\[\Delta = 1-(L_1+L_2+L_3)+L_1L_2 = 1+\frac{9}{s(s+1)}+\frac{9}{0.5s+1}+81+\frac{81}{s(0.5s+1)(s+1)}\]
\[p_1=\frac{9}{s(s+1)}, \quad L_2 与 p_1 不接触, \quad \Delta_1=1+\frac{9}{0.5s+1}\]
\[p_2=81, \quad \Delta_2=1\]
由梅森增益公式可得传递函数为
\[\frac{C_1(s)}{R_1(s)}=\frac{\sum p_i\Delta_i}{\Delta}=\frac{\dfrac{9}{s(s+1)}\left(1+\dfrac{9}{0.5s+1}\right)+81}{1+\dfrac{9}{s(s+1)}+\dfrac{9}{0.5s+1}+81+\dfrac{81}{s(0.5s+1)(s+1)}}\]
\[=\frac{81s(0.5s+1)(s+1)+9(0.5s+1)+81}{82s(0.5s+1)(s+1)+9s(s+1)+9(0.5s+1)+81}\]
\[=\frac{40.5s^3+121.5s^2+85.5s+90}{41s^3+132s^2+95.5s+90}\]
(2) 求 \(C_1(s)/R_2(s)\)。
与(1)相比,梅森增益公式中\(\Delta\)不变,只是\(\sum p_i\Delta_i\)发生变化,此时系统有一条前向通道,即
\[p_1=9, \quad \Delta_1=1\]
由梅森增益公式可得传递函数为
\[\frac{C_1(s)}{R_2(s)}=\frac{\sum p_i\Delta_i}{\Delta}=\frac{9}{1+\dfrac{9}{s(s+1)}+\dfrac{9}{0.5s+1}+81+\dfrac{81}{s(0.5s+1)(s+1)}}\]
\[=\frac{9s(0.5s+1)(s+1)}{82s(0.5s+1)(s+1)+9s(s+1)+9(0.5s+1)+81}\]
\[=\frac{4.5s^3+13.5s^2+9s}{41s^3+132s^2+95.5s+90}\]
(3) 求 \(C_2(s)/R_1(s)\)。
此时系统有一条前向通道,即
\[p_1=-9, \quad \Delta_1=1\]
由梅森增益公式可得传递函数为
\[\frac{C_2(s)}{R_1(s)}=\frac{\sum p_i\Delta_i}{\Delta}=\frac{-9}{1+\dfrac{9}{s(s+1)}+\dfrac{9}{0.5s+1}+81+\dfrac{81}{s(0.5s+1)(s+1)}}\]
\[=\frac{-9s(0.5s+1)(s+1)}{82s(0.5s+1)(s+1)+9s(s+1)+9(0.5s+1)+81}\]
\[=-\frac{4.5s^3+13.5s^2+9s}{41s^3+132s^2+95.5s+90}\]
(4) 求 \(C_2(s)/R_2(s)\)。
此时系统有两条前向通道,即
\[p_1=\frac{9}{0.5s+1}, \quad L_1 与 p_1 不接触, \quad \Delta_1=1+\frac{9}{s(s+1)}\]
\[p_2=81, \quad \Delta_2=1\]
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