最新自动控制原理习题课答案专业知识讲座课件

习题课一(2-2)求下列函数的拉氏变换。(1)f(t)=sin4t+cos4t解:∵L[sinwt]=ww2+s2sw2+s2∴L[sin4t+cos4t]=4s2+16ss2+16=s+4s2+16+L[coswt]=习题课一(2-2)求下列函数的拉氏变换。(1)f((2)f(t)=t3+e4t3!解:L[t3+e4t]=+=+3!s3+11s-4s41s-4(3)f(t)=tneat解:L[tneat]=n!(s-a)n+1(4)f(t)=(t-1)2e2t解:L[(t-1)2e2t]=e-(s-2)2(s-2)3(2)f(t)=t3+e4t3!解:L[t3+e4t]=2-3-1函数的拉氏变换。F(s)=s+1(s+1)(s+3)解：A1=(s+2)s+1(s+1)(s+3)s=-2=-1(s+1)(s+3)A2=(s+3)s+1s=-3=2F(s)=-2s+31s+2f(t)=2e-3t-e-2t2-3-1函数的拉氏变换。F(s)=s+1(s+1)(F(s)=s(s+1)2(s+2)2-3-2函数的拉氏变换。解:f(t)=est+lim[est]s(s+1)2s=-2ddsss+2s-1

=-2e-2t+lim(est+est)s-1

sts+22(s+2)2=-2e-2t-te-t+2e-t=(2-t)e-t-2e-2tF(s)=s(s+1)2(s+2)2-3-2函数的拉氏F(s)=2s2-5s+1s(s2+1)2-3-3函数的拉氏变换。解:F(s)(s2+1)s=+j=A1s+A2s=+jA1=1,A2=-5A3=F(s)s=1s=0∴f(t)=1+cost-5sintF(s)=++1ss2+1s-5s2+1F(s)=2s2-5s+1s(s2+1)2-3-3函数2-3-4函数的拉氏变换。(4)F(s)=s+2s(s+1)2(s+3)解:f(t)=est+ests+2(s+1)2(s+3)s=0s+2s(s+1)2s=-3+lims-1d[est]s+2s(s+3)ds=+e-3t+lim[+]23112s-1(-s2-4s-6)est(s2+3)2(s+2)tests2+3s=+e-3t-e-t-e-t2311234t22-3-4函数的拉氏变换。(4)F(s)=s+2s(2-4-1)求下列微分方程。d2y(t)dt2+5+6y(t)=6，初始条件：dy(t)dty(0)=y(0)=2。·解:s2Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)=1s′

A1=sY(s)s=0∴y(t)=1+5e-2t-4e-3tA2=(s+2)Y(s)s=-2A3=(s+3)Y(s)s=-3A1=1,A2=5,A3=-4

∴Y(s)=6+2s2+12ss(s2+5s+6)(2-4-1)求下列微分方程。d2y(t)dt2+5(2-4-2)求下列微分方程。d3y(t)dt3+4+29=29,d2y(t)dt2dy(t)dt初始条件:y(0)=0,y(0)=17,y(0)=-122···解:(2-4-2)求下列微分方程。d3y(t)dt3+42-5-a试画题2-1图所示电路的动态结构图,并求传递函数。＋－＋－Cuc＋R1R2uii1i2－＋－uoicC解:ui=R1i1+uo，i2=ic+i1UI(s)=R1I1(s)+UO(s)ducic=CdtI2(s)=IC(s)+I1(s)IC(s)=CsUC(s)即:=I1(s)UI(s)-UO(s)R1[UI(s)-UO(s)]Cs=IC(s)2-5-a试画题2-1图所示电路的动态结构图,并求传递函UO(s)UI(s)=1R1＋(sC)R21+1R1＋(sC)R2=R2+R1R2sCR1+R2+R1R2sC1R1sCR2UI(s)-UO(s)IC(s)I1(s)I2(s)＋＋1R1sCR2＋()UI(s)-UO(s)UO(s)UI(s)=1R1＋(sC)R212-5-b试画出题2-1图所示的电路的动态结构图,并求传递函数。

uo＋＋－－uiR1LR2C解：ui=R1I1+ucuc=uo+uLuL=LdiLdtiL=uoR2i1=iL+icic=CducdtUi(s)=R1I1(s)+UC(s)UC(s)=UO(s)+UL(s)UL(s)=sLIL(s)I1(s)=IL(s)+IC(s)∴2-5-b试画出题2-1图所示的电路的动态结构图,并求1R1CssLR2I1＋UOUiIC--UC=UO+ULILULI2(s)=UO(s)R2IC(s)=CsUC(s)I1(s)=UO(s)R2I1(s)=UI(s)+UC(S)R1即：IL(s)=I1(s)-IC(s)IC(s)=UC(s)Cs1R1CssLR2I1＋UOUiIC--UC=UO+ULIL解:电路等效为:2-6-a用运算放大器组成的有源电网络如图所示,试采用复数阻抗法写出它们的传递函数。UO=－R3＋SCR2R2＋1UIR1UOR3SCR2R21SC·＋1＋=－解:电路等效为:2-6-a用运算放大器组成的有源电网络R1+R3+R2R3CS=－R1(R2SC+1)R2R3=－(+)R1(R2SC+1)R1R1R2=－(+R3)(R2SC+1)1=R21R3R2SC＋＋R1－C(S)=UO(S)UI(S)∴－＋＋∞△CR1R2R3uiuo－＋＋∞△CR1R2R3uiuoR1+R3+R2R3CS=－R1(R2SC+1)R2R3=－－＋＋∞△CR1R2R3uiuoR4R52-6-b用运算放大器组成的有源电网络如力所示,试采用复数阻抗法写出它们的传递函数。－＋＋∞△CR1R2R3uiuoR4R52-6-b用运算=－R5R4+R5UO(R3SC+1)R2R3SC+R2+R3UOUI=(R2R3SC+R2+R3)(R4+R5)R1(R3SC+1)R5－=－(R4+R5)(R2+R3)(SC+1)R2R3R2+R3R1R5(R3SC+1)UIR1=R5R4+R5UOR2＋R3SCSCR3SC＋1－R5R4+R5UOR2＋R3R3SC＋1－=－R5R4+R5UO(R3SC+1)R2R3SC+R2+c(t)t0TKδ(t)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)。c(t)t0TKδ(t)

c(t)=

KTt-(t-T)KT

C(s)=

K(1-e)Ts2-TSC(s)=G(S)第二章习题课(2-8)解:c(t)t0TKδ(t)2-8设有一个初始条件为零的系统2-9若系统在单位阶跃输入作用时，已知初始条件为零的条件下系统的输出响应，求系统的传递函数和脉冲响应。r(t)=I(t)c(t)=1-e+e-2t-t解:R(s)=1sG(S)=C(s)/R(s)1s+21s-C(s)=1s+1+=s(s+1)(s+2)(s2+4s+2)

=(s+1)(s+2)(s2+4s+2)

C(s)=(s+1)(s+2)(s2+4s+2)

脉冲响应:2s+2=1+1s+1-c(t)=δ(t)+2e+e-2t-t第二章习题课(2-９)2-9若系统在单位阶跃输入作用时，已知初始条件为零的条件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。解:X1(s)=R(s)G1(s)-G1(s)[G7(s)-G8(s)]C(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)G1G2G3G5---C(s)-R(s)G4G6G8G7X1(s)={R(s)-C(s)[G7(s)-G8(s)]}G1(s)C(s)[G7(s)-G8(s)]G6(s)X3(s)X1(s)X2(s)C(s)G5(s)X3(s)G1G2G3G5---C(s)-R(s)G4G2G6G8G7G1G2G5-C(s)-R(s)G7-G81+G3G2G6G3G4-C(s)R(s)G7-G81+G3G2G6+G3G4G5G1G2G3G41+G3G2G6+G3G4G5+G1G2G3G4(G7-G8)G1G2G3G4R(s)C(s)=第二章习题课(2-10)2-10已知系统的微分方程组的拉氏变换式，试画出系统的动解:2-11(a)G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)H2(s)G1(s)G2(s)H1(s)__+R(s)C(s)H2(s)G3(s)求系统的传递函数1+G2H1G2

G1+G31+G1H21+G2H1G2

1+G2H1G2

=1+G2H1+G1G2H2G2

R(s)C(s)=1+G2H1+G1G2H2G2G1+G2G3G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)G1(s)H2(s)第二章习题课(2-11a)解:2-11(a)G1(s)G2(s)G3(s)H1(s)_2-11(a)G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)H2(s)求系统的传递函数解:L1L1=-G2H1L2L2=-G1G2H1P1=G1G2P2=G3G2Δ1=1Δ2=1R(s)C(s)=Σnk=1PkΔkΔΔ=1+G2H1+G1G2H21+G2H1+G1G2H2G2G1+G2G3

=第二章习题课(2-11a)2-11(a)G1(s)G2(s)G3(s)H1(s)__+解:2-11(b)G1(s)G2(s)G3(s)G4(s)_++R(s)C(s)H(s)求系统的传递函数G1(s)G2(s)G3(s)G4H_+R(s)C(s)H(s)1+G4G1HG1

G2(s)G3(s)_+R(s)C(s)H(s)1+G4HG1G1

G2G3_+R(s)C(s)1+G4HG1G1

HG1

1+G4HG1G1+G3

(1+HG1G4)1+G4HG1G2

(1+HG1G4)1+G4G1H+G1G2HR(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4H第二章习题课(2-11b)解:2-11(b)G1(s)G2(s)G3(s)G4(s)_解:2-11(b)G1(s)G2(s)G3(s)G4(s)_++R(s)C(s)H(s)求系统的传递函数R(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4HL1L1=-G1G2HL1=-G1G4HL2P1=G1G2Δ1=1P2=G3G2Δ=1+G4G2H+G1G2HΔ2=1+G1G4H第二章习题课(2-11b)解:2-11(b)G1(s)G2(s)G3(s)G4(s)_H1_+++G1+C(s)R(s)G3G22-11c求系统的闭环传递函数。解:H1_+G1+C(s)R(s)G3G2H1R(s)C(s)1+G1G2+G1H1–G3H1G1G2

(1–

G3H1)=_G1C(s)R(s)G2H1+G21-G3H11第二章习题课(2-11c)H1_+++G1+C(s)R(s)G3G22-11c求系统H_G1+C(s)R(s)G22-11d求系统的闭环传递函数。解:(1)_G1+C(s)R(s)G2HG21+G2H1(G1+G2)R(s)C(s)=(2)L1L1=-G2HP1=G1Δ1=1P2=G2Δ2=1第二章习题课(2-11d)H_G1+C(s)R(s)G22-11d求系统的闭环传递函-_G1+C(s)R(s)G2G3G42-11e求系统的闭环传递函数。解:(1)

_C(s)R(s)G1+G2G3-G4C(s)=R(s)1+(G1+G2)(G3-G4)(G1+G2)1+G1G3+G2G3–G1G4-G2G4=(G1+G2)第二章习题课(2-11e)L1L2L3L4L2=G1G4L3=-G2G3L4=G2G4(2)L1=-G1G3P1=G1Δ1=1P2=G2Δ2=11+G1G3+G2G3–G1G4-G2G4=(G1+G2)C(s)R(s)-_G1+C(s)R(s)G2G3G42-11e求系统的闭_G1+C(s)R(s)G22-11f求系统的闭环传递函数。_C(s)R(s)G11-G2G2C(s)=R(s)1+1-G2G1G1G21+G1G2–G2G1

(1–

G2)=第二章习题课(2-11f)解:(1)

(2)L1L1=-G1G2L2L2=G2P1=G1Δ1=1-G2Δ=1+G1G2-G2C(s)R(s)1+G1G2–G2G1

(1–

G2)=_G1+C(s)R(s)G22-11f求系统的闭环传递函数2-12(a)R(s)G1(s)G2(s)H2(s)_+C(s)H3(s)H1(s)_+D(s)解:求:D(s)C(s)R(s)C(s)D(s)=01-G2H2G2

G(s)=1-G2H2G1G2

C(s)=R(s)1+1-G2H2G1G2H31-G2H2G1G2

1-G2H2+G1G2H3G2G1=R(s)=0结构图变换成：G2(s)H2(s)_+C(s)G1H3G1H1_D(s)1-G2H2G2

1-G1H1C(s)=D(s)1+1-G2H2G21-G2H2G2

G1H3(1-G1H1)1-G2H2+G1G2H3G2(1-G1H1)=第二章习题课(2-12a)2-12(a)R(s)G1(s)G2(s)H2(s)_+C(2-12(b)求:D(s)C(s)R(s)C(s)R(s)Gn++D(s)解:D(s)=0G(s)=1+G1G2HG1G2

G1G2H__C(s)C(s)=R(s)1+1+G1G2HG1G2

1+G1G2HG1G2

1+G1G2H+G1G2G1G2=R(s)=0Gn++D(s)结构图变换成：G1G2H--C(s)Gn++D(s)G1G2H--C(s)Gn/G1++D(s)1+G1G2HG1G2

-C(s)+D(s)1+G1G2HG2Gn

1+G1G2HG2G1系统的传递函数:)C(s)=D(s)1+1(1+1+G1G2HG1G2

1+G1G2HGnG2

1+G1G2+G1G2H=1+GnG2+G1G2H第二章习题课(2-12b)2-12(b)求:D(s)C(s)R(s)C(s)R(s)G2-13(a)求:R(s)E(s)R(s)C(s)C(s)E(s)G1G2G3__+R(s)解:L1L1=-G2L2L2=-G1G2G3P1=G2G3P2=G1G2G3R(s)C(s)=1+G2+G1G2G3G2G3+G1G2G3Δ1=1Δ2=1E(s)结构图变换成：G1G2+-E(s)G3-R(s)G1+-E(s)R(s)1+G2G3G2

-E(s)-R(s)1+G2G2G3G1G2G31+G2系统的传递函数:)E(s)=R(s)1+1(1-G1G2G31+G2

1+G2G2G31+G2+G1G2G3=1+G2-G2G3第二章习题课(2-13a)2-13(a)求:R(s)E(s)R(s)C(s)C(s)ER(s)G4(s)+C(s)E(s)G1(s)G2(s)-+G3(s)第二章习题课(2-14)D(s)X(s)2-14求:R(s)C(s)解:D(s)=0结构图变换为

R(s)G4(s)+C(s)G1(s)G2(s)-+G3(s)G3(s)(G1+G2)(G3+G4)1+(G1+G2)G3G1+G2C(s)R(s)=1+G3(G1+G2)(G1+G2)(G3+G4)R(s)G4(s)+C(s)E(s)G1(s)G2(s)-+D(s)第二章习题课(2-14)R(s)+-E(s)G3G2G1E(s)R(s)=1+G3(G1+G2)1R(s)G4(s)+C(s)E(s)G1(s)G2(s)-+G3(s)X(s)求:R(s)E(s)2-14解:D(s)=0结构图变换为

G3(G1+G2)D(s)第二章习题课(2-14)R(s)+-E(s)G3D(s)C(s)D(s)R(s)C(s)G4(s)+E(s)G1(s)G2(s)-+G3(s)X(s)求:2-14解:R(s)=0D(s)C(s)=1E(s)X(s)=G2(s)E(s)X(s)第二章习题课(2-14)D(s)C(s)D(s)R(s)C(s)G4(s)+E(s)C1(s)R1(s)第二章习题课(2-15)求:2-15+G1G2G3C1(s)R1(s)+-H2H1G4G5-G6C2(s)R2(s)解:结构图变换为

+G1G2G3C1(s)R1(s)-H2H1G4G5-1+G4G4G5H1H21+G4G4G5H1H21+G1G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G1G2G3(1+G4)=C1(s)R1(s)1+G4+G1G4G5H1H2G1G2(1+G4)=C1(s)R1(s)第二章习题课(2-15)求:2-15C2(s)R2(s)G3C1(s)R1(s)++G1G2-H2H1G4G5G6-C2(s)R2(s)求:2-14解:结构图变换为

+G4G5G6C2(s)R2(s)-H1H2G2G1-第二章习题课(2-15)1+G4G4G5G1H1H21-G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G4G5G6(1-G1G2)==1+1+G4G6G4G5G1H1H21-G2G21+G4G4G5C2(s)R2(s)C2(s)R2(s)G3C1(s)R1(s)++G1G2-HC2(s)G4G5G6-++G1G2-H2H1R1(s)G3C1(s)R2(s)求:2-14解:结构图变换为

G4G5G6-+-G1G2H2H1R1(s)C2(s)第二章习题课(2-15)C2(s)R1(s)1+G4H2G4G5G11-G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G1G4G5G6H2==1+G6H1C2(s)R1(s)G11-G1G21+G4H2G4G5G11-G1G21+G4H2G4G5C2(s)G4G5G6-++G1G2-H2H1R1(s)G3C2(s)G6R1(s)G4G5-++G1G2-H2H1G3C1(s)R2(s)求:2-14解:结构图变换为

第二章习题课(2-15)G1G2G3++G5G4-H1R2(s)C1(s)H2-C1(s)R2(s)G1G2G3++G5G4-H1R2(s)C1(s)H2-G21+G4G4G51-G1G2-G1H11+G4+G1G4G5H1H2-G1G2-G1G2G4-G1G2G3G4G5H1==1+1+G4G4G5G1H11-G2G21+G4G4G5C1(s)R2(s)H2-G1H11-G2G2G2G3C2(s)G6R1(s)G4G5-++G1G2-H2H1G3G3C1(s)R1(s)+G1G2+-H2H1G4G5-G6C2(s)R2(s)C1(s)R1(s)求2-15第二章习题课(2-15)解:L1=G1G2L3=-G4L2=-G1G4G5H1H2P1=G1G2G3Δ=1-G1G2+G1G4G35H1H2+G4-G1G2G4Δ1=1+G41+G4+G1G4G5H1H2-G1G2-G1G2G4G1G2G3(1+G4)=C1(s)R1(s)G3C1(s)R1(s)+G1G2+-H2H1G4G5-G6G3C1(s)R1(s)+G1G2+-H2H1G4G5-G6C2(s)R2(s)C2(s)R2(s)求2-15解:L1=G1G2L3=-G4L2=-G1G4G5H1H2P1=G4G5G6Δ=1-G1G2+G1G4G5H1H2+G4-G1G2G4Δ1=1-G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G4G5G6(1-G1G2)=C2(s)R2(s)第二章习题课(2-15)G3C1(s)R1(s)+G1G2+-H2H1G4G5-G6G3C1(s)R1(s)+G1G2+-H2H1G4G5-G6C2(s)R2(s)C1(s)R2(s)求2-15解:L1=G1G2L3=-G4L2=-G1G4G5H1H2Δ=1-G1G2+G1G4G5H1H2+G4-G1G2G4Δ1=1P1=-G1G2G3G4G5H11+G4+G1G4G5H1H2-G1G2-G1G2G4-G1G2G3G4G5H1=C1(s)R2(s)第二章习题课(2-15)G3C1(s)R1(s)+G1G2+-H2H1G4G5-G6G3C1(s)R1(s)+G1G2+-H2H1G4G5-G6C2(s)R2(s)C2(s)R1(s)求2-15解:L1=G1G2L3=-G4L2=-G1G4G5H1H2Δ=1-G1G2+G1G4G5H1H2+G4-G1G2G4Δ1=1P1=G1G4G5G6H21+G4+G1G4G5H1H2-G1G2-G1G2G4G1G4G5G6H2=C2(s)R1(s)第二章习题课(2-15)返回G3C1(s)R1(s)+G1G2+-H2H1G4G5-G6习题课一(2-2)求下列函数的拉氏变换。(1)f(t)=sin4t+cos4t解:∵L[sinwt]=ww2+s2sw2+s2∴L[sin4t+cos4t]=4s2+16ss2+16=s+4s2+16+L[coswt]=习题课一(2-2)求下列函数的拉氏变换。(1)f((2)f(t)=t3+e4t3!解:L[t3+e4t]=+=+3!s3+11s-4s41s-4(3)f(t)=tneat解:L[tneat]=n!(s-a)n+1(4)f(t)=(t-1)2e2t解:L[(t-1)2e2t]=e-(s-2)2(s-2)3(2)f(t)=t3+e4t3!解:L[t3+e4t]=2-3-1函数的拉氏变换。F(s)=s+1(s+1)(s+3)解：A1=(s+2)s+1(s+1)(s+3)s=-2=-1(s+1)(s+3)A2=(s+3)s+1s=-3=2F(s)=-2s+31s+2f(t)=2e-3t-e-2t2-3-1函数的拉氏变换。F(s)=s+1(s+1)(F(s)=s(s+1)2(s+2)2-3-2函数的拉氏变换。解:f(t)=est+lim[est]s(s+1)2s=-2ddsss+2s-1

=-2e-2t+lim(est+est)s-1

sts+22(s+2)2=-2e-2t-te-t+2e-t=(2-t)e-t-2e-2tF(s)=s(s+1)2(s+2)2-3-2函数的拉氏F(s)=2s2-5s+1s(s2+1)2-3-3函数的拉氏变换。解:F(s)(s2+1)s=+j=A1s+A2s=+jA1=1,A2=-5A3=F(s)s=1s=0∴f(t)=1+cost-5sintF(s)=++1ss2+1s-5s2+1F(s)=2s2-5s+1s(s2+1)2-3-3函数2-3-4函数的拉氏变换。(4)F(s)=s+2s(s+1)2(s+3)解:f(t)=est+ests+2(s+1)2(s+3)s=0s+2s(s+1)2s=-3+lims-1d[est]s+2s(s+3)ds=+e-3t+lim[+]23112s-1(-s2-4s-6)est(s2+3)2(s+2)tests2+3s=+e-3t-e-t-e-t2311234t22-3-4函数的拉氏变换。(4)F(s)=s+2s(2-4-1)求下列微分方程。d2y(t)dt2+5+6y(t)=6，初始条件：dy(t)dty(0)=y(0)=2。·解:s2Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)=1s′

A1=sY(s)s=0∴y(t)=1+5e-2t-4e-3tA2=(s+2)Y(s)s=-2A3=(s+3)Y(s)s=-3A1=1,A2=5,A3=-4

∴Y(s)=6+2s2+12ss(s2+5s+6)(2-4-1)求下列微分方程。d2y(t)dt2+5(2-4-2)求下列微分方程。d3y(t)dt3+4+29=29,d2y(t)dt2dy(t)dt初始条件:y(0)=0,y(0)=17,y(0)=-122···解:(2-4-2)求下列微分方程。d3y(t)dt3+42-5-a试画题2-1图所示电路的动态结构图,并求传递函数。＋－＋－Cuc＋R1R2uii1i2－＋－uoicC解:ui=R1i1+uo，i2=ic+i1UI(s)=R1I1(s)+UO(s)ducic=CdtI2(s)=IC(s)+I1(s)IC(s)=CsUC(s)即:=I1(s)UI(s)-UO(s)R1[UI(s)-UO(s)]Cs=IC(s)2-5-a试画题2-1图所示电路的动态结构图,并求传递函UO(s)UI(s)=1R1＋(sC)R21+1R1＋(sC)R2=R2+R1R2sCR1+R2+R1R2sC1R1sCR2UI(s)-UO(s)IC(s)I1(s)I2(s)＋＋1R1sCR2＋()UI(s)-UO(s)UO(s)UI(s)=1R1＋(sC)R212-5-b试画出题2-1图所示的电路的动态结构图,并求传递函数。

uo＋＋－－uiR1LR2C解：ui=R1I1+ucuc=uo+uLuL=LdiLdtiL=uoR2i1=iL+icic=CducdtUi(s)=R1I1(s)+UC(s)UC(s)=UO(s)+UL(s)UL(s)=sLIL(s)I1(s)=IL(s)+IC(s)∴2-5-b试画出题2-1图所示的电路的动态结构图,并求1R1CssLR2I1＋UOUiIC--UC=UO+ULILULI2(s)=UO(s)R2IC(s)=CsUC(s)I1(s)=UO(s)R2I1(s)=UI(s)+UC(S)R1即：IL(s)=I1(s)-IC(s)IC(s)=UC(s)Cs1R1CssLR2I1＋UOUiIC--UC=UO+ULIL解:电路等效为:2-6-a用运算放大器组成的有源电网络如图所示,试采用复数阻抗法写出它们的传递函数。UO=－R3＋SCR2R2＋1UIR1UOR3SCR2R21SC·＋1＋=－解:电路等效为:2-6-a用运算放大器组成的有源电网络R1+R3+R2R3CS=－R1(R2SC+1)R2R3=－(+)R1(R2SC+1)R1R1R2=－(+R3)(R2SC+1)1=R21R3R2SC＋＋R1－C(S)=UO(S)UI(S)∴－＋＋∞△CR1R2R3uiuo－＋＋∞△CR1R2R3uiuoR1+R3+R2R3CS=－R1(R2SC+1)R2R3=－－＋＋∞△CR1R2R3uiuoR4R52-6-b用运算放大器组成的有源电网络如力所示,试采用复数阻抗法写出它们的传递函数。－＋＋∞△CR1R2R3uiuoR4R52-6-b用运算=－R5R4+R5UO(R3SC+1)R2R3SC+R2+R3UOUI=(R2R3SC+R2+R3)(R4+R5)R1(R3SC+1)R5－=－(R4+R5)(R2+R3)(SC+1)R2R3R2+R3R1R5(R3SC+1)UIR1=R5R4+R5UOR2＋R3SCSCR3SC＋1－R5R4+R5UOR2＋R3R3SC＋1－=－R5R4+R5UO(R3SC+1)R2R3SC+R2+c(t)t0TKδ(t)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)。c(t)t0TKδ(t)

c(t)=

KTt-(t-T)KT

C(s)=

K(1-e)Ts2-TSC(s)=G(S)第二章习题课(2-8)解:c(t)t0TKδ(t)2-8设有一个初始条件为零的系统2-9若系统在单位阶跃输入作用时，已知初始条件为零的条件下系统的输出响应，求系统的传递函数和脉冲响应。r(t)=I(t)c(t)=1-e+e-2t-t解:R(s)=1sG(S)=C(s)/R(s)1s+21s-C(s)=1s+1+=s(s+1)(s+2)(s2+4s+2)

=(s+1)(s+2)(s2+4s+2)

C(s)=(s+1)(s+2)(s2+4s+2)

脉冲响应:2s+2=1+1s+1-c(t)=δ(t)+2e+e-2t-t第二章习题课(2-９)2-9若系统在单位阶跃输入作用时，已知初始条件为零的条件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。解:X1(s)=R(s)G1(s)-G1(s)[G7(s)-G8(s)]C(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)G1G2G3G5---C(s)-R(s)G4G6G8G7X1(s)={R(s)-C(s)[G7(s)-G8(s)]}G1(s)C(s)[G7(s)-G8(s)]G6(s)X3(s)X1(s)X2(s)C(s)G5(s)X3(s)G1G2G3G5---C(s)-R(s)G4G2G6G8G7G1G2G5-C(s)-R(s)G7-G81+G3G2G6G3G4-C(s)R(s)G7-G81+G3G2G6+G3G4G5G1G2G3G41+G3G2G6+G3G4G5+G1G2G3G4(G7-G8)G1G2G3G4R(s)C(s)=第二章习题课(2-10)2-10已知系统的微分方程组的拉氏变换式，试画出系统的动解:2-11(a)G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)H2(s)G1(s)G2(s)H1(s)__+R(s)C(s)H2(s)G3(s)求系统的传递函数1+G2H1G2

G1+G31+G1H21+G2H1G2

1+G2H1G2

=1+G2H1+G1G2H2G2

R(s)C(s)=1+G2H1+G1G2H2G2G1+G2G3G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)G1(s)H2(s)第二章习题课(2-11a)解:2-11(a)G1(s)G2(s)G3(s)H1(s)_2-11(a)G1(s)G2(s)G3(s)H1(s)__+R(s)C(s)H2(s)求系统的传递函数解:L1L1=-G2H1L2L2=-G1G2H1P1=G1G2P2=G3G2Δ1=1Δ2=1R(s)C(s)=Σnk=1PkΔkΔΔ=1+G2H1+G1G2H21+G2H1+G1G2H2G2G1+G2G3

=第二章习题课(2-11a)2-11(a)G1(s)G2(s)G3(s)H1(s)__+解:2-11(b)G1(s)G2(s)G3(s)G4(s)_++R(s)C(s)H(s)求系统的传递函数G1(s)G2(s)G3(s)G4H_+R(s)C(s)H(s)1+G4G1HG1

G2(s)G3(s)_+R(s)C(s)H(s)1+G4HG1G1

G2G3_+R(s)C(s)1+G4HG1G1

HG1

1+G4HG1G1+G3

(1+HG1G4)1+G4HG1G2

(1+HG1G4)1+G4G1H+G1G2HR(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4H第二章习题课(2-11b)解:2-11(b)G1(s)G2(s)G3(s)G4(s)_解:2-11(b)G1(s)G2(s)G3(s)G4(s)_++R(s)C(s)H(s)求系统的传递函数R(s)C(s)=1+G1G2H+G1G4HG1G2+G2G3+G1G2G3G4HL1L1=-G1G2HL1=-G1G4HL2P1=G1G2Δ1=1P2=G3G2Δ=1+G4G2H+G1G2HΔ2=1+G1G4H第二章习题课(2-11b)解:2-11(b)G1(s)G2(s)G3(s)G4(s)_H1_+++G1+C(s)R(s)G3G22-11c求系统的闭环传递函数。解:H1_+G1+C(s)R(s)G3G2H1R(s)C(s)1+G1G2+G1H1–G3H1G1G2

(1–

G3H1)=_G1C(s)R(s)G2H1+G21-G3H11第二章习题课(2-11c)H1_+++G1+C(s)R(s)G3G22-11c求系统H_G1+C(s)R(s)G22-11d求系统的闭环传递函数。解:(1)_G1+C(s)R(s)G2HG21+G2H1(G1+G2)R(s)C(s)=(2)L1L1=-G2HP1=G1Δ1=1P2=G2Δ2=1第二章习题课(2-11d)H_G1+C(s)R(s)G22-11d求系统的闭环传递函-_G1+C(s)R(s)G2G3G42-11e求系统的闭环传递函数。解:(1)

_C(s)R(s)G1+G2G3-G4C(s)=R(s)1+(G1+G2)(G3-G4)(G1+G2)1+G1G3+G2G3–G1G4-G2G4=(G1+G2)第二章习题课(2-11e)L1L2L3L4L2=G1G4L3=-G2G3L4=G2G4(2)L1=-G1G3P1=G1Δ1=1P2=G2Δ2=11+G1G3+G2G3–G1G4-G2G4=(G1+G2)C(s)R(s)-_G1+C(s)R(s)G2G3G42-11e求系统的闭_G1+C(s)R(s)G22-11f求系统的闭环传递函数。_C(s)R(s)G11-G2G2C(s)=R(s)1+1-G2G1G1G21+G1G2–G2G1

(1–

G2)=第二章习题课(2-11f)解:(1)

(2)L1L1=-G1G2L2L2=G2P1=G1Δ1=1-G2Δ=1+G1G2-G2C(s)R(s)1+G1G2–G2G1

(1–

G2)=_G1+C(s)R(s)G22-11f求系统的闭环传递函数2-12(a)R(s)G1(s)G2(s)H2(s)_+C(s)H3(s)H1(s)_+D(s)解:求:D(s)C(s)R(s)C(s)D(s)=01-G2H2G2

G(s)=1-G2H2G1G2

C(s)=R(s)1+1-G2H2G1G2H31-G2H2G1G2

1-G2H2+G1G2H3G2G1=R(s)=0结构图变换成：G2(s)H2(s)_+C(s)G1H3G1H1_D(s)1-G2H2G2

1-G1H1C(s)=D(s)1+1-G2H2G21-G2H2G2

G1H3(1-G1H1)1-G2H2+G1G2H3G2(1-G1H1)=第二章习题课(2-12a)2-12(a)R(s)G1(s)G2(s)H2(s)_+C(2-12(b)求:D(s)C(s)R(s)C(s)R(s)Gn++D(s)解:D(s)=0G(s)=1+G1G2HG1G2

G1G2H__C(s)C(s)=R(s)1+1+G1G2HG1G2

1+G1G2HG1G2

1+G1G2H+G1G2G1G2=R(s)=0Gn++D(s)结构图变换成：G1G2H--C(s)Gn++D(s)G1G2H--C(s)Gn/G1++D(s)1+G1G2HG1G2

-C(s)+D(s)1+G1G2HG2Gn

1+G1G2HG2G1系统的传递函数:)C(s)=D(s)1+1(1+1+G1G2HG1G2

1+G1G2HGnG2

1+G1G2+G1G2H=1+GnG2+G1G2H第二章习题课(2-12b)2-12(b)求:D(s)C(s)R(s)C(s)R(s)G2-13(a)求:R(s)E(s)R(s)C(s)C(s)E(s)G1G2G3__+R(s)解:L1L1=-G2L2L2=-G1G2G3P1=G2G3P2=G1G2G3R(s)C(s)=1+G2+G1G2G3G2G3+G1G2G3Δ1=1Δ2=1E(s)结构图变换成：G1G2+-E(s)G3-R(s)G1+-E(s)R(s)1+G2G3G2

-E(s)-R(s)1+G2G2G3G1G2G31+G2系统的传递函数:)E(s)=R(s)1+1(1-G1G2G31+G2

1+G2G2G31+G2+G1G2G3=1+G2-G2G3第二章习题课(2-13a)2-13(a)求:R(s)E(s)R(s)C(s)C(s)ER(s)G4(s)+C(s)E(s)G1(s)G2(s)-+G3(s)第二章习题课(2-14)D(s)X(s)2-14求:R(s)C(s)解:D(s)=0结构图变换为

R(s)G4(s)+C(s)G1(s)G2(s)-+G3(s)G3(s)(G1+G2)(G3+G4)1+(G1+G2)G3G1+G2C(s)R(s)=1+G3(G1+G2)(G1+G2)(G3+G4)R(s)G4(s)+C(s)E(s)G1(s)G2(s)-+D(s)第二章习题课(2-14)R(s)+-E(s)G3G2G1E(s)R(s)=1+G3(G1+G2)1R(s)G4(s)+C(s)E(s)G1(s)G2(s)-+G3(s)X(s)求:R(s)E(s)2-14解:D(s)=0结构图变换为

G3(G1+G2)D(s)第二章习题课(2-14)R(s)+-E(s)G3D(s)C(s)D(s)R(s)C(s)G4(s)+E(s)G1(s)G2(s)-+G3(s)X(s)求:2-14解:R(s)=0D(s)C(s)=1E(s)X(s)=G2(s)E(s)X(s)第二章习题课(2-14)D(s)C(s)D(s)R(s)C(s)G4(s)+E(s)C1(s)R1(s)第二章习题课(2-15)求:2-15+G1G2G3C1(s)R1(s)+-H2H1G4G5-G6C2(s)R2(s)解:结构图变换为

+G1G2G3C1(s)R1(s)-H2H1G4G5-1+G4G4G5H1H21+G4G4G5H1H21+G1G1G21+G4+G1G4G5H1H2-G1G2-G1G2G4G1G2G3(1+G4)=C1(s)R1(s)1+G4+G1G4G5H1H2G1G2(1+G4)=C1(s)R1(s)第二章习题课(2-15)求:2-15C2(s)R2(s)G3C1(s)