\[= 0.4323z^{-2} + 0.9231z^{-3} + 1.2349z^{-4} + \cdots\]
\[c(0)=0,\quad c(1)=0,\quad c(2)=0.4323,\quad c(3)=0.9231\]
应用MATLAB,可得输出\(c(k)\)波形图如图7-57所示。
MATLAB文本:exe739.m
T=1;t=0:1:20;
sys=tf([0.4323],[1,-1.1353,0.5676],T);
step(sys,t);grid;
图7-57 系统单位阶跃响应(MATLAB)
7-40 采样系统的结构图如图7-58所示,两采样器同步采样,采样周期\(T=1\),并假设满足香农(Shannon)采样定理。要求:(1)求采样系统的误差脉冲传递函数\(E(z)/R(z)\);(2)判别系统的稳定性;(3)当\(r(t)=10\cdot1(t)\)时,求系统的稳态误差;(4)当\(r(t)=10\cdot1(t)\)时,求系统误差采样信号\(e^*(t)\)在前4个采样时刻的值\(e(kT)\),\(k=0,1,2,3\),并画出其波形图。
解 (1)求\(E(z)/R(z)\)。开环脉冲传递函数为
\[G_1(z)G_2(z) = \mathscr{Z}\left[\frac{1}{s}\right]\cdot\mathscr{Z}\left[\frac{1}{s+1}\right]\]
\[= \frac{z}{z-1}\cdot\frac{z}{z-\mathrm{e}^{-1}} = \frac{z^2}{z^2-1.3679z+0.3679}\]
图7-58 闭环采样系统结构图
误差脉冲传递函数为
\[\Phi_e(z) = \frac{E(z)}{R(z)} = \frac{1}{1+G_1(z)G_2(z)}\]
\[= \frac{z^2-1.3679z+0.3679}{2z^2-1.3679z+0.3679}\]
(2)系统稳定性。闭环特征方程为
\[D(z) = 2z^2 - 1.3679z + 0.3679 = 0\]
求得特征值为\(z_{1,2}=0.3420\pm \mathrm{j}0.2588\),因为\(|z_{1,2}|=0.4289<1\),故闭环系统稳定。
(3)系统稳态误差。输入\(z\)变换为
\[R(z) = \mathscr{Z}[10\cdot1(t)] = \frac{10z}{z-1}\]
采样误差函数为
\[E(z) = \Phi_e(z)R(z) = \frac{z^2-1.3679z+0.3679}{2z^2-1.3679z+0.3679}\cdot\frac{10z}{z-1}\]
稳态误差为
\[e_{ss}(\infty) = \lim_{z\to1}(1-z^{-1})E(z) = \lim_{z\to1}\frac{z-1}{z}\cdot\frac{z^2-1.3679z+0.3679}{2z^2-1.3679z+0.3679}\cdot\frac{10z}{z-1} = 0\]