推 linbanden :這張圖，讓我想到大一普物..... 03/26 17:10
以下是題目：
Find the steady-state current I(t) in the RLC-circuit in the Figure:
C
______││________
│ ││ │
│ │
│ │
< ξ
R > ξ L
< ξ
│ │
│ │
│ │
└───o o───┘
E(t)
R = 10 Ω L = 1 H C = 0.1 F
and with E(t) V as follows and periodic with period 2π.
Graph or sketch the first four partial sums.
Note that the coefficients of the solution decrease rapidly.
100( t - t^2 ) if -π < t < 0
問: E(t) = {
100( t + t^2 ) if 0 < t < π
1
Hint: LI" + RI' + ─── I = E'(t)
C
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題組內另外一題在同樣情況(R.L.C)下
-50 t^2 if -π < t < 0
E(t) = { 50 t^2 if 0 < t < π
的答案是:
I = 50 + A_1 cos t + B_1 sin t + A_3 cos (3t) + B_3 sin (3t) + ......
(10-n^2) a_n 10 n a_n
A_n = ────── B_n = ─────
D_n D_n
-400
a_n = ──── D_n = ( n^2 - 10 )^2 + 100n^2
π n^2
=======
我想問
1.a_n和D_n是啥?
4
前面的解說中只看到 C_n = √[(A_n)^2 + (B_n)^2] = ──────
πn^2 √(D_n)
為什麼要用這樣表示?
2. 將E'(t)轉換成傅利葉之後要怎麼再繼續解下去?
感覺卡住了......
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感謝版友們的幫助~
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