图 7-3(d)采样系统 \(C(z)=G(z)E(z)\)
\[E(s)=R(s)-C^*(s)H(s), \quad E(z)=R(z)-C(z)H(z)\]
则有
\[C(z)=G(z)[R(z)-C(z)H(z)], \quad C(z)=\dfrac{G(z)R(z)}{1+G(z)H(z)}\]
图 7-3(e)采样系统 \(C(s)=G(s)E(s)\)
\[E(s)=R(s)-H(s)C^*(s)\]
\[C(s)=G(s)[R(s)-H(s)C^*(s)], \quad C^*(s)=GR^*(s)-GH^*(s)C^*(s)\]
\[C^*(s)=\dfrac{GR^*(s)}{1+GH^*(s)}, \quad C(z)=\dfrac{GR(z)}{1+GH(z)}\]
图 7-3(f)采样系统 \(C(s)=G_2(s)E_2^*(s)\)
\[E_2(s)=G_1(s)E_1(s)=G_1(s)[R(s)-H(s)C(s)]\]
\[=G_1(s)[R(s)-H(s)G_2(s)E_2^*(s)]\]
\[E_2^*(s)=\dfrac{G_1R^*(s)}{1+G_1G_2H^*(s)}\]
\[C^*(s)=G_2^*(s)E_2^*(s)=\dfrac{G_2^*(s)G_1R^*(s)}{1+G_1G_2H^*(s)}\]
\[C(z)=\dfrac{G_2(z)G_1R(z)}{1+G_1G_2H(z)}\]
7-10 采样系统的结构图如图 7-4 所示,试求开环脉冲传递函数 \(G(z)\) 和闭环脉冲传递函数 \(\Phi(z)\)。
图 7-4 闭环采样系统结构图
解 开环脉冲传递函数为
\[G(z)=\mathscr{Z}\left[\dfrac{(1-\mathrm{e}^{-sT})K}{s^2(s+a)}\right]=(1-z^{-1})\mathscr{Z}\left[\dfrac{1}{s^2(s+a)}\right]\]
\[=(1-z^{-1})\cdot\dfrac{1}{a^2}\mathscr{Z}\left[\dfrac{a}{s^2}-\dfrac{1}{s}+\dfrac{1}{s+a}\right]\]
\[=(1-z^{-1})\cdot\dfrac{1}{a^2}\left[\dfrac{aTz}{(z-1)^2}-\dfrac{z}{z-1}+\dfrac{z}{z-\mathrm{e}^{-aT}}\right]\]
\[=\dfrac{(aT+\mathrm{e}^{-aT}-1)z+1-(aT+1)\mathrm{e}^{-aT}}{a^2(z-1)(z-\mathrm{e}^{-aT})}\]
闭环脉冲传递函数为
\[\Phi(z)=\dfrac{G(z)}{1+G(z)}=\dfrac{(aT+\mathrm{e}^{-aT}-1)z+1-(aT+1)\mathrm{e}^{-aT}}{a^2z^2+(aT-a^2-a^2\mathrm{e}^{-aT}+\mathrm{e}^{-aT}-1)z+1+(a^2-aT-1)\mathrm{e}^{-aT}}\]
7-11 已知 RC 电路图如图 7-5 所示,其中 \(r(t)=100\mathrm{e}^{-t}\),试求其输出 \(c(nT)\)。
解 RC 电路传递函数为