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1第5章5-1222212211arctanarctan1KTTTT222212211()sin(arctanarctan)1ssKTctRtTTT5-2(1)()arctan-arctanGjT(2)1212()arctanarctan-arctan-arctanGjaTbTTT5-3(1)(2)(3)(4)2(5)(6)(7)(8)5-4(1)3(2)(3)(4)4(5)(6)(7)(8)55-5(1)(2)(3)(4)6(5)(6)(7)(8)7(9)(10)5-6(a)12100()11(1)(1)Gsss(b)231221(1)()1(1)sGsss(c)112211()11csGsss(d)123()11(1)(1)sGsss5-7(a)稳定(b)稳定(c)稳定(d)不稳定(e)不稳定(f)不稳定(g)不稳定(h)稳定5-8(a)不稳定(b)稳定(c)稳定(d)稳定(e)稳定5-9(a)不稳定(b)稳定(c)稳定5-10稳定5-111:稳定2:不稳定3:稳定4:不稳定5-12不稳定5-13o73此题传递函数改为(0.561)()(1)(0.11)(0.0281)KsGsssss或幅值穿越频率5.13rad/sc,o46.5。5-14(1)(j)48.2cG,o131.8(2)(j)155.4cG,o24.685-15ooo(1)5520dB/dec(2)-15.840dB/dec(3)-52.860dB/dec,,,5-16s2s1s3p1p2p3ttt,5-17(1)o54.920lggK,,系统稳定。(2)o-46.520lg-gK,,系统不稳定。5-18o100(0.11)()1045(1)(0.011)csGssss,,rr2n123ζζθζθ1-ζζωMM曲线说明**大由图**小大,小**大小,大**大由nnn22nnζωζωζωζω1-ζω1-ζspp*大*由图*大小由及t*小大与反比σ**大小,σ大*大大由及9说明:以下各设计题目,一般都有多种解。5-19超前补偿法。固有部分(待补偿系统)0100()(0.041)Gsss超前补偿c0.041()0.0081sGss;设计后e100()(0.0081)Gsss滞后补偿法。固有部分010()(0.041)Gsss滞后补偿c110()10101sGss;设计后e100(1)()(0.041)(101)sGssss5-20K=10,c0.51()0.151sGss等等。提示:可利用二阶系统公式。5-21ce2(21)(0.21)100(21)()()(66.71)(0.11)(66.71)(0.11)sssGsGssssss;等等5-2211(1)100(1)1616()()11(1)(0.11)(1)160160cessGsGsssss;等等5-23(101)10(101)()()(1001)(1001)(1)(0.251)cessGsGssssss等等5-2411(1)(1)0.37.6()11(1)(1)0.067143cssGsss2111000(1)(1)0.37.6()1(1)(0.91)(0.0071)0.067essGsssss等等5-25根据二阶系统的计算公式。0.00635hK,反馈补偿后的开环传递函数为16000()(141.6)Gsss。5-26反馈补偿后系统的开环传递函数为215150()0.1(115)(10150)Gsshsssh0.0967h105-27110200KK,,c()200(0.11)sHss5-281171(1)710(1)3535()()11(1)(0.11)(1)(0.0011)400400cessGsGsssssss,等5-2901.41()11(1)(1)1010000Gsss;11(1)(1)2.510()10011(1)(1)0.26100cssGssss等5-30补偿装置的对数频率特性见图。)1201)(121()141(20)(sssssGe0201arctan41arctan1201)141(10)()()(0cecGsssGsGsG,补偿装置是超前网络,本身相位角为正,可提高相位裕度。图对数频率特性5-31K=2000,)14.01)(12001()131.0)(191()(sssssGc等等5-3211(1)(1)1.510()411(1)(1)0.36200cssGsss等等5-33K=200,1(1)10()1(1)3csGss115-3411(1)(1)350310()11300(1)(1)0.4140cssGsss等等5-3520.0167()0.21csHss5-3620.0020.91()0.005()0.11cccsHHssHss;;5-3720.63(0.11)()3.51cssHss等等5-380.283()0.251csHss5-39123222111ω01/1/1/(s)sTsTsTTTTTKTKTKKG2322221113s20lg(jω)20lg20lg20lg20lgωωωωKTKTKKGTTTT5-40(1)2n222nnn2222nnnn22nn2()()()212ssYsGsXsssKsKsss2n22nnn2(1)2Kss阻尼比n(1)2K,可见微分反馈可增加系统阻尼比。(2)2n2nn2222nnnnn(2)()()()21(1)(2)ssYsGsXsssKsKsss2n22nnn2(1)2Kss12阻尼比n(1)2K,可见微分反馈补偿系数K可增加系统阻尼比。
本文标题:自动控制原理(梅晓榕)习题答案5
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