山东大学：《生物医学信号处理 Biomedical Signal Processing》精品课程教学资源（PPT课件讲稿）Chapter 03 the Z-transform

EX, 3.3 Sum of two exponential sequences Determine the z-transform, including the roc pole-zero-plot, for sequence uln+ Solution 2 3 ul n ulna n=-0 n2“+ n n2 n=-00 2 ∑ n=0 2 n=-00 I Engineering_ shandong UniV

17 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. Ex. 3.3 Sum of two exponential sequences       1 1 2 3 n n x n u n u n     = + −         ( )     1 1 2 3 n n n n X z u n u n z  − =−         = + −                1 1 0 1 1 2 3 n n n n z z   − − = =−     = + −               1 1 2 3 n n n n n n u n z u n z   − − =− =−     = + −           ◆Determine the z-transform, including the ROC, pole-zero-plot, for sequence: Solution:

Example 3.3: Sum of two exponential sequences X(=2z1+∑|- n=-00 2z z 12 1+-z t-Z 2 R0c|>and|>l→Roc:|> 18 2021/1/31 Zhongguo Liu_ Biomedical Engineering_shandong Univ

18 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. Example 3.3: Sum of two exponential sequences 1 1 1 : 2 3 2 z and z ROC z     ( ) 1 1 0 1 1 2 3 n n n n X z z z   − − = =−     = + −           1 1 1 2 12 1 1 1 1 2 3 z z z z − −     −   =         − +     1 1 1 1 1 1 1 1 2 3 z z − − = + − + ROC:

手 m z-plane n z-plane e ORe 1+-z 2 12 Re 1+ Zhongguo Liu_ Biomedical Engineering_shandong Univ

19 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. 1 1 1 1 2 z − − 1 1 1 1 3 z − + 1 1 1 2 12 1 1 1 1 2 3 z z z z − −     −           − +    

Example 3.4: Sum of two exponential =吨 n+|-|a xin=a un Solution X(z) uIne> for z> uln> 1+-z l4n|+ ×人C 11+2z 20 ROC

20 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. Example 3.4: Sum of two exponential   2 1 2 1 1 1 2 1 1  −        − , z z u n Z n   3 1 3 1 1 1 3 1 1  +        − − , z z u n Z n     2 1 3 1 1 1 2 1 1 1 3 1 2 1 1 1  + + −         + −      − − , z z z u n u n Z n n       1 1 2 3 n n x n u n u n     = + −         xn a un n = ( ) z X z z a = − for z a  Solution: ROC:

Example 3.5 Two-sided exponential sequence n xn 啊n- n==aul-n Solution Ⅹt un> z-0 1+ for z<a l小-n-11 2< 2 Xz RoC: <z< 1+-z 2 +—2 21 2021/1/31 Zhongguo Liu_ Biomedical Engineering_shandong Univ

21 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. Example 3.5: Two-sided exponential sequence      1 2 1 3 1  − −       −      x n = − u n u n n n   3 1 3 1 1 1 3 1 1  +        − − , z z u n Z n   2 1 2 1 1 1 1 2 1 1  −  − −       − − , z z u n Z n ( ) 2 1 3 1 2 1 1 3 1 1 12 1 2 2 1 1 1 3 1 1 1 1 1 1 1          −      +       − = − + + = − − − − ROC : z z z z z z z X z ( ) z X z z a = − for z a  xn= −a u− n −1 n Solution: