\[
\frac{C(s)}{R(s)}=\frac{\dfrac{G_3G_5+G_2G_4}{1+G_4H_2+G_5H_2}\cdot\dfrac{G_1G_6}{1+G_1H_1}}{1+\dfrac{G_3G_5+G_2G_4}{1+G_4H_2+G_5H_2}\cdot\dfrac{G_1G_6}{1+G_1H_1}\cdot(H_4-H_3)}
\]
\[
=\frac{G_1G_6(G_2G_4+G_3G_5)}{(1+G_1H_1)(1+G_4H_2+G_5H_2)+G_1G_6(G_2G_4+G_3G_5)(H_4-H_3)}
\]
又解 可用信号流图方法对结果进行验证。
图2-40系统的信号流图如图2-43所示。由图可知,本系统有两条前向通道,七个单独回路,其中两对回路互不接触,即
\[
L_1=-G_1H_1,\qquad L_2=-G_4H_2,\qquad L_3=-G_5H_2,\qquad L_4=G_1G_2G_4G_6H_3
\]
\[
L_5=-G_1G_2G_4G_6H_4,\qquad L_6=G_1G_3G_5G_6H_3,\qquad L_7=-G_1G_3G_5G_6H_4
\]
其中,\(L_1\)与\(L_2\)不接触,\(L_1L_2=G_1G_4H_1H_2\);\(L_1\)与\(L_3\)不接触,\(L_1L_3=G_1G_5H_1H_2\)。
\[
\Delta=1-\sum_{i=1}^{7}L_i+L_1L_2+L_1L_3
\]
\[
=(1+G_1H_1)(1+G_4H_2+G_5H_2)+G_1G_6(G_2G_4+G_3G_5)(H_4-H_3)
\]
\[
p_1=G_1G_2G_4G_6,\qquad \Delta_1=1
\]
\[
p_2=G_1G_3G_5G_6,\qquad \Delta_2=1
\]
由梅森增益公式可得系统的传递函数为
\[
\frac{C(s)}{R(s)}=\frac{\sum p_i\Delta_i}{\Delta}=\frac{G_1G_6(G_2G_4+G_3G_5)}{(1+G_1H_1)(1+G_4H_2+G_5H_2)+G_1G_6(G_2G_4+G_3G_5)(H_4-H_3)}
\]
2-31 试求图2-44所示系统的输出\(C_1(s)\)及\(C_2(s)\)的表达式。
图2-43 题2-30系统信号流图 图2-44 题2-31系统信号流图
解 由图2-44信号流图可见,该系统有\(R_1(s)\)和\(R_2(s)\)两个源节点,\(C_1(s)\)和\(C_2(s)\)两个阱节点;有六个单独回路,其回路增益分别为
\[
L_1=G_1H_3,\qquad L_2=G_2H_4,\qquad L_3=G_3H_2
\]
\[
L_4=G_4H_1,\qquad L_5=G_1H_1G_2H_2,\qquad L_6=G_4H_3G_3H_4
\]
其中,\(L_1\)和\(L_2\)为不接触回路,\(L_3\)和\(L_4\)也是不接触回路。因此,流图特征式为
\[
\Delta=1-(L_1+L_2+L_3+L_4+L_5+L_6)+L_1L_2+L_3L_4
\]
\[
=1-G_1H_3-G_2H_4-G_3H_2-G_4H_1-G_1H_1G_2H_2
\]
\[
-G_4H_3G_3H_4+G_1H_3G_2H_4+G_3H_2G_4H_1
\]
(1) 求\(C_1(s)\)表达式;从源节点\(R_1(s)\)到阱节点\(C_1(s)\)有两条前向通路,其总增益为
\[
p_1=G_1,\qquad p_2=G_3H_4G_4
\]