中国石油大学(华东)《电力拖动自控系统》2015年春学期在线作业答案.

15春西交《电力拖动自动控制系统》在线作业满分答案Austrian Peng 15 spring West to "electric drive automatic control system" online homeworkTotal score: 100 test time: -Radio examinationA multiple-choice questionJudgment questionRadio examination questions (30 questions, 60 points)1. V-M system in four quadrant operation, forward braking ().A. the polarity of the armature terminal is negativeB. armature current polarity is positiveC. mechanical characteristics are in the second quadrantFull marks: 22. the following statement is correct.A. if the given voltage is constant, the partial pressure ratio that regulates the speed feedback voltage can not change the speedB. if the given voltage is constant, the partial pressure ratiothat regulates the speed feedback voltage does not necessarily change the speedC. if the given voltage is constant, the partial pressure ratio that regulates the speed feedback voltage can change the speedFull marks: 23. speed range means ().A. production machinery requires that the ratio of the maximum speed to the minimum speed provided by an electric motor is called the speed rangeB. production machinery requires that the ratio of the minimum speed to the maximum speed provided by an electric motor is called the speed rangeC. production machinery requires that the ratio of the average speed to the minimum speed provided by an electric motor is called the speed rangeD. production machinery requires that the ratio of the maximum speed to the average speed provided by an electric motor is called the speed rangeFull marks: 24. what are the basic characteristics of speed single closed loop speed regulation system?.A. uses only the proportional amplifier feedback control system, its modulation quantity still has no static difference.The function of B. feedback control system is to resist disturbance and obey the given. Disturbance performance is one of the most prominent characteristics of feedback control systemThe accuracy of the C. system depends only on the given accuracyThe accuracy of D. system depends only on the accuracy of feedback detectionFull marks: 25. when the constant voltage constant frequency sine wave power supply, the mechanical characteristic of the asynchronous motor is correct.A. when the S is very small, the torque is approximately proportional to the s, and the mechanical characteristics are a hyperbola that is symmetrical at the originB. when the S is very small, the torque is approximately inversely proportional to s, and the mechanical characteristic is a straight lineC. when the S is close to 1, the torque is approximately inversely proportional to s. At this point, Te = f (s) is a hyperbola that is symmetrical at the originD. when the S is close to 1, the torque is approximately proportional to s. At this point, Te = f (s) is a hyperbola that is symmetrical at the originFull marks: 2In the four quadrant operation of the 6. V-M system, the forward run time ().A. the polarity of the armature terminal voltage is positiveB. the polarity of the armature current is negativeC. mechanical characteristics are in the second quadrantFull marks: 27. a range of speed closed-loop speed control system is 1500~150r/min, static slip <2% system, static downhill open loop system is 100r/min, the closed loop system's open-loop amplification (what).A. 30B. 31.7C. 33.2D. 28Full marks: 28. a range of speed closed-loop speed control system is 1500~150r/min, requirement of static rate of <2% system, then the system is allowed (static downhill).A. 3.06r/minB. 2.8r/minC. 3.46r/minD. 4.00r/minFull marks: 29., the static frequency conversion device can be divided into (two) types.A.Indirect frequency conversion and direct frequency conversionB. only direct frequency conversionC. only indirect frequency conversionD. or more is incorrectFull marks: 2The advantage of 10. synchronous motors is thatA. speed and voltage, frequency strictly synchronous, power factor is as high as 1B. difficult to startThere are oscillations in C. overloading and even risk of missing stepFull marks: 211. power factor of synchronous motor ()A. can be adjusted by excitation current, which can be delayed or advancedB. can not be delayed or advancedC. can only lag behind, can not advanceFull marks: 212. when the speed and current double closed loop speed regulation system runs in a steady state, the input bias voltage and output voltage of the two regulators are wrong.A. when the two regulators are not saturated, their input bias voltage is zeroB. when the two regulators are not saturated, their input bias voltage is not zeroThe output limiting voltage of the C. speed regulator ASR determines the maximum value of the current given voltageThe output limiting voltage of the D. current regulator ACR limits the maximum output voltage of the power electronic converterFull marks: 2In the four quadrant operation of the 13. V-M system, the reverse run time ().A. the polarity of the armature terminal voltage is positiveB. armature current polarity is positiveC. mechanical characteristics are in the third quadrantFull marks: 214. direct start, also called cascade, allows the inverter to start in a fast manner. The rotor is connected with the AC network when the starting controller turns on the AC network, and then the rotor is ()A. high impedance stateB. closed stateC. open stateFull marks: 215. for the cascade speed regulation system, the starting should be large enough rotor current Ir or large enough rectifier DC current Id, therefore, the rotor rectifier voltage Ud and the inverter voltage Ui should ()A. larger differenceB. smaller differenceC. the same differenceFull marks: 216. the definition of static rate is ().A. when the system runs at a certain speed, the load falls from ideal no-load to rated value, and the speed drops at the ratio of ideal no-load speedB. when the system runs at a certain speed, the load falls from ideal no-load to the rated value, and the speed drops as compared with the rated valueFull marks: 217. which of the following is not a step-down starter?.A. star delta (Y- delta) startB. stator series resistance or reactance startingC. autotransformer step-up startFull marks: 218. when the parameters of the asynchronous motor are constant, the electromagnetic torque Te of the motor is equal to the square of the stator voltage U at a certain speed.A. is proportional toB. is inversely proportional toC. independentFull marks: 219. the maximum speed characteristic of a speed regulation system is 1500r/min, the minimum speed characteristic is 150r/min, the speed falls 15r/min with the rated load, and the constant speed drop at different speeds remains unchangedA. system can achieve speed range of 11, the system allows the static rate of 10%B. system can achieve speed range of 10, the system allows the static rate of 10%C. system can achieve speed range of 11,The allowable static slip is 8%D. system can achieve speed range of 10, the system allows the static rate of 12%Full marks: 2Calculation of steady-state parameters of 20. double closed loop speed regulation system and steady state calculation of non static differential system.A. is exactly the sameB. similarityC. has no contactFull marks: 2The feedback of the 21. position servo system is the position loop, and the feedback of the speed governing system is ().A. speed ringB. position ringC. time loopFull marks: 222. series speed regulation system has no brake stopping function. Can only by () gradually slow down, and rely on the role of load resistance torque free parking.A. increases beta angleB. reduces beta angleC. increases copper consumptionFull marks: 223. induction motor magnetic field () power generation.A. stator onlyB. rotors aloneC. consists of stator and rotorFull marks: 224. power supply motivation, because of the needs of the excitation, it must be drawn from the power lag reactive current, no-load power factor is very low. The synchronous motor can change the input power factor by adjusting the DC exciting current of the rotor, and it can lag behind or advance. When cos phi =1.0 ().A. armature has the greatest copper lossB. armature copper loss minimumC. armature has the greatest iron consumptionFull marks: 225. when the regulator saturation, the output is constant, the change of input output is no longer affected, unless there is a reverse input signal makes the regulator out of saturation; in other words, the regulator temporarily cut off saturation between the input and output connection, the equivalent of the adjusting ring ()A. ring openingB. closed loopC. short circuitFull marks: 226. the following options are not the characteristics of the starting process of the double closed loop DC speed regulation system.A. saturation nonlinear controlB. speed overshootC. quasi time optimal controlD. saturation linear controlFull marks: 227., when the load current reaches Idm, the speed regulator is saturated, the current regulator plays a major role in regulating, the system behaves as a current without static difference, and gets () automatic protection.A. overcurrentB. overvoltageC. zero driftFull marks: 228. in the DC speed control system, time constant small armature current, current loop must have a sufficiently high sampling frequency, and the current regulation algorithm is relatively simple, with high sampling frequency is possible. Therefore, the current regulator can generally be designed using () methodsA. directB. indirectC. combines direct and indirectFull marks: 2In the four quadrant operation of the 29. V-M system, the reverse braking is applied.A. the polarity of the armature terminal voltage is positiveB. armature current polarity is positiveC. mechanical characteristics are in the second quadrantFull marks: 230. as the synchronous motor rotor has independent excitation, it can operate at very low power frequency, so under the same conditions, the speed range of synchronous motor is greater than that of asynchronous motor.A. is narrowerB. widerSame as C.Full marks: 2Online operation of "electric drive automatic control system"Total score: 100 test time: -Radio examinationA multiple-choice questionJudgment question, (a total of 10 multiple-choice questions, a total of 20 points.)1. PWM - motor system has greater advantages in () aspect.The main circuit of the A. is simple and requires less power devices.B. switching frequency is high, the current is easy to continuous, harmonic less, motor loss and heating are small.C. low speed performance, stable speed, high precision, wide speed range, up to about 1:10000.If the D. is in line with the fast response motor, the system has wide frequency band, fast dynamic response and strong dynamic immunity.Full marks: 22., from the point of view of differential power, the asynchronous motor speed control system can be divided into several categories. ()A. slip power consumption type speed regulating systemB. slip power feed speed regulation systemC. variable power differential speed regulation systemFull marks: 23. the following statement is correct.The A. static rate is used to measure the speed stability of the speed governing system under varying loadsThe harder the mechanical properties of B. are, the smaller the static rate is, and the higher the stability of the rotational speedC. speed range and static rate of two indicators, collectively referred to as the steady-state performance of speed regulation systemFull marks: 2The starting process of 4. double closed loop DC speed regulation system is characterized byA. saturation nonlinear controlB. speed overshootC. quasi time optimal controlFull marks: 25., in the design of closed-loop speed regulation system, the method of dynamic correction is ().A. series correctionB. parallel correctionC. feedback correctionFull marks: 26., the following advantages of the PWM system are ().The main circuit of A. is complex and requires more power devicesB. switching frequency is high, the current is easy to continuous, harmonic less, motor loss and heating are smallThe C. DC power supply adopts uncontrolled rectifier, and the power factor of the grid is higher than that of the phase controlled rectifierFull marks: 2The type of the 7. PID regulator includes ()A. proportional differential (PD)B. proportional integral (PI)C. proportional integral derivative (PID)Full marks: 28. what are the dynamic performance indicators of the automatic control system?A. follow performance indicatorsB. anti-jamming performance indexC. current indexD. voltage indicatorFull marks: 2The essence of 9. closed-loop system can reduce thesteady-state (downhill).Mechanical characteristics of A. closed loop speed regulation systemMechanical characteristics of B. open loop systemC. blocking currentFull marks: 2The internal logic requirement of 10. DLC is ()A. converts the input signal and converts analog to switchingThe B. makes the correct logical decision based on the input signalC. requires two delay times to ensure reliable switching between the two thyristor devicesFull marks: 2Online operation of "electric drive automatic control system" Total score: 100 test time: -Radio examinationA multiple-choice questionJudgment questionJudgment questions (a total of 10 questions, a total of 20 points)OneThe magnetic field directional flux open-loop slip vector control system is determined by the flux and torque of the given signal by the vector control equation that no actual calculated rotor flux and phase, so it belongs to the indirect vector control. ()A. errorB. correctFull marks: 22. for speed single closed loop speed regulation system, the function of feedback control system is to resist disturbance and obey the given. Disturbance performance is one of the most prominent characteristics of feedback control system. ()A. errorB. correctFull marks: 23. voltage source inverter is a constant voltage source, the voltage control response is slow and difficult to fluctuate, so it is suitable for more than one power supply of synchronous motor operation. ()A. errorB. correctFull marks: 24. the AC frequency converter composed of thyristor is mainly used for rolling mill, main drive, ball mill, cement rotary kiln and other large capacity and low speed systems. ()A. errorB. correctFull marks: 2The 5. circulation can be divided into two major categories, static circulation and dynamic circulation.A. errorB. correctFull marks: 2The DC voltage waveform of the 6. voltage source inverter is relatively flat. Under ideal conditions, a constant voltage source with a resistance of 0 is used, and the rectangular wave or trapezoidal wave is generated when the AC voltage is output. ()A. errorB. correctFull marks: 27. for the same speed control system, Delta nN value, if the more stringent of static rate requirements, which requires the s value is smaller, speed range that allows is smaller. ()A. errorB. correctFull marks: 2The 8. PWM converter has two classes: irreversible and reversible.A. errorB. correctFull marks: 29. compared with the current model, the voltage model of the rotor flux is less affected by the change of motor parameters, and the algorithm is simple and easy to use. ()A. errorB. correctFull marks: 210. dynamic landing is generally greater than steady landing. ()A. errorB. correct Full marks: 2。

电机与拖动基础《电机与拖动基础》2015年春学期在线作业（二）一,单选题1. 由于磁饱和，计变压器铁心损耗，若铁心磁通为正弦波形，励磁电流的波形为（）。

A. 正弦B. 对称的尖顶C. 不对称的尖顶波?正确答案：C2. 题目和答案如下图所示：A.B.C.D.?正确答案：B3. 变压器的副边负荷电流产生的磁势和原边电流产生的磁势之间的关系（）。

A. 方向相反，起到去磁作用B. 方向相同，起到增磁作用?正确答案：A4. 三相对称绕组通入三相对称电流，产生（）磁势。

A. 脉振B. 椭圆形旋转C. 圆形旋转?正确答案：C5. 三相异步电动机其他条件不变，若电源电压降低，则电机转速、电机电流的变化（）。

A. 转速降低，电流减小B. 转速降低，电流增大C. 转速增加，电流增加D. 转速增加，电流减小?正确答案：B6. 题目和答案如下图所示：A.B.C.D.?正确答案：C7. 通常情况下，异步电机直接起动时，起动电流约为额定电流的（）。

A. 1-2倍B. 4-7倍C. 20倍?正确答案：B8. 电源电压不变，仅将降压变压器绕组反接，作升压变压器使用，则变压器的（）。

A. 励磁电流不变B. #励磁电流增加，甚至有可能烧坏变压器C. 励磁电流减小?正确答案：B9. 题目和答案如下图所示：5A.B.C.D.?正确答案：A10. 题目和答案如下图所示：A.B.C.D.?正确答案：A二,多选题1. 三相异步电动机的功率PM包括（）。

A. 定子铜耗B. 转子铜耗D. 总机械功率?正确答案：ABD2. 三相异步电动机的输入功率P1包括（）。

A. 定子铜耗B. 铁耗C. 电磁功率?正确答案：ABC3. 题目和答案如下图所示：A.B.C.D.E.F.?正确答案：CDEF4. 关于变压器空载实验，下列说法正确的是（）。

A. 目的是测变压器原边和副边的的漏阻抗参数B. 目的是测变压器激磁参数C. 可在高压侧测量，也可在低压侧测量，通常在低压侧测量D. 可在高压侧测量，也可在低压侧测量，通常在高压侧测量E. 折算后，在高压侧测量和在低压侧测量的参数值相同F. 折算后，在高压侧测量和在低压侧测量的参数值不同?正确答案：ACE5. 变压器副边的额定电压是指（）。

第一章1、请画出运动控制系统及其组成的框图。

2、请列出几种典型的生产机械负载转矩特性。

答：1.恒转矩负载负载转矩的大小恒定(TL=const)，称作恒转矩负载。

又分为（1）位能性恒转矩负载。

（2) 反抗性恒转矩负载。

2.恒功率负载负载转矩与转速成反比，而功率为常数，称作恒功率负载。

3.风机、泵类负载负载转矩与转速的平方成正比，称作风机、泵类负载。

3、请问如果构建一辆电动汽车，请问您认为必备哪些功能系统。

答：电动汽车系统应具备三大功能系统：电力驱动系统、主能源系统和辅助控制系统电力驱动系统：由电控单元、功率转换器、电动机、机械传动装置和驱动车轮组成主能源系统：有主电源、能量管理系统和充电系统构成辅助控制系统：具有动力转向、温度控制和辅助动力供给等功能=====================================================================第二章1、什么是调速范围和静差率？调速范围、静态速降和最小静差率之间有什么关系？为什么说“脱离了调速范围，要满足给定的静差率也就容易多了”？答：生产机械要求电动机提供的最高转速和最低转速之比叫做调速范围，用字母D 表示,即：min max n n D =当系统在某一转速下运行时，负载由理想空载增加到额定值时所对应的转速降落Δn N 与理想空载转速n 0之比，称作静差率s ，即0n n s N∆=。

调速范围、静态速降和最小静差率之间的关系是：)1(min mins n s n D N N -∆=按上述关系可得出：D 越小，s 越小，D 越大，s 越大；D 与s 相互制约，所以说“脱离了调速范围，要满足给定的静差率也就容易得多了”。

2、某一直流调速系统，测得的最高转速特性为=1500 r/min, 最低转速特性为=150 r/min 。

电动机额定转速为，带额定负载时的速度速降= 15r/min ，且在转速下额定速降如不变，试问系统能够达到的调速范围D 是多少？系统允许的静差率s 是多少？ 答：（1） n max =1500 r/min ，n min =150 r/min ，系统能够达到的调速范围是 ∆n N = 15r/min ，且不同转速下额定速降 ∆n N 不变， （2） 最大静差率=15/150=0.1， （3） 最小静差率=15/1500=0.01， 系统允许的静差率是0.01～0.1。

石油华东《电力拖动自控系统》2015年春学期在线作业一、单选题（共5 道试题，共30 分。

）1. 静止可控整流器的缩写是（）。

A. G-MB. V-MC. PWMD. SPWM。

正确答案：B2. 缩写术语PWM指的是（）。

A. 脉冲频率调制B. 脉冲幅值调制C. 脉冲宽度调制D. 混合调制。

正确答案：C3. 转速闭环控制中PD代表的是（）。

A. 微积分B. 比例积分C. 比例微分D. 转速闭环。

正确答案：C4. 下面属于静止可控整流器优点有（）。

A. 单向导电性B. 对过电压和过电流敏感C. 系统的动态响应比G-M系统快D. 低速时功率因数高。

正确答案：C5. 下面不属于直流斩波器控制方式的是（）。

A. PFMB. 两点式控制C. PWMD. PAM。

正确答案：《电力拖动自控系统》2015年春学期在线作业（一）二、多选题（共5 道试题，共40 分。

）1. 对于调速系统的转速控制要求有（）。

A. 调速B. 调速范围C. 加减速D. 稳速正确答案：2. 转速电流双闭环控制中，由于转速调节器经历了不饱和、饱和、退饱和三个阶段，分别对应下面中的几个阶段？（）。

A. 电流下降B. 转速调节阶段C. 电流上升阶段D. 恒流升速阶段正确答案：3. 直流电机常用的可控直流电源有（）。

A. 铅蓄电池B. 旋转变流机组C. 静止可控整流器D. 直线斩波器正确答案：4. 下列属于直流电动机调速方法有（）。

A. 定子串电阻B. 调节电枢两端电压C. 减弱励磁磁通D. 改变定子电压正确答案：5. 转速闭环系统看属于闭环可抵抗的扰动有（）。

A. 运算放大器放大倍数变化B. 反馈系数变化C. 供电电压扰动D. 电机负载变化正确答案：《电力拖动自控系统》2015年春学期在线作业（一）三、判断题（共5 道试题，共30 分。

）1. 在实际应用中，所谓的无静差调速系统是真正的转速无静差。

A. 错误B. 正确正确答案：2. 转速单闭环控制，采用PI控制可以实现无静差。

中国石油大学（华东）智慧树知到“电气工程及其自动化”《自动控制原理》网课测试题答案（图片大小可自由调整）第1卷一.综合考核(共15题)1.频率特性伯德图上，中频段特性反映了系统的()。

A.动态特性B.结构类型C.稳定裕度D.稳态特性2.传递函数只能描述系统的输入输出特性。

()A.错误B.正确3.二阶线性系统的阻尼比ξ决定了其振荡特性。

()A.错误B.正确4.一阶线性时不变系统模型参数包括()。

A.阶B.时间常数C.时间D.稳态增益5.稳态误差大小()。

A.与输入信号有关B.与系统结构类型有关C.与系统参数有关D.与输入信号、系统结构类型和参数均有关6.传递函数是一种()。

A.线性模型B.平衡点处变化量的动态模型C.稳态模型D.非线性模型7.一阶惯性环节的斜坡响应一定是无差跟踪输入的。

()A.错误B.正确8.根轨迹[S]右平面上有开环极点，控制系统一定是不稳定的。

()A.错误B.正确9.自动控制系统由()。

A.三个基本单元构成B.四个基本单元构成C.二个基本单元构成D.控制器和被控制对象10.人工控制与自动控制的原理是()。

A.不同的B.有一定相似之处C.存在本质区别D.相同的11.用开环频率特性研究闭环控制系统稳定性是有效的。

()A.错误B.正确12.被控对象的数学模型是()。

A.控制系统特性分析的基础B.自动控制系统运行的基础C.自动检测的基础D.自动操作调整的基础13.描述被控对象特性的数学模型包括()。

A.模型的参数B.模型的阶C.对象物理特性D.对象的结构特性14.传递函数方框图()。

A.是数学模型B.不是数学模型C.是控制系统的图示法D.是一种变换15.自动控制是指()。

A.人不参与操作调整的控制过程B.人部分参与操作调整的控制过程C.人参与的操作调整的控制过程D.有仪表参与的自动检测过程第2卷一.综合考核(共15题)1.一阶系统的时间常数代表了系统()。

A.稳态特性B.动态特性C.输入特性D.输出特性2.传递函数的频率特性()。

第二章双闭环直流调速系统2-1 在转速、电流双闭环调速系统中，若要改变电动机的转速，应调节什么参数？改变转速调节器的放大倍数行不行？改变电力电子变换器的放大倍数行不行？改变转速反馈系数行不行？若要改变电动机的堵转电流，应调节系统中的什么参数？答：改变电机的转速需要调节转速给定信号Un※；改变转速调节器的放大倍数不行，改变电力电子变换器的放大倍数不行。

若要改变电机的堵转电流需要改变ASR的限幅值。

2-2 在转速、电流双闭环调速系统中，转速调节器有哪些作用？其输出限幅值应按什么要求来整定？电流调节器有哪些作用？其输出限幅值应如何整定？答：转速调节器的作用是：（1）使转速n很快的跟随给定电压Un※变化，稳态时可减小转速误差，如果采用PI调节器，则可以实现无静差。

（2）对负载变化起抗扰作用。

转速调节器的限幅值应按电枢回路允许的最大电流来进行整定。

电流调节器作用：（1）使电流紧紧跟随给定电压Ui※变化。

（2）对电网电压的波动起及时抗扰的作用。

（3）在转速动态过程中，保证获得电动机允许的最大电流，从而加快动态过程。

（4）当电机过载甚至堵转时，限制电枢电流的最大值，起快速的自动保护作用。

电流调节器的最大值应该按变换电路允许的最大控制电压来整定。

2-3 转速、电流双闭环调速系统稳态运行时，两个PI调节器的输入偏差（给定与反馈之差）是多少？它们的输出电压是多少？为什么？答：若都是PI调节器，则两个调节器的输入偏差为0，即Ui※=Ui，Un※=Un；输出电压为：Ui※=βId=Ui，Uc=U d0/K s=RI d+C e n=(RUi※/β)+(CeUn※/α)。

2-4 如果转速、电流双闭环调速系统的转速调节器不是PI调节器，而是比例调节器，对系统的静、动态性能会有什么影响？答：若采用比例调节器可利用提高放大系数的办法使稳态误差减小即提高稳态精度，但还是有静差的系统，但放大倍数太大很有可能使系统不稳定。

2-5 在转速、电流双闭环系统中，采用PI调节器，当系统带额定负载运行时，转速反馈线突然断线，系统重新进入稳态后，电流调节器的输入偏差电压△Ui 是否为0，为什么？答：反馈线未断之前，Id=In，令n=n1，当转速反馈断线，ASR迅速进入饱和，Un※=Un※max，Uc↑，Id↑至Idm，Te＞T l，n↑，Id↓，△Ui出现，Id↑至Idm，n↑，Id↓，此过程重复进行直到ACR饱和，n↑，Id↓，当Id=In，系统重新进入稳态，此时的速度n2＞n1，电流给定为Un※max= Idmaxβ＞电流反馈信号Un= Inβ，偏差△Ui不为0。

《电力拖动自控系统》课程综合复习资料一、填空题1．位置随动系统一般由五个部分组成，这五个部分分别是：位置传感器、电压比较放大器、、和减速器与负载。

2．对调速系统转速控制的要求主要集中在三个方面，具体为、和。

其中和稳态性能指标相对应的是：和。

3．调压调速是直流调速系统的主要方法，调节电枢供电电压需要有专门的可控直流电源，常用的可控直流电源有三种，具体为：、和。

4．直流调速系统的主要形式是晶闸管－电动机调速系统，简称为系统。

5．常用的可逆PWM变换器的主电路的结构的结构型式有、型和等。

6．在随动系统动态校正中，常用的有、和。

7．在闭环控制系统中，常优先考虑串联校正方案。

由调节器可构成的超前校正，提高稳定裕度和改善快速性能；由调节器构成的滞后校正，可以保证系统的稳态精度；用调节器构成滞后－超前校正，兼具两者的优点。

8．在晶闸管—电动机系统中，抑制电流脉动可采取的措施是：、和。

二、选择题1．在直流电动机调速方法中，调速性能最好的是（）A、减弱励磁磁通B、调节电枢供电电压C、改变电枢回路电阻D、改变负载2．与开环系统相比，闭环系统增加了反馈电路，反馈电路对干扰（）不起作用。

A、电压扰动B、负载扰动C、反馈电路元件的本身影响D、负载冲击3．位置信号检测装置是构成位置闭环控制所必不可少的环节。

下列位置检测装置中，不属于电磁感应的是（）A、自整角机B、光电码盘C、旋转变压器D、感应同步器4．对闭环系统而言，控制精度依赖于给定和反馈检测精度，而反馈电路对干扰并不全起作用，主要对干扰（）起作用。

A、电压扰动B、给定电源扰动B、反馈电路元件的本身影响D、测速电机误差5．调速系统和随动系统在控制中十分常用，两者的主要性能不同。

对随动系统来说，主要是要求输出量准确地跟随给定量的变化，因此它的（）成了主要指标。

A、快速性B、灵活性C、跟随性能D、准确性三、简答题1．某双闭环调速系统，ASR、ACR均采用近似PI调节器，试问：①调试中怎么才能做到U im*=6V时，I dm=20A；如欲使U nm*=10V时，n=1000r/min，应调什么参数？②如发现下垂特性不够陡或工作段特性不够硬，应调什么参数？2．调速系统常用的稳态性能指标分为那几类？并写出相应的性能指标。

一、简答1、和V——M系统相比PWM系统有哪些优越性答:PWM系统的优点主电路线路简单，需用的功率器件少；（2）开关频率高，电流容易连续，谐波少，电机损耗及发热都较小；（3）低速性能好，稳速精度高，调速范围宽，可达1:10000左右；（4）若与快速响应的电机配合，则系统频带宽，动态响应快，动态抗扰能力强；（5）功率开关器件工作在开关状态，导通损耗小，当开关频率适当时，开关损耗也不大，因而装置效率较高；（6）直流电源采用不控整流时，电网功率因数比相控整流器高。

2、双闭环直流调速系统转速和电流调节器作用答：双闭环直流调速系统中转速调节器的作用（1）转速调节器是调速系统的主导调节器，它使转速n 很快地跟随给定电压变化，稳态时可减小转速误差，如果采用PI调节器，则可实现无静差。

（2）对负载变化起抗扰作用。

（3）其输出限幅值决定电机允许的最大电流。

电流调节器的作用（1）作为内环的调节器，在外环转速的调节过程中，它的作用是使电流紧紧跟随其给定电压（即外环调节器的输出量）变化。

（2）对电网电压的波动起及时抗扰的作用。

（3）在转速动态过程中，保证获得电机允许的最大电流，从而加快动态过程。

（4）当电机过载甚至堵转时，限制电枢电流的最大值，起快速的自动保护作用。

一旦故障消失，系统立即自动恢复正常。

这个作用对系统的可靠运行来说是十分重要的。

3、说明位置随动系统的组成答：1）交流伺服电动机2）主电路主电路为单相双向晶闸管交流调压电路。

交流伺服系统如今多采用微机控制及专用集成驱动模块。

调节双向晶闸管VTF(或VTR)的导通程度，今设导通角增大，则交流调压线路输出的电压便增大。

由角位移可知，伺服电动机的转速n愈大，或运转的时间t愈长，则角位移量θ将愈大。

3）触发电路在正、反两组触发电路中要增设互锁环节，以保证在正、反两组触发电路中，只能有一组发出触发脉冲(一组发出触发脉冲时，另一组将被封锁)。

4）控制电路4、从评价交流调速系统效率出发，异步电动机调速系统分为几类。

练习题一、交流异步电动机转速表达式是什么？160(1)(1)fn s n s p=-=- 其中1n 为同步转速，n 为电机转速，s 为转差率二、变频调速的协调控制原则是什么？1、恒磁通变频调速：Φ＝const —— 基频以下，即1U C f≈ 2、恒功率变频调速： P ＝const —— 基频以上 （弱磁调速）C三、根据下图，分别指出磁能和磁共能由哪一区域表示？并给出磁能和磁共能的关系。

磁能m W 由区域O-a-b 所表示，磁共能mW '由区域O-a-c 所表示。

mm W W i ψ'+=四、给出Clark 变换112210A B C V V V V V αβ⎡⎤--⎤⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦⎢⎥⎣⎦，121210A B C V V V V V αβ⎡⎤⎡⎤⎡⎤⎥⎢⎥=-⎢⎥⎥⎢⎥⎣⎦⎢⎥-⎣⎦⎣五、给出Park 变换M 11T 11cos sin sin cos V V V V αβθθθθ⎡⎤⎡⎤⎡⎤=⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦⎣⎦，11M 11T cos sin sin cos V V V V αβθθθθ-⎡⎤⎡⎤⎡⎤=⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦六、给出ABC 坐标系统异步电动机的动态数学模型SSSR S S RS RR R R L L i L L i ψψ⎡⎤⎡⎤⎡⎤=⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦ L U Ri LPi i ωθ∂=++∂ e 0120SR T RSL T pi i L θθ∂⎛⎫ ⎪∂=⎪∂ ⎪ ⎪∂⎝⎭2m me m 2L d d T T J R K dt dtθθθθΩ=+++ 其中：111111111111111111111112211221122l SS l l L L L L L L L L L L L L L ⎡⎤+--⎢⎥⎢⎥⎢⎥=-+-⎢⎥⎢⎥⎢⎥--+⎢⎥⎣⎦()()()()()()121212121212121212cos cos 120cos 120cos 120cos cos 120cos 120cos 120cos SR L L L L L L L L L L θθθθθθθθθ⎡⎤+-⎢⎥⎢⎥=-+⎢⎥⎢⎥+-⎣⎦222222222222222222222112211221122l RRl l L L L L L L L L L L L L L ⎡⎤+--⎢⎥⎢⎥⎢⎥=-+-⎢⎥⎢⎥⎢⎥--+⎢⎥⎣⎦T RS SR L L =，SSSR RSRR L L L L L ⎛⎫=⎪⎝⎭，m p θθ=七、以转子磁链定向，给出MT 坐标系的动态数学模型'M1M11M2'T1T11T2''''M21M1M22'''T21T1T20S M S M M R M R L i L i L i L i L i L i L i L i ψψψψψ⎧=+⎪=+⎪⎨=+=⎪⎪=+=⎩ 11111M1M111111T1T1'''12M2'''12T200000SS MM S SMM M RS MS Rr PL L PL L i u L r PL L PL i u PL r PL i L L r i ωωωωωω+--⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥+⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥+⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦ '12T1'M e RL T p i L ψ=， 22m m e L m d d T T J R K dt dt θθθθΩ=+++ 八、说明磁场定向控制思想以产生同样的旋转磁势为准则，在三相坐标系上的定子交流电流i A 、i B 、i C ，通过三相/两相变换和同步旋转变换，可以等效成同步旋转坐标系上的直流电流i m 和i t 。