过程控制-期末复习题

s200220240260280300320340360380400h/cm10.80.70.70.60.60.40.20.20.150.15已知矩形脉冲幅值AU=30%阀门开度变化,脉冲宽度At=20s,液位变送器量程为0〜200cm,且液位变送器和调节阀均为DDZ-II1型仪表。①试将该矩形脉冲响放曲线转换为阶跃响放曲线;②用一阶惯性环节加纯迟延近似该对象,试用两点法确定K、T和t值(t,=120.t2=180);③应用特性参数法选择PI调节器参数S、T|。4.解:将矩形脉冲响应曲线转换为阶跃响应曲线:%(0=沖)+>^-~),可得数据如下:t/s01020406080100120140160180h/cm000.20.823.65.47.49.31112.6t/s200220240260280300320340360380400h/cm13.614.415.115.816.41717.417.617.817.9518.118.1_^o"=0.6,v(^i)-y(Q)y(400)-y(0)=0.409/(G)=)_)’(0)y(400)-3;(0)=0.696=90.254⑴72.532(5)ln[l-A/(rJ]-ln[l-A/(r2)]ln[l-A/(f1)]-r1ln[l-A/(r2)]JZ系统数学模型为:G(s)=^^e^Ts+1根据Z-N公式:KKc=0.9(-)_,()=1.12=><=1.87/Tp*-^=3.33(-)LO=>r;=3.33r=241.532(s)T?T1因为等效调节器由调节器、变送器和调节阀组成,因此:其中变送器转换系数:20-4调节阀的转换系数.•w200-0100-020-4Kc=KcKmKv(%/mA)这样调节器的比例増益实际值为:/C.本1.8720-4xW0-03.74200-0?20-4相应的比例带实际值为.•f26.74%