自动控制原理及其应用 习题 答案 黄坚

《自动控制原理及其应用 习题 答案 黄坚》由会员上传分享，免费在线阅读，更多相关内容在教育资源-天天文库。

1、第二章习题课(2-1a)2-1(a)试建立图所示电路的动态微分方程。＋uc－i1=i2-ic解：CCicC＋uoRd1(ui-uo＋)＋i1R1＋uu1u=[]R-CR2dtuR1+uouiR2ioi2o－－2－－输入量为Ru=uRu-CRi，输出量为duiR+CRuduo。oRu+ui=uR1+uo2io1dt12dt12o2uoducd(ui-uo)u1=i1R1duoi2=Ric=Cduidt=dtuoR1+CRdt1R2+uoR22=R2ui+CRdt1R2第二章习题课(2-1b)2-1(b)试建立图所示电路的动态微分方程。ducCLd2uodu

2、oRLduoLic==+C2uL=1dtR2dtdt＋R2dt＋uoCCLd2uoduouiu=++CR2uo1iRRdt2dtoi2=Ru=u+u－22－2i1o输入量为ui，输出量为uo。u=iRducd(ui-uo)111ic=Cdt=dtdiLuu=Loi1=iL+icLdtiL=i2=R2习题课一(2-2)求下列函数的拉氏变换。(1)f(t)=sin4t+cos4tws解:∵L[sinwt]=L[coswt]=w2+s2w2+s24s∴L[sin4t+cos4t]=+s2+16s2+16s+4=s2+16(2)f(t)=t3+e4t3!13!1

3、解:L[t3+e4t]=+=+s3+1s-4s4s-4(3)f(t)=tneatn!解:L[tneat]=(s-a)n+1(4)f(t)=(t-1)2e2t2解:L[(t-1)2e2t]=e-(s-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=-1(s+1)(s+3)s=-2A2=(s+3)s+1=2(s+1)(s+3)s=-3F(s)=-21s+3s+2f(t)=2e-3t-e-2t2-3-2函数的拉氏变换。sF(s)=(s+1)2(s+2)解:f(t)=esst+limd[sest](s+1

4、)2s=-2s-1dss+2=-2e-2t+lim(estst+e2st)s-1s+2(s+2)2=-2e-2t-te-t+2e-t=(2-t)e-t-2e-2t2-3-3函数的拉氏变换。∴f(t)=1+cost-5sint2s2-5s+1F(s)=s(s2+1)解:F(s)(s2+1)s=+j=A1s+A2s=+jA1=1,A2=-5A3=F(s)s=s=011s-5F(s)=++ss2+1s2+12-3-4函数的拉氏变换。s+2213t(4)F(s)==+e-3t-e-t-e-ts(s+1)2(s+3)31242s+2s+2解:f(t)=est+es

5、t(s+1)2(s+3)s(s+1)2s=0s=-3d[s+2est]+lims(s+3)s-1ds21(-s2-4s-6)est(s+2)test=+e-3t+lim[+]312s-1(s2+3)2s2+3s(2-4-1)求下列微分方程。A2=(s+2)Y(s)s=-2A3=(s+3)Y(s)s=-3d2y(t)dy(t)+5+6y(t)=6，初始条件：dt2dty(0)=y(0)=2·。A1=1,A2=5,A3=-4∴y(t)=1+5e-2t-4e-3t解:s2Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)=′1s6+2s2+1

6、2s∴Y(s)=s(s2+5s+6)A1=sY(s)s=0(2-4-2)求下列微分方程。d3y(t)d2y(t)dy(t)+4+29=29,初始条件:dt3dt2dty(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。＋uc－解:ui=R1i1+uo，i2=ic+i1CCicduc＋R＋ic=Ci11udtuR2oii－－2UI(s)=R1I1(s)+UO(s)I2(s)=IC(s)+I1(s)IC(s)=CsUC(s)UI(s)-UO(s)即:=RI1(s)[UI(s)-UO(s)]Cs

7、=IC(s)1UI(s)UO(s)1()＋sCRR12-IC(s)sCUI(s)＋I2(s)UO(s)R2-1＋R1I1(s)1(sC)R＋2UO(s)R1R2+R1R2sCU(s)=1=I1+(sC)R＋2R1+R2+R1R2sCR12-5-b试画出题R1L2-1图所示的电路＋＋CRui2uo的动态结构图,并－－求传递函数。diL解：ui=R1I1+ucuc=uo+uLuL=Ldtuoduc∴iL=Ri1=iL+icic=Cdt2Ui(s)=R1I1(s)+UC(s)UC(s)=UO(s)+UL(s)UL(s)=sLIL(s)I1(s)=IL(s)+I

8、C(s)UO(s)IC(s)=CsUC(s)I2(s)=R2UI(s)+UC(S