推 atled:感謝你 03/04 14:32
※ 引述《atled (喬巴)》之銘言：
: http://www-o.ntust.edu.tw/~lib/pdf/Master/94/m940101.pdf
: 第一題指定用拉式
: 可是我取完拉式
: 上下兩式相減之後就沒得算了
: 有人可以給我點提示嗎?
[3 1]X'' = [ 1 3]X + [e^t ]
[2 1] [ 1 2] [e^-t]
X'' = [ 0 1]X + [e^t - e^(-t)] = AX + B
[ 1 0] [-2e^t + 3e^(-t)]
(A-λI) = 0 , λ= 1,-1 , s = [1 1] , s^(-1) = 1/2[1 1]
[1 -1] [1 -1]
let X=sY
Y'' = (S^-1AS)Y +S^(-1)B = [ 1 0]Y + 1/2 [-e^t +2e^(-t)]
[ 0 -1] [3e^t - 4e^(-t)]
=> Y = [c1e^t + c2e^(-t)+1/2*(-te^t+2te^-t)]
[c3sint + c4cost +1/4*(3e^t -4e^-t)]
X = SY = ..................
x(0) = y(0) = 1 , x'(0)=y'(0)=0
c1+c2+c4 = 1
c1+c2-c4 = 1
c1-c2+c3-1/2 + 1 = 0
c1-c2-c3-1/2 + 1 = 0
c1=c3 = 0 c1=1/4 , c2 = 3/4
X = [1/4e^t +3/4e^-t + 1/2(-te^t+2te^(-t)) +1/4(3e^t-4e^(-t))]
[1/4e^t +3/4e^-t + 1/2(-te^t+2te^(-t)) -1/4(3e^t-4e^(-t))]
= [e^t -1/4e^(-t) + 1/2(-te^t+2te^(-t))]
[-1/2e^t + 7/4e^-t + 1/2(-te^t+2te^(-t))]
--------------------
再來轉拉式給他(X)
(3s^2+1)X(s) + (s^2+3)Y(s) = 1/(s-1) + 4s
(2s^2+1)X(s) + (s^2+2)Y(s) = 1/(s+1) + 3s
1 1/4 1/2 1
X(s) = _______ - _______ - _______ + ______
4(s-1) (s+1) (s-1)^2 (s+1)^2
-1/2 7/4 1/2 1
Y(s) = _______ + ______ - ______ + ______
s-1 s+1 (s-1)^2 (s+1)^2
-1
=> X(t) = L{X(s)} = .........
過程錯了就算了Orz
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