技巧方法：应用留数法求部分分式的系数

• 本节课适用对象 对此方法感兴趣的学生
• 本节课所授知识点 应用留数法求部分分式的系数 1、针对单根的情况
• 公式 $F(s) = \frac{B(s)}{(s-s_1)(s-s_2)\cdots(s-s_n)} = \frac{c_1}{s-s_1} + \frac{c_2}{s-s_2} + \cdots + \frac{c_n}{s-s_n}$ $c_1 = \lim_{s \to s_1} [(s-s_1)F(s)]$ $\vdots$ $c_n = \lim_{s \to s_n} [(s-s_n)F(s)]$
• 例 $F(s) = \frac{s+2}{s^2+4s+3} = \frac{s+2}{(s+1)(s+3)} = \frac{c_1}{s+1} + \frac{c_2}{s+3}$ $c_1 = \lim_{s \to -1} [(s+1)F(s)] = \lim_{s \to -1} \frac{s+2}{s+3} = \frac{1}{2}$ $c_2 = \lim_{s \to -3} [(s+3)F(s)] = \lim_{s \to -3} \frac{s+2}{s+1} = \frac{1}{2}$ $F(s) = \frac{1}{2} \cdot \frac{1}{s+1} + \frac{1}{2} \cdot \frac{1}{s+3}$
• 例 $F(s) = \frac{s+3}{s^2+3s+2} = \frac{s+3}{(s+1)(s+2)} = \frac{a}{s+1} + \frac{b}{s+2}$ $a = \lim_{s \to -1} [(s+1)F(s)] = \lim_{s \to -1} \frac{s+3}{s+2} = 2$ $b = \lim_{s \to -2} [(s+2)F(s)] = \lim_{s \to -2} \frac{s+3}{s+1} = -1$ $F(s) = \frac{2}{s+1} + \frac{-1}{s+2}$ 2、针对重根的情况
• 公式 $F(s) = \frac{B(s)}{(s-s_1)^r} = \frac{c_r}{(s-s_1)^r} + \frac{c_{r-1}}{(s-s_1)^{r-1}} + \cdots + \frac{c_{r-j}}{(s-s_1)^{r-j}} + \cdots + \frac{c_1}{s-s_1}$ $c_r = \lim_{s \to s_1} \left[(s-s_1)^r F(s)\right]$ $c_{r-1} = \lim_{s \to s_1} \left\{\frac{d}{ds}\left[(s-s_1)^r F(s)\right]\right\}$ $\vdots$ $c_{r-j} = \lim_{s \to s_1} \left\{\frac{1}{j!} \cdot \frac{d^{(j)}}{ds^j}\left[(s-s_1)^r F(s)\right]\right\}$ $\vdots$ $c_1 = \lim_{s \to s_1} \left\{\frac{1}{(r-1)!} \cdot \frac{d^{(r-1)}}{ds^{r-1}}\left[(s-s_1)^r F(s)\right]\right\}$
• 例 $F(s) = \frac{s^2 + 2s + 3}{(s+1)^3} = \frac{a_3}{(s+1)^3} + \frac{a_2}{(s+1)^2} + \frac{a_1}{s+1}$ $a_3 = \lim_{s \to -1} \left[(s+1)^3 F(s)\right] = \lim_{s \to -1} (s^2 + 2s + 3) = 2$ $a_2 = \lim_{s \to -1} \left\{\frac{d}{ds}\left[(s+1)^3 F(s)\right]\right\} = \lim_{s \to -1} (2s+2) = 0$ $a_1 = \lim_{s \to -1} \left\{\frac{1}{2!} \cdot \frac{d^2}{ds^2}\left[(s+1)^3 F(s)\right]\right\} = \lim_{s \to -1} \frac{2}{2} = 1$ $F(s) = \frac{2}{(s+1)^3} + \frac{1}{s+1}$
• 例 $F(s) = \frac{s+2}{s(s+1)^2(s+3)} = \frac{a}{s} + \frac{b}{(s+1)^2} + \frac{c}{s+1} + \frac{d}{s+3}$ $a = \lim_{s \to 0} [s F(s)] = \lim_{s \to 0} \frac{s+2}{(s+1)^2(s+3)} = \frac{2}{3}$ $b = \lim_{s \to -1} \left[ (s+1)^2 F(s) \right] = \lim_{s \to -1} \frac{s+2}{s(s+3)} = -\frac{1}{2}$ $c = \lim_{s \to -1} \left\{ \frac{d}{ds} \left[ (s+1)^2 F(s) \right] \right\} = \lim_{s \to -1} \frac{s^2+3s-(2s+3)(s+2)}{(s^2+3s)^2} = -\frac{3}{4}$ $d = \lim_{s \to -3} \left[ (s+3) F(s) \right] = \lim_{s \to -3} \frac{s+2}{s(s+1)^2} = \frac{1}{12}$ $F(s) = \frac{2}{3} \cdot \frac{1}{s} - \frac{1}{2} \cdot \frac{1}{(s+1)^2} - \frac{3}{4} \cdot \frac{1}{s+1} + \frac{1}{12} \cdot \frac{1}{s+3}$