四、

\(N(s)=0\) 时,\(G(s) = \dfrac{10(2s+1)}{s^2}\)，2 型系统，故 \(e_{ssr} = 0\)

\(R(s) = 0,\dfrac{E(s)}{N(s)} = \dfrac{0-C(s)}{N(s)} = -\dfrac{10s}{s^2+20s+10}\)

\(e_{ssn} = \lim_{s \to 0} s \cdot \dfrac{1}{s} \cdot \left(-\dfrac{10s}{s^2+20s+10}\right) = 0\)

故误差为 0

五、

\(G(s)H(s) = \dfrac{K}{s^2(3s+1)}\)

相角范围：\(-180° \to -270°\)

令 \(s=j\omega\)

\(G(j\omega) = -\dfrac{k}{\omega^2}\dfrac{1-j3\omega}{1+9\omega^2}\) 只有第二象限

如图所示：

\(N = N_+ - N_- = 0÷1=-1\)

\(Z=P-2N=2\) 不稳定，有两个不稳定根

六、

(1)

\(N(A) = \dfrac{4M}{\pi A} \Rightarrow -\dfrac{1}{N(A)} = -\dfrac{\pi A}{4M}\)

页码: 67