z 反变换

Inverse Z-transform

采样信号 $e^{*} (t)$ 的 z 变换为：

E (z) = Z [e^{*} (t)]

则 $z$ 反变换可以记为

e^{*} (t) = Z^{- 1} [E (z)] = \sum_{n = 0}^{\infty} e (n T) δ (t - n T)

长除法得到洛朗级数

\begin{array}{r} F (z) = \frac{b_{0} z^{m} + b_{1} z^{m - 1} + \dots + b_{m}}{a_{0} z^{n} + a_{1} z^{n - 1} + \dots + a_{n}} = c_{0} + c_{1} z^{- 1} + c_{2} z^{- 2} + \dots \end{array}

\begin{array}{r} f^{*} (t) = c_{0} + c_{1} δ (t - T) + c_{2} δ (t - 2 T) + \dots \end{array}

\begin{array}{r} \frac{F (z)}{z} = \sum_{i = 1}^{n} \frac{c_{i}}{z - p_{i}} \Leftrightarrow F (z) = \sum_{i = 1}^{n} \frac{c_{i} z}{z - p_{i}} \end{array}

\begin{array}{r} f (k T) = Z^{- 1} [F (z)] f^{*} (t) = \sum_{k = 0}^{\infty} f (k T) δ (t - k T) \end{array}

\begin{aligned} Z^{- 1} [\frac{1 - a z^{- 1}}{z^{- 1} - a}] & = Z^{- 1} [- \frac{a (z^{- 1} - \frac{1}{a} + a - a)}{z^{- 1} - a}] = Z^{- 1} [- a + \frac{1 - a^{2}}{- a (1 - \frac{1}{a} z^{- 1})}] \\ = - a δ (k) + (a - \frac{1}{a}) (\frac{1}{a})^{k} \cdot 1 (k) \end{aligned}