六、
\[h'(t) =- 10e^{-20t} + 10e^{-20t}\]
\[\Phi(s) = \frac{-10}{s + 20}+\frac{10}{s + 10} = \frac{100}{(s + 20)(s + 10)}\]
显然闭环系统极点均为负实根，所以响应曲线无超调
七、
\[G = \frac{10}{s(s + 10)} \Rightarrow G(z) = z\left\{\frac{1}{s} + \frac{-1}{s + 10}\right\} = \frac{z}{z - 1} - \frac{z}{z - 0.368}\]
\[= \frac{0.632z}{(z - 1)(z - 0.368)}\]
\[K_v = \lim_{z \to 1}(z - 1)\frac{0.632z}{(z - 1)(z - 0.368)} = 1\]
\[\Rightarrow e = \frac{T}{K_v} = 0.1\]
八、
\[E = 10R - 10C \Rightarrow \frac{E}{R} = 10 - 10\Phi\]
\[\Phi = 10\frac{\dfrac{5}{s(s + 1)}}{1 +\dfrac{50}{s(s + 1)}} = \frac{50}{s^2 + s + 50}\]
\[\Rightarrow \frac{E}{R} = 10 - 10\Phi = 10 - \frac{500}{s^2 + s + 50} = \frac{10s^2 + 10s}{s^2 + s + 50}\]
\[\Rightarrow e = \lim_{x \to 0} s\frac{1}{s^2}\frac{10s^2 + 10s}{s^2 + s + 50} = 0.2\]
90