四、
\[G(z) = k(1-z^{-1})Z\left[\frac{1}{s^2(s+1)}\right] = k\frac{0.632z+0.264}{(z-1)(z-0.368)}\]
\[\Rightarrow D(z) = (z-1)(z-0.368) + k(0.632z+0.264) =\]
\[z^2 + z(0.632k-1.368) + 0.368+0.264k\]
朱莉判据:
\[D = z^2 + z(0.632k-1.368) + 0.368+0.264k\]
\[\begin{cases} D(1)>0 \Rightarrow k>0 \\ D(-1)>0 \Rightarrow k>0 \\ |0.368+0.264k|<1 \Rightarrow -1<0.368+0.264k<1 \Rightarrow k<2.4 \end{cases} \Rightarrow 0<k<2.4\]
五、
(1)
\[G(s) = \frac{k^*}{s(s+2)(s+4)}\]
开环极点:0, -2, -4;无零点
渐近线:\(\sigma_a = \dfrac{-6}{3-0} = -2, \varphi_a = \dfrac{(2k+1)\pi}{3-0} (k=0,1,2)\)
分离点:\(\dfrac{dG(s)}{ds} = 0 \Rightarrow s=-0.85, k=3.1\)
与虚轴的交点:
\[D(j\omega) = k - 6\omega^2 + j(8\omega-\omega^3) = 0 \Rightarrow \omega = \pm 2\sqrt{2}, k=48\]
如图所示:
60