祝守新-邢英杰-韩连英《机械工程控制基础》习题解答

1机械控制工程基础答案提示第二章系统的数学模型2-1试求如图2-35所示机械系统的作用力)(tF与位移)(ty之间微分方程和传递函数。)(tF)(tyf图2-35题2-1图解：依题意：22dytdytamFtfkytdtbdt故tFbatkydttdyfdttydm22传递函数:kfsmsbasFsYsG22-2对于如图2-36所示系统，试求出作用力F1(t)到位移x2(t)的传递函数。其中，f为粘性阻尼系数。F2(t)到位移x1(t)的传递函数又是什么？m2m1k1fk2F1(t)F2(t)x2(t)x1(t)图2-36题2-2图解：依题意：对1m：212121111dttxdmdttdxdttdxftxkF2对两边拉氏变换：sXsmssXssXfxksF12121111①对2m：222222212dttxdmtxkdttdxdttdxftF对两边拉氏变换：sXsmsxkssxssxfsF22222212②故：SFsxkfssmsfsxsFsfsxsxkfssm222221121121故得：22221212212122222121222211fskfssmkfssmkfssmsFsfsFsxfskfssmkfssmsfsFkfssmsFsx故求tF1到tx2的传递函数令：02sF2122211122432121212211212xsfsGsFsmsfskmsfskfsfsmmsfmmsmkmksfkkskk求tF2到tx1的传递函数令：01sF1122221122432121212211212xsfsGsFsmsfskmsfskfsfsmmsfmmsmkmksfkkskk2-3试求图2-37所示无源网络传递函数。iuou1R)(2ti)(1ti)(ti2Ra)uob）C1uiL1C2R1L2R2图2-37题2-3图解（a）系统微分方程为312121202121iitdtitRCuitRitRuitRititit拉氏变换得1122112012121iIsRIsCsUsIsRIsRUsIsRIsIsIs消去中间变量12,,IsIsIs得：210211212121121111iRRCsUsRCsRRRGsRRUsRRCsRCsRR（b）设各支路电流如图所示。uob）C1uiL1C2R1L2R2i1i2i3i4i5i6系统微分方程为1302131042502026234561213456iutRitutditRitLdtutitdtCditutLdtutRitititititit由（1）得：13ioUsRIsUs由（2）得：1312RIsLsIs由（3）得：421oUsisCs4由（4）得：25oUsLsIs由（5）得：26oUsRIs由（6）得：23456IsIsIsIsIs故消去中间变量123456,,,,,IsIsIsIsIsIs得：211212121212212121211oiLLsUsLLRLLRRUsLLCssLLLLRR2-4证明22cossstL证明：设cosftt由微分定理有22(1)200dftLsFssffdt（1）由于0cos01f，10sin00f，222cosdfttdt（2）将式（2）各项带入式（1）中得22cosLtsFss即22FssFss整理得22sFss2-5求21()2ftt的拉氏变换。解：2223001111222ststFsLttedtstedsts令stx，得23012xFsxedxs由于伽马函数01!nxnxedxn，在此2n所以33112!2Fsss2-6求下列象函数的拉氏反变换。（１）)3)(2)(1(35)(sssssX5（２）32)1(32)(ssssX（３）)2()1(1)(3ssssX解：（1）31253()(1)(2)(3)123AAAsXsssssss11513()(1)1(12)(13)sAXss同理27A，36A176()123Xssss拉式反变换得23()76tttxteee（2）22333122321()(1)(1)(1)1sssXsssss拉式反变换得2()ttxttee（3）351243321()(1)(2)(1)(1)(1)2AAAAAXsssssssss11111(2)1(12)sAss222112210(2)(2)sssdAdsssss2223232224411122111(2)(1)(4128)12(2)2(2)(2)ssssddssssssAdsssdsssss43011(1)(2)2sAss53211(1)2sAss所以31111()(1)122(2)Xsssss6拉式反变换得22111()222tttxtteee2-7绘制图2-38所示机械系统的方框图。xy1f2f图2-38题2-7图解依题意：2221dttydmdttdyfdttytxdftytxk两边拉氏变换得：sYmsssYfsYsXsfsYsXk221sYsYfsYsXsfkmssYmsssYfsYsXsfk2122211故得方块图：sf121mskX(s)sf2Y(s)--+++2-8如图2-39所示系统，试求（１）以X(s)为输入，分别以Y(s)、X1(s)、B(s)、E(s)为输出的传递函数。（２）以N(s)为输入，分别以Y(s)、X1(s)、B(s)、E(s)为输出的传递函数。7图2-39题2-8图解（1）sHsGsGsGsGsXsYsG212111sHsGsYsXsBsXsEsHsGsYsBsGsEcsY21211故sHsGsGsXsEsGsHsGsGsHsGsGsXsBsGsHsGsGsGsXsYsGsGsHsGsYsXsY2142121321111212111111（2）sHsGsGsHsGsNsEsGsHsGsGsHsGsNsBsGsHsGsGsHsGsGsNsYsGsHsGsGsGsNsYsG21242123212112212111112-9化简如图2-40所示各系统方块图，并求其传递函数。8)(sX)(sX)(sX)(sX)(sY)(sY)(sY)(sY图2-40题2-9图解：（a）第三回路传递函数：33321GHGsF2G()Xs()Ysa)-11G2H1H3331GGH第二回路传递函数：32233323323233323232211111GHGHGGGHGHGGHGGGsHsFGsFGsF()Xs()Ysa)-21G1H23332321GGGHGGH第一回路传递函数：1321332323212112111HGGGHGHGGGGGFHGFGsXsY故原图可化简为：9()Xs()Ysa)-31232323312311GGGGGHGHGGGH（b）2G()Ysb)-11G4G12HG+3G21HG+++()Xs234GGG()Ysb)-211211GGGH211HG+()Xs234GGG()Ysb)-211211GGGH211HG+()Xs()Ysb)-312341211GGGGGGH211HG+()Xs()Ysb)-4123412123212342141GGGGGGHGGHGGGGHGG()Xs10(c)23211GGGH()Ysc)-11G2H13HG++++()Xs4G123212321GGGGHGGH()Ysc)-213HG+++()Xs4G123212321211GGGGHGGHGGH()Ysc)-3+()Xs4G()Ysc)-41234122321211GGGGHGGGHGGH()Xs(d)3232322221113111FFHFFsXsYsFHGGsFHGGsF2-10解：112121213241232221111231HHGGLLGGLGGLHGLHGLGPGP△21213221221121432111HHGGGGGGHGHGLLLLLL而2L与21PP都不接触，所以222222211111HGLHGL故综上：212132212211223121213221221122122311111HHGGGGGGHGHGHGGGHHGGGGGGHGHGHGGHGGPsXsYsGPkkk第三章时间特性分析法3-1解：依题意，系统的闭环传递函数为：11112sssssG12%284%563%4.16%100%100%63.333242.293423333arctan1arctan23211121121133122222sWtsWteeMsWtssWtsrad3-2解：依题意：bkaksskbassksksXsY222111则10,1116.0%16%100%251021010331222baaeeMpaaWaakWbWbkannn3-3解：依题意：131.243.113213.11321.243.113223.11