1、( )( )( )stF sL f tf t edt*0( )() ()kftf kTtkT*0 ( )( )( )()kkZ f tZ ftF zf kT zsTze( )f t*( )ft( )f t*0( )()skTkFsf kT e*( )( )( )( ) ( )f tftFsF zF s)(zF( )f t)(zF( )f tsTze1lnszT( )F s( )F z1lnszT*( )Fs( )F z1lnszT *00kftf kTtkTftf TtTf kTtkT 012( )()0 12kkkF zf kT zff T zfT zf kT z *( )ft()1()1,
2、0,1,2f kTkTk0120( )()1111kkkF zf kT zzzzz1z112( )kz F zzzz1( )( )1F zz F z11( )11zF zzz)()()(sNsMsFniiiasAsF1)()(sM)(sN)(sF( )f t() ( )|iiisaAsa F s11iia TAezia tiAe11( ( )( )1inia TiAZ F sF zez)()(assasF11( )()aF ss sassa110,1aA22,1aa A 1111111(1)( )11(1)(1)aTaTaTezF zzezzezniiiasAsF1)(11( )1inia T
3、iAF zez( )f t( )f t1( )Re( )imis TizF zs F sze111Re( )()( )(1)!iiiiinniisTnsTs sizdzs F sssF szendsze0! 1)3)(1(1)(sssF11222,1,1,3,1msnsn 11( )(1)(1)(3)ssTzF zsssze31(3)(1)(3)ssTzsssze32 ()( 2)()TTzzz zeze33()2()()TTTTz eezeze21( )()F ssa111,2msa n 22211( )()2 1 !()()aTsTaTsadzTzeF zsadssazeze11( )(
4、)Z f tF z22( )( )Z f tF zab12( )( )( )f taf tbf t12( )( )( )F zaF zbF z1()( )()nnnjjZ f tnTzF zzfjT z0,( )0tf t()( )nZ f tnTzF z10()( )()nnnjjZ f tnTz F zzf jT z(0 )( )(1)0fTf TfnT()( )nZ f tnTz F z( )f t( )F zlim( )zF z(0)lim( )zfF z( )f t( )F z1(1) ( )zF z1111lim( )lim()lim(1) ( )(1) lim( )lim(1)
5、( )tkzzzf tf kTzF zzF zzF zz( )F z*( )ft1* ( )( )ZF zft*( )( )( )( ) f tftFsF z*0( )() ()kftf kTtkT0( )()kkF zf kT z111111()( )mmmmnnnnK zb zbzbF zmnza zaza( )F zkz120( )()(0)(1 )(2 )()kkkF zf kT zffT zfT zf kT z*( )(0)(1 ) ()(2 ) (2 ) () ()ftffTtTfTtTf kTtkT*0( )() ()kftf kTtkT25( )32zF zzz1z( )F z
6、1125( )1 32zF zzz123121123232343434545515351 325515101510154530353035105707570zzzzzzzzzzzzzzzzzzzzz123( )51535F zzzz*( )5 () 15 (2 )35 (3 )fttTtTtT( )F z10111( )( )( )mmmnnnb zb zbM zF zN zza za( )F z12( )()()()nN zzzzzzz12112( )nniiniA zAzA zA zF zzzzzzzzz( )()iiiz zF zAzzz1(), 0, 1,2,kiiiiiAzf kTZ
7、Azkinzz1212( )nnAAAF zzzzzzzz1()()niif kTf kT*0( )() ()kftf kTtkT例例2.6 求求 的的 z 反变换反变换 0.5( )(1)(0.5)zF zzz解：将解：将 除以除以z，并展开成部分分式，得，并展开成部分分式，得 ( )F z( )1110.5F zzzz上式两边乘以上式两边乘以z，得，得 1111( )10.511 0.5zzF zzzzz()1 (0.5) 0,1,2,kf kTk 于是得到于是得到 *00( )() ()1 (0.5) ()kkkftf kTtkTtkT()f kT1( )kF z z11()Res (
8、)imkz zif kTF z z11111Res ( )()( )(1)!iiiiinnkkz ziz znidF z zzzF z zndz*0( )() ()kftf kTtkT25( )32zF zzz11222,2,1,1,1mznzn12z 1111112Res ( )() ( )5 (2)5 2(2)(1)kkz zz zkkzF z zzz F z zzzzzz 21z 21115Res ( )(1)5 15(2)(1)kkkz zzzF z zzzzz 211()Res ( )5 255(21)ikkkz zif kTF z z *00( )() ()5(21) ()kkkftf kTtkTtkT( )F z12z 2( )(2)(1)zF zzz11222,2,1,1,2mznzn11122Res ( )(2)2(2)(1)kkkz zzzF z zzzzz21z 2,31212112111Res ( )(1)2 1 !(2)(1)(2)(2)(2)1kkz zzkkkzzdzF z zzzdzzzdzkzzzdzzzk ()21kf kTk*0( )(21) ()kkftktkT
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