考研851 自动控制原理
真题 · 真题答案

三、

(1)

\[G(s) = \dfrac{4}{s(s+1)^2}\]

相角范围:\(-90°\) \(\to\) \(-270°\)

\(s=j\omega\) 代入得

\[G(j\omega) = -4\dfrac{2\omega^2+j\omega(1-\omega^2)}{4\omega^4+(\omega-\omega^3)^2}\]

过第三象限和第二象限

与虚轴交点:\(G(j1)=-2\)

图:北方工业大学真题__北方工业大学真题及其答案__06--14,17,19--21真题及答案__06-14答案__扫描全能王_2024-07-10_13.47_07_fig1

(2)

\(N = N_+ - N_- = 0-1=-1\), \(Z = P-2N = 2\),系统不稳定

(3)

\[N(A) = \dfrac{4}{A\pi}\sqrt{1-\left(\dfrac{1}{A}\right)^2},\ A\ge1\]

\(A=1\) 时, \(-\dfrac{1}{N(A)}\to-\infty\),\(A\to+\infty\),\(-\dfrac{1}{N(A)}\to-\infty\)

\(-\dfrac{1}{N(A)}\)求导,得 \(A = 1.28\),\(-\dfrac{1}{N(A)}=-1.61\) 两曲线有交点从不稳定区到稳定区,系统有自振

令两曲线实部和虚部相等,可得:\(A = 2.29\),\(f=\dfrac{\omega}{2\pi}=0.16HZ\)

32