控制工程基础第二章参考答案

控制工程基础第二章参考答案第1页共32页第二章参考答案2-1（1）不是（2）是（3）不是（4）不是2-2(a))()()(3)(2222tutudttduRCdttudCRiooo=++(b))()()()()()()()(2211222121222111222121tudttduCRCRdttudCCRRtudttduCRCRCRdttudCCRRiiiooo+++=++++(c))()()()()()(33221312221tuRdttduCRRtuRRdttduCRRRRRiioo+=++++(d))()()()()()()()(1211222121211211222121tudttduCRCRdttudCCRRtudttduCRCRCRdttudCCRRiiiooo+++=++++(e))()()()()()()()(221222121211222222121tudttduRCCdttudCCRRtudttduCRCRCRdttudCCRRiiiooo+++=++++(f))()()()()()()(22121221tuRdttduLtuRRdttduLCRRdttudCLRiiooo+=++++2-3(a))]()([)()()(23213121tuRdttduCRRtuRdttduCRRRRiioo+=++－(b))()()()(4141232022213210tuRRtuRRdttduCRRRdttudCCRRRRiooo－=++(c))]()()([)(32321tuRRdttduCRRtuRiio++=－(d))()()()()(221122212121tudttduCRCRdttudCCRRdttduCRiiio+++=－(e))()()()(2412222142tudttduCRCRdttudCCRRooo+++)}()(])([)({21213224223221432132tudttduRRCCRRCRdttudRRCCRRRRRRiii+++++++=－2-4(a)dttdxfdttdxffdttxdmioo)()()()(12122=++(b)dttdxfktxkkdttdxfkkioo)()()()(12121=++(c))()()()()(121txkdttdxftxkkdttdxfiioo+=++(d))()()()()()(112121txkdttdxftxkkdttdxffiioo+=+++2-5(a))(1)()()()(1)()()(2112212221211211212221tuCCdttduCRCRdttudRRtuCCdttduCRCRCRdttudRRiiiooo+++=++++(b))()()()()()()()(2112212221211211212221txkkdttdxkfkfdttxdfftxkkdttdxkfkfkfdttxdffiiiooo+++=++++由(a)(b)两式可以看出两系统具有相同形式的微分方程，所以(a)和(b)是相似系统。2-6PLQΔ=Δ式中02PKL=，0QQQ－=Δ，0ΔPPP－=。线性化式子表示为：LPQ=2-7yFΔ=Δ12，式中73.2Δ－FF=，25.0Δy－y=控制工程基础第二章参考答案第2页共32页2-8设如右图(a)所示中间变量i，1i，2i，得方程组为21III+=1111IsCIRUi+=22221111IsCIRIsC+=221IsCUo=根据以上4式，按信号流动的方向绘制的系统方框图如右图(b)所示。传递函数1)(1)()(22211122121++++=CRCRCRsCCRRsUsUio2-998)1(5)()(2+++=ssssRsC9823)()(22++++=sssssRsE2-10(a)21212122132142112)()()()()(kkfskksmkmkfsmmsmmfssFsX+++++++=(b)212122132122)()()(kkfskksmksfmkfssXsF+++++=2-11)(2)(2)(3)(2)(2233trtcdttdcdttcddttcd=+++2-12(a)131222++RCssCR(b)1)(1)(22211122121221122121+++++++sCRCRCRsCCRRsCRCRsCCRR(c)21312221332)(RRCsRRRRRRCsRR+++++(d)1)(1)(21121122121121122121+++++++sCRCRCRsCCRRsCRCRsCCRR(e)1)(1)(2112222212122122121+++++++sCRCRCRsCCRRsRCCsCCRR(f)2121212)(RRsLCRRCLsRRLs+++++2-13(a)13121232)(RCsRRRRRCsRR+++－(b)412320221321041RRsCRRRsCCRRRRRR++－(c)13232RRRCsRR++－(d)sCRsCRCRsCCRR212211221211)(+++－(e)1)(1])([2412221422121322423221432132++++++++++sCRCRsCCRRsRRCCRRCRsRRCCRRRRRR－2-14(a)sffmssf)(2121++(b)21211)(kkfskkfsk++(c)211kkfskfs+++(d)212111)(kksffksf++++2-15(a)212121121GGGGGGG++++(b)7432164353243211GGGGGGGGGGGGGGG+++控制工程基础第二章参考答案第3页共32页(c)1212232211HGGHGGGGG+++(d)3211132121GGGGGGGG+++2-16(1)423211GGGGG+(2)321423211GGGGGGGGGRC++=3214211GGGGGGRE++=3214231GGGGGGGFC++=3214231GGGGGGGFE++=－(3))(1)(1)()()(32142332142321sFGGGGGGsRGGGGGGGGsFGsRGsCFCRC+++++=+=)]()([1)()()(321423sFsRGGGGGGsFGsRGsEFERE－++=+=2-17(a)211212121GGGGGGG++++(c)21213212211211HHGGHGGHGHGGG++++(b)211121211)1(HHHGHHGG+++(d)2132121432311433211HHGGGHHGGHGHGGGGGG+++++2-19：状态空间表达式为：Fmxxmfmkxx101021212101xxy系统的状态变量图为：题2-19状态变量图2-20：uxxxxxx020320032100321321321001xxxy控制工程基础第二章参考答案第4页共32页2-21：uTkxxxTTkkTkTTkxxx1132111412223332100101000321001xxxy2-22（1）uxxxxxx1006155100010321321321007xxxy（2）uxxxxxx800567100010321321321001xxxy（3）uxxxxxx2240164018192640100010321321321001xxxy2-23：（1）uxxxxxx10071481000103213213211815xxxy（2）uxxxxxx111400020001321321321612338xxxy第三章习题及答案3-1.假设温度计可用11Ts传递函数描述其特性，现在用温度计测量盛在容器内的水温。发现需要min1时间才能指示出实际水温的98%的数值，试问该温度计指示出实际水温从10%控制工程基础第二章参考答案第5页共32页变化到90%所需的时间是多少?解：41min,=0.25minTT1111()=1-e0.1,=ln0.9thttTT21T22()=0.9=1-eln0.1thttT，210.9ln2.20.55min0.1rtttTT2.已知某系统的微分方程为)(3)(2)(3)(tftftyty，初始条件2)0(,1)0(yy，试求：⑴系统的零输入响应yx(t)；⑵激励f(t)(t)时，系统的零状态响应yf(t)和全响应y(t)；⑶激励f(t)e3t(t)时，系统的零状态响应yf(t)和全响应y(t)。解：(1)算子方程为：)()3()()2)(1(tfptypp)()e25e223()()()()()e21e223()()()()()ee2()(2112233)()2(;0,e3e4)(34221ee)(2x2222x212121221xttytytyttthtytthppppppHttyAAAAAAAAtyttttttfftttt*)()e4e5()()()()()ee()(e)()()3(2x23ttytytyttthtytttttff*3.已知某系统的微分方程为)(3)(')(2)('3)(tftftytyty，当激励)(tf=)(e4tt时，系统的全响应)()e61e27e314()(42ttyttt。试求零输入响应yx(t)与零状态响应yf(t)、自由响应与强迫响应、暂态响应与稳态响应。解：控制工程基础第二章参考答案第6页共32页.,)();()e27e314(:);(e61:)()()e3e4()()()()()()e3221e61()()]e1(e21)e1(e32[)(]e2e2[e)(),()ee2()(,2112233)(242x242230)(2)(422}{不含稳态响应全为暂态自由响应强迫响应零状态响应零状态响应tyttttytytytettdtytthppppppHtttttttttttttttttff4.设系统特征方程为：0310126234ssss。试用劳斯-赫尔维茨稳定判据判别该系统的稳定性。解：用劳斯-赫尔维茨稳定判据判别，a4=1，a3=6，a2=12，a1=10，a0=3均大于零，且有3121001060031210010640610621011262051210110366101263015365123334所以，此系统是稳定的。5.试确定下图所示系统的稳定性.控制工程基础第二章参考答案第7页共32页解：