稳态误差分析例题

,单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,2011,年,4,月,11,日星期一,Sichuan University of Science and Engineering,#,2011,年,4,月,11,日星期一,Sichuan University of Science and Engineering,线性系统时域分析,稳态误差例题,2011,年,4,月,11,日星期一,Sichuan University of Science and Engineering,例题,s(,T,1,s+1)(,T,2,s+1),1,s+,2,s,2,s(,T,1,s+1)(,T,2,s+1),k,2,k,1,R(s),E(s),C(s),求图示系统中的,1,、,2,，,使系统由,一阶无差系统变为三阶无差系统。,解：,er,(s)=,s(,T,1,s+1)(,T,2,s+1),k,1,k,2,1+,1,s(,T,1,s+1)(,T,2,s+1)+k,1,k,2,因为,一阶无差,所以,系统稳定,，则当分子只有,s,3,项时，由终值定理可得,:,e,ss,=lim,s,er,(s)R(s),s0,=lim,s,s0,k,1,k,2,T,1,T,2,s,3,s,3,A,=,0,1,k,2,=1,2,k,2,=,T,1,+,T,2,即,:,1,=1/,k,2,2,=(,T,1,+,T,2,)/k,2,2011,年,4,月,11,日星期一,Sichuan University of Science and Engineering,例题,2,已知单位反馈系统开环传递函数为,G(s),，输入为,r(t),，试求稳态误差,e,ss,。,r,1,(t)=1(t),r,2,(t)=t,r,3,(t)=t,2,解,：,0,型,型,型,k=,10,k=,21/8,k=,8,e,ss,=1/11,e,ss,=8/21,e,ss,=1/8,系统,2,不稳定,，,系统,3,的,A=2,，,e,ss,e,ss,=1/4,2011,年,4,月,11,日星期一,Sichuan University of Science and Engineering,例题,3,求图示系统的,e,ssn,。,5,s,r=0,(0.1s+1)(0.5s+1),2,c(t),(1),n(t)=1(t),5,s,r=0,(0.1s+1)(0.5s+1),2,c(t),(2),n(t)=1(t),解,：,(1),C(s)=,s(0.1s+1)(0.5s+1)+10,5(0.1s+1)(0.5s+1),s,1,系统稳定,(2),C(s)=,s(0.1s+1)(0.5s+1)+10,2s,s,1,e,ssn,=,-,limsC(s)=,-,1/2,s0,e,ssn,=-limsC(s)=,0,s0,几点说明,增益,型别,差异,2011,年,4,月,11,日星期一,Sichuan University of Science and Engineering,例题,4,已知图示系统的调节时间,t,s,=0.3,秒,，,试求,r(t)=3t,时,输出端定义,的误差终值,e,ss,。,0.01s,1,k,h,R(s),C(s),t,s,=3T=0.03/k,h,=0.3,k,h,=0.1,e,ss,=3,2011,年,4,月,11,日星期一,Sichuan University of Science and Engineering,例题,5,设无零点的单位反馈二阶系统,h(t),曲线如图所示,1,、试求出该系统的开环传递函数及参数；,2,、确定串联校正装置的传递函数，使系统,对阶跃输入的稳态误差为零。,0,1,1.25,0.95,2011,年,4,月,11,日星期一,Sichuan University of Science and Engineering,0,1,1.25,0.95,s,k,),s,(,G,.,2,c,c,12,.,0,k,0,c,=,解,设,由,得,开环传递函数,1,、由于,