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作者:Han J, Wang P, Yang X.
文题:Tuning of PID controller based on fruit fly optimization algorithm
期刊来源:Mechatronics and Automation (ICMA), 2012 International Conference on. IEEE, 2012: 409-413.
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作者:Larsson P O, Hagglund T.
文题:Control signal constraints and filter order selection for PI and PID controllers
期刊来源:American Control Conference (ACC), 2011. IEEE, 2011: 4994-4999.
看看我以前回答过的一个问题,或许有帮助。评价一个控制系统是否优越,有三个指标:快、稳、准。所谓快,就是要使压力能快速地达到“命令值”(不知道你的系统要求多少时间)所谓稳,就是要压力稳定不波动或波动量小(不知道你的系统允许多大波动)所谓准,就是要求“命令值”与“输出值”之间的误差e小(不知道你的系统允许多大误差)对于你的系统来说,要求“快”的话,可以增大Kp、Ki值要求“准”的话,可以增大Ki值要求“稳”的话,可以增大Kd值,可以减少压力波动仔细分析可以得知:这三个指标是相互矛盾的。如果太“快”,可能导致不“稳”;如果太“稳”,可能导致不“快”;只要系统稳定且存在积分Ki,该系统在静态是没有误差的(会存在动态误差);所谓动态误差,指当“命令值”不为恒值时,“输出值”跟不上“命令值”而存在的误差。不管是谁设计的、再好的系统都存在动态误差,动态误差体现的是系统的跟踪特性,比如说,有的音响功放对高频声音不敏感,就说明功放跟踪性能不好。调整PID参数有两种方法:1、仿真法;2、“试凑法”仿真法我想你是不会的,介绍一下“试凑法”“试凑法”设置PID参数的建议步骤:1、把Ki与Kd设为0,不要积分与微分;2、把Kp值从0开始慢慢增大,观察压力的反应速度是否在你的要求内;3、当压力的反应速度达到你的要求,停止增大Kp值;4、在该Kp值的基础上减少10%;5、把Ki值从0开始慢慢增大;6、当压力开始波动,停止增大Ki值;7、在该Ki值的基础上减少10%;8、把Kd值从0开始慢慢增大,观察压力的反应速度是否在你的要求内;
PLC+ 风光变频器的小区恒压供水控制应用实例 The Example of PLC+ Fengguang Inverter Plot Constant Pressure Water Supply Control Application 山东新风光电子科技发展有限公司 杨国奎 孔亮 赵新军 郭培彬 Yang Guokui Kong Liang Zhao Xinjun Guo Peibin 摘 要: 本文介绍了风光变频器在某生活小区双恒压供水系统中的应用情况。 关键词: 风光变频器 PLC 恒压供水 Abstract: This article introduces Fengguang inverter in the some small-unit residential area double constant pressure water supply control application example situation. Key words: Fengguang inverter PLC Constant pressure water supply 1 引言 本文是针对某生活小区实际情况,结合用户生活 / 消防双恒压供水控制的要求,我们进行改造的一些心得。现将其中的改造情况介绍如下。 众所周知,恒压供水系统对于生活小区是非常重要的,例如在生活小区供水过程中,若自来水供水因故压力不足或短时断水,可能影响居民生活。又如当发生火警时,若供水压力不足或无水供应,不能迅速灭火,可能引起重大损失和人员伤亡。所以,生活小区采用生活 / 消防双恒压供水系统,具有较大的经济和社会意义。 基于上述情况,我公司对某生活小区供水系统进行改造,采用西门子 PLC 作为主控单元。利用风光供水变频器,根据系统状态可快速调整供水系统的工作压力,达到恒压供水的目的。改造提高了系统的工作稳定性,得到了良好的控制效果。 2 用户现场情况 如图 1 所示,市网自来水用高低水位控制器 EQ 来控制注水阀 YV1 ,自动把水注满储水水池,只要水位低于高水位,则自动向水箱注水。水池的高低水位信号也直接送给 PLC ,作为水位报警。为了保持供水的连续性,水位上、下限传感器高低距离较少。生活用水和消防用水共用二台泵,平时电磁阀 YV2 处于失电状态,关闭消防管网,二台泵根据生活用水的多少,按一定的控制逻辑运行,维持生活用水低恒压。当有火灾发生时,电磁阀 YV2 得电,关闭生活用水管网,二台泵供消防用水使用,并维持消防用水的高恒压值。火灾结束后,二台泵改为生活供水使用。 图 1 生活 / 消防双恒压供水系统示意图 现场设备参数如下 : 型号 65-315(I)A 流量 56m 3 /h 扬程 110m 效率 56% 转速 2900r/min 电机功率 37KW 3 系统控制要求 用户对二泵生活 / 消防双恒压供水系统的基本要求是: ⑴ 生活供水时,系统低恒压运行,消防供水时高恒压值运行。 ⑵ 二台泵根据恒压的需要,采取先开先停的原则接入和退出。 ⑶ 在用水量小的情况下,如果一台泵连续运行时间超过 1 天,则要切换下一台泵,系统具有倒泵功能,避免一台泵工作时间过长。 ⑷ 二台泵在启动时都要有软启动功能。 ⑸ 要有完善的报警功能。 ⑹ 对泵的操作要有手动控制功能;手动只在应急或检修时使用。 4 设备选型 ( 1 )风光 JD-BP32-XF 型供水变频器 JD-BP32-XF 型是山东新风光电子公司推出的专用于供水变频器,使用空间电压矢量控制技术适用于各类自控场合。在恒压供水中可以采用这类变频器。 JD-BP32-XF 型变频器除具有变频器的一般特性外,还具有以下特性:水压高、水压低输出接口,变频器运行上限、下限频率(可以任意设定),可以方便地进行双压力控制,内置智能 PI 控制,以上功能非常适用于供水控制要求。在本例中选用 JD-BP32-37F ( 37KW )风光供水变频器拖动用户水泵。 ( 2 ) PLC 选型 ① 控制系统的 IO 点及地址分配 根据图 1 所示及控制要求 , 统计控制系统的输入、输出信号的名称 , 代码及地址编号如下表 1 所示。水位上 、 下限信号分别为 、 。 输入输出点 / 代码及地址编号表 1 名 称 代 码 地址编码 输入信号 手动和自动消防信号 SA1 水池下限信号 SLL 水池上限信号 SLH 变频器报警信号 SU 消铃按钮 SB7 试灯按钮 SB8 水压低信号 SY1 水压高信号 SY2 输出信号 1# 泵工频运行接触器及指示灯 KM1,HL1 1# 泵变频运行接触器及指示灯 KM2,HL2 2# 泵工频运行接触器及指示灯 KM3,HL3 2# 泵变频运行接触器及指示灯 KM4,HL4 变频器开停机控制 KA2 生活 / 消防供水转换电磁阀、压力转换 YV2 KA1 水池水位下限报警指示灯 HL5 变频器报警指示灯 HL6 火灾报警指示灯 HL7 报警电铃 HA ② PLC 系统选型 系统共有开关量输入点 8 个,开关量输出点 10 个 , 选用西门子主机 CPU222 ( 8 入继电器出) 1 台,加上扩展模块 EM222 ( 8 继电器输出) 1 台。即可满足用户供水控制要求 ( 3 )压力传感器 在供水系统中,压力传感器既可以采用压力变送器,也可以采用远传压力表。在本例中采用远传压力表,压力表相应接线端子接到变频器主控板 3 脚、 4 脚、 5 脚即可。 5 电气控制系统原理图 电气控制系统原理图包括主电路图、控制电路图及 PLC 外围接线图三部分。 ( 1 )主电路图 如图 2 所示为电控系统主电路。二台电机分别为 M1 、 M2 。接触器 KM1 、 KM3 ,分别控制 M1 、 M2 的工频运行;接触器 KM2 、 KM4 ,分别控制 M1 、 M2 的变频运行; FR1 、 FR2 分别为二台水泵电机过载保护用的热继电器; QS1 、 QS2 和 QS3 分别为变频器和二台泵电机主电路的隔离开关; FU1 、 FU2 为主电路的熔断器; BPQ 为风光供水专用变频器。 图 2 主电路图 ( 2 )控制电路图 如图 3 所 示为电控系统电路。图中 SA 为手动 / 自动转换开关, SA 打在 1 的位置为手动控制状态,打在 2 的状态为自动控制状态。手动运行时,可用按钮 SB1~SB6 控制二台泵的起 / 停和电磁阀 YV2 的通 / 断;自动运行时,系统在 PLC 程序控制下运行。 图中的 HL8 为自动运行状态电源指示灯。 图 3 电控系统控制电路图 ( 3 ) PLC 接线图 下图 4 所示为 PLC 及扩展模块外围接线图。火灾时,火灾信号 SA1 被触动, 为 1 。 图 4 双恒压供水控制系统及扩展模块的外围接线图 6 系统程序设计 ( 1 )程序中使用的 PLC 内部器件及功能,如下表 2 所示: 表 2 器件地址 功 能 器件地址 功 能 VB400 变频工作泵的泵号 复位当前变频泵运行脉冲 VB401 工频运行泵的台数 当前泵工频运行启动脉冲 VD410 倒泵时间存储器 新泵变频启动脉冲 T33 工 / 变频转换逻辑控制 泵工 / 变频转换逻辑控制 T34 工 / 变频转换逻辑控制 泵工 / 变频转换逻辑控制 T37 工频泵增泵判断时间控制 泵工 / 变频转换逻辑控制 T38 工频泵减泵判断时间控制 故障信号汇总 T39 工 / 变频转换逻辑控制 水位下限故障逻辑 故障结束脉冲信号 水位下限故障消铃逻辑 泵变频启动脉冲 变频器故障消铃逻辑 - 火灾消铃逻辑 倒泵变频启动脉冲 生活 / 消防双恒压的两个恒压值是我公司生产的风光供水专用变频器直接设定的。在本实例中,根据用户要求,生活压力设定为 , 消防压力设定为 。 压力低、压力高信号分别由变频器内部主控板 14 脚、 15 脚给出。 供水运行下限频率、供水运行上限频率由变频器程序设定。在本系统中,运行下限频率设为 20Hz, 运行上限频率设为 50Hz 。 ( 2 )系统 PLC 控制程序如下: 7 结束语 随着变频调速技术的飞速发展,变频调速恒压供水技术在小区已普遍使用。用变频器来实现恒压供水,与其它供水方式相比较而言,其优点是非常明显的。节能效果十分显著,启动平稳,启动电流小,避免了电机启动时对电网的冲击,延长了泵和阀门等的使用寿命,消除了启动和停机时的水锤效应。供水控制系统提高了小区的供水质量。各项控制指标达到了用户的要求。风光变频器在小区恒压供水改造效果是明显的,改造是成功的
恒温室房间温度PID控制研究摘要:某恒温实验室的恒温精度为27±℃,但是由于实验室的非凡性,恒温室的内外扰量多且某些随机扰量的大小难于确定,而导致了其恒温精度很难达到预期效果。为了解决这个问题,通过建立恒温室被控对象的数学模型求出其传递函数,然后采用参数寻优方法确定PID控制器的参数,最后采用MATLAB仿真的方法,研究恒温室内外扰量对房间温度的影响。通过研究,可以得出,当设备散热干扰量为17℃以及送风温度干扰量为℃,渗透风干扰量不大于℃时,PID控制才能保证恒温室的恒温精度。关键词:恒温室,PID控制渗透风干扰量参数寻优温度1前言随着科学技术的发展,各类精密产品的生产制造以及特种科学实验都要求具有特定的工作环境,恒温就成为了不可缺少的条件之一。目前我国常见的恒温室的恒温精度为±1℃及±℃,也有±℃。而一些高精度的恒温室如光学仪器厂的刻线室恒温精度已达到了±℃。但是在某些非凡的科学实验室不仅恒温精度很高,而且干扰量多如渗透风、设备散热、送风温度波动以及电热器供电电压的波动等,且某些干扰量如渗透风其最大值难于确定而没有采用相应的措施控制渗透风扰量,导致了房间温度的波动过大,结果使恒温室的恒温精度很难达到要求。如何使这些非凡的科学实验室恒温精度达到使用要求,也成为了恒温室的空调系统和控制系统设计的一个巨大的难题。由于传统的PID控制算法,其运算简单、调整方便、鲁棒性强,在过程控制中,这种控制算法仍占据相当重要的地位.故目前恒温室的空调系统大部分采用PID控制。但PID控制的效果如何,在很大程度上是取决于控制器参数的正确整定。为此,人们提出了各种不同的参数整定方法,如误差积分最小、固定衰减比、极点配置等方法.这些方法主要是用经典控制理论中的一些设计方法或者依靠现场试验方法来进行PID控制器参数的计算与整定.显然,这就要求操作人员具有较高的理论基础和现场调试经验.而且,被控对象模型参数难以确定以及系统性能稳定性较差,则需频繁地进行参数整定,这必将影响系统的正常运行。对于这些非凡的空调房间温度的控制,由于被控对象具有较大的惯性和迟延,且受各种因素变化的影响,因此对象的传递函数具有非线性和时变特性,采用传统的PID控制难于取得较好的控制效果。本文采用单纯形法寻优PID参数,然后采用MATLAB仿真确定渗透风干扰量的最大值,PID控制才能保证恒温室的恒温精度。2工程概况恒温室建筑面积625m2,层高8m,总送风量27500m3/h,送风温度15℃,房间设计温度27±℃,设备散热量135KW,恒温室建筑墙体、地板采用绝热材料,渗透风来自外部房间其设计温度26±1℃。3恒温室空调过程建模1恒温室空调系统被控对象的数学模型要对一个恒温室空调系统被控对象进行控制,须为其建立一个合适的数学模型。使用数学语言对实际对象进行一些必要的简化和假设:由于该恒温室建筑墙体、地板采用绝热材料,故室内外墙体和地板热量传递忽略不计。恒温室顶棚由盖板组成,存在缝隙,考虑有一定的渗透风,其他地方如门窗的渗透风忽略不计。假如不考虑执行机构的惯性和室温调节对象的传递滞后,根据能量守恒定律,单位时间内进入对象的能量减去单位时间内由对象流出的能量等于对象内能量蓄存量的变化率,表达式和图1如下所示:图1室温自动调节系统数学表达式为:式中:Chrr——恒温室的热容;C——空气的比热;GS——送风量;θ0’——电加热器前的送风温度;θ1——室内空气温度,回风温度;QE——电加热器的热量;Qm——设备散热量;QI——渗透风带入的热量;由式QI=GIcit式中:GI——渗透风量;θIt——渗透风空气温度;cIt——渗透风空气的比热。把式代入式,整理得式中:T1——调节对象的时间常数,T1=Chrr/;K1——调节对象的放大系数,K1=GSc/;θE——电加热器的调节量,换算成送风温度的变化,θE=QE/GSC;θf——干扰量换算成送风温度的变化,;θf‘——送风温度干扰量,θf‘=θ0“θIf——渗透风的干扰量,θIf=QI/GSC;θMf——设备散热量的干扰量,θMf=QM/GSC。由式拉普拉斯变换,得假如考虑被控对象传递滞后,则恒温室空调过程的传递函数为:2感温元件和执行调节机构的传递函数感温元件采用热电阻,根据热平衡原理,其热量平衡方程式:式中:C2——热电阻的热容;θ2——热电阻温度;q2——单位时间内空气传给热电阻的热量;α2——室内空气与热电阻表面之间的换热系数;F2——热电阻的表面积;θ1——室内空气温度,回风温度。由式拉普拉斯变换,可得感温元件的传递函数:同样执行调节机构的传递函数:3恒温室特性参数及其他参数的确定恒温室特性即房间的特性,用传递滞后τ、时间常数T1和放大系数K1这三个参数来表示。时间常数T1和放大系数K1由式,η=4,GI=GS×3%,通过式,式计算可以得到,T1=18分,K1=。传递滞后τ由经验公式τ/T1=通过计算则得τ=35分由参考文献的附表6-可以得到感温元件的时间常数和不灵敏区为T3=50秒,2ε=℃。电加热器的比例系数K2=△θ/△N=,T2=50秒。4单纯形法寻优方法控制系统参数最优化是指对被控对象已知、控制器的结构和形式已确定,需要调整或寻找控制系统的某些参数使整个控制系统在某一性能指标下最佳。单纯形法的思想很简单,若要求一个函数的最大点,则可先计算若干点处的函数值,进行比较,并根据它们的大小关系确定函数的变化趋势作为搜索的参考方向,然后按参考方向搜索直到找到最小值为止。在三维空间内取不同一平面的四个点构成单纯形,如图3所示。图2三维空间的单纯形这四个点X0、X1、XX3对应的函数值为F0、FFF3,比较可看出最大者,则对应点X3作为差点,由此可以推测好点在差点XH的对称点XR处的可能性最大,然后计算XR处的函数值FR,若有FR≥max,说明从XH前进的步长太大,XR并不一定比XH好,因此可以压缩步长在XH与XR之间找一点XS为新点,然后X0,F1,F2中最大者说明情况有所改善,但前进和步长可能还不够,还可以加大步长得XH与XR延长线上的一点XE,若XE对应的函数FE小于FR则以XE作为新点,并以X0、X1、X2构成新的单纯形。最后比较构成新的单纯形的各点处的函数值,若其中最大者和最小者之间的相对差小于预先给定的数E,则说明最小值已经找到,否则继续重复上述步骤直到找到止。5恒温室控制系统仿真整个室温自动调节系统包括调节对象,调节器、感温元件以及PID控制器。根据参数计算结果,最后得到恒温室恒温控制系统如图3所示。图3恒温室恒温控制系统仿真框图?恒温室实验设备散热量相当稳定,由式计算可得,设备散热量干扰量θMf=17℃是稳定的扰量。而送风温度干扰量主要包括电加热器供电电压的波动和换热器冷冻水温度的波动以及管道温升等引起的送风温度的变化,其值为℃。渗透风干扰量是随机扰量,其随着恒温室外面的房间温度的变化和渗透风风量的变化而变化,它是影响恒温室的房间温度最重要的因数。当渗透风干扰量分别℃、℃、℃、℃时,PID控制的仿真曲线如图4-图7所示。图4θIf为℃时PID控制的仿真曲线图5θIf为℃时PID控制的仿真曲线图6θIf为℃时PID控制的仿真曲线图7θIf为℃时PID控制的仿真曲线分析图4-图可以得出:当渗透风扰量θIf不大于℃时,恒温室房间温度波动小于℃,满足恒温室的恒温精度要求。但是当渗透风扰量θIf为℃时,恒温室房间温度波动大于℃,超出答应的波动范围。6结论通过以上的仿真和分析,可以得出:恒温实验室的恒温精度为27±℃,但是由于实验室的非凡性,恒温室的内外扰量多,只有当设备散热干扰量为17℃以及送风温度干扰量为℃,渗透风干扰量不大于℃时,PID控制才能保证恒温实验室的恒温精度,达到使用的要求。=============Room temperature PID control of room temperatureAbstract: A temperature accuracy of laboratory temperature 27 ± ℃, but because of the extraordinary nature of the laboratory, constant temperature room and the volume inside and outside the interference of some random disturbance difficult to determine the size of the volume, which led to the accuracy of its temperature is very difficult to achieve the desired effect. To solve this problem, through the establishment of constant temperature room was charged with the mathematical model of the object to derive its transfer function, and then used to determine optimal parameters of PID controller parameters, and finally the use of MATLAB simulation method to study indoor and outdoor temperature the amount of room disturbance temperature. Through research, can be drawn, when the equipment cooling capacity of 17 ℃ interference and disruption of supply air temperature ℃, the volume of infiltration air interference at ℃ not more than, PID temperature control room in order to ensure the accuracy of the thermostat. Key words: constant temperature room, PID control parameters of the infiltration volume of wind interference temperature optimization 1 Introduction With the development of science and technology, various types of manufacturing precision products and the special requirements of scientific experiments are with a specific working environment has become a heated one of the conditions indispensable. At present, the temperature of our common room temperature accuracy of ± 1 ℃ and ± ℃, also ± ℃. And some, such as high-precision optical instrument factory room temperature of the engraved line has reached room temperature accuracy of ± ℃. However, in some extraordinary precision scientific laboratory is not only a high temperature, and interference, such as infiltration of the wind volume, equipment cooling, supply air temperature fluctuations, as well as electric heaters, such as supply voltage fluctuations, and some amount of interference, such as wind penetration of its difficult to determine the maximum value without the use of appropriate measures to control the amount of infiltration of the wind disturbance, leading to fluctuations in room temperature is too large, resulting in the constant temperature room thermostat accuracy requirement is very difficult to achieve. How to make these remarkable scientific accuracy of the use of constant temperature laboratory requirements, but also become a constant temperature room air-conditioning system and control system design of a huge problem. As a result of the traditional PID control algorithm, the computation is simple, convenient adjustment, robustness, and in process control, this control algorithm is still occupied a very important position. Therefore, the current room temperature most of the air-conditioning system using PID control. However, the effects of PID control to a large extent depends on the correct controller parameter tuning. To this end, the people made a variety of parameter tuning methods, such as minimum error integral, fixed attenuation ratio, pole placement and other methods. These methods are mainly used in classical control theory a number of design methods or testing methods rely on the scene to carry out PID control parameters of the calculation and setting. Obviously, this requires the operator has a higher theoretical basis and field testing experience. Moreover, the plant model parameters it is difficult to determine system performance, as well as less stable, it would take frequent tuning parameters, This will affect the normal operation of the system. For these extraordinary control of an air-conditioned room temperature, due to a larger plant with inertia and delay, and by the impact of changes in a variety of factors, so the object of the transfer function with nonlinear and time-varying characteristics, using the traditional PID control difficult to obtain a better control effect. Simplex method using PID parameter optimization, and simulation using MATLAB to determine the amount of infiltration of the maximum wind disturbance, PID control room temperature in order to ensure the accuracy of the thermostat. 2 Project Overview Constant temperature room floor area of 625m2, storey 8m, total air volume of 27500m3 / h, air temperature 15 ℃, room design temperature 27 ± ℃, heat dissipation equipment 135KW, constant temperature room building wall, floor insulation materials used, the wind penetration to Since the design of external room temperature 26 ± 1 ℃. 3 air-conditioned room temperature process modeling A constant temperature room air-conditioning systems charged with the mathematical model of the object To a constant temperature room air-conditioning system to control the object to be suitable for the establishment of a mathematical model. The use of mathematical language of the actual number of objects necessary to simplify and assumptions: Room temperature due to the building wall, floor insulation materials used, the indoor and outdoor wall and floor heat transfer is negligible. Room temperature by the flat roof of the existence of the gap, taking a certain degree of infiltration of wind, such as doors and windows in other parts of the wind penetration is negligible. If agencies do not consider the implementation of the inertia and temperature regulation of the transmission lags behind the target, according to the law of conservation of energy, per unit time into the object of energy per unit time minus the outflow of energy from the target object with the same energy with the rate of change of the stock, the expression and Figure 1 as follows: Figure 1 at room temperature automatic adjustment system Mathematical expression is: Where: Chrr - room temperature heat capacity; C - specific heat of air; GS - air traffic; θ0' - electric heater before the air temperature; θ1 - indoor air temperature, return air temperature; QE - the heat electric heater; Qm - equipment heat dissipation; QI - the infiltration of heat into the wind; QI = GIcit by type Type in: GI - the infiltration air flow; θIt - air temperature wind penetration; cIt - the air infiltration heat wind. Skill into the style, finishing a Where: T1 - the object of regulation time constant, T1 = Chrr /; K1 - adjust object magnification factor, K1 = GSc /; θE - regulation of electric heater volume, converted into changes in air temperature, θE = QE / GSC; θf - converted to interfere with the volume of supply air temperature changes ; θf' - interfere with the volume of supply air temperature, θf '= θ0 " θIf - interfere with the volume of wind penetration, θIf = QI / GSC; θMf - equipment heat dissipation capacity of the interference, θMf = QM / GSC. Laplace transform by the style, too If the accused objects to consider transmission lag, then the process of constant temperature air-conditioned room for the transfer function: 2 temperature components and the implementation of the transfer function of regulating agencies Temperature components using thermal resistance, according to the principle of heat balance, the heat balance equation: Where: C2 - thermal resistance of the heat capacity; θ2 - thermal resistance temperature; q2 - units of air time to the thermal resistance of heat; α2 - indoor air and thermal resistance between the surface heat transfer coefficient; F2 - thermal resistance of the surface area; θ1 - indoor air temperature, return air temperature. Laplace transform by type, available thermal transfer function components: Regulating agencies to implement the same transfer function: 3 Room temperature characteristics and other parameters to determine Room temperature properties of the characteristics of that room, with delivery lags τ, time constant T1 and the magnification factor of these three parameters K1 said. Time constant T1 and the magnification factor K1 By type, η = 4, GI = GS × 3%, through the ceremony, can be calculated, T1 = 18 minutes, K1 = . Transmission lag τ By the empirical formula by calculating the τ/T1 = is a τ = 35 minutes References schedule from 6 - temperature components can be time constant and dead-zone for the T3 = 50 seconds, 2ε = ℃. The ratio of electric heater coefficient K2 = △ θ / △ N = , T2 = 50 seconds. 4 simplex optimization method Control system parameters optimization refers to the object known, controls the structure and form have been identified, need to adjust the control system or to find some of the parameters so that the whole control system of performance indicators in a best. Simplex method of thinking is very simple, if the maximum point of a function may be a number of points to calculate the function values, for comparison, and in accordance with their function to determine the size of the relationship between changes in the trend of the reference as a search direction, and then by reference to the direction of the search until you find the minimum value so far. Three-dimensional space in a plane different from the four points constitute a simplex, as shown in Figure 3. Figure 2 Three-dimensional simplex space These four points X0, X1, XX3 corresponding value is a function of F0, FFF3, comparison can be seen the largest, then the corresponding point as almost X3, it can be a good point that in the nearly symmetric point XH Department is most likely to XR , and then calculating the value function XR Department FR, if FR ≥ max, from the XH big step forward, XR is not necessarily better than the XH, the compression step can be in between the XH and XR point to find the new XS point X0, F1, F2 largest note in the situation has improved, but step forward and might not be enough, you can increase the step-by-step look XH and XR point to extend the online XE, if XE corresponding FE function of small while in the FR as a new point of XE, and X0, X1, X2 constitute a new simplex. Finally, constitute a new comparison of the simplex function value point, if one of the largest and the smallest relative difference between the pre-less than a given number of E, then the minimum has been found, otherwise continue to repeat the above steps until you find the only. 5 Room temperature control system simulation Automatic adjustment of the temperature regulation system includes objects, regulators, and the PID temperature controller components. The calculation results based on parameters, and finally to be room temperature thermostat control system shown in figure 3. Figure 3 Room temperature thermostat control system simulation block diagram? Room temperature a considerable amount of laboratory equipment, heat stability, available from the computing equipment interfere with heat dissipation capacity θMf = 17 ℃ traffic disturbance is stable. Interfere with the volume of supply air temperature and the main electric heater supply voltage fluctuations and the chilled water heat exchanger, as well as fluctuations in temperature, such as pipeline temperature rise caused by changes in air temperature, its value is ℃. Interfere with the volume of air infiltration is the amount of random disturbance, the room with the outside temperature changes in room temperature and the amount of infiltration风风changes, it is the impact of room temperature room temperature the most important factor. When the infiltration of wind interference respectively ℃, ℃, ℃, ℃, when, PID control of the simulation curve in Figure 4 - as shown in Figure 7. Figure ℃ when 4θIf for the simulation curve of PID control Figure ℃ when 5θIf for the simulation curve of PID control Figure ℃ when 6θIf for the simulation curve of PID control Figure ℃ when 7θIf for the simulation curve of PID control Analysis of Figure 4 - plans can be drawn: When the volume of infiltration θIf wind disturbance is not more than ℃, the temperature fluctuations in room temperature of less than rooms ℃, room temperature thermostat to meet accuracy requirements. However, when the volume of infiltration air disturbance θIf for ℃, the temperature fluctuations in room temperature greater than room ℃, beyond the agreed scope of the fluctuations. 6 Conclusion Through the above simulation and analysis, can be drawn: Temperature accuracy of laboratory temperature 27 ± ℃, but because of the extraordinary nature of the laboratory, both inside and outside the room temperature volume disturbance, interference only when the cooling equipment of air temperature of 17 ℃ and the interference of ℃, wind penetration interfere with the volume of not more than ℃, when, PID temperature control to ensure the accuracy of laboratory temperature, to achieve the requirements of use.【这是我在网上找的,希望可以帮助你】
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