Z轉換
性质

|   | 时域      | Z域      | 证明       | 收敛域   |
| --- | --- | --- | --- | --- |
| 记法      |  x[n]={\mathcal {Z}}^{-1}\{X(z)\}  |  X(z)={\mathcal {Z}}\{x[n]\}  |   |  r_{2}<|z|<r_{1}  |
| 線性  |  a_{1}x_{1}[n]+a_{2}x_{2}[n]  |  a_{1}X_{1}(z)+a_{2}X_{2}(z)  |  {\begin{aligned}X(z)&=\sum _{n=-\infty }^{\infty }(a_{1}x_{1}(n)+a_{2}x_{2}(n))z^{-n}\\&=a_{1}\sum _{n=-\infty }^{\infty }x_{1}(n)z^{-n}+a_{2}\sum _{n=-\infty }^{\infty }x_{2}(n)z^{-n}\\&=a_{1}X_{1}(z)+a_{2}X_{2}(z)\end{aligned}}  | 包含 ROC1 ∩ ROC2  |
| 时间膨胀  |  x_{K}[n]={\begin{cases}x[r],&n=rK\\0,&n\not =rK\end{cases}} r: 整数  |  X(z^{K})  |  {\begin{aligned}X_{K}(z)&=\sum _{n=-\infty }^{\infty }x_{K}(n)z^{-n}\\&=\sum _{r=-\infty }^{\infty }x(r)z^{-rK}\\&=\sum _{r=-\infty }^{\infty }x(r)(z^{K})^{-r}\\&=X(z^{K})\end{aligned}}  |  R^{\frac {1}{K}}  |
| 降采样  |  x[nK]  |  {\frac {1}{K}}\sum _{p=0}^{K-1}X\left(z^{\tfrac {1}{K}}\cdot e^{-i{\tfrac {2\pi }{K}}p}\right)  | ohio-state.edu （页面存档备份，存于）  或  ee.ic.ac.uk （页面存档备份，存于）  |   |
| 时移      |  x[n-k]  |  z^{-k}X(z)  |  {\begin{aligned}Z\{x[n-k]\}&=\sum _{n=0}^{\infty }x[n-k]z^{-n}\\&=\sum _{j=-k}^{\infty }x[j]z^{-(j+k)}&&j=n-k\\&=\sum _{j=-k}^{\infty }x[j]z^{-j}z^{-k}\\&=z^{-k}\sum _{j=-k}^{\infty }x[j]z^{-j}\\&=z^{-k}\sum _{j=0}^{\infty }x[j]z^{-j}&&x[\beta ]=0,\beta <0\\&=z^{-k}X(z)\end{aligned}}  | ROC，除了 k > 0 时 z = 0 和 k < 0 时 z = ∞  |
| Z域的  尺度性质  |  a^{n}x[n]  |  X(a^{-1}z)  |  {\begin{aligned}{\mathcal {Z}}\left\{a^{n}x[n]\right\}&=\sum _{n=-\infty }^{\infty }a^{n}x(n)z^{-n}\\&=\sum _{n=-\infty }^{\infty }x(n)(a^{-1}z)^{-n}\\&=X(a^{-1}z)\end{aligned}}  |  |a|r_{2}<|z|<|a|r_{1}  |
| 时间反转      |  x[-n]  |  X(z^{-1})  |  {\begin{aligned}{\mathcal {Z}}\{x(-n)\}&=\sum _{n=-\infty }^{\infty }x(-n)z^{-n}\\&=\sum _{m=-\infty }^{\infty }x(m)z^{m}\\&=\sum _{m=-\infty }^{\infty }x(m){(z^{-1})}^{-m}\\&=X(z^{-1})\\\end{aligned}}  |  {\tfrac {1}{r_{1}}}<|z|<{\tfrac {1}{r_{2}}}  |
| 共轭复数  |  x^{*}[n]  |  X^{*}(z^{*})  |  {\begin{aligned}{\mathcal {Z}}\{x^{*}(n)\}&=\sum _{n=-\infty }^{\infty }x^{*}(n)z^{-n}\\&=\sum _{n=-\infty }^{\infty }\left[x(n)(z^{*})^{-n}\right]^{*}\\&=\left[\sum _{n=-\infty }^{\infty }x(n)(z^{*})^{-n}\right]^{*}\\&=X^{*}(z^{*})\end{aligned}}  |   |
| 实部  |  \operatorname {Re} \{x[n]\}  |  {\tfrac {1}{2}}\left[X(z)+X^{*}(z^{*})\right]  |   |   |
| 虚部  |  \operatorname {Im} \{x[n]\}  |  {\tfrac {1}{2j}}\left[X(z)-X^{*}(z^{*})\right]  |   |   |
| 微分      |  nx[n]  |  -z{\frac {dX(z)}{dz}}  |  {\begin{aligned}{\mathcal {Z}}\{nx(n)\}&=\sum _{n=-\infty }^{\infty }nx(n)z^{-n}\\&=z\sum _{n=-\infty }^{\infty }nx(n)z^{-n-1}\\&=-z\sum _{n=-\infty }^{\infty }x(n)(-nz^{-n-1})\\&=-z\sum _{n=-\infty }^{\infty }x(n){\frac {d}{dz}}(z^{-n})\\&=-z{\frac {dX(z)}{dz}}\end{aligned}}  |   |
| 卷积  |  x_{1}[n]*x_{2}[n]  |  X_{1}(z)X_{2}(z)  |  {\begin{aligned}{\mathcal {Z}}\{x_{1}(n)*x_{2}(n)\}&={\mathcal {Z}}\left\{\sum _{l=-\infty }^{\infty }x_{1}(l)x_{2}(n-l)\right\}\\&=\sum _{n=-\infty }^{\infty }\left[\sum _{l=-\infty }^{\infty }x_{1}(l)x_{2}(n-l)\right]z^{-n}\\&=\sum _{l=-\infty }^{\infty }x_{1}(l)\left[\sum _{n=-\infty }^{\infty }x_{2}(n-l)z^{-n}\right]\\&=\left[\sum _{l=-\infty }^{\infty }x_{1}(l)z^{-l}\right]\!\!\left[\sum _{n=-\infty }^{\infty }x_{2}(n)z^{-n}\right]\\&=X_{1}(z)X_{2}(z)\end{aligned}}  | 包含 ROC1 ∩ ROC2  |
| 互相关  |  r_{x_{1},x_{2}}=x_{1}^{*}[-n]*x_{2}[n]  |  R_{x_{1},x_{2}}(z)=X_{1}^{*}({\tfrac {1}{z^{*}}})X_{2}(z)  |   | 包含  X_{1}({\tfrac {1}{z^{*}}}) 与  X_{2}(z) 的ROC的交集   |
| 一阶差分      |  x[n]-x[n-1]  |  (1-z^{-1})X(z)  |   | 包含 X1(z) 与 z ≠ 0 的ROC的交集   |
| 累积      |  \sum _{k=-\infty }^{n}x[k]  |  {\frac {1}{1-z^{-1}}}X(z)  |  {\begin{aligned}\sum _{n=-\infty }^{\infty }\sum _{k=-\infty }^{n}x[k]z^{-n}&=\sum _{n=-\infty }^{\infty }(x[n]+\cdots +x[-\infty ])z^{-n}\\&=X[z]\left(1+z^{-1}+z^{-2}+\cdots \right)\\&=X[z]\sum _{j=0}^{\infty }z^{-j}\\&=X[z]{\frac {1}{1-z^{-1}}}\end{aligned}}  |   |
| 乘法  |  x_{1}[n]x_{2}[n]  |  {\frac {1}{j2\pi }}\oint _{C}X_{1}(v)X_{2}({\tfrac {z}{v}})v^{-1}\mathrm {d} v  |   | - |
帕塞瓦尔定理
初值定理：如果 x[n] 为因果的，那么
终值定理：如果 (z−1)X(z) 的极点在单位圆内，则