133-拉普拉斯反變換的部分分式展開-資料下載頁
2025-07-25 14:18本頁面
【正文】 )!1(11111 psqqqq sFpsdsdqK ??????…… ( sp1 )qF(s)|s = p1 1)]()[( 1 psq sFpsdsd ??1)]()[(21122psq sFpsdsd??例：求 F(s)的原函數(shù) 32 )1(1)(?? sssF解： D(s)=0的根為 p1=1為三重根 p2=0為二重根 2212231121213)1()1(1)( sKsKsKsKsKsF ????????首先以 (s+1)3乘以 F(s)得 )(1)()1(1123 KssFs ???121???ssK11 = ( sp1 )3F(s)|s = p1 =1 122213121???ssdsdK14621???ss=3 121???ssdsd132????ss1)]()[( 3112 pssFpsdsdK ???=2 32 )1(1)(?? sssF32 )1(1)(?? sssF下面計算 K21和 K22，首先以 S2乘 F(s)得 )()1( 1)( 2132 KssFs ???3)1(10322 ?????ssdsdK1)1(10321 ????ssK23213)1(1)1(213)(ssssssF ????????所以 相應(yīng)的原函數(shù)為 f(t)= 3et +2tet + 3 +t 2212231121213)1()1(1)( sKsKsKsKsKsF ????????f(t)= tpeK 113 tpteK 112? tpetK 121121? ???nitpiieK
環(huán)評公示相關(guān)推薦
鄂ICP備17016276號-1