[理學(xué)]7二階電路-wenkub
2023-02-03 14:35:57 本頁面
【正文】 ???????? ?? jj epep0201 , ?????)s i n ()s i n (00 ?????? ?? ???? ?? tKeteU ttteLUtuCi tC ?? ? s i ndd 0 ????)s i n (dd 00 ???? ? ???? ? teUtiLu tL??????s i nc o s00??)(2 )(12012120 21tjtjttptpCepepejUepepppUu????????????)s i n (00 ???? ? ?? ? teUu tcteLCUtuteLUtuCitCtC??????s i ndds i ndd 0 0???????))s i n (s i nc o s()c o s ())s i n (()c o s ()s i n (0000000000????????????????????????????????????????????teUteUteUteUdtduteUuttttctc??????s i nc o s00?????????????????????????????????ttttteULCeUeUKtKttKttKteUteU?????????????????????????00200000001s i n)s i n ())s i n ()c o s ()c o s ()( s i n ())s i n (c o s)c o s (( s i n))s i n (s i nc o s()c o s (??????s i nc o s00?? 2 LR??22022 )2(1 ??? ??? LRLC)s i n (00 ???? ? ?? ? teUu tcteLUi t ?? ? s i n 0 ??)s i n ( 00 ???? ? ??? ? teUu tLuL uC ?? 2?? uc t U0 teU ??? ?00teU ??? ??000 ? 2? i ? ?+? )s i n ( 00 ???? ? ?? ? teUu tCuc 是幅值按指數(shù)規(guī)律衰減的正弦函數(shù)。 設(shè)：方程的解為 代入上式得 ptc Aeu ?LCLCCRRCp24222,1???? LCLRLR 1)2(2 2 ????p有三種情況 二個(gè)不等負(fù)實(shí)根 2 CLR ?二個(gè)相等負(fù)實(shí)根 2 CLR ?二個(gè)共軛復(fù)根 2 CLR ?過阻尼 臨界阻尼 欠阻尼 LCLRLRpLCLRLRp1)2(21)2(22221????????tptpC eeupp212121AA ??? 時(shí)0210 AA)0( UUu C ?????0AA0)0(dd)0( 2211 ?????? ?? pptuCi C0121201221 AA UpppUppp?????)( 21 12120 tptpC epepppUu ???R L C + i uc uL + (t=0) uC(0)=U0 i(0)=0 )ee()pp(LU )ee()pp(ppCUdtduCitptp120tptp12210C2121??????????)epep(ppUu tp1tp2120C21 ???)epep()pp(UdtdiLu tp2tp1120L21 ?????p1p2=1/LC 2 . CLR ?一特征根為兩個(gè)不等的負(fù)實(shí)數(shù) LC1)L2R(L2RpLC1)L2R(L2Rp2221????????12212pp0p,0pL2RLC1)L2R(?????) ( 2 1 1 2 1 2 0 t p t p C e p e p p p U u ? ? ? )epep()pp(U)epep()pp(Uutp2tp2120tp1tp2120c1121????????2tm uL tm i t U0 uc ) ( 2 1 1 2 1 2 0 t p t p C e p e p p p U u ? ? ? ,0)(,)0( 0 ???? CC uUu0)( ?tu c 且 uc(t)單調(diào)下降 )0( ??? t)ee()pp(LU )ee()pp(ppCUdtduCitptp120tptp12210C2121??????????2tm uL tm i t U0 uc LCpp 121 ?)0(0)(,0)(,0)0( ???????? ttiii令 di / dt =0 , 求得 i 的極值點(diǎn) t=0+ , i=0 t = tm 時(shí) i 最大 t0 i0 2tm uL tm i t U0 uc R L C + i uc uL + (t=0) t=? , i=0 由 di/dt=0,即 uL= 0 可計(jì)算 tm 0epep tp2tp1 21 ??2112lnppppt m??tpptptpeeepp )(12 2121???2tm uL tm i t U0 uc R L C + i uc uL + (t=0) )epep()pp(UdtdiLu tp2tp1120L21 ?????2tm uL tm i t U0 uc R L C + i uc uL + (t=0) 0 t tm , i 增加 , uL0 t tm , i 減小 , uL 0 t=0+ ， uL=U0 t= ? ， uL=0 t = 2tm 時(shí) uL 極小 tpptptpeeepp )(212 2121)( ???由 duL / dt =0,可確定 uL為極小值的時(shí)間 t 021 2221 ?? tptp epepmtpp
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