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10第二章控制系统的数学模型练习题及答案2-1试建立图2-27所示各系统的微分方程。其中外力)(tF,位移)(tx和电压)(tur为输入量;位移)(ty和电压)(tuc为输出量;k(弹性系数),f(阻尼系数),R(电阻),C(电容)和m(质量)均为常数。解(a)以平衡状态为基点,对质块m进行受力分析(不再考虑重力影响),如图解2-1(a)所示。根据牛顿定理可写出22)()(dtydmdtdyftkytF整理得)(1)()()(22tFmtymkdttdymfdttyd(b)如图解2-1(b)所示,取A,B两点分别进行受力分析。对A点有)()(111dtdydtdxfxxk(1)对B点有ykdtdydtdxf21)((2)联立式(1)、(2)可得:dtdxkkkykkfkkdtdy2112121)(11(c)应用复数阻抗概念可写出)()(11)(11sUsIcsRcsRsUcr(3)2)()(RsUcsI(4)联立式(3)、(4),可解得:CsRRRRCsRRsUsUrc212112)1()()(微分方程为:rrccuCRdtduuRCRRRdtdu121211(d)由图解2-1(d)可写出CssIsIsIRsUcRRr1)()()()((5))()(1)(sRIsRICssIcRc(6)CssIsIRsIsUcRcc1)()()()((7)联立式(5)、(6)、(7),消去中间变量)(sIC和)(sIR,可得:1312)()(222222RCssCRRCssCRsUsUrc微分方程为rrrcccuRCdtduCRdtduuRCdtduCRdtdu2222222212132-2试证明图2-28中所示的力学系统(a)和电路系统(b)是相似系统(即有相同形式的数学模型)。解(a)取A、B两点分别进行受力分析,如图12解2-2(a)所示。对A点有)()()(1122yyfyxfyxk(1)对B点有1111)(ykyyf(2)对式(1)、(2)分别取拉氏变换,消去中间变量1y,整理后得)()(sXsY=21212121221212212121()1()1ffffsskkkkfffffsskkkkk(b)由图可写出sCRsUc221)(=sCRsCRsCRsUr111112111)(整理得)()(sUsUrc=1)(1)(21221122121221122121sCRCRCRsCCRRsCRCRsCCRR比较两系统的传递函数,如果设112211221,1,,,RkRkCfCf则两系统的传递函数相同,所以两系统是相似的。2-3假设某容器的液位高度h与液体流入量rQ满足方程rQShSdtdh1,式中S为液位容器的横截面积,为常数。若h与rQ在其工作点),(00hQr附近做微量变化,试导出h关于rQ的线性化方程。解将h在0h处展开为泰勒级数并取一次近似hhhhdthdhhh00021|0(1)代入原方程可得)(1)21()(0000rrQQShhhSdthhd(2)在平衡工作点处系统满足13000rQhdtdh(3)式(2),(3)相减可得h的线性化方程rQhhdthdS022-4试求图2-29所示各信号)(tx的象函数)(sX。解(a))(2)(0tttx)(sX=stess0212(b))())(())(()(321ttcttcbttabatx)(sX=])()([1321stststceecbeabas(c))(tx=)(4)2(4)2(442222TtTTtTTtTtT)21(4)(222TssTeesTsX2-5求下列各拉氏变换式的原函数。(1)1)(sesXs(2))3()2(1)(3ssssX(3))22(1)(2sssssX解(1)1)(tetx14(2)原式=)3(31241)2(83)2(41)2(2123sssssx(t)=24131834432222tttteeetet(3)原式=1)1(1211)1(12121222121222ssssssss)(tx=)cos(sin2121ttet2-6已知在零初始条件下,系统的单位阶跃响应为tteetc221)(,试求系统的传递函数和脉冲响应。解单位阶跃输入时,有ssR1)(,依题意ssssssssC1)2)(1(2311221)()2)(1(23)()()(ssssRsCsGtteessLsGLtk21142411)()(2-7已知系统传递函数232)()(2sssRsC,且初始条件为1)0(c,0)0(c,试求系统在输入)(1)(ttr作用下的输出)(tc。解系统的微分方程为)(2)(2)(3)(22trtcdttdcdttcd(1)考虑初始条件,对式(1)进行拉氏变换,得ssCssCssCs2)(23)(3)(2(2)22141)23(23)(22sssssssssCtteetc2241)(2-8求图2-30所示各有源网络的传递函数)()(sUsUrc。15解(a)根据运算放大器“虚地”概念,可写出12)()(RRsUsUrc(b)22112211111122)1)(1(111)()(sCCRsCRsCRsCRsCRsCRsUsUrc(c))1(11)()(212122CsRRRRCsRCsRsUsUrc2-9某位置随动系统原理框图如图2-31所示,已知电位器最大工作角度mQ=3300,功率放大器放大系数为3k。(1)分别求出电位器的传递函数0k,第一级和第二级放大器的放大系数1k,2k;(2)画出系统的结构图;(3)求系统的闭环传递函数)()(sQsQrc。16解(1)电位器的传递函数11180180330300000mQEK根据运算放大器的特性,可分别写出两级放大器的放大系数为310101030331K,210101020332K(2)可画出系统结构如图解2-9所示:(3))1(11)1()()(3210323210sTsKKKKKsTKKKKsTsKKKKKsQsQmmmtmmmrc11132103223210sKKKKKKKKKsKKKKKTmtmmm2-10飞机俯仰角控制系统结构图如图2-32所示,试求闭环传递函数)()(sQsQrc。解经结构图等效变换可得闭环系统的传递函数68.0)42.018.1()7.09.0()6.0(7.0)()(23sKsKsssQsQrc172-11已知系统方程组如下:)()()()()]()()([)()]()()()[()()()]()()[()()()(3435233612287111sXsGsCsGsGsCsXsXsXsGsXsGsXsCsGsGsGsRsGsX试绘制系统结构图,并求闭环传递函数)()(sRsC。解系统结构图如图解2-11所示。利用结构图等效化简或梅逊增益公式可求出系统的闭环传递函数为843217432154363243211)()(GGGGGGGGGGGGGGGGGGGGsRsC2-12试用结构图等效化简求图2-32所示各系统的传递函数)()(sRsC。18解(a)所以:432132432143211)()(GGGGGGGGGGGGGGsRsC(b)所以:HGGGsRsC2211)()((c)19所以:32132213211)()(GGGGGGGGGGsRsC(d)所以:2441321232121413211)()(HGGGGGGHGGHGGGGGGGsRsC(e)所以:2321212132141)()(HGGHGHGGGGGGsRsC202-13已知控制系统结构图如图2-34所示,求输入()31()rtt时系统的输出)(tc。解由图可得)3)(1(2)1(1221122)()(22SsssssssRsC又有ssR3)(则311323)3)(1(2)(ssssSssC即tteesssLtc313231132)(2-14试绘制图2-36所示系统的信号流图。解2-15试绘制图2-36所示信号流图对应的系统结构图。21解2-16试用梅逊增益公式求2-12题中各结构图对应的闭环传递函数。解(a)图中有1条前向通路,3个回路,有1对互不接触回路,,,2111432111GGLGGGGP,,,21321323432)(1LLLLLGGLGGL43213243214321111)()(GGGGGGGGGGGGGGPsRsC(b)图中有2条前向通路,1个回路,,,,,HGLGPGP212221111111LHGGGPPsRsC22122111)()((c)图中有1条前向通路,3个回路,,,211132111GGLGGGP,,,)(13213213322LLLGGGLGGL3213221321111)()(GGGGGGGGGGPsRsC(d)图中有2条前向通路,5个回路22,,,,11241213211GGPGGGP,,,,414321323221211GGLGGGLHGGLHGGL,,)(154321245LLLLLHGL24413212321214132122111)()(HGGGGGGHGGHGGGGGGGPPsRsC(e)图中有2条前向通路,3个回路,,,,242132111GPGGGP,,,,)(132123231221211LLLHGGLHGLHGGL2321212132141122111)()(HGGHGHGGGGGGPPPPsRsC2-17试用梅逊增益公式求图2-37中各系统的闭环传递函数。解(a)图中有1条前向通路,4个回路1143211,GGGGP23)(143212434443213332121321LLLLHGGLHGGGGLHGGGLHGGL,,,则有2434432133211324321111)()(HGGHGGGGHGGGHGGGGGGPsRsC(b)图中有2条前向通路,3个回路,有1对互不接触回路,,,,111243213211111HGLGGPGGGP,,,3213213332111HHHGGGLHGLHGL,21321)(1LLLLL则有33113213213311114332122111)1()()(HGHGHHHGGGHGHGHGGGGGGPPsRsC(c)图中有4条前向通路,5个回路,,,,1242321211GGPGPGGPGP,,,,,2151242321211GGLGGLGLGGLGL,,)(1143214321LLLL则有44332211)()(PPPPsRsC21212121211222111222113121GGGGGGGGGGGGGGGGGGGGGG(d)图中有2条前
本文标题:第二章习题及答案
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