PPT《第二章习题课 （2-1a）》

第二章习题课 (2-1a) uo i2= R2 C + - + - ui uo R1 R2 2-1(a) 试建立图所示电路的动态微分方程.

解: 输入量为ui,输出量为uo. ui=u1+uo u1=i1R1 ic=C duc dt = dt d(ui-uo) i1=i2-ic u1=R1+uo uo -C d(ui-uo) dt R2 R2ui=uoR1-C R1R2+C R1R2+uoR2 dui dt dt duo uoR1+C R1R2+uoR2=R2ui+C R1R2 duo dui dt dt + - + - C uc + R1 R2 ui i1 i2 - + - uo ic C

第二章习题课 (2-1b) 2-1(b) 试建立图所示电路的动态微分方程. uo + + - - ui R1 L R2 C i1=iL+ic uL=L diL dt uo iL=i2= R2 uL= L R2 duo dt ic=C duc dt CL R2 d2uo dt2 duo dt + uo R2 CL R2 d2uo dt2 duo dt i1= +C uo i2= R2 输入量为ui,输出量为uo. ui=u1+uo u1=i1R1 ic=C duc dt = dt d(ui-uo) 习题课一 (2-2) 求下列函数的拉氏变换. (1) f(t)=sin4t+cos4t 解:∵L[sinwt] w w2+s2 s w2+s2 ∴L[sin4t+cos4t]

4 s2+16 s s2+16 = s+4 s2+16 + L[coswt]= (2) f(t)=t3+e4t 3! 解:L[t3+e4t] 3! s3+1

1 s-4 s4

1 s-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 2-3-1 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+1 s=-3 =

2 F(s)

2 s+3

1 s+2 f(t)=2e-3t-e-2t 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)2 s=-2 d ds s s+2 s -1 =-2e-2t+lim( est+est) s -1 st s+2

2 (s+2)2 =-2e-2t-te-t+2e-t =(2-t)e-t-2e-2t F(s)= 2s2-5s+1 s(s2+1) 2-3-3 函数的拉氏变换. 解:F(s)(s2+1) s=+j =A1s+A2 s=+j A1=1, A2=-5 A3=F(s)s =1 s=0 ∴ f(t)=1+cost-5sint F(s)

1 s s2+1 s -5 s2+1 2-3-4 函数的拉氏变换. (4) F(s)= s+2 s(s+1)2(s+3) 解:f(t)est est s+2 (s+1)2(s+3) s=0 s+2 s(s+1)2 s=-3 + lim s -1 d[est ] s+2 s(s+3) ds e-3t+lim[

2 3

1 12 s -1 (-s2-4s-6)est (s2+3)2 (s+2)test s2+3s e-3t- e-t- e-t

2 3

1 12

3 4 t

2 (2-4-1) 求下列微分方程. d2y(t) dt2 +5 +6y(t)=6 ,初始条件: dy(t) dt y(0)=y(0)=2 . ・ 解:s2Y(s)-sY(0)-Y(0)+5sY(s)-5Y(0)+6Y(s)=

1 s ′ A1=sY(s) s=0 ∴ y(t)=1+5e-2t-4e-3t A2=(s+2)Y(s) s=-2 A3=(s+3)Y(s) s=-3 A1=1 , A2=5 , A3=-4 ∴ Y(s)= 6+2s2+12s s(s2+5s+6) (2-4-2) 求下列微分方程. d3y(t) dt3 +4 +29 =29, d2y(t) dt2 dy(t) dt 初始条件: y(0)=0 , y(0)=17 , y(0)=-122 ・ ・ ・ 解: 2-5-a 试画题2-1图所示电路的动态结构图,并求传递函数. + - + - C uc + R1 R2 ui i1 i2 - + - uo ic C 解:ui=R1i1+uo ,i2=ic+i1 UI(s)=R1I1(s)+UO(s) duc ic=C dt I2(s)=IC(s)+I1(s) IC(s)=CsUC(s) 即:I1(s) UI(s)-UO(s) R1 [UI(s)-UO(s)]Cs=IC(s) UO(s) UI(s) =

1 R1 + sC)R2 1+

1 R1 + sC)R2 = R2+R1R2sC R1+R2+R1R2sC

1 R1 sC R2 UI(s) - UO(s) IC(s) I1(s) I2(s) + +

1 R1 sC R2 + UI(s) - UO(s) 2-5-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 + + - - ui R1 L R2 C 解:ui=R1I1+uc uc=uo+uL uL=L diL dt iL= uo R2 i1=iL+ic ic=C duc dt Ui(s)=R1I1(s)+UC(s) UC(s)=UO(s)+UL(s) UL(s)=sLIL(s) I1(s)=IL(s)+IC(s) ∴

1 R1 Cs sL R2 I1 + UO Ui IC - - UC=UO+UL IL UL I2(s)= UO(s) R2 IC(s)=CsUC(s) I1(s)= UO(s) R2 I1(s)= UI(s)+UC(S) R1 即: IL(s)=I1(s)-IC(s) IC(s)= UC(s) Cs 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 + SC R2 R2 +

1 UI R1 UO R3 SC R2 R2

1 SC ・ +

1 + =- R1+R3+R2R3CS = - R1(R2SC+1) R2 R3 R1(R2SC+1) R1 R1 R2 R3 ) (R2SC+1)

1 = R2

1 R3 R2 SC + + R1 - C(S)= UO(S) UI(S) ∴ - + + ∞ C R1 R2 R3 ui uo - + + ∞ C R1 R2 R3 ui uo - + + ∞ C R1 R2 R3 ui uo R4 R5 2-6-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. = - R5 R4+ R5 UO(R3SC+1) R2R3SC+R2+R3 UO UI = (R2R3SC+R2+R3)(R4+R5) R1(R3SC+1)R5 - = - (R4+R5)(R2+R3)SC+1) R2R3 R2+R3 R1R5(R3SC+1) UI R1 = R5 R4+ R5 UO R2 + R3 SC SC R3 SC +