4. 解:(1)a 22()()s a
H s s a b +=++
设采样周期为TH a (s)的极点为:s 1=-a+j b , s 2=-a -jb 将H a (s)部分分式展开(用待定系数法) 12a 2212122122
1
2122122()()()()
()()()A A s a H s s a b s s s s A s s A s s s a b A A s A s A s s a b +=
=+++---+-=+++--=++比较分子各项系数可知,A 1,A 2应满足方程
解之得: A 1=1/2, A 2=1/2, 所以
a 1/21/2()(j )(j )H s s a
b s a b =+--+---
2
1(j )1(j )111/21/2()1e 1e 1e k k T s a b T a b T k A H z z z z --+----===+---∑ 1122
1e cos()
()12e cos()e aT aT
aT z bT H z bT z z -------=-+ a 22()()s a
H s s a b +=++a 22()()b
H s s a b =++
(2)H a (s)的极点为:s 1=-a+jb,s 2=-a -jb 将H a (s)部分分式展开 a j j 22()(j )(j )H s s a b s a b -=+-----+ (j )1(j )1j j 22()1e 1e a b T a b T H z z z ----+--=-+-- 通分并化简整理, 得
1122e sin()
()12e cos()e aT aT
aT z bT H z bT z z ------=-+
5. 解:
H a (s)的极点为
将T=2代入上式,得
或通分合并两项得