(接上题，第六题解答续)

(1) K=10，20lgK=20，转折频率1，20，斜率变化-20dB/dec

前一个积分环节，低频段斜率为-20dB/dec。

(2) 由图 \(\dfrac{-20}{\lg\omega_c - \lg 1} = -40\)，\(\lg\omega_c = \dfrac{1}{2}\)，\(\omega_c = \sqrt{10}\)

\(\gamma = 180° - 90° - tg^{-1}\omega_c - tg^{-1}0.05\omega_c = 90° - 72.45° - 8.9° = 8.56°\)

(3) \(G_c(s) = \dfrac{0.1s+1}{0.125s+)}\)

(4) 超前校正. wc个 快速性 改善动态性能.

七. (10分)

离散系统如图所示，其中采样周期 T=1，试求单位阶跃响应前四次的采样值。

图3

已知：\(Z[1(t)] = \dfrac{z}{z-1}\)，\(Z\left[\dfrac{1}{s+a}\right] = \dfrac{z}{z-e^{-aT}}\)

甲：\(G(z) = Z\left[\dfrac{1}{s+1}\right] = \dfrac{z}{z-e^{-1}}\)

\(\Phi(z) = \dfrac{G(z)}{1+G(z)} = \dfrac{z}{2z-e^{-1}}\)

\(R(z) = \dfrac{z}{z-1}\)

\(D(z) = 2z - e^{-1} = 0\)，\(z = \dfrac{e^{-1}}{2} =\)

\(|z| < 1\)（系统稳定）

\(C(z) = \Phi(z)R(z) = \dfrac{z}{2z-0.368} \cdot \dfrac{z}{z-1} = \dfrac{z^2}{2z^2-1.368z+0.368} = \dfrac{1}{2-1.368z^{-1}+0.368z^{-2}}\)

乙：\(\Phi(z) = \dfrac{G(z)}{1+G(z)} = \dfrac{\dfrac{z}{z-e^{-1}}}{\dfrac{2z-e^{-1}}{z-e^{-1}}} = \dfrac{z}{2z-e^{-1}}\)

\(C(z) = \dfrac{z}{2z-e^{-1}} \cdot \dfrac{z}{z-1} = \dfrac{z^2}{2z^2-\cdots}\)