线性定常系统的差分方程求解

差分方程
非齐次线性定常前向差分方程、非齐次线性定常后向差分方程

设离散系统的输入为 $e (k)$ ，输出为 $u (k)$ ，（此时采样周期为 1 ，则 $t = k T \Rightarrow k$ ）

\begin{aligned} \sum_{j = 0}^{n} a_{j} u (k - j) = \sum_{i = 0}^{m} b_{i} e (k - i) \\ \Rightarrow & \sum_{j = 0}^{n} a_{j} z^{- j} U (z) = \sum_{i = 0}^{m} b_{i} z^{- i} E (z) \\ \Rightarrow & U (z) = \frac{\sum_{i = 0}^{m} b_{i} z^{- i}}{\sum_{j = 0}^{n} a_{j} z^{- j}} E (z) \\ \Rightarrow & u (k) = Z^{- 1} [U (z)] \end{aligned}

$u (k) + a u (k - 1) = e (k), e (k) = 1 (k), u (k) = 0 (k \leq - 1)$ , 求输出 $u (k)$

\begin{aligned} U (z) + a z^{- 1} [U (z) + u (- 1)] = E (z) \\ \Rightarrow & U (z) = \frac{1}{1 + a z^{- 1}} E (z) = \frac{1}{1 + a z^{- 1}} \frac{1}{1 - z^{- 1}} \\ \Rightarrow & \frac{U (z)}{z} = \frac{z}{(z + a) (z - 1)} = \frac{a}{a + 1} \frac{1}{z + a} + \frac{1}{a + 1} \frac{1}{z - 1} \\ \Rightarrow & u (k) = \frac{a}{a + 1} (- a)^{k} + \frac{1}{a + 1} 1 (k) \end{aligned}

\begin{array}{r} y (k + 2) - 1.2 (k + 1) + 0.32 y (k) = 1.2 u (k + 1) y (0) = 1, y (1) = 2.4, u (0) = 1 \end{array}

\begin{aligned} z^{2} [Y (z) - y (0) - y (1) z^{- 1}] - 1.2 z [Y (z) - y (0)] + 0.32 Y (z) = 1.2 z [U (z) - u (0)] \\ \Rightarrow & (z^{2} - 1.2 z + 0.32) Y (z) - z^{2} - 2.4 z + 1.2 z = 1.2 z U (z) - 1.2 z \\ \Rightarrow & Y (z) = \frac{1.2 z U (z)}{z^{2} - 1.2 z + 0.32} + \frac{z^{2}}{z^{2} - 1.2 z + 0.32} \\ \Rightarrow & y (k) = Z^{- 1} [Y (z)] \end{aligned}

\begin{array}{r} y (k) - 5 y (k - 1) + 6 y (k - 2) = u (k) \end{array}

\begin{aligned} Y (z) - 5 z^{- 1} Y (z) + 6 z^{- 2} Y (z) = U (z) \\ \Rightarrow & Y (z) = \frac{z^{2}}{z^{2} - 5 z + 6} \times \frac{z}{z - 1} = \frac{z^{3}}{(z - 2) (z - 3) (z - 1)} \\ \Rightarrow & \frac{Y (z)}{z} = \frac{1}{2} \frac{1}{z - 1} - 4 \frac{1}{z - 2} + \frac{9}{2} \frac{1}{z - 3} \\ \Rightarrow & y (k) = \frac{1}{2} - 4 \cdot 2^{k} + \frac{9}{2} \cdot 3^{k} \end{aligned}