考研851 自动控制原理
真题 · 真题/手写解答

(左上角,撕角/透影处,字迹残缺,见 uncertain)

\[\approx \dfrac{X(s)}{N(s)}\]

(左侧竖排,疑似相邻页透影,非本页正文,未转写)


拉氏变换 时间函数 Z变换
\(\dfrac{1}{s+a}\) \(e^{-at}\) \(\dfrac{z}{z-e^{-aT}}\)
\(\dfrac{a}{s(s+a)}\) \(1-e^{-at}\) \(\dfrac{(1-e^{-aT})z}{(z-1)(z-e^{-aT})}\)
\(\dfrac{a}{s^{2}(s+a)}\) \(t-\dfrac{1}{a}(1-e^{-aT})\) \(\dfrac{Tz}{(z-1)^{2}}-\dfrac{(1-e^{-aT})z}{a(z-1)(z-e^{-aT})}\)

(表格左侧手写批注,曲线指向表格)

\[\lim_{s\to 0} sE(s)\]

① 令 \(N(s)=0.\)

\[C(s)=\dfrac{\dfrac{1}{s(s+1)}}{1+\dfrac{1}{s(s+1)}}\times (s+1)\cdot R(s) = \dfrac{s+1}{s^{2}+s+1}\cdot R(s)\]
\[\underline{E(s)}=R(s)-C(s)\]
\[=R(s)\left[\dfrac{s^{2}+s+1-s-1}{s^{2}+s+1}\right]=\dfrac{s^{2}}{s^{2}+s+1}R(s)\]
\[\dfrac{E(s)}{R(s)} \qquad\qquad e_{ssr}=\lim_{s\to 0} sE(s)\]

② 令 \(R(s)=0\)

\[\dfrac{C(s)}{N(s)}=\dfrac{1}{1+\dfrac{1}{s(s+1)}}\cdot(\ \ )\quad\checkmark\]
\[\underline{E(s)}=R(s)-C(s)=-C(s)\ \cdots\]
\[e_{ssn}=\lim_{s\to 0}sE(s)\]

(右下角,零散笔画 "lin / lu / kw" 等,字迹不成文,未转写)

\[e_{ss}=e_{ssr}+e_{ssn}\]