0.268s+33 Example 2: F(s) ,fnd∫(t +50s+10 Solution: N(S=0.268S+33 and D(=s+50s+10 from D(s=0, we have P1=-25+j315 P2=-25-j315 N(S) K 0.268(-25+j315)+33 D(s)s=-25+j3152(-25+j315)+50 26.3+184.4 =0.14∠-17.3° j630 f()=0.28c-25cos(315t-1,3°)Example 2: , ( ). 50 10 0.268 33 ( ) 2 5 find f t s s s If F s + + + = N(s)=0.268s+33 and D(s)=s 2+50s+105 from D(s)=0, we have    = − − = − + 25 315 25 315 2 1 p j p j  0.14 17.3 630 26.3 84.4 =  − + = j j ( ) 25 315 ( ) 1 ' D s s j N s k = − + = 2( 25 315) 50 0.268( 25 315) 33 − + + − + + = j j ( ) 0.28 cos(315 17.3 ) 25   = − − f t e t t Solution: