86-1 The source-free parallel circuit i(0)=I R C (0-)= Re∫ KCL:-+ odt -i(to)+C doo RLJto 十 +-b=0 r dt Let v= aest CAs2e”+ As,st e+— 0 R Ae(Cs+s/R+1/l=0
§6-1 The source-free parallel circuit 0 0 (0 ) (0 ) V i I = = − − 0 2 + + = st st st e L A e R As CAs e ( ) 0 1 : 0 0 + − + = dt d dt i t C R L KCL t t 0 1 1 2 2 + + = dt L d dt R d C st Let = Ae ( / 1/ ) 0 2 Ae Cs + s R+ L = st R i(t) L C Re f
Ae (Cs+S/R+1/L)=0 Let: CS+S/R+1/L=0 RVR-4L 1,2 2C 2RC 2RC LC or S= 十 2RC 2RC LC 2 2RC V 2RC LC St e = S?t 2 1922 two arbitrary constants SL S2----real numbers or conjugate complex Since the st and s,t must be dimensionless, S, S2----(1/s)
LC 1 ) 2RC 1 ( 2RC 1 s LC 1 ) 2RC 1 ( 2RC 1 or s 2 2 2 1 = − + − = − − − : / 1/ 0 2 Let Cs + s R + L = C RC RC LC L C R R s 1 ) 2 1 ( 2 1 2 ) 4 1 ( 1 2 2 1,2 = − − − − = s t s t A e A e 1 2 = 1 + 2 A1 , A2 ----two arbitrary constants s1, s2 ----real numbers or conjugate complex Since the s1 t and s2 t must be dimensionless, s1, s2 ---- (1/s). ( / 1/ ) 0 2 Ae Cs + s R+ L = st
Let 0 LC resonant(谱谐振) frequency a neper frequency or exponential damping coefficient 2RC 十 ZRO ZRO LC 、6 2RC LO -√-0 ZRO a>00 c-0 real-number c<00 imaginary-number 0,V c-0 zero
-- neper frequency or exponential damping coefficient. 2RC 1 = 2 0 2 2 2 2 0 2 2 1 1 ) 2 1 ( 2 1 1 ) 2 1 ( 2 1 = − − − = − − − = − + − = − + − RC RC LC s RC RC LC s zero imaginary number real number = − − − − − − − − − − − − − − 2 0 2 0 2 0 2 0 2 0 2 0 , , , LC 1 Let resonant( 0 = 谐振) frequency
1. The over-damped parallel rlc circuit a>00---( ZRO 2 0<C 00<-a+a <0 The S, and s, are negative real numbers, thus, the response can be expressed as the sum of two decreasing exponential terms, both of which approach zero as time increases without limit. The term containing S,(T2 <T has the more rapid rate of decrease
1. The over-damped parallel RLC circuit ) 1 2 1 ( 0 RC LC − − − The S1 and S2 are negative real numbers, thus, the response can be expressed as the sum of two decreasing exponential terms, both of which approach zero as time increases without limit. 2 1 s s The term containing s2 ( ) has the more rapid rate of decrease. 2 1 1 2 t 0 2 0 2 2 0 2 2 0 2 − − − − − + −
Example: Find u(t) 3.5 R i(t) 2RC 2×6× 42 607H 0o=√LC 7× Re∫ 42 D(0)=0 ∴> i(0)=10A c+√ 3.5+√3.52-6=-3.5+2.5=-1 2 3.5-25=-6 D(t)=Ae t+Ae-6t
Example: Find (t). i(0) 10A (0) 0 = = 6 42 1 7 1 1 0 = = = LC 3.5 2.5 6 2 0 2 s2 = − − − = − − = − 0 3 5 3 5 6 3 5 2 5 1 2 2 0 2 s1 = − + − = − . + . − = − . + . = − t t t A e A e 6 1 2 ( ) − − = + 3.5 42 1 2 6 1 2 1 = = = RC R i c i 6 i(t) 7H F 42 1 Re f
(t)=412e+A2 R i(t) U(0)=0→>0=41+42 6g71F do i(0)=10A一 e-642 dt Re f dudu d t dt t=0 Cs←(0)+i(0)i(0) 420 or-41-642=420∴41=84,A2=-84 N84e- U(t)=84(e--e" 84 d let 84(-e-+6e)=0 +6e 0→e 6 At time t the voltage becomes maximum 0.358s→m=48.9 5
t t A e A e dt d i A 6 10 1 6 2 (0) − − = → = − − 420 (0) (0) (0) (0) 0 = = + = = = = C i C i i C i dt t d dt d i C c R c t e e V t t ( ) 84( ) − −6 = − A1 = 84, A2 = −84 : 84( 6 ) 0 6 = − + = −t − t e e dt d let t e − 84 t e 6 84 − − t m t At time tm the voltage becomes maximum. t s V e e e m m t t t m m m 0.358 48.9 5 ln 6 6 0 6 6 5 = = → = − + = → = − − 0 0 1 2 (0) = → = A + A or − A1 − 6A2 = 420 R i c i 6 i(t) 7H F 42 1 Re f t t t A e A e 6 1 2 ( ) − − = +
R1 7H b 0238 0 50V 25V 5.0s 口V(R1:2) Time
V L1 7H 1 2 C 1 0.0238 R 1 6 0 Time 0s 1.0s 2.0s 3.0s 4.0s 5.0s 6.0s V(R1:2) 0V 25V 50V
2. Critical damping a=ooa-O=0 →R 2 =35√692) 2RC√LC i(t) 0 LC 35607H Ref =e (Ait+A2)=Ate vt me v6t U(0)=0→A2=0u()=41te dU=A1t(-、6e√6t+heV6r dv dt t=0 du do i2(0)i(0)+i(0)i(0 =420= dt t=0 C U(t)=420te √6t =422454
2. Critical damping 3.5 6 ) 2 1 1 2 1 , 0 ( 2 0 2 = 0 − = − − − = → = = C L R RC LC ( ) t A t A te 6 0 2 0 1 (0) − = → = = 6 6 1 1 2 0 = = − = − = = = s s LC ( ) e e e t t t t A t A A t A 6 2 6 1 2 1 ( ) − − − = + = + t te te V 6 t 2.45t ( ) 420 420 − − = = 1 6 1 6 1 0 ( 6) A dt t d A t e A e dt d t t = = = − + − − 420 1 (0) (0) (0) (0) 0 A C i C i i C i dt t d dt d i C c R c = = = + = = = = R i c i 3.5 6 i(t) 7H F 42 1 Re f
U(t)=420te t 420te45t 420t lim u(t)=lim =420im 0 t→0 t→0 2.45t t->∞2.45e 2.45t du(t) 0,420e0(-√6t+1)=0 dt rm=、6=04090m=4208=03 63.1
t t e t e V 6 t 2.45t ( ) 420 420 − − = = t . s . e . V . . m 0 408 m 420 0 408 63 1 6 1 2 4 5 0 408 = = → = = − m = 63.1V t 0 t m 0 2.45 1 420lim 420 lim ( ) lim 2.45 2.45 = = = → → → t t t t t e e t t 0, 420 ( 6 1) 0 ( ) 6 = − + = − e t dt d t t
C1 7H 0238 8.573 0 “… “““ 口V(L1:1)◇V(L1:1)v(L1:1)
V C 1 0.0238 0 L1 7H 1 2 V R 1 8.573 Time 0s 1.0s 2.0s 3.0s 4.0s 5.0s 6.0s V(L1:1) V(L1:1) V(L1:1) 0V 40V 80V
