五、闭环采样系统如下图所示，采样周期 \(T=0.5\)，试求：

(1) 开环脉冲传递函数 \(G_hG_0(Z)\)；

(2) 采样系统的单位阶跃响应 \(C(nT)\)。

注：\(e^{-0.5}=0.61\)，计算保留两位小数。

Z变换公式考试会给出。

（右上角推导，图中位置在题目右侧空白处，先于正式解答书写）

\(C(Z) = \bar{e}(Z) \cdot G_hG_0(Z)\)

\(C(S) = \bar{e}^*(S) \cdot G_hG_0(S)\)

\(\bar{e}(S) = R(S) - C(S)\)

\(\bar{e}(Z) = R(Z) - C(Z)\)

\(C(Z) = G_0G_h(Z)\left[R(Z) - C(Z)\right]\)

\(C(Z) + G_0G_h(Z)C(Z) = G_0G_h(Z)R(Z)\)

\(\dfrac{C(Z)}{R(Z)} = \dfrac{G_0G_h(Z)}{1+G_0G_h(Z)}\)

\(C(Z) = R(Z) \cdot \dfrac{G_0G_h(Z)}{1+G_0G_h(Z)}\)

\(= \dfrac{Z}{Z-1} \cdot \dfrac{0.22Z+0.17}{2Z^2-3Z+1.19}\)

(1)

\(G_hG_0(Z) = (1-Z^{-1})Z\left(\dfrac{1}{S^2(S+1)}\right)\)

\(= \dfrac{T}{Z-1}(-1) + \dfrac{Z-1}{Z-e^{-T}}\)

\(= \dfrac{(2e^{-\frac{1}{2}}-1)Z+2-3e^{-\frac{1}{2}}}{2(Z-1)(Z-e^{-\frac{1}{2}})}\)

\(\approx \dfrac{0.22Z+0.17}{2(Z-1)(Z-0.61)+2}\)

(2)

\(C(Z) = \dfrac{G_hG_0(Z)}{1+G_hG_0(Z)}R(Z)\)

其中 \(R(Z) = \dfrac{Z}{Z-1}\)

（以下使用长除法，具体内容略）。