Re: [問題] RLC混聯電路
※ 引述《consley (je ne sais quoi)》之銘言：
: ------^^^--------------------------------------------------
: | R1=5 | | | | |
: | | ) > | + )
: + c1=0.1F = L1=3H ) R2=5/3 > C2=0.1F = ) L2=1.5H
: _10V | ) > | Vc )
: | | | | | _ |
: | | | i3 | | i1 | | i4 | | | i2
: | V | V | V | | V |
: |----------------------------------------------------------
: DC電壓源再t=0的時候接上，所以我知道是step response。T<=0的時候
: 電感沒存有電荷。電感初始電流是i1(0)=1/3A，i2(0)=2/3A
: 求是哪種damping? i1(t)? Vc(t)? i3(t)? i4(t)?
: 如果用拉式變換會不會太複雜了?
: 感謝
C_tot = 0.2F
L_tot = 1H
R_p = R2 = 5/3
R_s = R1 = 5
分壓公式=> 落在C1, L1, R2, C2, L2上面的電壓如下
Vc = Vin*(R2||C_tot||L_tot)/(R1+ R2||C_tot||L_tot)
R2||C_tot||L_tot = 1/[1/R2 + sC_tot + 1/sL_tot]
= sR2*L/[s^2LCR2 + sL + R2]
= sL/[s^2LC + sL/R2 + 1]
= s/[s^2/5 + s*3/5 + 1]
= 5s/(s^2 + 3s + 5)
Vc = Vin*5s/[5s^2 + 20s + 25] = Vin*s/[s^2 + 4s + 5]
由laplace transform得
Vin(s) = 10/s
Vc(s) = 10/[(s+2)^2 + 1]
Vc(t) = 10*exp(-2t)sin(t) + Vo(0-) = 10exp(-2t)sin(t)#
under damping (因為還有震盪項)
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