资源描述
4-1,已知系统的零、极点分布如图，大致绘制出系统的根轨迹。,第四章习题课,(4-1),解,:,j,0,(1,),(2,),j,0,(3,),j,0,(4,),j,0,60,0,90,0,60,0,第四章习题课,(4-1),(5,),j,0,(6,),j,0,j,0,(7,),(8,),j,0,60,0,45,0,135,0,36,0,108,0,4-2,已知开环传递函数，试用解析法绘制出系统的根轨迹，并判断点,(-2+j0),(0+j1),(-3+j2),是否在根轨迹上。,第四章习题课,(4-2),解,:,K,r,(s+1),G(s,)=,K,r,(s)=,s+1+K,r,K,r,=0,s=-1-K,r,j,0,系统的根轨迹,s=-1,-1,K,r,=,s=-,s=-2+j0,-2,s=0+j1,0+j1,-3+j2,s=-3+j2,j,0,4-3,已知系统的开环传递函数，试绘制出根轨迹图。,第四章习题课,(4-3),解,:,s(s+1)(s+5),(1),G(s,)=,K,r,(s+1.5)(s+5.5),1,）开环零、极点,p,1,p,1,=0,p,2,p,2,=-1,p,3,p,3,=-5,2,）实轴上根轨迹段,p,1,p,2,z,1,z,1,=-1.5,z,2,z,2,=-5.5,z,1,p,3,z,2,-,3,）根轨迹的渐近线,n-m,=,1,=,+,180,o,4,）分离点和会合点,A,(,s,),B,(,s,)=,A,(,s,),B,(,s,),A(s,)=s,3,+6s,2,+5s,B(s,)=s,2,+7s+8.25,A(s,),=3s,2,+12s+5,B(s,),=2s+7,解得,s,1,=-0.63,s,2,=-2.5,s,3,=-3.6,s,4,=-7.28,第四章习题课,(4-3),s(s+1)(s+4),(2),G(s,)=,K,r,(s+1.5),1,）开环零、极点,p,1,p,1,=0,p,2,p,2,=-1,p,3,p,3,=-4,j,0,2,）实轴上根轨迹段,p,1,p,2,z,1,z,1,=-1.5,p,3,z,1,3,）根轨迹的渐近线,n-m,=,2,=,+,90,o,2,=,-1-4+1.5,=-1.75,-1.75,4,）分离点和会合点,A(s,)=s,3,+5s,2,+4s,B(s,)=s+1.5,A(s,),=3s,2,+10s+4,B(s,),=1,解得,s=-0.62,5,）系统根轨迹,第四章习题课,(4-3),s(s+1),2,(3),G(s,)=,K,r,1,）开环零、极点,p,1,=0,p,2,=-1,p,3,p,3,=-1,2,）实轴上根轨迹段,p,1,p,2,p,3,-,3,）根轨迹的渐近线,n-m,=3,j,0,p,1,p,2,3,=,-1-1,=-0.67,-0.67,4,）根轨迹与虚轴的交点,=,+,180,o,+,60,o,闭环特征方程为,s,3,+2s,2,+s+K,r,=0,K,r,=0,K,r,=2,2,3,=1,1,=0,1,-1,5,）分离点和会合点,A(s,)=s,3,+2s,2,+s,B(s,)=1,A(s,),=3s,2,+4s+1,B(s,),=0,解得,s=-0.33,6,）系统根轨迹,第四章习题课,(4-3),1,）开环零、极点,p,1,=0,p,2,=-3,p,3,=-7,j,0,2,）实轴上根轨迹段,p,1,p,2,p,4,p,4,=-15,z,1,z,1,=-8,p,3,z,1,p,4,-,3,）根轨迹的渐近线,n-m,=3,s(s+3)(s+7)(s+15),(4),G(s,)=,K,r,(s+8),p,1,p,2,p,3,3,=,-3-7-15+8,=-5.67,=,+,180,o,+,60,o,-5.67,4,）根轨迹与虚轴的交点,闭环特征方程为,s,4,+25s,3,+171s,2,+323s+8K,r,=0,K,r,=0,1,=0,K,r,=638,2,3,=6.2,6.2,-6.2,5,）分离点和会合点,A(s,)=s,4,+25s,3,+171s,2,+315s,B(s,)=s+8,A(s,),=4s,3,+75s,2,+342s+315,B(s,),=2s+7,解得,s=-1.4,6,）系统根轨迹,第四章习题课,(4-4),4-5,已知系统的开环传递函数。,(1),试绘制出根轨迹图。,(2),增益,K,r,为何值时，复数特征根的实部为,-2,。,第四章习题课,(4-5),解,:,，,s(s+1),K,r,(s+2),G,(,s,)=,j,0,p,1,=0,p,1,p,2,=-1,p,2,z,1,=-2,z,1,p,1,p,2,z,1,-,分离点和会合点,s,2,+4s+2=0,s,1,=-3.41,s,2,=-0.59,s,2,+s+K,r,s+2K,r,=0,闭环特征方程式,系统根轨迹,s=-2+j,(-2+j,),2,+(-2+j,)(1+K,r,)+2K,r,=0,-4,+(1+K,r,),=0,4-,2,-2(1+K,r,)+2K,r,=0,K,r,=3,=,1.41,4-6,已知系统的开环传递函数，试确定闭环极点,=0.5,时的,K,r,值。,第四章习题课,(4-6),解,:,p,1,=0,p,2,=-1,p,3,=-3,p,1,p,2,p,3,-,8,+,60,o,=,+,180,o,3,=,-1-3,=,-1.3,根轨迹的分离点：,A,(,s,),B,(,s,)=,A,(,s,),B,(,s,),3s,2,+8s+3=0,s,1,=-0.45,s,2,=-2.2,s,2,没有位于根轨迹上，舍去。,j,s(s+1)(s+3),K,r,G,(,s,),H,(,s,)=,与虚轴交点,s,3,+4s,2,+3s+K,r,=0,2,=0,K,r,-4,3,+,3,=0,-,K,r,=0,K,r,=12,2,3,=,1.7,1,=0,1.7,-1.7,p,3,-3,0,p,1,p,2,-1,系统根轨迹,s,1,=0.5,得,s,1,=-0.37+j0.8,K,r,=,|s,3,|s,3,+1|s,3,+3|,s,3,=-4+0.37,2=-3.26,=3.26,2.26,0.26=1.9,s,3,4-7,已知系统的开环传递函数，,(1),试绘制出根轨迹图。,第四章习题课,(4-7),解,:,p,1,=0,p,2,=-2,p,3,=-4,p,1,p,2,p,3,-,8,+,60,o,=,+,180,o,3,=,-2-4,=,-2,根轨迹的分离点：,A,(,s,),B,(,s,)=,A,(,s,),B,(,s,),3s,2,+12s+8=0,s,1,=-0.85,s,2,=-3.15,s,2,没有位于根轨迹上，舍去。,j,(3),与虚轴交点,s,3,+6s,2,+8s+K,r,=0,2,=0,K,r,-6,3,+,8,=0,-,K,r,=0,K,r,=48,2,3,=,2.8,1,=0,2.8,-2.8,p,3,-4,0,p,1,p,2,-2,系统根轨迹,s,1,s,3,s(s+2)(s+4),K,r,G,(,s,),H,(,s,)=,(2),阻尼振荡响应的,K,r,值范围,s=-0.85,K,r,=0.85,1.15,3.15=3.1,s=,j2.8,K,r,=48,(4),=0.5,s,1,=-0.7+j1.2,s,3,=-6+0.7,2=-4.6,K,r,=4.6,2.6,0.6=7.2,第四章习题课,(4-8),第四章习题课,(4-9),第四章习题课,(4-10),第四章习题课,(4-11),
展开阅读全文