线性系统理论+课后答案+(程兆林+马树萍+著)+科学出版社

    课后答案网，用心为你服务！  大学答案---中学答案---考研答案---考试答案 最全最多的课后习题参考答案，尽在课后答案网（）！Khdaw团队一直秉承用心为大家服务的宗旨，以关注学生的学习生活为出发点，旨在为广大学生朋友的自主学习提供一个分享和交流的平台。 爱校园（）课后答案网（）淘答案（） 线性系统理论习题答案习题11.解：a)由电路学知识得∫=++=idtcuiLuiRteCc1,)(.设iyteuixuxC====),(,,21,则,1121xCiCx==,11))((1122uLxLxLRuiRteLxC+−−=−−=即[]⎪⎪⎪⎩⎪⎪⎪⎨⎧⎥⎦⎤⎢⎣⎡=⎥⎥⎦⎤⎢⎢⎣⎡+⎥⎦⎤⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−=⎥⎦⎤⎢⎣⎡.10,10110212121xxyuLxxLRLCxxb)∫=idtCuC111,11xuC=,∫=idtCuC212,22xuC=,ccCCuuiRte21)(++=,)(teu=,ccCCuuu21+=,Cuy=.则[]⎪⎪⎪⎩⎪⎪⎪⎨⎧⎥⎦⎤⎢⎣⎡=⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡+⎥⎦⎤⎢⎣⎡⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−=⎥⎦⎤⎢⎣⎡.11,111111212121221121xxyuRCRCxxRCRCRCRCxx2.解：由电路学知识得.1,)(∫=++=idtCVViLiRteCC设CCVyteuVxix====),(,,21,则,11)(11211uLxLxLRteLVLiLRixC+−−=+−−==1211xCiCx==.即[]⎪⎪⎪⎩⎪⎪⎪⎨⎧⎥⎦⎤⎢⎣⎡=⎥⎥⎦⎤⎢⎢⎣⎡+⎥⎦⎤⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−−=⎥⎦⎤⎢⎣⎡.10,01011212121xxyuLxxCLLRxx3.解：由电路学知识得213Ciii+=①132Ciii+=②2113311CCCVRidtiCV+==∫③1112221uRidtiCVCC+==∫④1222CVRiu+=⑤由①-⑤得)(1)(1211132232CCCCVVRVuRiii−−−=−=,11)11(2233221uRVRVRRCC++−−=)(1)(1113132212uVRVVRiiiCCCC−−−=−=,1)11(11131321uRVRRVRCC+−−+=,1111CCiCV=,1222CCiCV=设,11xVC=,22xVC=,21⎥⎦⎤⎢⎣⎡=CCVVy,21⎥⎦⎤⎢⎣⎡=uuy则,01101111112112212112323231312121⎥⎦⎤⎢⎣⎡⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡+⎥⎦⎤⎢⎣⎡⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡−−−−=⎥⎦⎤⎢⎣⎡uuRCRCxxRCRCRCRCRCRCxx⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡=211001xxy.4.解：对方程uuyy+=−两边作拉氏变换得：)()1()()1(2sussys+=−令),()1()(),()1()(2szssyszssu+=−=则.)(,)(zztyzztu+=−=设,,21zxzx==则[].11y,100110212121⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡+⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡xxuxxxx5.解：uyyyy66116=+++，令,,,321yxyxyx===则()⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛+⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛−−−=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛.001,6006116100010321321321xxxyuxxxxxx6.解：(1)⎟⎟⎠⎞⎜⎜⎝⎛=1011A,2)1(1011−=−−−=−sssAsI,⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎝⎛−−−=⎟⎟⎠⎞⎜⎜⎝⎛−−−=−−110)1(1111011)1(1)(221ssssssAsI.(2)⎟⎟⎠⎞⎜⎜⎝⎛=1001A,⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎝⎛−−=−−1111)(1ssAsI.(3)⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛=221131132A,410722113113223−+−=−−−−−−−−−=−ssssssAsI,⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛+−−−−+−−−+−−+−=−−351211341434541071)(222231sssssssssssssssAsI.7．解：(1)⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛−−−=311032010A，⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛=210B，()100=C.()⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛−−+−=−=−−21031103201100)()(11sssBAsICsG().6737221023150362039100671322223−−++=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛++−−−−+−−−−−=sssssssssssssss.(2)⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛−−−=213100010A,⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛=111001B,()111=C.()()()sssssssssssssssssssssBAsICsG3212321111001332312121113211110012131001111)()(223222311+−+++=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛−−−+−++++++=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛+−−=−=−−8.解：.)(0111111CasasasCsCsCsGnnnnnnn++++++++=−−−−9.证明类似定理1.4,此处略.10.解：]321[])[(1111−−−−⎟⎟⎠⎞⎜⎜⎝⎛+−=−=ssLAsILeAt]2211221221112112[1⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎝⎛+++−+++−+−++−+=−ssssssssL⎟⎟⎠⎞⎜⎜⎝⎛+−+−−−=−−−−−−−−tttttttteeeeeeee22222222.τττdBuexetxttAAt)()0()(0)(∫−+=⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛+−+−−−=−−−−−−−−102222222tttttttteeeeeeeeττττττττττdeeeeeeeeett)(02222022222−−−−−−−−−−⎟⎟⎠⎞⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛−+−−−+∫τττdeeeeeeetttttt∫⎟⎟⎠⎞⎜⎜⎝⎛+−−+⎟⎟⎠⎞⎜⎜⎝⎛−−=−−−−−−−02244242⎟⎟⎠⎞⎜⎜⎝⎛−−+−=−−−−tttteeteet222)43()14(.11.解：200000ttAttteteeee−−−−⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦，()1,0,utt=≥.()00()()tAtAtxtexeBudτττ−=+∫=00()tAtAexeButdτττ+−∫=200000tttteteee−−−−⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦121⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦+∫t0200000eeeeτττττ−−−−⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦014⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦τd=2112tttteee−−−⎡⎤+⎢⎥+⎢⎥⎢⎥−⎣⎦.12.解：0140A−⎡⎤=⎢⎥⎣⎦，21||44ssIAss−==+−22122211144()44444sssssIAsssss−−⎡⎤⎢⎥−⎡⎤++−==⎢⎥⎢⎥+⎣⎦⎢⎥⎢⎥++⎣⎦11111()()22242211111()()22222isisisisiisisisisi⎡⎤+−⎢⎥−+−+=⎢⎥⎢⎥−−+⎢⎥−+−+⎣⎦111cos2sin2()22sin2cos2AttteLsIAtt−−⎡⎤−⎢⎥⎡⎤=−=⎣⎦⎢⎥⎣⎦1cos2sin2()22sin2cos2ttxttt⎡⎤−⎢⎥=⎢⎥⎣⎦010(())01tAedτδττ−⎡⎤⎡⎤+⎢⎥⎢⎥⎣⎦⎣⎦∫110cos2sin2()2012sin2cos2tttt⎡⎤−⎡⎤⎡⎤⎢⎥=+⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦=1cos2sin222sin2cos2tttt⎡⎤−⎢⎥⎢⎥+⎣⎦.13.解：(1)1001000Atttee−⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦()()()AtHtCeBDtττδ−−=+[][]()100100110010ttetτττ−−−⎡⎤⎡⎤⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦=−(2)sin()0atutω⎡⎤=⎢⎥⎣⎦,00()()tAtAxytCeeBudτττ−=∫[]01100sin1001110001tttadeeττωττ−−−⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦∫[][]022sin10sin0cossin10cos0sin.tatadaatttaattaattτωτωττωωωωωωωωωω−⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦⎡⎤−⎢⎥⎢⎥⎢⎥=−+⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦=−∫14.解：解法一做z变换：2()(1)zZkz=−，求得()xz.解法二递推方法：(2)2(1)()(1)(3)2(2)(1)1(2)xkxkxkkxkxkxkk++++=⎧⎨+++++=+⎩(2)(1)−⇒(3)(2)(1)()1,xkxkxkxk+++−+−=即2(3)(2)(1)()22kkxkxkxkxk++++−=++−.2(21)(2)(1)(0)00,(3)11121(22)(21)(2)(1).(4)222klxlxllxxklxlxllxx=++−=+−==++++−−=+−=−(3)(4)(2)0xl−⇒=.由(3)得：(23)(22)10,(5)xlxll+++−−=(5)(4)(21)0xl−⇒+=.15解：解法一设123(1)()(1)(1)(1)(1)(2)xkykxkxkykxkyk+⎡⎤⎡⎤⎢⎥⎢⎥+=+=+⎢⎥⎢⎥⎢⎥⎢⎥++⎣⎦⎣⎦,12()()()()(1)ukukukukuk⎡⎤⎡⎤==⎢⎥⎢⎥+⎣⎦⎣⎦,则1223323212(1)(),(1)(),(1)2()3()2()3(),()().xkxkxkxkxkxkxkukukykxk+=+=+=−−++=01000(1)001()00()02332xkxkuk⎡⎤⎡⎤⎢⎥⎢⎥+=+⎢⎥⎢⎥⎢⎥⎢⎥−−⎣⎦⎣⎦,[]()010()ykxk=.解法二作z变换：2()3()2()2()3(),zyzzyzyzzuzuz++=+2()23()32yzzuzzz+=++.令()(23)()yzzxz=+,2()(32)()uzzzxz=++,12()(),(1)(),xkxkxkxk=+=则1122(1)()010(),(1)()231xkxkukxkxk+⎡⎤⎡⎤⎡⎤⎡⎤=+⎢⎥⎢⎥⎢⎥⎢⎥+−−⎣⎦⎣⎦⎣⎦⎣⎦[]12()()32()xkykxk⎡⎤=⎢⎥⎣⎦.16.解:1,01Atte⎡⎤=⎢⎥⎣⎦则12,01ATGe⎡⎤==⎢⎥⎣⎦222000102,01112AtttHeBdtdtdt⎡⎤⎡⎤⎡⎤⎡⎤====⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦∫∫∫则122(1)()().012xkxkuk⎡⎤⎡⎤+=+⎢⎥⎢⎥⎣⎦⎣⎦17.解:010,,021AB⎡⎤⎡⎤==⎢⎥⎢⎥⎣⎦⎣⎦1T=,21112111111()122()221002tAttessseLsIALes−−−⎡⎤−−⎡⎤⎢⎥−+−⎢⎥⎡⎤=−==⎢⎥⎣⎦⎢⎥⎢⎥⎣⎦⎢⎥−⎣⎦,22220022113.19451(1),207.38910111(21)1.09734,2213.1946(1)21ATTtTTAttTeGeeTeeHeBdtdteeT⎡⎤−⎡⎤⎢⎥===⎢⎥⎢⎥⎣⎦⎣⎦⎡⎤−+−⎡⎤⎢⎥−+⎡⎤⎢⎥====⎢⎥⎢⎥⎢⎥⎣⎦⎢⎥−⎣⎦⎢⎥⎣⎦=∫∫[](1)()(),()()01().xkGxkHukykCxkxk+=+==习题23.设CXx∈0，证明在任意控制)(tu作用下，自0x出发的轨线)(tx上的任一点均属于.CX证：τττdBuexetxttAAt)()(0)(0∫−+=，因为XcxAtxeIjnjjAt∈==∑−=0100)(α∫∑∫−=−−==tnjjjttAdBuAtdBueII0110)()()()(τ