2-1试建立如图 所示电路的动态微分方程。

(a) 输入ui 输出uo u1 C i2 i1=i-i2 u1=ui-uo ＋ + uo u1 ui-uo i1 R1 ui R2 i uo i= i1= R1 = R1 R2 - - du1 d(ui-uo) ui-uo uo d(ui-uo) i2=C dt =C - dt R1 = R2 C dt du du R2(ui-uo )=R1u0-CR1R2( dt i- dt o) duo du CR1R2 dt +R1uo+R2u0=CR1R2 dt i+R2ui

解：(b) ＋ ui 1 (ui-u1) i=i1+i2 i= Ri R1 u1 L C i1 Ri2 2 + uo - uo duui-u1 uo du1 i1= R2 i2=C dt 1 R1 = R2 +C dt L duo L duo u1-uo= u=u+ 1oR2 dt R2 dt ui uo L duo uo duo CL d2uo - - dt2 R1 R1 R1R2 dt = R2 +C dt + R2 2ui CL duo duo 1 1 L ) + R)uo 2 +(C+ = +( R1 R1R2 dtR1 R2 dt2

2-2 求下列函数的拉氏变换。 (1) f(t)=sin4t+cos4tsωL[sinωt]= L[cosωt]=2+s2解：22ωω+s4ss+4L[sin4t+cos4t]= s2+16+s2+16=s2+16(2) f(t)=t3+e4t解：3!+ 1= 6s+24+s4L[t3+e4t]= 4s4(s+4)ss-4-

(3) f(t)=tneatn!解：L[tneat]=(s-a)n+1(4) f(t)=(t-1)2e2t

解：L[(t-1)2e2t]=e-(s-2)23(s-2)

2-3求下列函数的拉氏反变换。

s+1(1) F(s)=(s+2)(s+3)s+1解：A1=(s+2)(s+2)(s+3)s=-2=-1s+1A2=(s+3)(s+2)(s+3)s=-3=22-1F(s)= s+3s+2f(t)=2e-3t-e-2t

2-5s+1As+A12A3(3) F(s)=2s= 22ss(s+1)s+1+ 解：F(s)(s2+1)s=+j=A1s+A2s=+j2s2-5s+1=As+A2 s s=j1s=j-5j-1=-A1+jA2 1s-5F(s)= s+s2+1+s2+1A1=1A2=-5A3=F(s)ss=0=1

f(t)?1?cost?5sint

d2y(t)dy(t)?(0)?2 ?5?6y(t)?6 初始条件：y(0)?y2-4 解下列微分方程 2dtdt6解：s2Y(s)-sy(0)-y'(0)+5sY(s)-5y(0)+6Y(s)= s6s2Y(s)-2s-2+5sY(s)-10+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

2-8设有一个初始条件为零的系统，当其输入端作用一个脉冲函数δ(t)时，它的输出响应c(t)如图所示。试求系统的传递函数。

解:δ(t)c(t)K0T

t

-TSKt-K(t-T)K(1-e )c(t)=TC(s)=TTs2C(s)=G(s)

2-9 若某系统在阶跃输入作用r(t)=1(t)时，系统在零初始条件下的输出响应为：

c(t)?1?e?2t?e?t，试求系统的传递函数。

解:(s2+4s+2)G(s)=C(s)/R(s)=(s+1)(s+2)(s2+4s+2)=1+2-1脉冲响应:C(s)=(s+1)(s+2)s+2s+1c(t)=δ(t)+2e-2t-e-t

2-10 已知系统的微分方程组的拉氏变换式，试画出系统的动态结构图并求传递函数

C(s)。 R(s)解:X1(s)=R(s)G1(s)-G1(s)[G7(s)-G8(s)]C(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)]-G1-X1(s)R(s)G6(s)X3(s)G2X2(s)G6-G3X3(s)G4C(s)C(s)[G7(s)-G8(s)]C(s)G5(s)-G5G7G8 R(s)-R(s)G1--G1G2G3G4C(s)C(s)G4GGG231+GGG+GG326 -345G7GG1G2G3G48-G6G7-G8G5C(s)R(s)=1+G3G2G6 +G3G4G5+G1G2G3G4(G7 -G8)R(s)-G1G2-G3G4C(s)1+G3G2G6G5G7-G8

2-11 已知控制系统结构图如图所示，试分别用结构图等效变换和梅逊公式求系统传递函数

C(s)。 R(s)解：(a)

(a)求系统的传递函数G1+G3R(s)R(s)G1(s)G3(s)__G3(s)+G1(s)G2G3(s)解:1+G2HG21=R(s)G+2H1+G1G2C(s)1+GH22G1(s)1+G1H2__G2(s)1+G2H1G2G1+G2HG1(s)C(s)3=R(s)1+G2H1H+G1G2H22(s)+G2(s)_G(s)C(s)2_H(s)H1(s)1H(s)G 1(s)H2(s)2G21+G2H1C(s)(b)求系统的传递函数R(s)_G3(s)G1(s)L2+L1G2(s)C(s)+G4(s)H(s)解:L1=-G1G2HL2=-G1G4HΔ=1+G1G4H+G1G2HP1=G1G2P2=G3G2Δ2=1+G1G4HΔ1 =1C(s)G1G2+G2G3+G1G2G3G4 HR(s)=1+G1G2H+G1G4H

R(s)(c)解: _G1G3G2C(s)R(s)_G1G2C(s)(d) 解: (1)R(s)R(s)G1G2+_R(s)C(s)C(s)_G1G2HC(s)+R(s)_++C(s)GG23G1+++H1H11+1-G3H1H1HG2L1H1+G2C(s)(G+G)1R(s)=12 1+G2H(2)L1=-G2HP1=G1P2=G2C(s)=G1G2(1–G3H1)R(s)1+G1G2+G1H1–G3H1C(s)(G+G)1R(s)=12 1+G2HΔ1 =1Δ2 =1

(f) (e)R(s)解: (1)_R(s)C(s)解: (1)_R(s)C(s)-G1L3L4G2G3G4C(s)+R(s)_G1G2C(s)G1G21-G2L1L2G1+G2G3-G4+(G1+G2)C(s)=R(s)1+(G1+G2)(G3-G4)G1G1(1–G2)=G1G21+G1G2–G21+1-G2(2)L1=-G1G3L2=G1G4L3=-G2G3(2)L1=-G1G2L2=G2P1=G1Δ1=1-G2L4=G2G4P1=G1Δ1=1P2=G2Δ2=1C(s)G1(1–G2)(G1+G2)C(s)Δ=1+G1G2-G2=R(s)=1+G1G2–G2R(s)1+G1G3+G2G3–G1G4-G2G4

L2_L1C(s)R(s)=2-12求图所示系统的传递函数 解：(a)

C(s)R(s)L1=G2H2L2=-G1G2H3R(s)__H1(s)G1(s)L2C(s)C(s)，。 R(s)D(s)D(s)+C(s)G2(s)+L1H2(s)H3(s)(b)求:C(s)C(s)R(s)D(s)R(s)Gn__G1H+G2+D(s)C(s)P1=G1G2Δ1=1G2G1C(s)=R(s)1-G2H2+G1G2H3C(s)L=GH122L2=-G1G2H3P1=G2Δ1=1D(s)P2=-G1G2H1C(s)G2(1-G1H1 )=D(s)1-G2H2+G1G2H3Δ2=1解:L1=-G1G2L2=-G1G2HP1=G1G2Δ1=1G1G2C(s)=R(s)1+G1G2H+G1G2P1=GnG2Δ1=1P2=1Δ2=1+G1G2HC(s)=1+GnG2+G1G2HD(s)1+G1G2+G1G2H

2-13求图所示系统的传递函数

C(s)E(s)，。 R(s)R(s)E(s)_G1L2_+L1G2G3C(s)(a)求:C(s)E(s)R(s)R(s)R(s)L1=-G2L2=-G1G2G3解:P1=G2G3Δ1=1P2=G1G2G3Δ2=1C(s)G2G3+G1G2G3R(s)=1+G2+G1G2G3P1=-G2G3Δ1=1P2=1Δ2=1+G2C(s)-G2G3+1+G2R(s)=1+G2+G1G2G32-14求图所示系统的传递函数

C(s)E(s)C(s)X(s)，,，。 R(s)R(s)D(s)E(s)