作者iyenn (晓风)
看板Grad-ProbAsk
标题Re: [理工] [工数]-拉式转换
时间Mon Feb 1 14:08:25 2010
※ 引述《mdpming (阿阿 要加油)》之铭言:
: 1.
: 3
: -1 s
: L [ ---------- ]
: 4 4
: s + 4a
3
-1 s
L [ ---------- ]
4 4
s + 4a
1 a+i∞ s^3 e^st
=------∫ -----------ds
2πi a-i∞ s^4+4a^4
z^3e^zt
F(z)=--------
z^4+4a^4
∮F(z)dz=∫F(z)dz+∫F(z)dz+∫F(z)dz
c L1+L2 c' L
c:L1+L2+c'+L
=0+∫F(z)dz
L
=2πiRezF(z)
pole:z^4=-4a^4
z=√2ae^(j(π+2nπ)/4) n=0,1,2,3
=a(1+i),a(-1+i),a(-1-i),a(1-i)
=bi,i=1,2,3,4
z^3e^zt
->Rez(Fz,bi)=lim (z-bi)---------
z->bi z^4+4a^4
e^zt
=lim ---------
z->bi 4
1
RezF(z)=---(e^(a(1+j)t+e^(a(-1+j)t+e^(a(-1-j)t+e^(a(1-j)t)
4
1
=---(e^at+e^-at)(e^jat+e^-jt)
4
=coshatcosat
1 a+i∞ s^3 e^st
∴------∫ -----------ds
2πi a-i∞ s^4+4a^4
=coshatcosat
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※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 123.193.214.165
※ 编辑: iyenn 来自: 123.193.214.165 (02/01 14:27)
1F:推 bjackal:之前这题用因式分解的 强 用留数的....更猛= = 02/01 15:35
2F:→ bjackal:中间POLE bi那边怎麽弄得阿 有点看不懂 02/01 15:37
3F:→ mdpming:谢谢i大...有比较正统的解法吗...................... 02/01 17:06
4F:推 CRAZYAWIND:这很正统阿 这是最正统的解法了= = 02/01 17:10
5F:推 mdpming:..=.=.. 02/01 17:23
6F:→ iyenn:应该只有这两种方法,选一个吧= = 02/01 20:36
7F:推 CRAZYAWIND:越看这个方法越看越喜欢= = 02/01 21:04