8 15-6 The partial-fraction-expansion method N(S) aos s 0 …+a F(S) +∴+b 1.n>m D(s) bos"+b, (The roots of D(s) are unequal Assume Eg. D(=0, to be n simple-roots: P,p2,.,p 十 Multiply two sides by(s-P1 PI s-p (s-p1)F(S)=k1+(s-p1( P S-Pn PPI k1=(s-p1)F(s) In a similar manner we may obtain (S-PZ)F(S kn=[s-n)F(③ S f(t)=L{F(s)}=(k1e+k2+…+k,e"川u§15-6 The partial-fraction-expansion method n n n m m m b s b s b a s a s a D(s) N(s) F(s) + + + + + + = = − −   1 0 1 1 0 1 1.n>m (1)The roots of D(s) are unequal: Assume Eq. D(s)=0, to be n simple-roots: p1 ,p2 ,…,pn . n n s p k s p k s p k F(s) − + + − + − =  2 2 1 1 ) s p k s p k (s p )F(s) k (s p )( n n − + + − − = + −  2 2 1 1 1 Let s=p1   1 1 1 ( ) ( ) s p k s p F s = = − In a similar manner we may obtain   2 2 2 ( ) ( ) s p k s p F s = = −   n n n s p k (s p )F(s) = = − f (t ) L F(s) ( k e k e k e )u(t ) p t n p t p t n  = = + + + −1 1 1 2 2  Multiply two sides by (s-p1 )