C2⒉C(S)G1G2 G3 G1G4R(S) 1 G1G2H1G1G2G3 G1G4 G2G3H 2 G4H 2⒊ ① 0<K<6 ② K ≤0 ③ K ≥6 ④ no answer⒋⒌① the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2 ③⒍31.62( S1)⒎ GH (S)S S SS1)(1)((1)(1)AUTOMATIC CONTROL THEOREM(2)⒈Derive the transfer function and the differential equation of the electric network shown in Fig.1. (12% )R1R1V1(S)C1C2V2(S)⒉Consider the equation group shown in Equation.1. Draw block diagram and obtainthe closed-loop transfer function C (S). （16%）R(S)X1 (S) G1 (S)R(S)G1(S)[G7 (S) G8 (S)]C(S)X2 (S)G2(S)[ X1( S)G6 (S)X 3(S)][ X2(S)C(S)G5 (S)]G3(S)X3(S)C(S)G4(S)X3 (S)⒊Use Routh’s criterion to determine the number of roots in the right-half S plane forthe equation 1GH (S) S53S428S3226S2600S 400 0 .Analyze stability.（12% ）⒋ Determine the range of K value ,when r(1 t t 2 ) , e SS0.5 .（12%）RE3S2 4S KS2(S 2)C⒌Fig.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%）R K CS(S24S 5)R E⒍Sketch root-locus diagram（.18% ）Im Im ImRe Re ReIm Im ImRe Re Re⒎Determine the transfer function. Assume a minimum-phase transfer function. （10% ）L(dB)0200–20–40 -604030205ωω1ω2ω3ω4⒈V2(S)1V1 ( S) R1C1R2 C2 S2(R1C1 R2 C2 R1C2 )S 1⒉C(S)G1 G2G3G 4R( S) 1 G2G3G6G3G4 G5 G1G2G3G4 (G7 G8 )⒊There are 4 roots in the left-half S plane, 2 roots on the imaginary axes, 0 root in the RSP. The system is unstable.⒋8K20⒌K=20⒍31.62( S1)⒎GH (S)S S SS1)(1)((1)(1)AUTOMATIC 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 network shown in Fig.1.（ 16% ）C1R1U2U1R2C2⒊Consider the system shown in Fig.2. Obtain the closed-loop transferC(S)E(S)C(S)function,,.（12%）E G5R P CG1G2G3G4H2H1H3⒋ The characteristic equation is given 1GH (S) S33S22S 200 . Discuss the distribution of the closed-loop poles. (16%)5. Sketch the root-locus plot for the systemK. (The gain K is GH (S)S( S1)assumed to be positive.)④Determine the breakaway point and K value.⑤Determine 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.G1K4. Suppose ,G2S2S(S3)r (2t ) , n 1 . Determine the value of K to ensure e SS 1 .(15%)RNEG2C G17. Consider the system with the following open-loop transfer function:GH (S)K① Draw Nyquist diagrams. ② Determine the.S(T1S1)(T2 S 1)stability of the system for two cases, ⑴ the gain K is small, ⑵ K is large. (15%)⒈Solution: The advantages of open-loop control systems are as follows:①Simple construction and ease of maintenance②Less expensive than a corresponding closed-loop system③ There is no stability problem④ Convenient when output is hard to measure or economically not feasible. (Forexample, 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:①Disturbances and changes in calibration cause errors, and the output may bedifferent from what is desired.②To maintain the required quality in the output, recalibration is necessaryfrom time to time.U2(S)R1C1 R2 C2 S2(R1C1 R2C2 )S 1⒉U1(S)R1C1 R2 C2 S2(R1C1 R2 C2 R1C2 )S 1⒊ C(S)G1G2 G3G4 G1G5R(S) 1 G1G2H 1 G2G3H 2G1G2G3G4 H 3G1G5H 3C( S)G3G4 (1 G1G2 H 1) G3G5 H 2P( S) 1 G1G2H 1 G2G3H 2G1G2G3G4H 3G1G 5H 3⒋R=2, L=1⒌S:① the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2⒍ 3.5 KAUTOMATIC CONTROL THEOREM(4)⒈Find the poles of the following F ( s) :F ( s)1(12%)e s1⒉ Consider the system shown in Fig.1,where0.6 and n 5 rad/sec. Obtain the rise time t r , peak time t p , maximum overshoot M P , and settling time t s when the system is subjected to a unit-step input. （ 10%）2ns(s2n )C(s)R(s)⒊Consider the system shown in Fig.2. Obtain the closed-loop transferC(S)E(S)C(S)function,,.（12%）G5ERP CG1G2G3G4H2H1H3⒋ The characteristic equation is given 1GH (S) S33S22S 200 . Discuss the distribution of the closed-loop poles. (16%)5. Sketch the root-locus plot for the systemK. (The gain K is GH (S)S( S1)assumed to be positive.)⑦Determine the breakaway point and K value.⑧Determine 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.G1K4. Suppose ,G2S2S(S3)r (2t ) , n 1 . Determine the value of K to ensure e SS 1 .(12%)RNEG2C G17. Consider the system with the following open-loop transfer function:GH (S)K① Draw Nyquist diagrams. ② Determine the.S(T1S1)(T2 S 1)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%)R(S)S 2S3C(S) (S2)(S 5)(S10)⒈ Solution: The poles are found from e s1or e(j ) e (cosj sin ) 1From this it follows that0,2n(n0,1,2, ) . Thus, the poles are located at s j 2n⒉ Solution: rise timet rt0.785 sec, , peak time pmaximum overshoot M P0.095 ,and settling time t s 1.33 sec for the 2%criterion, settling time t s1sec for the 5% criterion.⒊C(S)G1G2 G3G4G1G5R(S) 1 G1G2H 1G2G3H 2G1G2G3G4 H 3G1G5H 3C( S)G3G4 (1 G1G2 H 1) G3G5 H 2P(S) 1 G1G2H 1G2G3H 2G1G2G3G4H 3G1G 5H 3⒋R=2, L=15.S:①the breakaway point is –1 and –1/3; k=4/27 ② The imaginary axis S=±j; K=2⒍ 3.5 KAUTOMATIC CONTROL THEOREM(5)⒈Consider the system shown in Fig.1. Obtain the closed-loop transferfunction C (S),E( S).（18%）R(S) R(S)H2R E CG1G2G3H1H3H4⒉The characteristic equation is given1GH (S) S5 3S4 12S3 24S2 32S 48 0 . Discuss the distribution of the closed-loop poles. (16%)⒊ Sketch the root-locus plot for the system GH (S)K. (The gain S( S1)(0.5S 1)K is assumed to be positive.)①Determine the breakaway point and K value.②Determine 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. G1K 1, G2K 2.①T1S 1T2S 1Suppose r 0 , n 1 . Determine the value of e SS.②Suppose r 1 , n 1. Determine the value of e SS . (14%)RNEG2C G1⒌ Sketch the Bode diagram for the following transfer function. GH (s)K, s(1Ts)K 7 , T 0.087. (10%)⒍ A system with the open-loop transfer function GH (S)K is inherently2 (TSs1)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%)⒎Draw the block diagram and determine the transfer function. (10%)U1(s)R C U2(s)⒈C (S) G 1G 2 G 3 R(S)⒉ R=0, L=3,I=2⒋① e ssK 21 K2 K 1K 2② e ssK 1K 21 1 ⒎U 2 ( s) 1U 1 ( s)RCs 1AUTOMATIC CONTROL THEOREM(6)⒈Consider the system shown in Fig.1. Obtain the closed-loop transferfunction C (S),E( S).（18%）R(S) R(S)R E CG1G2H1H2H3⒉The characteristic equation is given1 GH (S) 25S5105S4120S3122S220S 1 0 . Discuss the distribution of the closed-loop poles. (12%)⒊ Sketch the root-locus plot for the system GH ( S)K (S 1). (The gain K isS( S3) assumed to be positive.)①Determine the breakaway point and K value.②Determine 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. G11, G210. Suppose S1)r 1t , n 0.1 . Determine the value of e SS .(12%)R E N CG1G2⒌ Calculate the transfer function for the following Bode diagram of the minimum phase. (15%)dB14816w-40-200dB/dec 200⒍ For the system show as follows, G(s)4,(16%), H (s) 1s(s5)①Determine the system output c(t ) to a unit step, ramp input.② Determine the coefficient K P , K V and the steady state error to r (t )2t .⒎Plot the Bode diagram of the system described by the open-loop transfer functionelements G(s)10(1s), H (s) 1.(12%)s(10.5s)⒈C(S)G 1G 2 (1 G 2H 2)R(S) 1 G 1H 1G 2H 2 G 1G 2H 3H 2 G 1G 2H 1H 2 G 2H 2H 3⒉ R=0, L=50.05(10s 1)( s 1)(s1)⒌ G(s)41s)s 2 (116⒍c(t ) 1 4e t 1 e 4 tc(t) t5 4 e t 1 e 4t K P, K V ,334 312essAUTOMATIC CONTROL THEOREM(7)⒈Consider the system shown in Fig.1. Obtain the closed-loop transferfunction C (S),E( S).（16%）R(S) R(S)R E CG1G2G3⒉The characteristic equation is given1 GH (S) S64S54S44S37S28S 10 0 . Discuss the distribution ofthe closed-loop poles. (10%)⒊Sketch the root-locus plot for the system GH ( S)K(S 1). (The gain K is assumed to be positive.)S3①Determine the breakaway point and K value.②Determine the value of K at which root loci cross the imaginary axis.③ Discuss the stability.(15%)⒋Show that the steady-state error in the response to ramp inputs can be made zero, if the closed-loop transfer function is given by:C( s)a n 1s a n; H ( s) 1（12%）R( s)s n a1s n 1a n 1 s a n⒌ Calculate the transfer function for the following Bode diagram of the minimum phase.dB-40-20dB/decw（15%）w1 w2w3-40⒍Sketch the Nyquist diagram (Polar plot) for the system described by the open-looptransfer function GH (S)1, and find the frequency and phase such that 1)magnitude is unity.（16%）⒎ The stability of a closed-loop system with the following open-loop transferfunction GH (S)K (T2 s1)s2 (T1s depends on the relative magnitudes of T1 and T2 .1)Draw Nyquist diagram and determine the stability of the system.（ 16%）( K 0T1 0T20 )⒈C (S)G1G1G2G 3R(S) 2 G1G1G2G2G3 G1G2 G3⒉R=2, I=2,L=222 (s1)⒌ G(s)1ss2 ()31⒍0.986rad / s oAUTOMATIC CONTROL THEOREM (8)⒈ Consider the system shown in Fig.1. Obtain the closed-loop transfer functionC (S) , E(S) . （16%）R( S) R(S)G1G2RECG3G4⒉ The characteristic equation is given 1 GH ( S) S 33KS 2 (2 K)S 4 0.Discuss the condition of stability. (12%)⒊ Draw the root-locus plot for the system GH ( S)K2 ; H (s) 1.1)2( S 4)( SObserve that values of K the system is overdamped and values of K it is underdamped. (16%)K (1 0.5s) . Determine the⒋ The system transfer function is G (s), H ( s) 1s(1 2s)(1 s)steady-state error e SS when input is unit impulse (t) 、unit step 1(t ) 、unit ramp t and unit parabolic function1 t2 . (16%)2⒌ ① Calculate the transfer function (minimum phase);② Draw the phase-angle versusdB-40-20dB/decw(12%)w1 w2 w3-40⒍Draw the root locus for the system with open-loop transfer function.K (1 s)(14%)GH (s)s( s 2)(s3)⒎ GH ( s)K Draw the polar plot and determine the stability of system.3 (Tss1)(14%)⒈ C(S)G 1G 2 G 3G 4 1 G 1G 2G 3G 4 R(S)G 1G 2 G 3G 4 G 1G 2 G 3G 4⒉ 0.528 K⒊ S:0<K<0.0718 or K>14 overdamped ;0.0718<K<14 underdamped ⒋ S:(t)ess0 ; 1(t ) e ss0 ; t e ss1 ; 1 t2 e ssK 21 2 (s1)⒌S:K12 ;G( s)1ss 2 ( )31AUTOMATIC CONTROL THEOREM(9)⒈Consider the system shown in Fig.1. Obtain the closed-loop transferfunction C (S),E(S) .（12%）R(S)N(S)G5RE N C G1G2G3G4H1H2H3⒉The characteristic equation is given1 GH (S) S327500S 7500K 0 . Discuss the condition of stability. (16%)⒊Sketch the root-locus plot for the system(s a)4a is GH (S)2 (s. (The gains1)assumed to be positive.)①Determine the breakaway point and a value.②Determine 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. G1(s) K i s1, G 2 ( s)K. Assume s(Ts 1)that the input is a ramp input, or r (t)at where a is an arbitrary constant. Show that by properly adjusting the value of K i, the steady-state error e SS in the response to ramp inputs can be made zero.(15%)R(s)E(s)C(s)G1(s)G2(s)⒌ Consider the closed-loop system having the following open-loop transfer function:K . ① Sketch the polar plot ( Nyquist diagram). ② Determine theGH (S)S(TS1)stability of the closed-loop system. (12%)⒍ Sketch the root-locus plot. (18%)Im Im ImRe Re ReIm Im ImRe Re Re⒎ Obtain the closed-loop transfer functionC (S). （15%）R( S)G4RCG1G2G3H2H1⒈C(S)G1G2G3G4G1G2 G4G5 (1 G3 H 1 )R(S) 1 G3H1G2G3H 2G1G2G3 G4 H 2G1G2 G4G5 H 3G3 H 1G1G2 G4 G5 H 3 E (S)G4H 3H 2G2G4G5 H 3N(S) 1 G3H1G2G3H 2G1G2G3G4H 2G1G2 G4G5 H 3G3 H 1G1G2 G4G5 H 3⒉⒌S: N=1 P=1 Z=0; the closed-loop system is stable⒎C(S)G1G2 G3 G1G4R(S) 1 G1G2H 1G1G2G3 G1G4 G2G3H 2 G4H 2AUTOMATIC CONTROL THEOREM(10)⒈Consider the system shown in Fig.1. Obtain the closed-loop transfer functionC(S) ,C(S). （16%）R( S)N (S)G3NR CG1G2G4G5H⒉The characteristic equation is given1 GH (S) S420KS 35S210S 15 0 . Discuss the condition of stability. (14%)⒊Consider a unity-feedback control system whose open-loop transfer function is1G(S). Obtain the response to a unit-step input. What is the rise time for S( S0.6)this system? What is the maximum overshoot?（ 10%）⒋ Sketch the root-locus plot for the system GH (S)K (10.5s). (The gain K isS(10.25s) assumed to be positive.)③Determine the breakaway point and K value.④Determine the value of K at which root loci cross the imaginary axis. Discuss the stability.(15%)⒌4, H (s) 1 .① Determine the The system transfer function is G ( s)s(s5)steady-state output c(t ) when input is unit step1(t )、unit ramp t .②Determine the K P、K V and K a, obtain the steady-state error e SS when input is r (t ) 2t .（12%）⒍Consider the closed-loop system whose open-loop transfer function is given by:K K K. Examine the stability ① GH (S); ② GH(S); ③GH(S)1 TS 1 TS TS1of the system.（15%）⒎Sketch the root-locus plot。