内容发布更新时间 : 2024/5/21 19:33:55星期一 下面是文章的全部内容请认真阅读。
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]= =22s+16s+16s2+16
(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= -12A3= 32s+1A4= 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
(2-4)求解下列微分方程。2(t)dy(t)·(1)dy2y(0)=y(0)=2 +6y(t)=6+5dtdt62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)c(t)K0T解:δ(t)c(t)Kt
0Tt
-TSKt-K(t-T)K(1-e )c(t)=TC(s)=G(s)C(s)=2TTs
2-9 若系统在单位阶跃输入作用时，已知初始-t-2t条件为零的条件下系统的输出响应，求系统的r(t)=I(t)c(t)=1-e +e1R(s)=s传递函数和脉冲响应。 1-1+1=(s2+4s+2)解:C(s)=ss+2s+1s(s+1)(s+2)(s2+4s+2)G(s)=C(s)/R(s)=(s+1)(s+2)(s2+4s+2)C(s)=(s+1)(s+2)
c(t)=δ(t)+2e-2t-e-t
G6(s)X3(s)G2X2(s)2-10 已知系统的拉氏变换式，试画出系统的动态结构图并求传递函数。解:X1(s)=R(s)G1(s)-G1(s)[G7(s)-G8(s)]C(s)R(s)-G1-X1(s)G6-G3X3(s)G4C(s)={R(s)-C(s)[G7(s)-G8(s)]}G1(s)X2(s)=G2(s)[X1(s)-G6(s)X3(s)]X3(s)=G3(s)[X2(s)-C(s)G5(s)]C(s)=G4(s)X3(s)C(s)[G7(s)-G8(s)]C(s)G5(s)-G5G7G8
R(s)-G1G2G3G4C(s)1+G3G2G6G5G7-G8G1G2G3G4C(s)=R(s)1+G3G2G6 +G3G4G5+G1G2G3G4(G7 -G8)-
2-11求系统的传递函数
(a)R(s)_G3(s)G1(s)_+G2(s)H1(s)H2(s)C(s) 解:L1=-G2H1Δ=1+G2H1+G1G2H2L2=-G1G2H2P1=G1G2Δ1 =1P2=G3G2Δ2 =1nPkΔkG2G1+G2G3C(s)Σk=1==R(s)1+G2H1+G1G2H2Δ
(b)R(s)_G3(s)G1(s)+G2(s)C(s)+解:L1=-G1G2HL2=-G1G4HP1=G1G2Δ1 =1Δ=1+G1G4H+G1G2HP2=G3G2Δ2=1+G1G4HC(s)G1G2+G2G3+G1G2G3G4 H=R(s)1+G1G2H+G1G4H
G4(s)H(s)R(s)_G1G3++G2C(s)(c)R(s)_G1G3G2C(s)++H1 ++H1H1 C(s)=G1G2(1–G3H1)R(s)1+G1G2+G1H1–G3H1
(d) R(s)_G1G2HC(s)+解: L1=-G2H C(s)P1=G1Δ1 =1P2=G2Δ2 =1C(s)(G+G)1R(s)=12 1+GH2
(e)R(s)-G1G2G3G4+_L3=-G2G3解: L1=-G1G3L2=G1G4L4=G2G4P1=G1Δ1=1P2=G2Δ2=1(G1+G2)C(s)=R(s)1+G1G3+G2G3–G1G4-G2G4
(f) R(s)_G1G2C(s)L1L2+解: L1=-G1G2L2=G2P1=G1Δ1=1-G2G(1–G2)Δ=1+G1G2-G2C(s)=1R(s)1+G1G2–G2 D(s)+C(s)G2(s)+L1H2(s)H3(s)2-12 (a)R(s)__H1(s)G1(s)L2L1=G2H2L2=-G1G2H3P1=G1G2Δ1=1G2G1C(s)=R(s)1-G2H2+G1G2H3P1=G2Δ1=1P2=-G1G2H1Δ2=1G2(1-G1H1 )C(s)=D(s)1-G2H2+G1G2H3
(b)R(s)__G1HG2解:L1=-G1G2L2=-G1G2HP1=G1G2Δ1=1G1G2C(s)=R(s)1+G1G2H+G1G2C(s)P1=GnG2Δ1=1P2=1Δ2=1+G1G2HC(s)=1+GnG2+G1G2HD(s)1+G1G2+G1G2H
解:L1=-G2L2=-G1G2G3P1=G2G3Δ1=1GG+GGGP2=G1G2G3Δ2=1C(s)=23123R(s)1+G2+G1G2G3P1=-G2G3Δ1=1P2=1Δ2=1+G2C(s)-G2G3+1+G2=1+G+GGGR(s)2123 2-13 (a)R(s)E(s)_G1L2_+L1G2G3C(s)
(b)R(s)G1E(s)G2--G3G4+G5C(s)解:L1=-G3G4L2=-G2G3G5P1=G1G5Δ1=1P2=G2G3G5Δ2=1C(s)=G1G2G5+G1G5R(s)1+G2G3G5+G3G4P1=G1G5Δ1=1P2=1Δ2=1+G3G4E(s)G1G5+(1+G1G5 )R(s)=1+G2G3G5+G3G4