根据系统的结构图可绘制系统的信号流图，如图 2-39 所示。
图 2-39 系统(2)的信号流图
① 求 \(C(s)/R(s)\)。由图可知，此时系统有两条前向通道，一个单独回路，即
\[L_1 = -\frac{K}{(s+1)(Ts+1)}, \quad \Delta = 1 - L_1 = 1 + \frac{K}{(s+1)(Ts+1)}\]
\[p_1 = \frac{K}{(s+1)(Ts+1)}, \quad \Delta_1 = 1\]
\[p_2 = \frac{\tau s}{(s+1)(Ts+1)}, \quad \Delta_2 = 1\]
由梅森增益公式可得系统的传递函数为
\[\frac{C(s)}{R(s)} = \frac{\sum p_i \Delta_i}{\Delta} = \left(\frac{K+\tau s}{(s+1)(Ts+1)}\right) \Bigg/ \left(1 + \frac{K}{(s+1)(Ts+1)}\right) = \frac{\tau s + K}{(s+1)(Ts+1)+K}\]
② 求 \(C(s)/N(s)\)。由图可知，此时系统有两条前向通道，一个单独回路，即
\[L_1 = -\frac{K}{(s+1)(Ts+1)}, \quad \Delta = 1 - L_1 = 1 + \frac{K}{(s+1)(Ts+1)}\]
\[p_1 = 1, \quad \Delta_1 = 1; \qquad p_2 = 1, \quad \Delta_2 = 1\]
由梅森增益公式可得系统的传递函数为
\[\frac{C(s)}{N(s)} = \frac{\sum p_i \Delta_i}{\Delta} = 2 \Bigg/ \left(1 + \frac{K}{(s+1)(Ts+1)}\right) = \frac{2(s+1)(Ts+1)}{(s+1)(Ts+1)+K}\]
又解 可用消去法对结果进行验证。
(1) 对系统(1)的方程作拉普拉斯变换，整理后可得
\[C(s) = X_1(s) + K_3 R(s)\]
\[X_1(s) = \frac{1}{s+T_1}X_2(s) + \frac{K_2}{s+T_1}R(s), \quad X_2(s) = -\frac{T_2}{s}X_1(s) + \frac{K_1}{s}R(s)\]
则
\[\frac{s(s+T_1)+T_2}{s(s+T_1)}X_1(s) = \frac{K_1+sK_2}{s(s+T_1)}R(s)\]
即
\[X_1(s) = \frac{K_1+sK_2}{s(s+T_1)+T_2}R(s)\]
将 \(X_1(s)\) 代入 \(C(s)=X_1(s)+K_3R(s)\)，可得
\[\frac{C(s)}{R(s)} = K_3 + \frac{K_1+K_2s}{T_2+T_1s+s^2}\]
(2) 方法同上，可解得系统(2)的传递函数
\[\frac{C(s)}{R(s)} = \frac{\tau s + K}{(s+1)(Ts+1)+K}, \qquad \frac{C(s)}{N(s)} = \frac{2(s+1)(Ts+1)}{(s+1)(Ts+1)+K}\]
·41·