二阶电路课件.ppt

§7—1二階電路的零輸入回應RC+–uciLS(t=0)uRLuL已知:uC(0–)=U0,iL(0–)=0,t=0時將開關S閉合，求t≥0時uC(t)、iL(t)、uL(t)。解：由KVL,t0時:–uC+uR+uL=0uR=iLR=–RCduCdtuL=L=–LCd2uCdt2diLdtLC+RC+uC=0d2uCdt2duCdt設:uC=Aept代入上式LCAP2ept+RCAPept+Aept=0LCP2+RCP+1=0特徵方程特徵根LCP2+RCP+1=0P=–RC±(RC)2–4LC2LC=–±2LR2LR()2–LC12LR2LR()2–LC1P1=–+P2=––2LR2LR()2–LC1可以看出:P1、P2只與組成電路的元件的參數有關。當電路中的R、L、C確定之後，便可求出P1、P2。uC(t)=A1ept+A2ept12由初始條件uC(t)=A1ept+A2ept12可解出:iL(0+)=iL(0–)=–C=0dtduCt=0A1+A2=U0A1P1+A2P2=0A1=P2–P1P2U0A2=–P2–P1P1U0uC(0–)=uC(0+)=U0討論：2LR2LR()2–LC1P1=–+P2=––2LR2LR()2–LC1A1=P2–P1P2U0A2=–P2–P1P1U0uC(t)=A1ept+A2ept12uC(t)=(P2ep1t–P1ep2t)P2–P1U0P1、P2均小於0，|P2||P1|；A1ep1t比A2ep2t放電慢。uCA1A2tuC1、0,R2時，P1、P2為不相等的兩個實數根。2LR()2–LC1CL無震盪衰減電感的電壓、電流uC(t)=(P2ep1t–P1ep2t)P2–P1U0iL=–CduCdt=–(ep1t–ep2t)L(P2–P1)U0uL=LdiLdt=–(P1ep1t–P2ep2t)(P2–P1)U0iLuLttmt10ttmCRLuCuLiLt?uC?uL??uR=uC–uL?i?t=tmuL=0,uR=uC,iL=iLmaxttmuL0,uC?i?能量轉換關係:0ttmCRLWCWLCRLttmWLWCWCWRWR2、P=–±2LR2LR()2–LC1令:?=?2=–()22LRLC12LR則:P1=–?+–?2=–?+j?P2=–?–j?uC(t)=(P2ep1t–P1ep2t)P2–P1U0=[(–?–j?)e(–?+j?)t–(–?+j?)e(–?–j?)t](–?–j?)–(–?+j?)U0=e–?t[–?(ej?t–e–j?t)–j?(ej?t+e–j?t)]j2?–U0=U0e–?t(??ej?t–e–j?tj2+2ej?t+e–j?t)=U0e–?t(sin?t+cos?t)??2LR()2–LC1CL0,R2時，P1、P2為一對共軛複根令：?=?2=–()22LRLC12LRuc(t)=U0e–?t(sin?t+cos?t)??LC1?0=uc(t)=U0e–?t(sin?t+cos?t)??0?0??0?=?0–?22????0??0?=cos??0?=s