Re: [理工] [工數]
看板Grad-ProbAsk作者ntust661 (Enstchuldigung~)時間13年前 (2010/09/16 21:30)推噓5(5推 0噓 17→)留言22則, 5人參與討論串8/12 (看更多)
※ 引述《SS327 (土豆人)》之銘言:
: http://tw.myblog.yahoo.com/jw!SHTR_hWBA0SAiAiBeiVw/photo?pid=1
: 上面那2張圖可以用左邊函數代表??為什麼阿??如果可以的話單位步階函數在空心點跟實
: 心點
: 有什麼差啊??
: http://ezproxy.lib.ncu.edu.tw:8080/~arhui/cexamn/exam/EC01_90_03.pdf
: 第4題r(t)的表示不是在正負無限大都有...那麼為什麼可以用拉式解??題目需要再加
: t>0嗎???
第四題 您也可以使用Fourier Transform
y'' + 2y' + 10y = r(t)
┌ 1 0 < t < π
│
r(t) = ─┤
│
└ -1 π < t < 2π
r(t) is a periodic function , and period = 2π
y(0) = c1
y'(0) = c2
假設 y 是 piecewise 並且至少符合 Laplace 存在性定理
L {y'' + 2y' + 10y} = L{r(t)}
2π -st
2 ∫0 r(t) e dt
s Y(s) - sy(0) - y'(0) + 2[sY(s) - y(0)] + 10Y(s) = ─────────
1 - e^(-2sπ)
2 1 π -st 2π -st
[s + 2s + 10 ]Y(s) + c1(- s - 2 ) - c2 = ────── [∫ e dt -∫ e dt]
1 - e^(-2sπ) 0 π
2 -1/s -πs -2πs -πs
[s + 2s + 10 ]Y(s) + c1(- s - 2 ) - c2 = ────── (e - 1 - e + e )
1 - e^(-2sπ)
2 1/s -πs 2
[s + 2s + 10 ]Y(s) + c1(- s - 2 ) - c2 = ────── ( 1 - e )
1 - e^(-2sπ)
-πs
2 (1 - e )
[s + 2s + 10 ]Y(s) + c1(- s - 2 ) - c2 = ────────
s ( 1 + e^(-sπ))
-πs
1 1 - e ∞ n -nπs
Y(s) = ─────── (c2 + c1(s+2) + ──── Σ (-1) e )
s^2 + 2s + 10 s n=0
-πs
c2 ( s + 1 ) +1 1 - e ∞ n -nπs
= ─────── + c1 ─────── + ─────── Σ (-1) e
(s+1)^2 + 9 (s + 1)^2 + 9 s((s+1)^2 + 9) n=0
C1 = c2 + 1
C2 = c1
-πs
-t -t -1 1 - e ∞ n -πns
y(t) = C1 sin(3t) e u(t) + C2 cos(3t) e u(t) + L{───────Σ(-1) e }
s((s+1)^2+9)) n=0
我們來觀看後面的級數
-πs
1 - e -πs -2πs -3πs
───────── ( 1 - e + e - e + ... )
s ((s+1)^2 + 9)
第一可拆,第二要積分= ="
-πs
先拆 1 與 e
(1)
1 -πs -2πs -3πs
──────── ( 1 - e + e - e + ... )
s((s+1)^2 + 9)
(2)
1 -πs -2πs -3πs -4πs
──────── ( - e + e - e + e - ....)
s((s+1)^2 + 9)
考慮(1)反轉換後
1 t ∞ n
── ∫ Σ (-1) sin3(t - nπ) u(t-nπ) dt
3 0 n=0
1 ∞ n t
── Σ (-1) ∫ sin3(t-nπ) dt
3 n=0 nπ
1 ∞ n 1
── Σ (-1) ── [cos(0) - cos3(t-nπ)]
3 n=0 3
1 ∞ n 1
── Σ (-1) ── [1 - cos3(t-nπ)] u(t - nπ)
3 n=0 3
1 ∞ n
── Σ (-1) [1 - cos3(t-nπ)] u(t - nπ)
9 n=0
考慮(2)反轉換
1 ∞ n 1
── Σ (-1) [1 - cos3(t-nπ) ] u(t - nπ) - ── ( 1 - cos3t ) u(t)
9 n=0 9
(1) + (2) = =
2 ∞ n 1
── Σ (-1) [ 1 - cos3(t-nπ) ] u(t - nπ) - ── ( 1 - cos3t )u(t)
9 n=0 9
so...
-t -t
y(t) = C1 sin(3t) e u(t) + C2 cos(3t) e u(t) + (1) + (2)
--
解完就飽啦XD
--
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