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 )
Example I 4s+5 If F(s) ,Find∫(t) s2+5s+6 Solution: N(s)=45+5 and D()=s2+55+6=(5+2)(5+3) e have p 4s+5 4s+5 k1=(s+2) 3 (+2)+3)s=-2L(s+3)s=-2 4s+5 4s+5 k2=(s+3) 7 (s+2s+3)s=-3L(+2)s=-3 f()=(-3e2+7e1)u()
Example 1: , ( ). 5 6 4 5 ( ) 2 Find f t s s s If F s + + + = Solution: N(s)=4s+5 and D(s)=s2+5s+6=(s+2)(s+3) We have p1= -2,p2= -3 3 ( 3) 2 4 5 ( 2)( 3) 2 4 5 ( 2) 1 = − = − + + = = − + + + = + s s s s s s s k s 7 ( 2) 3 4 5 ( 2)( 3) 3 4 5 ( 3) 2 = = − + + = = − + + + = + s s s s s s s k s ( ) ( 3 7 ) ( ) 2 3 f t e e u t − t − t = − +
By means of Nopital's rule may determine k, that is ki=lim (S-p;)N(s) =lin(s-Pi)N()+N(s)=im D(s)D(pi) N(s) N(Pi) s→Pi D(S) D'(S) s→ i=1,2, f(t)=[F(s)=(k,ePit+k2 ePat+.+knep,n )u(t) 4s+5 :F()=-2 2 s2+5s+6s+2s+3 4s+5 4s+5 K -3k 7 2s+5s 2 2s+5J=-3 ∴f()=(-3e-+7e)u(t)
By means of Nopital' s rule may determine k, that is i , ,...n D'( p ) N( p ) D (s ) N(s) lim D'(s ) (s p )N'(s) N(s ) lim D(s ) (s p )N(s) k lim i i ' s p i s p i s p i i i i = 1 2 = = − + = − = → → → 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 5 6 2 3 4 5 : ( ) 1 2 2 + + + = + + + = s k s k s s s If F s 3 2 5 2 4 5 1 = − + = − + = s s s k 7 2 5 3 4 5 2 = + = − + = s s s k ( ) ( 3 7 ) ( ) 2 3 f t e e u t − t − t = − +
()The roots of D(s)are conjugate complex: Assume Eq. D(S=0, to contain conjugate roots PI=a+ya 2 k1=(s-n1)(F(s) N(S) s=P, D(s)s=a+ja s-n2)(F()】 N(S) s=P2 D(s)s=a-ja ku,k2 must be conjugate complex number, j 9 K k 1e9 f(t)=kiese(ation +kle j9(a-jo)t (at+91) 十C j(at+9) keat cos(at+9
(2)The roots of D(s) are conjugate complex : Assume Eq., D(s)=0, to contain conjugate roots p1 = + j p2 = − j D s s j N s s p k s p F s = + = = = − ( ) ( ) ( )( ( ) ' 1 1 1 D s s j N s s p k s p F s = − = = = − ( ) ( ) ( )( ( ) ' 2 2 2 k1 ,k2 must be conjugate complex number, 1 1 1 j k = k e 1 2 1 j k k e − = j j t j j t f t k e e k e e ( ) 1 ( ) 1 1 1 ( ) + − − = + ( ) ( ) 1 = +1 + − +1 t j t j t k e e e 2 cos( ) 1 1 = k e t + t
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:
()The roots of D(s)are equal In D(s) there are factors(S-P1 F(ky、+k2+ (s-p1)(s-n1)2a(s-p1) k1=(s-P1)2F(s) k12 (s-P1)2F(s S= PI S=PI (m-1)!ds m-(s-p1)"F(s) f(t)=k12e+knte+∑ken i=2 n-1 e pIt (s+a)"(m-1)!
(3)The roots of D(s) are equal: In D(s) there are factors (s-p1 ) 2 . = − + − + − = n i i i s p k s p k s p k F s 2 2 1 11 1 12 ( ) ( ) ( ) ( ) 1 2 11 1 ( ) ( ) s p k s p F s = = − 1 2 1 2 1 ( ) ( ) s p s p F s ds d k = = − 1 1 1 1 1 ( ) ( ) ( 1)! 1 s p s p F s ds d m k m m m m = − − = − − } ( ) ( 1)! 1 { 1 1 p t m m e m t s − + − = = + + n i p t i p t p t i f (t ) k e k t e k e 2 1 2 1 1 1 1
s+4 Example 3: If F(s) find f(t). (S+2)(s+1) Solution: (+2)(+12)(+2)3、气 3 12 k FO 十 十 (S+1) k1=(s+2)F(s) s=-2s+1s=-2 S+1-(S+4) 3 12 S+2)F(s) 3 ds =-2(+1)2s=-21 1 d2s+4 1-(-3)2(s+1) 32d2L+1=-22(s+1)4s==-3 3 k2=(s+1)F(S 3 ∫(t)=(-3e2-3te21-te-2+3e)u(t)
Example 3: , ( ). ( 2) ( 1) 4 ( ) 3 find f t s s s If F s + + + = Solution: ( 2) ( 2) ( 2) ( 1) ( ) 2 3 1 1 2 1 3 1 2 + + + + + + + = s k s k s k s k F s 2 1 2 4 2 ( 2) ( ) 3 1 1 = − + = − + = = − = + s s s s k s F s 3 1 3 ( 1) 2 1 ( 4) 2 ( 2) ( ) 2 3 1 2 = − − = + = − + − + = = − = + s s s s s s F s ds d k 3 ( 1) 2 ( 3)2( 1) 2 1 1 2 4 2 1 2 4 2 1 3 = − + = − − − + = = − + + = s s s s s s ds d k 3 1 3 1 ( 1) ( ) 2 = = = − = + s k s F s ( ) ( 3 3 3 ) ( ) 2 2 2 2 f t e t e t e e u t − t − t − t −t = − − − +
2.n≤nt Example 4 s3+62+15s+11 ff F(s) , find j∫(t) s2+5s+6 Solution: 4s+5 37 F(3)=s+1+ s+1+ s2+5s+6 s+2s+3 f(t)=8()+(t)+(-3e-+7e)u() 5()}=L d6(t)
2. n m Example 4: , ( ). 5 6 6 15 11 ( ) 2 3 2 find f t s s s s s If F s + + + + + = 3 7 2 3 1 + + + − = + + s s s ( ) ( ) ( ) ( 3 7 ) ( ) ' 2 3 f t t t e e u t − t − t = + + − + s dt d t L t L = = ( ) ( ) ' Solution: 5 6 4 5 ( ) 1 2 + + + = + + s s s F s s
