自动控制原理 英文版
自动控制原理 英语
自动控制原理英语Automatic control principleAutomatic control principle refers to the theory and techniques used in controlling mechanical, electrical, or other systems without human intervention. It involves monitoring the system's output and adjusting it based on predefined criteria or input signals.The basic principle of automatic control is to minimize the error between the desired output and the actual output. This is achieved by continuously measuring the output and comparing it to the desired value. If there is a deviation, a control system takes corrective action to bring the system back to the desired state.There are several key components and concepts in automatic control systems. The first is the sensor, which measures the system's output. This can be a physical device that measures variables such as temperature, pressure, or position, or it can be a software component that processes digital signals.The second component is the controller, which receives the measurement from the sensor and determines the appropriate action to take. The controller uses algorithms and mathematical models to calculate the desired output based on the current conditions.The third component is the actuator, which translates the control signal from the controller into physical action. This can be a motor, valve, or any device that manipulates the system to achieve the desired output.Feedback is another important concept in automatic control. It involves continuously monitoring the system's output and feeding it back to the controller. This allows the controller to make adjustments and fine-tune its output to minimize errors.There are different types of control systems, including open-loop and closed-loop systems. In open-loop systems, the controller doesn't receive feedback and operates solely based on the input signal. Closed-loop systems, on the other hand, use feedback to continuously adjust the output.Automatic control principles are used in a wide range of applications, including industrial processes, robotics, aerospace, and automotive systems. They improve efficiency, accuracy, and reliability by reducing human intervention and ensuring consistent performance.In conclusion, the automatic control principle is the theory and techniques used to control systems without human intervention. It involves measuring the system's output, comparing it to the desired value, and taking corrective action. Key components include sensors, controllers, actuators, and feedback. Automatic control systems are used in various industries to improve performance and reliability.。
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。
自控原理课件3(英文版)
r(t) 1 t 2 1(t)
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3-1. Introduction
Function imagine
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image
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Image function
Unit-impulse response of first-order system
For the unit-impulse input, R(s)=1, so
(s) C(s) 1 R(s) Ts 1
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3-2 Time response of first-order system
Figure 5-4 Unit-impulse response of the system shown
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3-2 Time response of first-order system
r(t)§R3(s).2C.3(s)=一F(阶s) R系(s)统的典c型(t) 响应
Given an absolute stable system, we will discuss further the relative stability of the system.
If the output of a system at steady state does not exactly agree with the input, the system is said to have steady-state error. This error is indicative of accuracy of the system.
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。
自控原理英文版分析
Desired temperature
thermometer
operator
gas
valve
air
furnace
mixer
1-3 What is an automatic control system?
Definition of automatic control system
A control system without people involved is called automatic control system.
2. According to whether the system is satisfied the superposition principle
Linear Systems versus Nonlinear System
A linear system satisfies the properties of superposition and Homogeneity: (Principle of Superposition).
1-3 What is an automatic control system?
water level automatic control system
1-4 types and Basic components of control systems.
1. According to the form of given signal Regulator Systems versus Servo Systems (恒值系 统和随动系统）
Open-loop system, closed-loop system (feedback control systems), Compound Control systems
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。
《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。
英文版自动控制原理教学大纲,参考
PRINCIPLES OF AUTOMATIC CONTROL SYLLABUSCourse name：principles of automatic control(Ⅱ) Chinese name:自动控制原理Course code：0243003 Course times：48 hours （Laboratory：6 hours ）Credit hour：3.5Applicable major：Thermal energy and powerplant engineeringPenner：Lingling Zhong Checker：1. Course GoalPrinciples of automatic control is a compulsory course for students majoring thermal energy and powerplant engineering. The course deals with introduction to design of feedback control systems, properties and advantages of feedback systems, time-domain and frequency-domain performance measures, stability and degree of stability. It also covers root locus method, nyquist criterion and frequency-domain design. By learning this course, the students are expected and required to understand the difficulties, to master the basic concepts and principles of the control system analysis and design, to develop the problem-solving skills and feedback control thinking, to lay massive foundation for further study and future career.2. Course ContentsThe course includes six chapters which consist of Introduction to Control Systems, Mathematic Models of Control Systems, Time-Domain Analysis of Control Systems, Root Locus Method, Frequency Response Method and Compensation of Control Systems. The break-down details in each chapter make step by step understanding.A fter learning this course, the students will be able to obtain a basic understanding of feedback control systems theory, the ability to perform analysis and design of linear feedback control systems, using both time and frequency domain techniques and hands on experience analyzing and designing control system.3. Grading Policy and SchedulingThe final grade is based on homework performances, midterm test and a final exam. The proportion is set like this: homework performances (20%) , midterm test (20%) , final exam score (60%).Scheduling4. Detailed ContentsPart 1: LectureChapter 1 Introduction to control systems1) Main contents•Introduction•Basic idea and history of automatic control•Basic types of automatic control systems•Basic requirements of automatic control systems2) Emphases and DifficultiesEmphases: The basic concept of automatic control and the principle of feedback (closed-loop) control3) Teaching requirements•To understand the history, application fields and development direction of automatic control•To comprehend the basic concept of automatic control and principles of feedback (closed-loop) controlChapter 2 Mathematical models of control systems1) Main contents•Introduction to system modeling•Laplace transform•Differential Equations of simple Physical Systems•The Transfer Function of Linear Systems•Block Diagram Model•Signal-Flow Graph Models•Equivalence among Models and Summary2) Emphases and Difficulties•Emphases: The method of getting the linear differential equation and transfer function of simple physical systems. deducing and calculating the transfer function of closed-loop system using block diagram and signal flow graph•difficulties: Lapalce transform; The equivalence transform and simplification of block diagram3) Teaching requirements•To understand the basic principle and method of mathematical modeling, such as approximation, complication and similar systems•To comprehend some basic concepts such as similar systems, transfer function, typical elements, block diagram, signal-flow graph•To master the method of getting linear differential equation and transfer function of simple physical systems. Equivalence transform and simplification of block diagram.Mason’s gain formula of signal-flow graphChapter 3 Time-Domain Analysis of Control Systems1) Main contents•Basic concepts of time-domain analysis•Typical input responses and specifications•The performance of 1-order systems•The performance of 2-order systems•The performance of high-order systems•Stability analysis for linear systems•Steady-state error of linear systems•Disturbance rejection2) Emphases and Difficulties•Emphases: The concept of time response；Transient response analysis and calculation of 2-order Systems. The concept of stability; The Routh—Hurwitz stability criterion. Analysis and calculation of steady-state error•Difficulties: Transient response analysis of high-order systems. The correlation between the location of closed-loop poles and zeros and the system performance.Definitions of stability in different meanings.3) Teaching requirements•To comprehend the concepts of typical input signal, the specifications of step-response, the effects of addition of closed-loop poles and zeros •To understand the concept of stability and the sufficient and necessary condition for stability•To understand the causes of steady-state error and the ways to reduce or eliminate the steady-state error•To master the characteristics of the step response of first-order and second-order systems and their relationship with system parameters•To master the Routh—Hurwitz stability criterion•To master calculation method of steady-state errorChapter 4 Root Locus Method1) Main contents•The root locus concept•Rules for Plotting root locus•Typical root locus and extended root locus•Control system analysis and design using root locus2) Emphases and Difficulties•Emphases: The root locus concept. The root locus rules•Difficulties: To analyze system’s performance using root locus3) Teaching requirements•To comprehend the root locus concept and the method of drawing root locus•To master the rules of drawing root locus by hand and how to analyze system’s performance using root locusChapter 5 Frequency Response Methods1) Main contents•The concept of frequency response•Bode diagrams of elementary factors•Open-Loop frequency response•Nyquist stability criterion•Relative stability•Closed-Loop frequency-domain analysis•Open-Loop frequency-domain analysis2) Emphases and Difficulties•Emphases: The concept of frequency response. Frequency characteristics of typical elements. Drawing of frequency characteristics. The concept of relative stability. The Nyquist criterion. The calculation of gain and phase margins.•Difficulties: Drawing the open-loop Bode diagram of general systems. The relation between frequency performance indices and time performance indices3) Teaching requirements•To understand the measurements of frequency response, the frequency performance indices•To comprehend the concept of frequency response and the concept of relative stability •To master the frequency characteristics of typical elements and the methods of drawing the open-loop Bode diagram for general systems•To master the Nyquist stability criterion; the calculation of gain and phase marginsChapter 6 Compensation of Control System1) Main contents•Concept of compensation•Compensation networks•Phase-lead design using the Bode diagram•Phase-lag design using the Bode diagram•PID controller•Feedback compensation2) Emphases and Difficulties•Emphases: The design of phase-lead and phase-lag networks•Difficulties: The selection of phase-lead and phase-lag networks3) Teaching requirements•To understand the feedback compensation•To comprehend the concept and method of control system design•To master the procedure of phase-lead and phase-lag compensationPart 2: Laboratories1) RequirementsThere will be three lab assignments during the term. Before a laboratory, every student should make preparations for it. Each laboratory will be carried out by teams of twostudents. A lab report is required for each lab. This report is due one week after your laboratory session.2) EquipmentsLabact experiment box and PC4) ContentsLab 1：The virtual oscillograph use and experiment box verification programMain contents•Be familiar with the structure and function of every section in Labact experiment box and building experiment circuit.•Start the application of virtual oscillograph, configure communication interface, and be familiar with the UI of virtual oscillograph and the testing method of parameters.•Test the function of every section of experiment box by verification program.Teaching requirements•Master the method of building experiment circuit in Labact box•Master the usage of virtual oscillograph software•Be familiar with the box verification program and record the test resultLab 2: Time-Domain analysis of linear systemMain contents•Verify the characteristics of the step response of elementary factors (gain element, inertial element, PID). Record the result of step response by adjusting the value of resistances, capacitors, voltage or other parameter.•Verify the characteristics of transient response and stability of 2-order systems. The test system will be in underdamped, overdamped or critical damped state by adjusting the input resistance. Observe the stability of this system and record the result of step response.Teaching requirements•Mater the structure of test circuit of elementary factors and 2-order systems•Observe the result of step response of elementary factors of gain, inertial and PID. Record the experiment result.•Observe the result of step response of 2-order systems of underdamped, overdamped and critical damped. Record the experiment result.Lab 3: Frequency-Domain analysis of linear systemMain contentsResearch the affection to frequency response of the circuit structure of 2-order closed-loop systems. Calculate the natural undamped oscillatory frequency nω, the damping ratioξ,the resonant frequency mωand the resonant magnitude mM of the 2-order test system and compare with corresponding theoretical value.Teaching requirementsMater the method of calculating the natural undamped oscillatory frequency nω,the damping ratioξ,the resonant frequency mωand the resonant magnitude mM of the 2-order test system. Compare with corresponding theoretical value and think of the reason of error.5) Lab reportA lab report is required for each lab. The report should be written in the unified report paper, which includes lab name, target, theory, steps, record, data processing and result analysis. This report is due one week after your laboratory session.6) GradingThe grade will be based on the hands on ability during the experiment and lab report after the experiment, which is marked by centesimal grade. The laboratory grade accounts for 20% in the final grade.5. Course Texts and References:Course texts:[1]自动控制原理（中英文对照），李道根主编，哈尔滨：哈尔滨工业大学出版社，2007.8[2] 张德银、魏鳞、钟玲玲编.自动控制原理实验.2006Reference:[1] Modern Control Engineering: Fourth Edition, （美）Katsuhiko Ogata, 影印本，北京：清华大学出版社，2006.2[2] Modern Control Systems, 8th ed. by Richard C. Dorf, Robert H. Bishop. Translated by Xie Hongwei. Higher Education Press（高等教育出版社）, 2001.06.[3] 胡寿松主编.自动控制原理简明教程.科学出版社，2003[4] 吴麒主编.自动控制原理.北京,清华大学出版社,1990。
完整word版,《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(K =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)()(Λ ;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∞ππK 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 --+++--= ⒉45.30ππK⒌S: 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.。
自控原理课件(英文版)分析
Linear Systems versus Nonlinear System
A linear system satisfies the properties of superposition and Homogeneity: (Principle of Superposition).
1-4 types and Basic components of control systems.
Example:
Desired output response (Command input)
Actuating device Controller
process plant
Output
1-4 types and Basic components of control systems.
Example: Water Level control system
Tank
Valve 1
Valve 2
Disturbance(valve 2)
Lever
Valve 1
Tank
1-4 types and Basic components of control systems.
Example: Temperature control system
Continuous Systems versus Discrete Systems 5. According to the number of I/O
SISO Systems versus MIMO Systems 6. According to the structure of the system
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.。
自动控制原理2008(双语)chapter5
The Principle of Automatic Control 2008
5-1 Frequency characteristics
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The Principle of Automatic Control 2008
Obtaining steady-state outputs to sinusoidal inputs
a AG( j) a AG( j)
2 j
2j
G(-jw) is a complex quantity, G( j) G( j) e jG( j)
A G( j) e j
a 2j
A G( j) e j
a 2j
G( j)
css (t)
A G( j) e j(t ) A G( j) e j(t )
2j
2j
A G( j) e j(t ) e j(t )
2j
css (t) A G( j) sin(t )
r(t) Asint
12
frequency characteristic / frequency response
The Principle of Automatic Control 2008
RC网络只允许低频信号通过,具 有低通滤波器(low-pass filter) 的性质。
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The Principle of Automatic Control 2008
A few conclusions from the example
• Physical meaning of FC:表示系统或环节对不同频率正弦 信号的跟踪能力或复现能力。确定了在不同频率正弦信号作 用下,系统或环节输入与输出的单值对应关系。 • Properties of FC: G(jw)只与系统或环节本身的结构参数 有关,是系统或环节本身的属性,与输入信号和初始条件无关。 • 频率法就是利用稳态的频率响应，间接地研究系统的动态特 性，从而避免了求解高阶微分方程的困难。
5.9.题解
■SolutionsP5.1 The open-loop transfer function of a unity feedback system is 125)(+=s s G .Determine the steady-state output of the closed-loop system due to the following input signals: (a) )30sin()( +=t t r , (b) )452cos(2)( -=t t rSolution: The closed-loop transfer function and frequency response are 625)(1)()(+=+=s s G s G s Φ226)2(5)(+=ωωΦj , 62arctan)(ωωΦ-=∠jrespectively.(a) In the case of )30sin()( +=t t r , since 1=ωand 300=ϕ, we have()0.794j ωΦ===, 2()arctan 18.436j ω∠Φ=-=-and the steady-state output is()0.79c o s (3018.43)0.79s ic t t t =+-=+(b) When )452cos(2)( -=t t r , since 2=ω and 450-=ϕ, we have()0.69j ωΦ===,69.33622arctan )(-=⨯-=∠ωΦjand the steady-state output is()1.39c o s (24533.69)1.39c oc t t t =--=-P5.2 The unit-step response of a system is t t e e t c 948.08.11)(--+-=, 0≥t Find the frequency response of the system.Solution: The impulse response and the transfer function are tteett c t k 942.72.7d )(d )(---==, )9)(4(3692.742.7)(++=+-+=s s s s s ΦRespectively. Hence, we have )9)(4(36)(++=ωωωΦj j j22229436)(+⋅+=ωωωΦj , 9arctan4arctan)(ωωωΦ--=∠jP5.3 Plot the asymptotic log-magnitude curves and phase curves for the following transferfunctions (a) )15.0)(12(1)()(++=s s s H s G , (b) 215.0)()(ss s H s G +=, (c) )1.0()2.0(10)()(2++=s s s s H s G ,(d) )8()2(32)()(2++=s s s s H s G ,(e))254)(1()1.0(8)()(22+++++=s s s s s s s H s GSolution: (a))15.0)(12(1)()(++=s s s H s G (b)215.0)()(ss s H s G +=(c))1.0()2.0(10)()(2++=s s s s H s G (d))8()2(32)()(2++=s s s s H s G)110()15(202++=s s s)1125.0()15.0(82++=s s s---------------(e))254)(1()110(252)254)(1()110(50)()(2222+++++⨯=+++++=s ss ss s s ss ss s s H s GP5.4 The asymptotic log-magnitude curves of some systems are given in Fig. P5.4. Determine the transfer function and sketch the corresponding asymptotic phase curves for each system. Assume that the systems have minimum phase transfer functions.Figure P5.4----Solution:(a) This a 0-pype system. (b) This a 2-pype system. 16.3 10lg 20=⇒=K K 1.0 20lg 20=⇒-=K K)205.0)(15.0(16.3)()(++=s s s H s G )1()110(1.0)()(2++=s s s s H s G(c) There is a differential factor.1.0 20lg 20=⇒-=K K (d) This a 1-pype system. 102.01.0)()(+=s s s H s G 1000 602080lg 20=⇒=-=K K)101.0)(110()12.0(1000)()(+++=s s s s s H s G(e) This a 1-pype system. 100 =K )101.0)(1100(100)()(++=s s s s H s G]--]---(f) There is a differential factor. (g) This a 0-pype system.11 ω=K 10 2020lg =⇒=K K)11)(11(1)()(321++=s s s s H s G ωωω55.2215.0 2025.21121lg2022=⇒=-==⇒-=-n n m ωςωωςςς255250)()(2++=s ss H s G(h) This a 1-pype system. (i) This a 1-pype system. 10 2020lg =⇒=K K 100 =K52.0 202821lg20==⇒-=n ωςς50 3.45213.0 85.4121lg2022=⇒=-==⇒=-n n m ωςωωςςς)25.6()1(5.62)()(2+++=s ss s s H s G)250030(250000)()(2++=s ss s H s G--90---0---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, υ is the number of the integral factors in the open-loop transfer function.Solution: The stability of each system will be determined by sketching half complete Nyquist plot and considering the difference between positive and negative crossovers as ω varies from zero to infinite.(a) 0=υ, 1 =P (b) 0=υ, 1 =P221021 p N N ==-=--+221210 p N N ≠-=-=--+The system is stable. The system is unstable.(c) 2 =υ, 0 =P (d) 2=υ,=P2000 p N N ==-=--+2110 p N N ≠-=-=--+The system is stable. The system is unstable.Figure P5.5(e) 0=υ, 1 =P (f) 1 =υ, 1 =P221021 p N N ==-=--+ 221210 p N N ≠-=-=--+The system is stable. The system is unstable.(g) 3=υ, 0 =P (h) 1 =υ, 2=P2011 p N N ==-=--+2101 p N N ==-=--+The system is stable. The system is unstable.(i) 0=υ, 2 =P (j) 0=υ, 1 =P2010 p N N ≠=-=--+221210 p N N ≠-=-=--+The system is unstable. The system is unstable.P5.6 Sketch the polar plots of the following open-loop transfer functions. Sketch only the portion that is necessary to determine the stability of the closed-loop systems. Determine the stability of the systems by using the Nyquist criterion. (a))1)(25.0)(12.0(10)()(+++=s s s s H s G , (b))14.0)(21.0(100)()(++=s s s s H s G ,ω(c))22)(1(10)()(2+++=s ss s s H s G (d))4)(2(50)()(2++=s s s s H s G(e) ss s H s G 2.01)()(-=Solution: (a))1)(25.0)(12.0(10)()(+++=s s s s H s G . By inspection, the phase-angle variesfrom 0 to 180-, and we have10)()(lim 0=→ωωωj H j G ,2700)()(lim j e j H j G -∞→⋅=ωωωLetting180)()(-=∠g g j H j G ωω, i.e.180arctan 5.0arctan 2.0arctan -=---g g g ωωωresults in sec /rad 17=g ω and 794.01171175.01172.010)()(22=+⋅+⨯⋅+⨯=g g j H j G ωωThe polar plot is sketched as shown.Since 0=p and 000=-=--+N N , the closed-loop system is stable.(b))14.0)(21.0(100)()(++=s s s s H s G . By inspection, the phase-angle varies from 90- to270-, and we have90)()(lim j ej H j G -→⋅∞=ωωω2700)()(lim j ej H j G -∞→⋅=ωωωLetting180)()(-=∠g g j H j G ωω, i.e.1804.0a r c t a n 1.0a r c t a n 90-=---g g ωωresults in sec /rad 5=g ω and 81215.05100)()(22=+⋅+⋅=g g j H j G ωωThe polar plot is sketched as shown.Since 0=p and 110-=-=--+N N , the closed-loop system is unstable.(c))22)(1(10)()(2+++=s ss s s H s G . By inspection, the phase-angle varies from 90- to360-, and we have90)()(lim j e j H j G -→⋅∞=ωωω3600)()(lim j e j H j G -∞→⋅=ωωωLetting180)()(-=∠g g j H j G ωω, i.e.18022arctanarctan 902-=----gg g ωωωresults in sec /rad 32=g ω and5.43223221323210)()(22=⨯+⎪⎭⎫ ⎝⎛-⋅+⋅=g g j H j G ωωThe polar plot is sketched as shown.Since 0=p and 110-=-=--+N N , the closed-loop system is unstable.(d))4)(2(50)()(2++=s s s s H s G . By inspection, the phase-angle varies from 90- to360-. It should be noted that412+sis an oscillatory element with 0=ςand its phase anglevaries from 0 to 180- suddenly in the case of 2=ω. To draw the polar plot, someThen, the polar plot is sketched as shown. Since=p and 110-=-=--+N N , the closed-loop system is unstable.(e) ss s H s G 2.01)()(-=. By inspection, the phase-angle, varies from 90 to 180,because the phase-angle of s2.011- is12.0arctanω-- and varies from 0 to 90.Considering that90)()(lim j e j H j G +→⋅∞=ωωω1805)()(lim j ej H j G +∞→⋅=ωωωthe polar plot can be plotted as shown.Since1=pand221021p N N ==-=--+, the closed-loop system is stable.P5.7 Sketch the polar plots of the following open-loop transfer functions, and find the maximum value for the open-loop gain so that the system is stable by using the Nyquist criterion. (a))4()()(2++=s s s k s H s G ,(b))4()2()()(2++=s s s k s H s GSolution: (a))4()()(2++=s s s ks H s G . By inspection, the phase-angle varies from 90-to 270-, and we have900)()(lim j e j H j G -→⋅∞=ωωω2700)()(lim j ej H j G -∞→⋅=ωωωnoting that the corner frequency of the oscillatory element just is the phase crossing frequency , we have s e c /r a d 2==n g ωω and422)()(k k j H j G g g =⨯=ωωThe polar plot is sketched as shown.As shown in the polar plot, the system is stable if and only if 1)()(<g g j H j G ωω, i.e.4<k . Since the open-loop gain is 4k K <, the system is stable if and only if 1<K .(b))4()2()()(2++=s s s k s H s G . There are two integral elements.Considering that the corner frequency of the first-order differential element is less than that of the inertial element, the polar plot can be drawn as shown.Obviously, the system is always unstable for any given open-loop gain.P5.8 A negative feedback system has an open-loop transfer function)10)(2()5.0()(2+++=s s s s k s GPlot the Bode diagrams with asymptotic curves and determine whether the system is stable using the Nyquist criterion for 10=k and 1000=k , respectively.Solution: The open-loop gain is given by 40)(lim 20k s G s K s ==→Hence, db 1225.0lg 20lg 2010-===k Kdb2825lg 20lg 201000===k Kand the Bode diagram can be plotted as shownIn the case of 10=k , there is no crossover in the phase-angle plot and the system is stable. In the case of 1000=k , there is a negative crossover and the system is unstable.P5.9 A unity negative feedback system has the open-loop transfer function)11.0)(105.0(7.11)(++=s s s s GDetermine the crossover frequency and the phase margin.Solution: Letting 1)(=c j G ω yields22462227.110125.0005.0 7.111)1.0(1)05.0(=++⇒=++c c c c c c ωωωωωω To find c ω, let224627.110125.0005.0)(-++=c c c c f ωωωω we have09.135)1(<-=f , 01.113)10(>=f , 07.103)5(<-=f , 01.15)8(<-=f 00.10)5.8(>=f , 05.0)3.8(<-=f , 01.2)35.8(>=f Taking sec / 3.8rad c =ω yields)3.81.0a r c t a n ()3.805.0arctan(90180)(180⨯-⨯--=∠+= c j G ωγ 8.277.395.2290=--=Or, considering that 101<<c ω, we have22427.1101.0 7.111)1.0(≈+⇒≈+⋅c c c c ωωωω s e c / 79.8rad c ≈ω)79.81.0arctan()79.805.0arctan(90180)(180⨯-⨯--≈∠+≈ c j G ωγ 0.253.417.2390=--=P5.10 A closed-loop system has the open-loop transfer functionsKes G sτ-=)((a) Determine the gain K so that the phase margin is 60 when s2.0=τ. (b) Plot the phasemargin versus the time delay τ for K as in part (a).Solution: (a) Letting 1)(=c j G ω yieldsK Kc c=⇒=ωω 1hence, we have632)(πτωπτωππωπγ=⇒=--=∠+=c c c j G62.26===τπωc K(b) In the case of62.2=K , the open-loop transfer function isses G s2.062.2)(-=.Then,τπτωπγ62.222-=-=cand the required plot is drawn as shown.P5.11 A time-delay system has the open-loop transfer function)1()(+=-s s es G sτ(a) Determine the time delay τ to maintain stability.Solution: Solving the crossover frequency yields215 01 111)(2242-=⇒=-+⇒=+=c c c ccc j G ωωωωωωr a d /s79.0=c ω The system is stable if and only if 0>γ, i.e.14.1 0arctan 2<⇒>---=ττωωππγc cHence, when14.1<τ the system is stable.P5.12 The polar plot of a conditionally stable system, for a specific gain 50=K , is shown in Fig. P5.12. (a) Determine whether the system is stable. Assume that the open-loop characteristic has the minimum phase. (b) Find the range of K so that the system is stable.Solution: (a) It is assumed that the open-loop transfer function has the minimum phase, i.e. 0=p . A half Nyquist plot is completed as shown. By inspection, in the case of 50=K ,2011p N N ==-=--+the system is stable.(b) The magnitude-phase curve will pass the point )0 ,1(j - when5001.0501==K , 252502==K ,105503==KThe system is stable if and only if 011=-=--+N N or 000=-=--+N N . Therefore, the system is stable when 10<K or50025<<K .P5.13 Consider a unity feedback system with the open-loop transfer function21)(ss s G +=τDetermine the value of τ that results in the system with a phase margin of 45.Solution: Letting 45=γ, i.e. 45arctan 180180=+-c τω resulting in 1=c τω. Letting 1)(=c j G ω getsFigure P5.1212 112222=⇒=+τωωτcc84.0=τP5.14 Consider a unity feedback system with the open-loop transfer function3)101.0()(+=s Ks G(a) Determine the value of K that results in the system with a phase margin of 45.(b) Determine the gain margin corresponding to the gain obtained in (a). Solution: (a) Solving the crossover frequency, we have4501.0arctan 4501.0arctan 3180=⇒=-=c c ωωγ r a d /s 100=c ω Letting 1)(=c j G ω yields221)01.0(32=⎪⎭⎫ ⎝⎛+=c K ω(b) Solving the phase crossover frequency, we have6001.0arctan 18001.0arctan 3)(=⇒-=-=∠g g g j G ωωω r a d /s3100=g ω Therefore, the corresponding gain margin is 22221)01.0()(132=⎪⎭⎫ ⎝⎛+==g g g j G K ωωP5.15 The open-loop transfer function of a unity feedback system is)100()(2++=s s s ks G(a) Determine the value of open-loop gain so that the system has a gain margin ofdb20.(b) Determine the phase margin corresponding to the gain obtained in (a). Solution: (a) The corner frequency of the oscillatory element is rad/s 10=n ω and its damping ratio is 05.0=ς. Considering that1<<ς,the corrected value of magnitude isdb2021lg20=ςThen, the log-magnitude plot can be drawn as shown. It can be seen that, in the case that the system has a gain margin of db 20, the asymptotic magnitude should be db 40- at rad/s 10=ω. By inspection, the crossover frequency is rad/s 1.0=c ω and the corresponding open-loop isdb),(ωL-----1.0==c K ω(b) Considering 1<<ς and n c ωω<<, we have 90)(-≈∠c j G ω. Hence, the phase margin is90)(180≈∠+=c j G ωγP5.16 The asymptotic logarithmic magnitude curve of a minimum-phase system is shown in Fig. P5.16. Estimate the phase margin and gain margin of the system.Solution: The corner frequency of the oscillatory element is rad/s 100=n ω and its damping ratio is 1.0=ς. Considering that 1<<ς, the corrected value of magnitude isdb1421lg20=ςhence, the gain margin is db 61420=-=g K Since 1<<ς, n c ωω<<=10and 180)(-=∠n j G ω, we have 90)(-≈∠c j G ω and90)(180≈∠+=c j G ωγP5.17 Consider a unity feedback system with the open-loop transfer function 1001000)(+=s s GFind the bandwidths of the open-loop system and the closed-loop system and compare the results.Solution: For the open-loop system, by inspection, we have 2510001000221001000)0(22)(222..=+⋅=+⇒=openb open b j G j G ωωr a d /s 100.=o p e n b ω As for the closed-loop system, we have 11001000)(+=s s Φ463.01100010002211001000)0(22)(222.=+⋅=+⇒=⋅closeb close b j j ωΦωΦr a d /s4.1099.=close b ω It can be seen that the bandwidth of the closed-loop system is wider than that of the open-loop system.-Figure P5.16P5.18 The open-loop transfer function of a unity feedback system is)2(16)(+=s s s G(a) Determine the crossover frequency c ω and the phase margin γ. (b) Determine the resonant frequency r ω and the relative resonant peak r M of the closed-loop system.Solution: (a) Letting 1)(=c j G ω yields12.14 0164 1216)(222422=⇒=-+⇒=+=c c c ccc j G ωωωωωωr a d /s 76.3=c ω01.28276.3arctan90180)(180=--=∠+=c j G ωγ(b) The closed-loop transfer function is given by16216)(2++=s ss GThis is a second-order system with rad/s 4=n ω and 25.0=ς. Hence, we have07.225.0125.02112122=-⨯=-=ςςr Mr a d /s3.740.52-1 42122=⨯=-=ςωωn r P5.19 The asymptotic logarithmic magnitude curve of a unity feedback system is shown in Fig. P5.19. Assume that the system has the minimum-phase. (a) Determine the open-loop transfer function. (b) Determine whether the system is stable. (c) Discuss the effect on the performance specifications p σ, s t , and ss e if the magnitude curve is translated right by a decade.Solution: (a) By inspection, the open-loop transfer function is in the form of)125.0)(110()15()(+++=s s s s K s GNoting that 0)(=c L ω and rad/s 1=c ω, we have, for the asymptotic characteristic,21100)10(10)5()(222=⇒=+⋅+⨯+⋅=K K j G c c c ωωωTherefore, the open-loop transfer function is)125.0)(110()15(2)(+++=s s s s s G(b) Estimating the phase margin yields4.7025.0arctan 10arctan 905arctan 180)(180=---+=∠+=c j G ωγFigure P5.19It can be seen thatω<. Hence, the system is stable.∠ωjG as cω>180)(-(c) If the magnitude curve is translated right by a decade, the phase curve is translated right by a decade too. Obviously, the phase margin will not be changed. Hence, the overshootσp will not be changed. Meanwhile, the setting timet will be reduced, because the crossoversfrequency will become larger and the phase margin will not be changed. As fore, thess conclusion is that the steady-state error to a ramp input will be reduced due to larger open-loop gain.。
《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（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Error 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Mass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。
The Principle of Automatic Control 自动控制原理
本课程涉及的是数字仿真（或称计算机仿真），仿真应包含 两个过程：建立模型及模型试验。计算机仿真包含三要素：系 统、模型与计算机；与之相联系的三个基本活动：模型建立、 仿真模型建立及仿真试验。它们之间的关系可由图表示：
系统
模型建立
仿真实验
模型
仿真模型建立
计算机
图1 计算机仿真的基本要素与基本活动
•系统仿真语言MATLAB
胡寿松主编. 自动控制原理（第3版）. 北京： 国防工业出版社，1994
薛定宇著. 控制系统计算机辅助设计---MATLAB语言. 北京：清华大学出版社. 1996
Significance
（课程意义）
Control engineering is an exciting and challenging field. It is a multidisciplinary subject, and a core course in the
系统仿真：广义上讲，为了系统的分析和设计， 首先建立系统的模型，然后在模型上进行实验这 一过程就称为系统仿真。根据模型的种类不同， 系统仿真可分为三种：
物理仿真（实物，如飞机风洞实验）真实直 观；投资大、周期长，试验受限制
数字仿真（基于系统数学模型的仿真）经济 、方便、灵活；真实性要依赖模型
物理---数学仿真（半实物仿真，如飞机、导 弹等运动体的转台实验）
The Principle of Automatic Control 自动控制原理
References
(参考教材）
Richard C.Dorf and Robert H.Bishop. Modern Control Systems. 9th edition, Science Press.
自动控制原理 英文版
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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.