《自动控制原理》黄坚课后习题答案

2-1试建立图所示电路的动态微分方程C＋-+-uiuoR1R2i1ii2C＋-+-uiuoR1R2i1ii2C＋-+-uiuoR1R2i1ii2Lu1C＋-+-uiuoR1R2i1ii2LC＋-+-uiuoR1R2i1ii2Lu1解：u1=ui-uoi2=Cdu1dti2=Cdu1dti1=i-i2uoi=R2uoi=R2u1i1=R1u1i1=R1=ui-uoR1=ui-uoR1dtd(ui-uo)=Cdtd(ui-uo)=C(a)Cd(ui-uo)dtuo-R2=ui-uoR1Cd(ui-uo)dtuo-R2=ui-uoR1i=i1+i2i2=Cdu1dti2=Cdu1dtuoi1=R2uoi1=R2u1-uo=LR2duodtu1-uo=LR2duodtR1i=(ui-u1)R1i=(ui-u1)(b)解：)-R2(ui-uo)=R1u0-CR1R2(duidtdtduo)-R2(ui-uo)=R1u0-CR1R2(duidtdtduoCR1R2duodtduidt+R1uo+R2u0=CR1R2+R2uiCR1R2duodtduodtduidt+R1uo+R2u0=CR1R2+R2ui=R1ui-u1uo+CR2du1dt=R1ui-u1uo+CR2du1dtu1=uo+LR2duodtu1=uo+LR2duodtduodtR1R2Lduodt+CLR2d2uodt2=--uiR1uoR1uoR2+CduodtR1R2Lduodt+CLR2d2uodt2=--uiR1uoR1uoR2+C)uoR1R2Lduodt)CLR2d2uodt2=++(uiR11R11R2+(C+)uoR1R2Lduodt)CLR2d2uodt2=++(uiR11R11R2+(C+2-5试画题图所示电路的动态结构图，并求传递函数。(1)ii2+-uruc+-R2R1c+-uruc+-R2R1ci1解：I2(s)I1(s)+Uc(s)Ur(s)_Cs1R11R1+R2Uc(s)I(s)Ur(s)Uc(s)=1R1(1+(+sC)R21R1+sC)R2Ur(s)Uc(s)=1R1(1+(+sC)R21R1+sC)R2=R2+R1R2sCR1+R2+R1R2sC=R2+R1R2sCR1+R2+R1R2sC(2)C＋-+-urucR1R2Lu1C＋-+-urucR1R2Lu1I(s)Ur(s)_1R11R1U1(s)解：I1(s)-I2(s)L31Cs1CsU1(s)Uc(s)-1Ls1LsR2I1(s)Uc(s)L1L2L1=-R2/LsL2=-/LCs2L3=-1/sCR1Δ1=1L1L3=R2/LCR1s2P1=R2/LCR1s2=R1CLs2+(R1R2C+L)s+R1+R2Ur(s)Uc(s)R2=R1CLs2+(R1R2C+L)s+R1+R2Ur(s)Uc(s)R22-11求系统的传递函数已知控制系统结构如图，试分别用结构图等效变换和梅森公式求系统传递函数(a)G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)H2(s)G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)H2(s)解:L1=-G2H1L2=-G1G2H2P1=G1G2P2=G3G2Δ1=1Δ2=1R(s)C(s)=Σnk=1PkΔkΔR(s)C(s)=Σnk=1PkΔkΔΔ=1+G2H1+G1G2H21+G2H1+G1G2H2G2G1+G2G3=1+G2H1+G1G2H2G2G1+G2G3=(b)G1(s)G2(s)G3(s)G4(s)_++R(s)C(s)H(s)G1(s)G2(s)G3(s)G4(s)_++R(s)C(s)H(s)解:R(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4HR(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4HL1=-G1G2HL2=-G1G4HP1=G1G2Δ1=1P2=G3G2Δ=1+G1G4H+G1G2HΔ2=1+G1G4H2-14,。系统结构图如图，求传递函数R(s)G4(s)+C(s)E(s)G1(s)G2(s)-+G3(s)D(s)X(s)R(s)G4(s)+C(s)E(s)G1(s)G2(s)-+G3(s)D(s)D(s)X(s)2-14C(s)R(s)=1+G3(G1+G2)(G1+G2)(G3+G4)C(s)R(s)=1+G3(G1+G2)(G1+G2)(G3+G4)解:L1=-G1G3L2=-G2G3Δ1=1P1=G1G3P2=G2G3Δ2=1P3=G1G4Δ3=1P4=G2G4Δ4=1E(s)R(s)=1+G3(G1+G2)1E(s)R(s)=1+G3(G1+G2)1D(s)C(s)=1D(s)C(s)D(s)C(s)=1=G2(s)E(s)X(s)=G2(s)E(s)X(s)E(s)X(s)1s(s+1)G(s)=1s(s+1)G(s)=解:=s2+s+1C(s)R(s)1=s2+s+1C(s)R(s)12=1ωn2=1ωn2ωnζ=12ωnζ=1ζ=0.5=1ωn=1ωn=0.866dω=ωn2ζ1-=0.866dω=ωn2ζ1-=60o-1ζ=tgβ21-ζ=60o-1ζ=tgβ21-ζ-1ζ=tgβ21-ζ21-ζtr=dωπβ-=3.14-3.14/30.866=2.42tr=dωπβ-tr=dωdωπβ-=3.14-3.14/30.866=2.42tp=dωπ3.140.866==3.63tp=dωπtp=dωdωπ3.140.866==3.63σ%=100%e-ζζπ1-2σ%=100%e-ζζπ1-2-ζζπ1-2=16%-1.8e-1.8ets=ζ3ωn=6ts=ζ3ωnts=ζ3ωnζ3ωnωn=6ts=ζ4ωn=8ts=ζ4ωnts=ζ4ωnζ4ωnωn=83-11闭环系统的的特征方程如下，试用劳斯判据判断系统的稳定性(1)s3+20s2+9s+100=0解：劳斯表如下：s1s0s3s219201004100系统稳定。(3)s4+8s3+18s2+16s+5=01185s4s3816劳斯表如下：s2165s12161621616s05系统稳定。3-17已知系统结构如图。-R(s)-K1τs１s2C(s)-R(s)-K1τs１s2C(s)(1)单位阶跃输入:确定K1和τ值。σ%=20%ts=1.8(5%)解:G(s)=s2+K1sτK1G(s)=s2+K1sτK1Φ(s)=s2+K1s+K1τK1Φ(s)=s2+K1s+K1τK12ωnζ=K1τ2ωnζ=K1τ2=K1ωn2=K1ωn=0.2e-ζζπ1-2=0.2e-ζζπ1-2e-ζζπ1-2-ζζπ1-2ts=ζ3ωn=1.8ts=ζ3ωnts=ζ3ωnζ3ωnωn=1.8ζ=0.45ωn31.8*0.45=ωn31.8*0.45==3.72ωnK1=2ωnK1==13.7τ=0.24(2)求系统的稳态误差：r(t)=I(t),t,12t2r(t)=I(t),t,12t2解:G(s)=s2+K1sτK1G(s)=s2+K1sτK1=s(s+1)τK11τ1=s(s+1)τK11τ1υ=1Kp=∞ess1=0R(s)=1sR(s)=1sR(s)=s21R(s)=s21Kυ=KKυ=Kess2=τess2=τ=0.24R(s)=s31R(s)=s31Ka=0ess3=∞解:jσ0ωjσ0ωσ0ω(1)(2)jσ0ωjσ0ω600(3)jσ0ωjσ0ω(4)jσ0ωjσ0ω9006003-4已知单位负反馈系统的开环传递函数，求系统的上升时间tr、峰值时间tp、超调量σ%和调整时间ts。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。(5)jσ0ωjσ0ω(6)jσ0ωjσ0ω600jσ0ωjσ0ω(7)(8)jσ0ωjσ0ω45013503601080解:，s(s+1)Kr(s+2)G(s)=，s(s+1)Kr(s+2)G(s)=p1=0p2=-1z1=-2p1~p2z1~-∞分离点和会合点s2+4s+2=0s1=-3.41s2=-0.59s2+s+Krs+2Kr=0闭环特征方程式s=-2+jωs=-2+jω(-2+jωω)2+(-2+j)(1+Kr)+2Kr=0(-2+jωω)2+(-2+j)(1+Kr)+2Kr=0jσ0ωjσ0ωp1p2z1ω-4+(1+Kr)ω=0ω-4+(1+Kr)ω=0ω4-2-2(1+Kr)+2Kr=0ω4-2-2(1+Kr)+2Kr=0Kr=3ω=±1.41s(s+5)(s+15)(1)G(s)=750s(s+5)(s+15)(1)G(s)=750解：ωω-2002040-20020400-180-90-2700-180-90-270dBL(ω)dBL(ω))(ωφ)(ωφs(G(s)=1051s+1)s+1)(151s(G(s)=1051s+1)s+1)(151ω1=5ω1=5ω2=15ω2=15-20dB/dec20lgK=20dB15-40dB/dec15-60dB/decω=0ω=∞-90oφ(ω)=-90oφ(ω)=-270oφ(ω)=-270oφ(ω)=s(s-1)(5)G(s)=10s(s-1)(5)G(s)=10解：ωω-2002040-20020400-180-90-2700-180-90-270dBL(ω)dBL(ω))(ωφ)(ωφ20lgK=20dB1-20dB/decω1=1ω1=1-40dB/decω=0ω=∞-270oφ(ω)=-270oφ(ω)=-180oφ(ω)=-180oφ(ω)=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。5-2已知单位负反馈系统开环传递函数，试绘制系统开环对数频率特性曲线。5-17已知系统开环幅频率特性曲线(1)写出传递函数。(2)利用相位裕量判断稳定性(3)将对数幅频特性向右平移十倍频程，讨论对系统性能的影响。解：s10G(s)=(10s+1)K=10(0.05s+1)cω1010≈12cωcω1010≈12=1cω=1cωcω=180o-90o-tg-110-tg-10.05γ=180o+)(ωφcγ=180o+)(ωφc=90o-84.3o-2.9o=2.8o0.120-20dB/dec-60dB/dec-40dB/decωdBL(ω)dBL(ω)ωcωc010-180-900-180-900ω)(ωφ)(ωφγG0(s)=s(s+5)500KoG0(s)=s(s+5)500Ko解：K=Kv=100G0(s)=s(0.2s+1)100G0(s)=s(0.2s+1)10020lgK=40dBγ=12.6°取Δ=5.6o=45o–12.6o+5.6o=38°1+sinφm1–sinφma=1+sinφm1–sinφma=φm=γ-′γ+Δφm=γ-′γ+Δ=4.2cω0.2100≈12cωcω0.2100≈12=22.4cω=22.4cωcωωω90-900-18090-900-180-2002040-2002040L05-20dB/dec-40dB/decγ408224LcLγ'γ'dBL(ω)dBL(ω))(ωφ)(ωφcφcφ0φ0φφωc'ωc'ωcωc=6.2dBaT=0.04Gc(s)=1+0.04s1+0.01sGc(s)=1+0.04s1+0.01sG(s)=G0(s)Gc(s))=10lgaL0(ωm)=10lgaL0(ωm=ωm=40ωc'=ωm=40ωc'a=82=ω2ωma=82=ω2ωmaT1=23.8=ω1aT1=23.8=ω11=23.8=ω1821T==0.01821T==0.016-1已知单位负反馈系统开环传递函数，采用超前校正满足Kυ=100,γ≥45o。