练习：求下列各函数的原函数K.KK(s + 1)(s + 3)单实根(1)F(s) =s+ 4S+ 2ss(s + 2)(s + 4)3(s+1)(s+3)K, =[sF(s)]=0 =[8s(s +2)(s+4)(s +1)(s +3)K, =[(s + 2)F(s)]=-2 =[(s + 2)4s(s +2)(s +4)3(s+1)(s +3)K, =[(s+4)F(s)]s=-4=[(s+4s(s +2)(s + 4)833原函数4tf(t)88
练习：求下列各函数的原函数 ( 2)( 4) ( 1)( 3) (1) ( ) + + + + = s s s s s F s 2 4 1 2 3 + + + = + s K s K s K 1 0 [ ( )] = s= K sF s t t f t e e 2 4 8 3 4 1 ε(t) 8 3 ( ) - - 原函数 = + + 8 3 = 2 2 [( 2) ( )] = + s= - K s F s 3 4 [( 4) ( )] = + s= - K s F s 0 ] ( 2)( 4) ( 1)( 3) [ = + + + + = s s s s s s s 2 ] ( 2)( 4) ( 1)( 3) [( 2) = - + + + + = + s s s s s s s 4 ] ( 2)( 4) ( 1)( 3) [( 4) = - + + + + = + s s s s s s s 4 1 = 8 3 = 单实根
1单实根1(2)F(s) :s(s2 + 2s+ 2)(s-pi)(s-P21)(s P22)共轭复根分母有三个根:Pi=0，P21=-1+j，P22=-1-jKKK2221F(sss-P21S-P21K, = sF(s) s=0 = ? + 2s+2三2112135°K, =(s- Pa)F(s)2(1 + j)2V2S=P21T135°K, =(s- P,)F(s)2V2=P22原函数f(t) ==e(t) + 2K21leαt cos(ot + 0))cos(t +135°)2/2217
17 分母有三个根：p1=0， (s-p1)(s-p21)(s-p22) 1 = 1 s 0 (s) = = K sF 2 1 = 22 22 21 1 21 ( ) s p K s p K s K F s - + - = + 21 ( ) ( ) 21 21 s p K s p F s = - = 22 ( ) ( ) 22 22 s p K s p F s = - = 2(1 j) 1 + = - 原函数 2 s 0 2 2 1 = + + = s s = 135 2 2 1 ε(t) 2 cos( ) 2 1 ( ) 21 w 1 a f t = + K e t + t cos( 135 ) 2 2 1 ε(t) 2 2 1 = + + - e t t ( 2 2) 1 (2) ( ) 2 + + = s s s F s 共轭复根 单实根 p21=-1+j，p22=-1-j = - 135 2 2 1
2.D(s)=0有重根一三重根若D(s)=0具有相等的实根，则D(s)中应含有(s-pi)n的因式。 设n=3，F(s)表示为KK,K.F(s) =(s-p)+(s-p)+s- PiS-pzPi为D(s)=0的三重根，确定Ki1、Ki2、K13(1)把Ku单独分离出来设T(s)= (s- P)"F(s) = (s - pi)"K13 +(s - pI)K12 +Kμ + ..K =[(s- p)°F(s)]s=P18
18 若D(s)=0具有相等的实根，则D(s)中应含有(s-p1)n 的因式。设n=3，F(s)表示为 + - + - + - + - = . ( ) ( ) ( ) 2 2 1 13 2 1 12 3 1 11 s p K s p K s p K s p K F s p1为D(s)=0的三重根，确定K11、K12、K13。 (1)把K11单独分离出来 ( ) ( ) 3 1 s - p F s 1 [( ) ( )] 3 11 1 s p K s p F s = - = 2. D(s)=0有重根—三重根 设T(s)= ( ) ( ) . 13 1 12 11 2 = s - p1 K + s - p K + K +
KiK2K13(2)把Kz单独分离出来，F(s).(s-p)3(s- p)s-piT(s)= (s - p)F(s)=(s- p)"K3 +(s- p)K12 + Kμ +..对式中s进行一次求导，dT(s) _ d(s- p.)"F(s)l = 2(s - P,)K13 + Ki2dsdsd[(s - p)"F(s)]K12=[(s- p,)"F(s)]S=P1S=P1ds1 d同理：K[(s - p.)"F(s)]s=p132 ds..d9-11Kia[(s - pi)"F(s)](q -1)! ds--119
19 (2)把K12单独分离出来， 1 13 12 3 1 2( ) ( ) [( ) ( )] s p K K ds d s p F s ds dT s = - + - = 1 d d[( ) ( )] 3 1 12 s p s s p F s K = - = 同理： 1 [( ) ( )] d d 2 1 3 13 2 1 s p s p F s s K = - = 对式中s进行一次求导， 1 13 2 1 12 3 1 11 ( ) ( ) ( ) s p K s p K s p K F s - + - + - = ( ) ( ) ( ) ( ) . 13 1 12 11 2 1 3 s - p1 F s = s - p K + s - p K + K + 1 [( ) ( )] 3 1 s p s p F s = = - T(s)= 1 [( ) ( )] d d ( 1)! 1 1 1 1 1 s p q q q q s p F s q s K - = - - - = . .
KKK行F(s)=(s- p)3(s-p)s--pP原函数f(t)= K1(tepr')+Ki2(tept) +Kirseprt +ZK,epri=2其中Kn =[(s-p)"F(s)S=P1d[(s - pr)"F(s)]K12S=P1dsd1K[(s-p)"F(s)]13S=P2 ds1f(t)=1epirF(s) =(s-p,)n+1n!20
20 + - + - + - + - = . ( ) ( ) ( ) 2 2 1 13 2 1 12 3 1 11 s p K s p K s p K s p K F s ) 2! 1 ( 2 1 11 p t K t e 原函数 f(t)= 1 d d[( ) ( )] 3 1 12 s p s s p F s K = - = 1 [( ) ( )] 3 11 1 s p K s p F s = - = 1 [( ) ( )] d d 2 1 3 13 2 1 s p s p F s s K = - = ) 1! 1 ( 1 12 p t + K te p t K e 1 + 13 p t n i i K e 1 2 = + n p t t e n f t 1 ! 1 ( ) = 其中 1 1 ( ) 1 ( ) + - = n s p F s