201141595312稳态误差分析例题
线性系统时域分析,稳态误差例题,例题,s(T1s+1)(T2s+1),1s+ 2s2,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(T1s+1)(T2s+1),k1k2,1+,1,s(T1s+1)(T2s+1)+k1k2,因为一阶无差所以系统稳定,则当分子只有s3项时,由终值定理可得:, 1k2=1,2k2=T1+T2,即:,1 = 1/k2,2=(T1+T2)/k2,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ess。,r1(t)=1(t),r2(t)=t,r3(t)=t2,解:,0型,型,型,k=10,k=21/8,k=8,ess=1/11,ess= 8/21,ess=1/8,×,×,系统2不稳定,,系统3的A=2,, ess, ess=1/4,例题3,求图示系统的essn。,解:,(1),C(s)=,s(0.1s+1)(0.5s+1)+10,5 (0.1s+1)(0.5s+1),系统稳定,(2),C(s)=,s(0.1s+1)(0.5s+1)+10,2s,例题4,已知图示系统的调节时间ts= 0.3秒,,试求r(t)=3t时输出端定义的误差终值ess。,0.01s,1,kh,R(s),C(s),ts=3T=0.03/kh=0.3,kh=0.1,ess= 3,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;,0,1,1.25,0.95,解,设,由,得,开环传递函数,1、由于,