自动控制课后习题答案

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1、课后答案网www.khdaw.com《自动控制理论第2版》习题参考答案第二章U()sRRCS+RRRCS+12122212-1(a)==⋅U1()sR1R2CS+R1+R2R1+R2R1R2CS+1R+R12U()s11=(b)()2UsRRCCs+(RC+RC+RC)s+121212111222U2()sRCs+1U2()sR11U2(s)R1⎛R1⎞2-2(a)=(b)=−⋅(c)=−⎜Cs+1⎟U1()sRCsU1()sRRCs+1U1()sR⎝4⎠42-3设激磁磁通φ=Ki恒定ffΘ()sCφ=mUa()s⎡2()60⎤sLJs+Lf+RJs+Rf+C

2、φCφ⎢aaaaem⎥⎣2π⎦C()sKCφ=Am2-4R()s3()2⎛60⎞iLJs+iLf+RJs+i⎜Rf+CφCφ⎟s+KCφaaaaemAm⎝2π⎠−32-5i−.219×10=.0084()u−2.0ddC()s()G1+G2G3C()sG1(G2G3+G4)2-8(a)=(b)=R()s1+()G+HGR()s1+GGH+()GG+G(H+GH)1131212342132-9框图化简中间结果如图A-2-1所示。R(s)++1C(s)0.72s+3.0s+1_+.0162.1+2sKs图A-2-1题2-9框图化简中间结果C()s7.0s+.042

3、=()3()2()Rss+9.0+7.0ks+.118+.042ks+.052C()sG1G2G32-10=+G()4Rs1+GH−GGH+GGH211212322-11系统信号流程图如图A-2-2所示。课后答案网www.khdaw.com图A-2-2题2-11系统信号流程图C()sGGG1=123R()s1+GG+G−GGGGHH124124512C()sGGGGGH2=124562R()s1+GG+G−GGGGHH124124512C()s12-12(a)=(abcdef+agdef+abcdi+adgi)(b)R()s1−cdhC()sR=2()2()R

4、sRCRCs+RC+RC+RCs+111221121222-13由选加原理,可得1C()s=[]GGR()s+GD()s−GD()s−GGHD()s()12212212131+HG+HG1122第三章3-1分三种情况讨论(a)当ζ>1时(2)(2)s=−ζ−ζ−1ω,s=−ζ+ζ−1ω1n2n−⎛ζ−ζ2−1⎞⎟ωt−⎛ζ+ζ2−1⎞ωt⎡⎜n⎜⎟n⎤2ζ1e⎝⎠e⎝⎠c()t=t−+⎢−⎥222⎢()2()2⎥ωn2ζ−1ωnζ−ζ−1ζ+ζ−1⎣⎦(b)当0<ζ<1时(2)(2)s=−ζ−j1−ζω,s=−ζ+j1−ζω1n2n22ζ2ζ−ζωnt21−2

5、ζ−ζωnt2c()t=t−−ecos1−ζωt+esin1−ζωtnnωω2nn1−ζωn2ζ1⎛2ζ1−ζ2⎞=t−+e−ζωntsin⎜1−ζ2ωt−arctg⎟n22⎜1−2ζ⎟ω1−ζωnn⎝⎠(c)当ζ=1时s=−ω2,1nc()t=t−2+2e−ωnt⎛1+ωnt⎞⎟⎜ωnωn⎝2⎠课后答案网www.khdaw.com设系统为单位反馈系统,有s()s+2ζωE()s=R()()s−cs=R()snr22s+2ζω+ωnn1s(s+2ζω)2ζn系统对单位斜坡输入的稳态误差为e=lims⋅⋅=sr222s→0ss+2ζωs+ωωnnn3-2(1)K

6、=50,K=,0K=0(2)K=∞,K=K,K=0pvapvaKK(3)K=∞,K=∞,K=(4)K=∞,K=,K=0pvapva102003-3首先求系统的给定误差传递函数E(s)1s1.0(s+)1Φ()s===e2R(s)1+G(s)1.0s+s+10误差系数可求得如下s1.0(s+)1C=limΦ()s=lim=00s→0es→01.0s2+s+10d102.0(s+)1C=limΦ()s=lim=1.01s→0dses→01.0(s2+s+10)2222d1.0(2s+s+10)−202.0(s+)1C=limΦ()s=lim=02s→0ds2es→

7、01.0(s2+s+10)3(1)rt)(=R，此时有rt)(=R,r&t)(=r&&t)(=0，于是稳态误差级数为0s0sse(t)=Crt)(=0，t≥0sr0s(2)rt)(=R+Rt，此时有r(t)=R+Rt,r&(t)=R,r&&(t)=0，于是稳态误差级数为01s01s1se()t=Crt)(+Cr&(t)=1.0R，t≥0sr0s1s11212(3)r(t)=R+Rt+Rt，此时有r(t)=R+Rt+Rt,r&(t)=R+Rt，r&&t)(=R，于是稳态误差级数012s012s12s222为Ce()t=Crt)(+Cr&(t)+2r&&(t)=(

8、1.0R+Rt)，t≥0sr0s1ss