求以下序列的z變換

1、習(xí)題習(xí)題五Z變換變換數(shù)字信號(hào)數(shù)字信號(hào)處理精品理精品課1習(xí)題五Z變換1求以下序列的z變換，并畫(huà)出零極點(diǎn)圖和收斂域。分析：Z變換定義，????????nnznxzXnxZ)()()]([n的取值是的有值范圍。Z變換的收斂域)(nx是滿足?????????Mznxnn)(的z值范圍。解：解：(1)由Z變換的定義可知：nnnzazX????????)(nnnnnnzaza?????????????01nnnnnnzaza?????????01)

2、)(1()1()1)(1(1111212azazazaazazazaazaz?????????????)(21)()2(nunxn???????)1(21)()3(??????????nunxn)1(1)()4(??nnnx為常數(shù)）00(0)sin()()5(????nnnnx10)()cos()()6(0????rnunArnxn??)1||()()1(??aanxn習(xí)題習(xí)題五Z變換變換數(shù)字信號(hào)數(shù)字信號(hào)處理精品理精品課3解：(4)??

3、????11)(nnznzX，??????????11)(1)(nnznndzzdX21)(11zzznn?????????1||?z。的收斂域?yàn)楣实氖諗坑蛳嗤?，的收斂域和因?yàn)?||)()()(1ln)1ln(ln)(???????zzXdzzdXzXzzzzzX????z10零點(diǎn)為：極點(diǎn)為：zz解：解：(5)設(shè))()sin()(0nunny???則有1||cos21sin)()(20101???????????????zzzzznyz