\[\Rightarrow D(z)=z-0.368+0.632K\]
\[\Rightarrow |0.368 - 0.632K| < 1 \Rightarrow 1 < K < 2.165\]
(2)
\[G = \frac{0.632}{z - 0.368} \Rightarrow K_p = \lim_{r\to 1}(1+G) = 2 \Rightarrow e = 0.5 \Rightarrow C(\infty) = 0.5\]
七、解答
(1)
\[G = \frac{20K}{s(s + 20)} \Rightarrow \begin{cases} 2\xi\omega_n = 20 \\ \omega_n^2 = 20K \end{cases} \Rightarrow K = 13.89\]
(2)
判别稳定性:
\[D=s^2 +20s+20K\]
| \(s^2\) | 1 | \(20K\) |
| \(s^1\) | 20 | |
| \(s^0\) | \(2K\) |
\[\Rightarrow K > 0\]
求 \(\xi=0.6\) 时的超调量为 0.09478, 故 c(t) 最大值为: \(3 \times 1.09478 = 3.28434\)
(3)
由于系统是 4 型，故可以实现有差跟踪速度输入信号
八、解答
