7电路分析第8章二阶

1、当uC 0时，i= i=+LCuC=电场能量全部转成磁场能量|uC |，|i当uC 0时，i= i=+LCuC=电场能量全部转成磁场能量|uC |，|i当uCU0 时i=i=+CLuC=LC=i+u (0)=i(0)=LCCLuLit0i= = LC=i+u (0)=i(0)=LCCLuLit0i= = CC=i=su =LC0ti = uCuCLC电路的正弦振i+i =CLCu =u =LiL(0)=LLC电路的正弦振i+i =CLCu =u =LiL(0)=LCi设 uC(0)t0RLC电i=sRLC0tuC7-RLC串联电路的零输入i=C =L uLiLRu+ u =+LRCS-C L+

2、Ri+u =7-RLC串联电路的零输入i=C =L uLiLRu+ u =+LRCS-C L+Ri+u =CS+RCduC +u =LC CSuC(0)=uS = i(0)= C+u =CCCC设解为uC(t)=+RCduC +u =LC 设解为uC(t)=+RCduC +u =LC CLCs2Kest+RCsKest +Kest=(LCs2 +RCs+1)Kest=LCs2 +RCs+1=特征方程的根（固有频率(RC)2-RC -=- s=()21取值不同，根号里的值有四种不同情况。(RC)2- -=- 2L()2s 特征方程的根（固有频率取(RC)2- -=- 2L()2s 特征方程的根（

3、固有频率取值不同，根号里的值有四种不同情况。1.( )2 s1,s2为两个不相等的负实数。即 R 2.( )2 3.( 2L)2 4.R=s1,s2为共轭虚1.( )2 s s12 =-( )2s =-+即 R1121s =- )2- ( =-1.( )2 s s12 =-( )2s =-+即 R1121s =- )2- ( =-22通解的形ttss+K-u (t)=K+K=K21212C112 C=2 t=012K1+K2 =-K -K uC(t)=K1e1t+K2e例已知图示电路中t 0R=L= 2LuS=iR+C= 1-Cu (0)=i(0)=4求uC(t)及C例已知图示电路中t 0R=

4、L= 2LuS=iR+C= 1-Cu (0)=i(0)=4求uC(t)及CL若以uC(t)为求解变=2.828阻尼电阻 Rs2+6s+8=+RC 36-2= -LC +u =sC18=-+u =24Cs =-s =-+8u =+12CuC(t)= K1e-2t + K2e-K1+K2=-2K1 uC(t)= K1e-2t + K2e-K1+K2=-2K1 -4K2 =K1=K2 =-u (0)=K +K =C12|=-2K 4K =C12iL(t)=iC(t)=uC(t)=6e-2t-4e-4t 6= -3e-i+ 4e-4210tt0例已知图示电路中t 0R=L= 2LuS=iR+C= 1-

5、Cu (0)=i(0)=4求uC(t)及例已知图示电路中t 0R=L= 2LuS=iR+C= 1-Cu (0)=i(0)=4求uC(t)及CL若以iL(t)uR+uL+uC=d2i +6 di +8i=Ri+Ltidt=+u (0)0CCs2+6s+8=s =-s =-LCd2i +RC +i=12iL(t)= K1e-2t + K2e-183 + 4 dt +i=例已知图示电路中t 0R=L= 2uS=LiRC=1+-u (0)=i(0)=C4CL求u (t)及i CL解iL(t)= K1e-2例已知图示电路中t 0R=L= 2uS=LiRC=1+-u (0)=i(0)=C4CL求u (t)

6、及i CL解iL(t)= K1e-2t + K2e-iL(0)=K1+K2 =+u (0)+Lu di-L=-2K -4K L12uL(0)=-31-2=-得K1 =-3-2K -4K =-K2=12iL(t)=-3e-2t-4e-4t |tt-=2+u (t)=u (0)2C00CC=2+4(3 e-2t-e-4t1)=6e-2t-4e-4t 22例已知图示电路中t 0R=L= 2uS=LiR+C=1 -Cu (0)=i(0)=4求uC(t)及例已知图示电路中t 0R=L= 2uS=LiR+C=1 -Cu (0)=i(0)=4求uC(t)及CL解: （3）=2.828阻尼电阻 RuC(t)=

7、 K1e-2t + K2e-=6e-2t-4e-4t =- ( )2- s=-9-i (t)=CL=-=-3e-2t +4e-s =-s =-12例已知图示电路中t 0R=L= 2LuS=iR+C=1 -Cu (0)=i(0)=例已知图示电路中t 0R=L= 2LuS=iR+C=1 -Cu (0)=i(0)=4求uC(t)及CLuC(t)=6e-2t-4e-4t iL(t)=iC(t)=C= -3e-i+ 4e-uC64210tt0s,ss s = R =-2.( )2 12即 RuC(t)= K1e-t + K2te-=(K1 s,ss s = R =-2.( )2 12即 RuC(t)=

8、K1e-t + K2te-=(K1 +K2t)e-K1=duC |(K +K=K K -=K tt21221i i +u K K LLK i (0)=i CCC212CLi +u u (t)=u (0)e-+tLCCCCi +u -=u (0)+LCCC无振荡衰减，临界阻RLC串联电路中t0iLuS=C=iL(0)=uC(0)=R+-L=R=C求u RLC串联电路中t0iLuS=C=iL(0)=uC(0)=R+-L=R=C求u C解=- ( )2- s=-1-1=-uC(t) =(K1 +K2t )e-K =1i |=K -K L=CC21t0K2=uC(t)=te-t 3.( )2 s =-

9、 -( =-R3.( )2 s =- -( =-Rd1 1 -( R s=-=-d2uC(t)=e-t K1cosd t+K2sind uC(0)=|=-e-t(Kcost+K sin1d2d+e-t(dK1sindt + =-K +K 1ui CK L+2duC(t) =e-tK1cosdt + sin tcos t=K2+K2e-tdduC(t) =e-tK1cosdt + sin tcos t=K2+K2e-tdd12K2+KK2+K12cos12K+2212sin2KK2+K112 uC(t)=K12 +K22 e-t coscosdt +t-+212d=Ke-t cos(dt=-ar

10、ctg K=K2+KK121uC(t)=Ke-t cos(dt用初始条件确定K和=-arctg uC(t)=Ke-t cos(dt用初始条件确定K和=-arctg K=K2+KK121=Ke-i K +L2ddtt0-Ke-R比较小，称为欠阻尼uC(t)是衰减振d 例RLC串联电路的零输入响应为3tcos(t)= C已知R=4，求L和CuC(t)=e-t 例RLC串联电路的零输入响应为3tcos(t)= C已知R=4，求L和CuC(t)=e-t K1cosd t+=1()2= sd1、=1( ) =23d解得：L=1H， CF74.R= =j =-js =-s =j0012特征根 s2uC(t

11、4.R= =j =-js =-s =j0012特征根 s2uC(t)=K1=t+ti |=K LCi LK 2uC(t)= Kcos(0t =-arctgK=K2+KK121uC(t)=K1cosO t+K2sinO sintcost=K2+KOO12KuC(t)=K1cosO t+K2sinO sintcost=K2+KOO12K2+KK2+K1212cos+22KK1222sin2K+22112 uC(t) =K12 +K22 coscosOt +=K12 +K22 cos(Ot -=K cos(Ot=-arctgK=K2+KK121 7-RLC串联电路的全响+RC duC +u =LC

12、L+ -uC(0)=RCS+ 7-RLC串联电路的全响+RC duC +u =LC L+ -uC(0)=RCS+-C|t=0=uC(t)=uch +设(t)=+=若为直流激励，则 Qtt-+Ku (t)=K+12s =-C12S11uch(t)=K1e-1t+K2e-s2 =-K1，K2求图示电路中i+已知uC(0) = iL(0)=-US=-+u =CS设ucp求图示电路中i+已知uC(0) = iL(0)=-US=-+u =CS设ucp(t)= Q=+=C12-tu (t)=Kt2C1s2+s+1=-11-2122=-js1,22u (0)=K +2=C1-uch(t)=e2 K1cos

13、2t+12i K +|=-K L2C12K =- 2 K =-312112i K +|=-K L2C12K =- 2 K =-3121t- 23sin3t+-t-u(t)=e232C1-t=-t-cos( )+22GCL并联电路的分7-iC+iG+iL =CdtC+i =CLSd 2L+L+i =LS-(GL)2GCL并联电路的分7-iC+iG+iL =CdtC+i =CLSd 2L+L+i =LS-(GL)2-i =s=SL +L+i =L -=-2C()2|t=0=iL(0)=LCs2 +GLs+1=G=dL-LGCLLCd2iL+i =LG=LSdL=- G ( )2- GCLLCd2i

14、L+i =LG=LSdL=- G ( )2- sRLC+u =CCR=CSdC=- ( )2- s例图示电路中，欲使电路产生临界阻尼响应，则C解CG=dL欲使电路产生临界阻尼响应，应满足故因：G2=L得例图示电路中，欲使电路产生临界阻尼响应，则C解CG=dL欲使电路产生临界阻尼响应，应满足故因：G2=L得C=0.5例：RLC并联电路的零输入响uc(t) = 100e-iR1+-CRL求R，L，C以及电感的初始电1wC(0)=例：RLC并联电路的零输入响uc(t) = 100e-iR1+-CRL求R，L，C以及电感的初始电1wC(0)=uC(0)=1 Cu 2(0 )= C=6.67230u 2

15、(0 +C+CuC(t)=e-t K1cosd t+d K1 K2 )2- =- G ( =-d)2- =- G ( =-=d= CRLC-G)2- =- G ( =-=d= CRLC-G=60026.6710-6 =80.0410-R=1 = 1 -=400G 1 =4002+dL=iL(0+)=-iR(0+)-=- uC(0+) -C duC Rd(100e-=- - 6.6710-=-0.8+0.4=-二阶电路分析方法总dX+a =Ad2X +a102dXX(0)=X二阶电路分析方法总dX+a =Ad2X +a102dXX(0)=X(t)=Xh(t)+Xp(tXh(t)=a0s2 +a1

16、s+a2 =-2s0=- ( )2- RLCs=- G ( )2- GCLss1,s2为两个不相等的负实s2 =-s1 =-Xh(t) = s1,s2为两个不相等的负实s2 =-s1 =-Xh(t) = K1e-1t+ K2e-s1,s2为两个相等的负实s1 =s2=-Xh(t)=(K1 +K2t)e-(3) s1,s2s1 =-s2=- Xh(t) =e-tK1cosdt + s1,s2为共轭虚s1s2 =-Xh(t)=K1cos0t +求设Xp(求设Xp(t) = Q QA如果是直流激励的渐近稳定电路X(t)=Xh(t)+用初始条件确定K1和例(1求固有频率s及uC(t)(2若并联C1=3/4F，求uC(t解R+CC1L(