自动控制原理第二章习题课答案教学文案.ppt

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1、第二章习题课　(2-1a)uoi2=R2C＋－＋－uiuoR1R22-1(a)试建立图所示电路的动态微分方程。解：输入量为ui，输出量为uo。ui=u1+uou1=i1R1ic=Cducdt=dtd(ui-uo)i1=i2-icu1=[]R1+uouo-Cd(ui-uo)dtR2R2ui=uoR1-CR1R2+CR1R2+uoR2duidtdtduouoR1+CR1R2+uoR2=R2ui+CR1R2duoduidtdt＋－＋－Cuc＋R1R2uii1i2－＋－uoicC第二章习题课　(2-1b)2-1(b)试建

2、立图所示电路的动态微分方程。uo＋＋－－uiR1LR2Ci1=iL+icuL=LdiLdtuoiL=i2=R2uL=LR2duodtic==+CducdtCLR2d2uodt2duodt+uoR2CLR2d2uodt2duodti1=+Cuoi2=R2输入量为ui，输出量为uo。ui=u1+uou1=i1R1ic=Cducdt=dtd(ui-uo)(2)f(t)=t3+e4t3!解:L[t3+e4t]=+=+3!s3+11s-4s41s-4(3)f(t)=tneat解:L[tneat]=n!(s-a)n+1(4)

3、f(t)=(t-1)2e2t解:L[(t-1)2e2t]=e-(s-2)2(s-2)32-3-1函数的拉氏变换。F(s)=s+1(s+1)(s+3)解：A1=(s+2)s+1(s+1)(s+3)s=-2=-1(s+1)(s+3)A2=(s+3)s+1s=-3=2F(s)=-2s+31s+2f(t)=2e-3t-e-2tF(s)=s(s+1)2(s+2)2-3-2函数的拉氏变换。解:f(t)=est+lim[est]s(s+1)2s=-2ddsss+2s-1=-2e-2t+lim(est+est)s-1sts+22

4、(s+2)2=-2e-2t-te-t+2e-t=(2-t)e-t-2e-2tF(s)=2s2-5s+1s(s2+1)2-3-3函数的拉氏变换。解:F(s)(s2+1)s=+j=A1s+A2s=+jA1=1,A2=-5A3=F(s)s=1s=0∴f(t)=1+cost-5sintF(s)=++1ss2+1s-5s2+12-3-4函数的拉氏变换。(4)F(s)=s+2s(s+1)2(s+3)解:f(t)=est+ests+2(s+1)2(s+3)s=0s+2s(s+1)2s=-3+lims-1d[est]s+2s(s

5、+3)ds=+e-3t+lim[+]23112s-1(-s2-4s-6)est(s2+3)2(s+2)tests2+3s=+e-3t-e-t-e-t2311234t2(2-4-1)求下列微分方程。d2y(t)dt2+5+6y(t)=6，初始条件：dy(t)dty(0)=y(0)=2。·解:s2Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)=1s′A1=sY(s)s=0∴y(t)=1+5e-2t-4e-3tA2=(s+2)Y(s)s=-2A3=(s+3)Y(s)s=-3A1=1,A2=5,A3

6、=-4∴Y(s)=6+2s2+12ss(s2+5s+6)(2-4-2)求下列微分方程。d3y(t)dt3+4+29=29,d2y(t)dt2dy(t)dt初始条件:y(0)=0,y(0)=17,y(0)=-122···解:2-5-a试画题2-1图所示电路的动态结构图,并求传递函数。＋－＋－Cuc＋R1R2uii1i2－＋－uoicC解:ui=R1i1+uo，i2=ic+i1UI(s)=R1I1(s)+UO(s)ducic=CdtI2(s)=IC(s)+I1(s)IC(s)=CsUC(s)即:=I1(s)UI(s)

7、-UO(s)R1[UI(s)-UO(s)]Cs=IC(s)UO(s)UI(s)=1R1＋(sC)R21+1R1＋(sC)R2=R2+R1R2sCR1+R2+R1R2sC1R1sCR2UI(s)-UO(s)IC(s)I1(s)I2(s)＋＋1R1sCR2＋()UI(s)-UO(s)2-5-b试画出题2-1图所示的电路的动态结构图,并求传递函数。uo＋＋－－uiR1LR2C解：ui=R1I1+ucuc=uo+uLuL=LdiLdtiL=uoR2i1=iL+icic=CducdtUi(s)=R1I1(s)+UC(s)U

8、C(s)=UO(s)+UL(s)UL(s)=sLIL(s)I1(s)=IL(s)+IC(s)∴1R1CssLR2I1＋UOUiIC--UC=UO+ULILULI2(s)=UO(s)R2IC(s)=CsUC(s)I1(s)=UO(s)R2I1(s)=UI(s)+UC(S)R1即：IL(s)=I1(s)-IC(s)IC(s)=UC(s)Cs解:电路等效为:2-6-a用运算放大