图 7-5 RC 电路图
\[
G_c(s) = \frac{1}{1+RCs} = \frac{1}{1+s}
\]
因为
\[
C(s) = G_c(s)R^*(s)
\]
则有
\[
C(z) = G_c(z)R(z)
\]
\[
= \frac{z}{z-\mathrm{e}^{-T}} \cdot \frac{100z}{z-\mathrm{e}^{-T}} = \frac{100z^2}{(z-\mathrm{e}^{-T})^2}
\]
采用反演积分法,可得
\[
c(nT) = \mathrm{Res}\left[\frac{100z^2}{(z-\mathrm{e}^{-T})^2}\right]_{z\to \mathrm{e}^{-T}} = \frac{1}{1!}\lim_{z\to \mathrm{e}^{-T}}\frac{\mathrm{d}}{\mathrm{d}z}\left[\frac{(z-\mathrm{e}^{-T})^2 \cdot z^{n-1} \cdot 100z^2}{(z-\mathrm{e}^{-T})^2}\right]
\]
\[
= \lim_{z\to \mathrm{e}^{-T}}\left[100(n+1)z^n\right] = 100(n+1)\mathrm{e}^{-nT}
\]
幂级数法验证
\[
C(z) = \frac{100z^2}{(z-\mathrm{e}^{-T})^2} = \frac{100}{1-2\mathrm{e}^{-T}z^{-1}+\mathrm{e}^{-2T}z^{-2}}
\]
\[
= 100(1+2\mathrm{e}^{-T}z^{-1}+3\mathrm{e}^{-2T}z^{-2}+\cdots)
\]
\[
c(nT) = 100(n+1)\mathrm{e}^{-nT}
\]
7-12 已知采样系统结构图如图 7-6 所示,试判断系统的稳定性。
图 7-6 采样系统结构图
解 开环脉冲传递函数为
\[
G_hG_0(z) = \mathscr{Z}\left[\frac{1-\mathrm{e}^{-Ts}}{s}\cdot\frac{1.5}{s(0.1s+1)(0.05s+1)}\right]
\]
\[
= (1-z^{-1})\mathscr{Z}\left[\frac{1.5}{s^2(0.1s+1)(0.05s+1)}\right]
\]
\[
= 1.5(1-z^{-1})\mathscr{Z}\left[-\frac{0.15}{s}+\frac{0.2}{s+10}-\frac{0.05}{s+20}+\frac{1}{s^2}\right]
\]
\[
= 1.5(1-z^{-1})\left[-\frac{0.15z}{z-1}+\frac{0.2z}{z-\mathrm{e}^{-10}}-\frac{0.05z}{z-\mathrm{e}^{-20}}+\frac{z}{(z-1)^2}\right]
\]
\[
= \frac{1.275z+0.225}{z^2-z}
\]
闭环脉冲传递函数为
\[
\Phi(z) = \frac{G_hG_0(z)}{1+G_hG_0(z)} = \frac{1.275z+0.225}{z^2+0.275z+0.225}
\]
闭环特征方程为
\[
D(z) = z^2+0.275z+0.225 = 0
\]
求得特征根为
\[
z_{1,2} = -0.1375\pm0.454\mathrm{j}
\]
显然,系统的特征根全部在单位圆内,所以离散系统是稳定的。
・382・