Z[F(s)]?(1?z?1)Z[?11]2s(s?1)11111Te?Tz?1?1?(1?z)[??]?(1?z)[??]
ss?1(s?1)21?z?11?e?Tz?1(1?e?Tz?1)2z(?e?T?Te?T?1)?e?T?Te?T?e?2T?(?z?e?T)21?e?sT(6)F(s)?2
s(s?1)留数法:
Z[F(s)]?(1?z?1)Z[?11]2s(s?1)2d1z1zz?1?z?z?Tz?(1?z)[()?()]?(1?z)[?] s?0s??1dss?1z?esTs2z?esT(z?1)2z?e?Tz?1(?1?T?e?T)?z?2(?e?T?Te?T?1)?(1?z?1)(1?e?Tz?1)部分分式法：
Z[F(s)]?(1?z?1)Z[?11]2s(s?1)111Tz?111?1?(1?z)[2??]?(1?z)[??]
ss?1s(1?z?1)21?e?Tz?11?z?1z?1(?1?T?e?T)?z?2(?e?T?Te?T?1)?(1?z?1)(1?e?Tz?1)2.13 用级数求和法求下列函数的Z变换 (1) f(k)?ak 解答：
F(z)?Z[a]??akz?kkk?0??1?az?1?a2z?2?a3z?3???(2) f(k)?ak?1 解答：
11?az?1
F(z)?Z[a?k?1]??ak?1z?kk?0?1?1?1a1?z?az?2?a2z?3????a1?a?1z?1a?a2z?1
(3) f(t)?tak?1
16
解答：
由于Z[tf(t)]??Tzd F(z)，dzf(k)?ak?1的z变换为F(z)?Ta0z?1?2Ta1z?2?3Ta2z?3?...
F1(z)?aF(z)?Taz?1?2Ta2z?2?...
2?2所以az?1F?2Ta3z?3?... 1(z)?Taz(1?az?1)F1(z)?Taz?1?Ta2z?2?... Taz?1 F1(z)??12(1?az)1Tz?1 F(z)?F1(z)??12a(1?az)(4) f(t)?t2e?5t 解答：
由于Z[tf(t)]??TzdF(z)，dzf(kT)?e?5kT的z变换为?5t?k?0
(Fz)?Z[e]??e?5kTz?k?1?e?5Tz?1?e?10Tz?2?e?15Tz?3??将两边同时乘以e?5Tz?1，得：e?5Tz?1F(z)=e?5Tz?1?e?10Tz?2?e?15Tz?3??将上两式相减，得：1?e?5Tz?1dTe?5Tz?1?5T [Zte]??TzF(z)?dz(1?e?5Tz?1)2Z[te2?5TF(z)?1
d?5TT2e?5Tz?1(1?e?5Tz?1)]??Tz[te]?dz(1?e?5Tz?1)32.14 用长除法、部分分式法、留数法对下列函数进行z反变换
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z?1(1?e?aT)(1)F(z)?
(1?z?1)(1?e?aTz?1)解答：长除法
z?1(1?e?aT)原式=1?(1?e?aT)z?1?e?aTz?21?(1?e?aT)z?1?e?aTz?2(1?e?aT)z?1?(1?e?2aT)z?2?(1?e?3aT)z?3??z?1(1?e?aT)(1?e?aT)z?1?(1?e?2aT)z?2?(1?e?aT)e?aTz?3
(1?e?2aT)z?2?(1?e?aT)e?aTz?3(1?e?2aT)z?2?(1?e?2aT?e?aT?e?3aT)z?3?(1?e?2aT)e?aTz?4?f*(t)?(1?e?aT)?(t?T)??部分分式法：
z(1?e?aT)F(z)?(z?1)(z?e?aT)F(z)11?? ?aTzz?1z?ef*(t)?1?e?at留数法：
z1?1,z2?e?aTRes[F(z)zRes[F(z)z??e?atf*(t)?1?e?atk?1z(1?e?aT)k?1]z?1?[(z?1)z]z?1?1?aT(z?1)(z?e)]z?e?aT?[(z?e?aTk?1z(1?e?aT))zk?1]z?e?aT ?aT(z?1)(z?e)z(1?e?aT)(2) F(z)?
(z?1)(z?e?aT)解答：长除法
F(z)?2z
(z?1)(z?2)18
2z?1原式=1?3z?1?2z?22z?1?6z?2??1?3z?1?2z?22z?12z?1?6z?2?4z?3 ?2?36z?4z6z?2?18z?3?12z?4?f*(t)?2?(t?T)?6?(t?2T)?部分分式法：
F(z)22??zz?2z?1f(kT)?2*2k?2f(t)??(2k?1?2)?(t?kT)*k?0?
留数法：
z1?1,z2?2Res[F(z)zk?1]z?1?[(z?1)Res[F(z)zk?1]z?2?2k?1f(kT)?2k?1?2f(t)??(2k?1?2)?(t?kT)*k?0?2zzk?1]z?1??2(z?1)(z?2)2z?[(z?2)zk?1]z?2(z?1)(z?2)
?6?2z?1(3)F(z)? ?1?21?2z?z解答：长除法
1?2z?1?z?2?6?10z?1???6?2z?1?6?12z?1?6z?2?10z?1?6z?2
?10z?1?20z?2?10z?3?f*(t)??6?(t)?10?(t?T)??部分分式法：
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F(z)?64??zz?1(z?1)2f(kT)??6?4kf(t)??(?6?4k)?(t?kT)*k?0?
留数法：
z1,2?12d2?6z?2zk?1Res[F(z)z]z?1?[(z?1)z]z?1??6?4kds(z?1)2
f(kT)??6?4kk?1f(t)??(?6?4k)?(t?kT)*k?0?0.5z?1(4) F(z)?
1?1.5z?1?0.5z?2解答：长除法
1?1.5z?1?0.5z?20.5z?1?0.75z?2?0.5z?10.5z?1?0.75z?2?0.25z?3 0.75z?2?0.25z?30.75z?2?1.125z?3?0.375z?4?f*(t)?0.5?(t?T)?0.75?(t?2T)?部分分式法：
F(z)?0.5z(z?1)(z?0.5)F(z)11??zz?1z?0.5f(kT)?1?(0.5)kf(t)??(1?(0.5)k)?(t?kT)*k?0?
留数法：
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