《自动控制原理》黄坚课后习题答案解析
2-1试建立图所示电路的动态微分方程Ci2
＋uii1R1R2-+iuo-
＋uiu1LiR1Ci1i2R2+uo-
-(a)解：i1=i-i2u1=ui-uouou1ui-uoi=Ri1=R=R211du1d(ui-uo)i2=Cdt=Cdt(b)解：(ui-u1)i=i1+i2i=Rudui1=Roi2=Cdt12Lduou1-uo=R2dt1ui-uouod(ui-uo)-C=R1R2dt
duduR2(ui-uo )=R1u0-CR1R2(dti-dto)duoduiCR1R2dt+R1uo+R2u0=CR1R2dt+R2uiduoCLd2uouiuoLduouo--R2dt2R1R1R1R2dt=R2+Cdt+ uiCLd2uoduo11L) +(C+=R1R2dt+(R1+R2)uoR1R2dt2
du1ui-u1uo+C=dtR1R2Lduou1=uo+Rdt2
2-2 求下列函数的拉氏变换。(1) f(t)=sin4t+cos4tωL[cosωt]=2s2解：L[sinωt]= 2ω+sω+s24ss+4+L[sin4t+cos4t]= =222s+16s+16s+16
《自动控制原理》黄坚课后习题答案解析
(2) f(t)=t3+e4t43!16s+24+s34t解：L[t+e]= s-4= s4(s+4)s4+ (3) f(t)=tneat解：L[tneat]=(s-a)n!n+1(4) f(t)=(t-1)2e2t解：L[(t-1)2e2t]=e-(s-2)(s-2)23
2-3求下列函数的拉氏反变换。(1) F(s)=(s+2)(s+3)s+1= As+21+ As+32解：A1=(s+2)(s+2)(s+3)s+1s=-2=-1A2=(s+3)s+1F(s)= s+32(s+2)(s+3)s=-3=2-s+21f(t)=2e-3t-e-2t(2) F(s)=AA(s+1)s2(s+2)= (s+1)12+ s+12+ As+23解：A1=(s+1)2(s+1)2s(s+2)s=-1=-1Ads2= ds[s+2]s=-1=2A3=(s+2)(s+1)2s(s+2)s=-2=-2f(t)=-2e-2t-te-t+2e-t(3) F(s)=2s2s(s-5s+1A2+1)= s1s+A2+12+ As3解：F(s)(s2+1)s=+j=A1s+A2s=+j2s2-5s+1s s=j=A1s+A2 s=j-2-5j+1j =jA1+A2 -5j-1=-A1+jA2 A1=1A2=-5A3=F(s)ss=0=1F(s)= s1+s2+1s+s2-5+1f(t)=1+cost-5sint
(4) F(s)=s+2解：=As(s+1)2(s+3)(s+1)12+A2+As3+s+3A4A1= -12A2s+13= 3A4= 121A2= -34Ad[2= s(s+3) (s+2)]dss=-1= [s(s+3)-(s+2)(2s+3)] [s(s+3)]2s=-1 = -3f(t)=2-t4e-t-43e-t+23+121e-3t
《自动控制原理》黄坚课后习题答案解析
(2-4)求解下列微分方程。2y(t)dy(t)d·(1)y(0)=y(0)=2 +6y(t)=6+52dtdt62解：sY(s)-sy(0)-y'(0)+5sY(s)-5y(0)+6Y(s)= sA1A2A36+2s2+12s= Y(s)=s(s2+5s+6)s+ s+2+ s+3A1=1A2=5 A3=-4y(t)=1+5e-2t-4e-3t
并求传递函数。2-5试画题图所示电路的动态结构图，(1)+
ci1i2R1解：+iR2Ur(s)CsI1(s)++I(s)ur-uc-
_R2Uc(s)Uc(s)R1I2(s)1
1+sC)R( 2R1R2+R1R2sCUr(s)==R+R+RRsCUc(s)11+(R+sC)R212121
(2)＋uru1R1CLR2+uc-
解：Ur(s)_-U1(s)L3R1L1I(s)c(s)I2(s)U1(s)UI(s)-1111RI1(s)-CsUc(s)2L2Ls
L1=-R2 /LsL2=-/LCs2L3=-1/sCR1L1L3=R2/LCR1s2R2P1=R2/LCR1s2Δ1=1Ur(s)=Uc(s)R1CLs2+(R1R2C+L)s+R1+R2
2-8 设有一个初始条件为零的系统，系统的输入、输出曲线如图，求G(s)。δ(t)K0Tc(t)解:δ(t)c(t)Kt
0Tt
-TSKt-K(t-T)K(1-e )c(t)=TC(s)=G(s)C(s)=2TTs
《自动控制原理》黄坚课后习题答案解析
