《自动控制原理》黄坚课后习题答案VIP专享

2-1试建立图所示电路的动态微分方程 +C＋--u iu oR1 R2i1ii2 +C＋--u iu oR1R2i1ii2Lu1 解：u1=ui-uoi2=Cdu1dti1=i-i2uoi=R2u1i1=R1= ui-uoR1dtd(ui-uo)=C(a)uCd(ui-uo)dtuo-R2=i-uoR1 i=i1+i2i2=Cdu1dtuoi1=R2u1-uo= LR2duodtR1i= (ui-u1)(b) 解：)-R2(ui-uo )=R1u0-CR1R2( duidtdtduoCR1R2duodtduidt+R1uo+R2u0=CR1R2+R2ui u=R1i-u1uo+CR2du1dtu1=uo+ LR2duodtuduodtR1R2L duodt+ CLR2d2uodt2=--iR1uoR1uoR2+C)uoR1R2Lduodt) CLR2d2uodt2=++(uiR11R11R2+(C+ 2-2 求下列函数的拉氏变换。(1) f(t)=sin4t+cos4tL[sinωt]= ωω2+s2= s+4s2+16L[sin4t+cos4t]= 4s2+16ss2+16+sω2+s2L[cosωt]=解： (2) f(t)=t3+e4t解：L[t3+e4t]= 3!s41s-4+ 6s+24+s4s4(s+4)= (3) f(t)=tneatL[tneat]=n!(s-a)n+1解： (4) f(t)=(t-1)2e2tL[(t-1)2e2t]=e-(s-2)2(s-2)3解： 2-3 求下列函数的拉氏反变换。A1=(s+2)s+1(s+2)(s+3) s=-2=-1=2f(t)=2e-3t-e-2t(1) F(s)=s+1(s+2)(s+3)解：A2=(s+3)s+1(s+2)(s+3) s=-3F(s)= 2s+31s+2-= A1s+2 s+3+ A2 (2) F(s)=s(s+1)2(s+2)f(t)=-2e-2t-te-t+2e-t解：= A2s+1 s+2+ A3+ A1(s+1)2A1=(s+1)2s(s+1)2(s+2) s=-1A3=(s+2)s(s+1)2(s+2) s=-2ddsss+2 ][A2= s=-1=-1=2=-2 (3) F(s)=2s2-5s+1s(s2+1)F(s)(s2+1)s=+j=A1s+A2 s=+jA2=-5A3=F(s)s s=0f(t)=1+cost-5sint解：= s+ A3s2+1A1s+A2=12s s 2-5s+1=A1s+A2 s=js=jj -2-5j+1=jA1+A2 -5j-1=-A1+jA2 A1=1F(s)= 1ss2+1s-5s2+1++ (4) F(s)=s+2s(s+1)2(s+3)解：=+ s+1A1s+3A2(s+1)2+ sA3 + A4-12A1= 23A3= 112A4= A2= d[s=-1ds](s+2)s(s+3) -34= -34A2= +- 43+f(t)=e-t32e-3t2-te-t121= s=-1 [s(s+3)]2[s(s+3)-(s+2)(2s+3)] (2-4) 求解下列微分方程。y(0)=y(0)=2 ·+6y(t)=6+5d2y(t)dt2dy(t)dt(1)解：s2Y(s)-sy(0)-y'(0)+5sY(s)-5y(0)+6Y(s)= 6s A1=1y(t)=1+5e-2t-4e-3tA2=5 A3=-4Y(s)= 6+2s2+12ss(s2+5s+6) = A1s+2s+3+ A3s + A2 2-5 试画题图所示电路的动态结构图，并求传递函数。 (1)ii2+-uruc+-R2R1ci1 解：I2(s)I1(s)+Uc(s)Ur(s)_Cs1R1+R2Uc(s)I(s) ( Ur(s)Uc(s)=1R11+(+sC)R21R1+sC)R2=R2+R1R2sCR1+R2+R1R2sC (2)+C＋--urucR1R2Lu1 I(s)Ur(s)_1R1U1(s)解：I1(s)-I2(s)L31CsU1(s) Uc(s)-1LsR2I1(s)Uc(s)L1L2 L1=-R2 /Ls L2=-/LCs2 L3=-1/sCR1Δ1=1L1L3=R2/LCR1s2P1=R2/LCR1s2...