1）开环零、极点p1=0p2=-1p3=-5z1=-1.5z2=-5.52）实轴上根轨迹段p1~p2z1~p3z2~-∞3）根轨迹的渐近线n-m= 1θ= +180o4）分离点和会合点A(s)B'(s)=A'(s)B(s)A(s)=s3+6s2+5sB(s)=s2+7s+8.25A(s)'=3s2+12s+5B(s)'=2s+7s1=-0.63s2=-2.5 s3=-3.6s4=-7.28
(2) G(s)=s(s+1)(s+4)Kr(s+1.5)1）开环零、极点p1=0p2=-1pz3=-41=-1.52）实轴上根轨迹段p1~p2p3~z13）根轨迹的渐近线n-m= 2θ= +90oσ=-1-4+1.52=-1.75(3) G(s)=s(s+1)Kr21p）开环零、极点1=0p2=-1p3=-12）实轴上根轨迹段p1~p2p3~-∞3）根轨迹的渐近线n-m=3θ= +60o, +180oσ=-1-14）根轨迹与虚轴的交点3=-0.67s3+2s2+s+KKr=0r=0 ω1=0Kr=2 ω2,3=±1jωp3-1.75p2p1z10σ4）分离点和会合点A(s)=s3+5s2+4sB(s)=s+1.5A(s)'=3s2+10s+4B(s)'=1s=-0.62jω1p2p1p3-0.670σ-15）分离点和会合点A(s)=s3+2s2+sB(s)=1A(s)'=3s2+4s+1B(s)'=0s=-0.33
Kr(s+8)(4) G(s)=s(s+3)(s+7)(s+15)1）开环零、极点p1=0p2=-3p3=-7p4=-15z1=-82）实轴上根轨迹段p1~p2p3~z1p4~-∞3）根轨迹的渐近线n-m=3σ=-3-7-15+8=-5.673oo++, θ= 601804）根轨迹与虚轴的交点s4+25s3+171s2+323s+8Kr=0ω2,3=±6.2Kr=0 ω1=0Kr=638
p4p3p2z1-5.67jω6.2p10σ-6.25）分离点和会合点A(s)=s4+25s3+171s2+315sA(s)'=4s3+75s2+342s+315B(s)'=2s+7B(s)=s+8s=-1.4
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Kr为何值时，
复数特征根的实部为-2。
Kr(s+2)G(s)=s(s+1)，解:p1=0p2=-1z1=-2p1~p2z1~-∞分离点和会合点s2+4s+2=0s1=-3.41s2=-0.59jωp2z1p10σ-4ω=0ω+(1+Kr )闭环特征方程式4-ω2-2(1+Kr )+2Kr=0s=-2+js2+s+Krs+2Kr=0ωω=±1.41Kr=3(-2+jω)2+(-2+jω)(1+Kr)+2Kr=0
4-6 已知系统的开环传递函数，试确定闭环极点ζ=0.5时的Kr值。
Kr(1)G(s)H(s)=s(s+1)(s+3)解:p1=0p2=-1p3=-3p1~p2p3~ -+60o, +180o-1.3σ= -1-3=θ= 3根轨迹的分离点：A(s)B'(s)=A'(s)B(s)3s2+8s+3=0s1=-0.45s2=-2.2舍去与虚轴交点s3+4s2+3s+Kr=0-ω1=0ω=0Kr=0 ω3+3ω2,3=±1.7Kr-4ω2=0Kr=12 s1s3p3-3p2-1jω1.78p10σ-1.7ζ=0.5得s1=-0.37+j0.8s3=-4+0.37×2=-3.26Kr=|s3||s3+1||s3+3|=3.26×2.26×0.26=1.9
Kr(2)G(s)H(s)=s(s+3)(s2+2s+2)解：p1=0p2=-3p3.4=-1±jp1~p2+135o+45o, =-1.25σ=-3-1-1θ= 4根轨迹的出射角-1θ-2θ-4θ3=+πθ-90o-26.6oπ-135o=+=-71.6o与虚轴的交点s(s+3)