故得系统的模频特性为
H?ej??=1?1?0.36cos2????0.36sin2??22
系统的相频特性为
??????arctan五．解： （1）H?z??0.36sin2?
1?0.36cos2?z?1
z2?2z?16z2?17z?19（2）H?z??H(E)|E?z?3
z?8z2?17z?102z3?z2?2z?2（3）H?z??2z?z?2z?2z? 4z?1?2?3?4六．解：由单位序列响应得：
z2?4z?4?z?2??1?2 H?z??1?4z?4z??22zzY?z??9z z?12??Y?z?9zz29z39z2H?z??????z??2F?z?z?1?z?2?2?z?2?2?z?1????z?2??z?1????kk12k2?11?z????2z?2z?1????z?2??
k11?9z2?z?2??z?1?2?z?2?2z??2??12?d?9z22k12???z?2??2dz????z?2??z?1??k2?9z2?8z??2
?z?2??z?1?2?z?1?z?1?1故得
??1281??12z8zzF?z??z?????? ?22z?2z?1z?2z?1????z?2???z?2?故得其激励序列
kkkf?k????12k??2??8??2???1??U?k?
??七．解：引入
sin2tf1(t)?2??G4???
2tF1?j????G4???
F1?j??的图形如图6-（c）所示
?F1?j???2026??c??/?rad?s?1?
f(t)?2sin2tcos2000?t 2tF?j???1F1?j?????????2000???????2000?????2?1???G4???????????2000???????2000???? 2???2G4???2000????2G4???2000??F?j??的图形如图6（d）所示
F?j???2?2000??206??d?2000??22000??2000?引入
?
f0?t??GT?t?
2F0?j???T?T?Sa??? 2?4?故
An?式中??22T?T????F0?j????Sa?n???Sa?n? TT2?4??2???n?2?2???2000?rad/s T1?10?3S?j???n?????An????n????n???????Sa??2n?????n2000??
???S?j??的图形如图6（e）所示
S?j?????????1?2000?????06??e?1?2000???n2000??
H?j???H?j??ej????的图形如图6（f）所示
H?j??????2??101?6??f???0?
X?j???1F?j???S?j?? 2?Y?j???X?j??H?j???G2???e?j2?
故得
y?t??八．解：
1?Sa?t?2?,t?R
（1）因有
1??s?1?1??(s)??sI?A???11?22???s?1s?3?0?? ?1?s?3??故得系统状态转移矩阵
?e?t?(t)??1?t1?3t?e?e??220??U?t? ?3t?e??（2）因有
1????H?s??C?(s)B?D??1?2?
?s?3???故得冲激矩阵
1??h?t?????t??e?3t?U?t?
2??（3）因有
F?s??1 s?510??6??6Y?s??C?(s)x?0??H?s?F?s????? ss?3????故得系统输出
y?t??九．解：
51?2e?3t?U?t? ?6（1）因有
H?z??故得系统的直接形式信号流图如图7所示
z1z2?z?2
y?k??f?k?DD1?12
图7
（2）系统函数
H?z??（3）因有
z1z2?z?2
H?ej???ej?ej2??ej??ej?1212??11?1?12?0.4?180o
H?ej???故得系统正弦稳态响应为
ej2??ej??y?k??0.4?100cos??k?90o?180o??40cos??k?90o?