控制工程基础第三版习题答案_清华大学出版社

发布时间:2023-03-09 01:45:21   来源:文档文库   
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各位童鞋仅供参考第二章2-11:F(SL[(4t(t]L[5(t]L[t1(t]L[21(t]052:F(S1212522SSSS3s522(s251es3:F(S2s14:F(SL{[4cos2(ts6]1(te5t1(t}66s64Se14Se1s222S5s24S5e2se2s65:F(S006SS6:F(SL[6cos(3t45901(tS44]S46Se6SeL[6cos3(t1(t]22244S3S97:F(SL[e6tcos8t1(t0.25e6tsin8t1(t]S62S822222(S68(S68S12S1008:F(S22-2解:1:f(tL1(259es20(s202s29s612(e2t2e3t1(tS2S312:f(tsin2t1(t213:f(tet(cos2tsin2t1(t2eset11(t14:f(tL(S11
5:f(t(tet2et2e2t1(t81515t6:f(tL1(152815e2sin15t1(t(S121521522(27:f(t(cos3t13sin3t1(t2-3解:1对原方程取拉氏变换,得:S2X(SSx(0x(06[SX(Sx(0]8X(S1S将初始条件代入,得:S2X(SS6SX(S68X(S1S(S26S8X(S1SS6177X(SS26S18S(S26S8S4S28S4取拉氏反变换,得:x(t17e2t748e4t82t=0时,将初始条件x(050代入方程,得:50+100x(0300x(0=2.5对原方程取拉氏变换,得:sx(s-x(0+100x(s=300/sx(0=2.5代入,得:SX(S-2.5100X(S300SX(S2.5S300S(S1003s0.5s100取拉氏反变换,得:x(t3-0.5e-100t2-4解:该曲线表示的函数为:u(t61(t0.0002则其拉氏变换为:-1-

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