各章习题及答案第一章绪论1 .举例说明什么是测控？答：(1) 测控例子：为了确定一端固定的悬臂梁的固有频率，我们可以采用锤击法对梁进行激振，再利用压电传感器、电荷放大器、波形记录器记录信号波形，由衰减的振荡波形便可以计算出悬臂梁的固有频率。

（2）结论：由本例可知：测控是指确定被测对象悬臂梁的属性—固有频率的全部操作，是通过一定的技术手段—激振、拾振、记录、数据处理等，获取悬臂梁固有频率的信息的过程。

2. 测控技术的任务是什么？答：测控技术的任务主要有:通过模型试验或现场实测，提高产品质量；通过测控，进行设备强度校验，提高产量和质量；监测环境振动和噪声，找振源，以便采取减振、防噪措施；通过测控，发现新的定律、公式等；通过测控和数据采集，实现对设备的状态监测、质量控制和故障诊断。

3. 以方框图的形式说明测控系统的组成，简述主要部分的作用。

测控系统方框图如下：（2）各部分的作用如下：●传感器是将被测信息转换成某种电信号的器件；●信号的调理是把来自传感器的信号转换成适合传输和处理的形式；●信号处理环节可对来自信号调理环节的信号，进行各种运算、滤波和分析；●信号显示、记录环节将来自信号处理环节的信号显示或存贮。

●模数（A/D）转换和数模（D/A）转换是进行模拟信号与数字信号相互转换，以便用计算机处理。

4.测控技术的发展动向是什么?传感器向新型、微型、智能型方向发展；测控仪器向高精度、多功能、小型化、在线监测、性能标准化和低价格发展；参数测量与数据处理向计算机为核心发展；5. A precise optional signal source can control the output power level to within 1%. A laser is controlled by an input current to yield the power output. A microprocessor controls the input current tothe laser. The microprocessor compares the desired power level with a measured signal proportional to the laser power output obtained from a sensor. Complete the block diagram representing thisclosed-loop control system shown in Fig E1.1, identifying the output, input, and measured variables and the control device.答：6. many luxury automobiles have thermostatically controlled air-conditioning system（恒温空调系统）for the comfort of the passengers. Sketch a block diagram of an air-condition temperature on a dashboard panel（仪表盘）. Identify the function of each element of the thermostatically controlled cooling system.答：7. In the past, control systems used a human operator as part of a closed-loop control system. Sketch the block diagram of the valve control system shown in Fig. P1.2.答：8. The student-teacher learning process is inherently a feedback process intended to reduce the system error back model of the learning process and identify each block of the system.答：9. Automatic control of water level using a float level was used in the Middle East for a water clock. The water clock was used from sometime before Christ until the seventeenth century. Discuss the operation of the water clock, and establish how the float provides a feedback control that maintains the accuracy of the clock. Sketch a block diagram of the feedback system.答：第二章信号与系统分析基础1求周期方波的傅立叶级数（复指数函数形式），画出|c n|-ω和ϕ-ω图。

解：(1)方波的时域描述为：(2) 从而：2 . 求正弦信号的绝对均值和均方根值。

解(1)(2)3.求符号函数和单位阶跃函数的频谱。

解：(1)因为不满足绝对可积条件，因此，可以把符合函数看作为双边指数衰减函数：其傅里叶变换为：(2)阶跃函数：4. 求解下列微分方程(1)初始条件，式中均为常数，解：set(2)假设解：(3)初始条件解：(4)初始条件解：5. 一阶微分方程组为已知，求解解：6. 已知一液压控制系统的运动方程为式中参数：。

试求在单位阶跃输入作用下，系统的输出。

已知解：7. An electric circuit is shown in Fig. P2.1. Obtain a set of simultaneous integrodifferential equations representing the network.8. A dynamic vibration absorber is shown in Fig. P2.2. This system is representative of many situations involving the vibration of machines containing unbalanced components. The parameters M2 and k12may be chosen so that the main mass M1 does not vibrate in the steady-state when . Obtain the differential equations describing the system.For mass 1the differential equations:isFor mass 2the differential equations：isresults:9. A couple spring-mass system is shown in Fig. P2.3. The masses and springs are assumed to be equal. Obtain the differential equations describing the system.for the left massThe differential equations:for the right mass:the differential equations:results:第三章第四章测控系统的时域分析1.说明线性系统的频率保持性在测量中的作用。

答：（1）线性系统的频率保持性，在测试工作中具有非常重要的作用。

因为在实际测试中，测试得到的信号常常会受到其他信号或噪声的干扰，这时依据频率保持特性可以认定测得信号中只有与输入信号相同的频率成分才是真正由输入引起的输出。

（2）同样，在故障诊断中，根据测试信号的主要频率成分，在排除干扰的基础上，依据频率保持特性推出输入信号也应包含该频率成分，通过寻找产生该频率成分的原因，就可以诊断出故障的原因。

2.在使用灵敏度为80nC/MPa的压电式力传感器进行压力测量时，首先将他与增益为5mV/nC的电荷放大器相连，电荷放大器接到灵敏度为25mm/V的笔试记录仪上，试求该压力测试系统的灵敏度。

当记录仪的输出变化30mm时，压力变化为多少？2 解：（1）求解串联系统的灵敏度。

（2）求压力值。

4.把灵敏度为的压电式力传感器与一台灵敏度调到的电荷放大器相接，求其总灵敏度。

若要将总灵敏度调到，电荷放大器的灵敏度应作如何调整？解：4.用一时间常数为2s的温度计测量炉温时，当炉温在200℃—400℃之间，以150s为周期，按正弦规律变化时，温度计输出的变化范围是多少？解：（1）已知条件。

（2）温度计为一阶系统，其幅频特性为（3）输入为200℃、400℃时，其输出为：y=A(w)×200=200.7(℃) y=A(w)×400=401.4(℃)5.用一阶系统对100Hz的正旋信号进行测量时，如果要求振幅误差在10%以内,时间常数应为多少？如果用该系统对50Ｈz的正旋信号进行测试时，则此时的幅值误差和相位误差是多少？解：（1）一阶系统幅频误差公式。

幅值误差为：2.9%，相位差为:-67.540第五章第六章第七章第八章测控系统的频域设计与校正1. the frequency response for a process of the formis shown in Fig. E8.4. Determine K and a by examining the frequency response curves.The transfer function:When ,whenopen-loop gain:from the magnitude relationship:2. Several studies have proposed an extravehicular robot that could move about a NASA space station and perform physical tasks at various worksites. The arm is controlled by a unity feedback control withDraw the Bode diagram for K=100, and determine the frequency when 20log|G| is 0dB.Low frequency: -90°High frequency: -270°Determine the frequency when20log|G| is 0dB:3. A feedback system has a loop transfer function(a) Determine the corner frequencies (break frequencies) for the Bode plot.(b) Determine the slope of the asymptotic plot at very low frequencies and at high frequencies.(c) Sketch the Bode magnitude plot.(a) the corner frequencies:(b) the slope of the asymptotic plot at very low frequencies is 0d B/decthe slope of the asymptotic plot at high frequencies is -20d B/de(c) the Bode magnitude plotthe phase: 180°~ -90°4. A control system for controlling the pressure in a closed chamber is shown in FigP8.4.The transfer function for the measuring element isand the transfer function for the valve isThe controller transfer function isObtain the frequency response characteristics for the loop transfer functionthe transfer function:the corner frequencies:K=1the phase: -90°~-360°5. The robot industry in the United States is growing at a rate of 30% a year. A typical industrial robot has six axes or degrees of freedom. A position control system for a force-sensing joint has a transfer functionwhere H(s)=1 and K=10. Sketch the Bode diagram of this system.the phase: 0°~-360°6. The asymptotic log-magnitude curves for two transfer function are given in FigP8.6. Sketch the corresponding asymptotic phase shift curves for each system. Determine the transfer function for each system. Assume that the systems have minimum phase transfer functions.from the low frequency asymptotic, we havewhere W=1 , K=3.98The transfer function:orthe phase: -90°~- 180°the transfer function:the phase: 90°~- 90°第九章测控传感器1.应变片的灵敏系数与电阻丝（敏感栅）的灵敏系数有何不同？为什么？答：(1)一般情况下，应变片的灵敏系数小于电阻丝的灵敏系数。