HH sY(S)-1+2sY(s)-3Y(S) s+1 解出Y(s) s2Y(s)-1+2s(s)-3Y(s)1 S+1 (s2+2s-3)Y(s) S+/1S+2 s+1 s+2 Y(S (S+1)s2+2s-3) 2士√4+12「1 s2+2s-3=0的解为s
6 由 解出Y(s) 1 1 ( ) 1 2 ( ) 3 ( ) 2 + - + - = s s Y s sY s Y s - = - + + - = = + + - + = + + + = + + - = + - + - = 3 1 2 2 4 12 2 3 0 ( 1)( 2 3) 2 ( ) 1 2 1 1 1 ( 2 3) ( ) 1 1 ( ) 1 2 ( ) 3 ( ) 2 2 2 2 s s s s s s s Y s s s s s s Y s s s Y s sY s Y s 的解为
即Y(s)有三个单极点为-1,1,3 s+2 s+2 Y(S) (s+1)s2+2s-3)s3+2s2-3s+s2+2s-3 s+2 A(S) s3+3s2-s-3B(s) B(s)=3s2+6s-1,因此 1+2 1+2 3+2 v(t) e+ e+ 3-6-1 3+6-127-18-1 3,1 e+=e
7 即Y(s)有三个单极点为-1,1,-3 t t t t t t y t 3 3 e 8 1 e 8 3 e 4 1 e 27 18 1 3 2 e 3 6 1 1 2 e 3 6 1 1 2 ( ) - - - - = - + - - - - + + + - + + - - - + = B'(s)=3s 2+6s-1, 因此 ( ) ( ) 3 3 2 2 3 2 3 2 ( 1)( 2 3) 2 ( ) 3 2 2 3 2 2 B s A s s s s s s s s s s s s s s s Y s = + - - + = + - + + - + = + + - + =
例2求解方程组 -x"+x'-y=e′-2 2y"-x"-2y+x=-t 满足初始条件 ∫y()=y(0)=0 x(0)=x(0)2=0 的解
8 例2 求解方程组 - - + = - - + - = - y x y x t y x x y t 2 2 e 2 = = = = (0) (0) 0 (0) (0) 0 x x y y 满足初始条件 的解
y”-x"+x'-y=e-2 2 "-2y+x=-t 对两个方程取拉氏变换,设[y(O刀Y(s [x()]=X(s),并考虑到初始条件,得 s Y(S)SX(S)+sX(s)-Y(S) 2s2Y(s)-S2X(S)-2.(S)+X(S)= 整理得(s+)y(s)-sX() s+2 S(S-1 2sY()-(S+1)X(s) SS
9 对两个方程取拉氏变换, 设L [y(t)]=Y(s), L [x(t)]=X(s), 并考虑到初始条件, 得 - - + = - - + - = - y x y x t y x x y t 2 2 e 2 - - + = - - - - + - = 2 2 2 2 2 1 2 ( ) ( ) 2 ( ) ( ) 2 1 1 ( ) ( ) ( ) ( ) s s Y s s X s sY s X s s s s Y s s X s sX s Y s 整理得 - - + = - - - + + - = ( 1) 1 2 ( ) ( 1) ( ) ( 1) 2 ( 1) ( ) ( ) 2 2 s s sY s s X s s s s s Y s sX s
解此线性方程组 s+2 (S+1)Y(s)-sX(s)= (S-1) 2sY(S)-(+1)X(s)= (S-1) s+1 S D -(S+1)2+2S 2 (s+1) 2s-1+2s2=s2-2s-1
10 解此线性方程组 2 1 2 2 1 ( 1) 2 2 ( 1) 1 ( 1) 1 2 ( ) ( 1) ( ) ( 1) 2 ( 1) ( ) ( ) 2 2 2 2 2 2 2 = - - - + = - - = - + + = - + + - = - - + = - - - + + - = s s s s s s s s s s s D s s sY s s X s s s s s Y s sX s