填空题解析
1、解析：上升时间，超调量，稳态误差，调节时间，超调量，稳态误差
2、解析：\(\Phi(s) = \dfrac{1}{10s(3s+1)}\)
3、解析：0
4、解析：\(\dfrac{1}{4}\) \(\infty\)
5、解析：左侧
6、解析：\(4e^{-2t}\) \(2t+e^{-2t}-1\) \(\dfrac{4}{s+2}\)
7、解析：复变函数
8、解析：\(r=\begin{cases}t & t\geqslant 0\\0 & t<0\end{cases}\)
9、解析：稳态误差
10、解析：\(\xi, w_n\)
11、解析：过阻尼系统
12、解析：
\[G(s) = \frac{1}{s^2}+\frac{0.9}{s}-\frac{0.9}{s+10} = \frac{s+10+0.9s(s+10)-0.9s^2}{s^2(s+10)} = \frac{10s+10}{s^2(s+10)}\]
\[R(s) = \frac{1}{s}+\frac{1}{s^2} = \frac{s+1}{s^2}\]
故 \(\Phi(s) = \dfrac{c(s)}{R(s)} = \dfrac{10s+10}{s^2(s+10)}\times\dfrac{s^2}{s+1} = \dfrac{10}{s+10}\) 单位负反馈\(H(s)=1\)
故
\[\Phi(s) = \frac{G(s)}{1+G(s)} \rightarrow (1+G)\Phi = G \rightarrow G(\Phi-1) = -\Phi \rightarrow G = \frac{\Phi}{1-\Phi}\]
\[G = \frac{\dfrac{10}{s+10}}{1-\dfrac{10}{s+10}} = \frac{\dfrac{10}{s+10}}{\dfrac{s+10-10}{s+10}} = \frac{10}{s}\]
且 \(\Phi(s) = \dfrac{1}{1+\dfrac{10}{s}} = \dfrac{s}{s+10}\)，故 \(\lim\limits_{s\to\infty} s\cdot\left[\dfrac{s}{s+10}\cdot\dfrac{s+1}{s^2}\right] = \lim\limits_{s\to\infty}\dfrac{s-1}{s+10} = 0.1\)
13、解析：\(T=0.01, K=1\)
\[G(s) = \frac{K(1-0.5s)}{s(Ts+1)} \Rightarrow \phi(s) = \frac{G(s)}{G(s)+1} = \frac{K(1-0.5s)}{s(Ts+1)+K(1-0.5s)}\]
\[= \frac{K(1-0.5s)}{Ts^2+(1-0.5K)s+K}\]
由题干可知，\(\omega = 10\)