5.4 解答

5.4

1.

(a) 逆変換公式を用いる．

$${\cal L}^{-1}\{\frac{5}{(s-2)^{6}}\} = 5\frac{t^5 e^{2t}}{5!} = \frac{t^{5}e^{2t}}{4!} \ \ \framebox{終}$$

(b)

$${\cal L}^{-1}\{\frac{s}{(s+2)^{2}+1}\} = {\cal L}^{-1}\{\frac{s+2-2}{(s+2)^{2}+1}\}$$

$$= {\cal L}^{-1}\{\frac{s+2}{(s+2)^{2}+1}\} - {\cal L}^{-1}\{\frac{2}{(s+2)^{2}+1}\}$$

$$= e^{-2t}\cos{t} - 2e^{-2t}\sin{t} \ \ \ \framebox{終}$$

(c)部分分数分解を行なうと

$$\frac{1}{s^2 + 3s +2} = \frac{1}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}$$

$$A = (s+1)\frac{1}{(s+1)(s+2)}\mid_{s=-1} = 1, \ B = (s+2)\frac{1}{(s+1)(s+2)}\mid_{s = -2} = -1$$

よって

$${\cal L}^{-1}\{\frac{1}{s^2 + 3s +2}\} = {\cal L}^{-1}\{\frac{1}{s+1}\} + {\cal L}^{-1}\{\frac{-1}{s+2}\} = e^{-t} - e^{-2t} \ \ \framebox{終}$$

(d)

$${\cal L}^{-1}\{\frac{s+1}{s^2 + 2s +2}\} = {\cal L}^{-1}\{\frac{s+1}{(s+1)^2 + 1}\} = e^{-t}\cos{t} \ \ \framebox{終}$$

(e) 部分分数分解を行なうと

$$\frac{s^2}{(s+1)(s+2)(s+3)} = \frac{A}{s+1} + \frac{B}{s+2} + \frac{C}{s+3}$$

$$A = (s+1)\frac{s^2}{(s+1)(s+2)(s+3)}\mid_{s=-1} = \frac{1}{2}$$

$$B = (s+2)\frac{s^2}{(s+1)(s+2)(s+3)}\mid_{s=-2} = -4$$

$$C = (s+3)\frac{s^2}{(s+1)(s+2)(s+3)}\mid_{s=-3} = \frac{9}{2}$$

よって

$${\cal L}^{-1}\{\frac{s^2}{(s+1)(s+2)(s+3)}\} = {\cal L}^{-1}\{\frac{1/2}{s+1} + \frac{-4}{s+2} + \frac{9/2}{s+3} \}$$

$$= \frac{1}{2}{\cal L}^{-1}\{\frac{1}{s+1}\} -4{\cal L}^{-1}\{\frac{1}{s+4}\} + \frac{9}{2}{\cal L}^{-1}\{\frac{1}{s+3}\}$$

$$= \frac{e^{-t}}{2} - 4e^{-2t} + \frac{9e^{-3t}}{2} \ \ \ \framebox{終}$$

(f) 部分分数分解を行なうと

$$\frac{3s^2 +4}{s^4 + s^2} = \frac{3s^2 + 4}{s^2(s^2 + 1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs +D}{s^2 + 1}$$

$$B = s^2(\frac{3s^2 +4}{s^2(s^2 + 1)}\mid_{s=0} = 4$$

$$A = (\frac{3s^2 + 4}{s^2 +1})^{\prime}\mid_{s=0} = \frac{6s(s^2 + 1) - (3s^2 + 4)(2s)}{(s^2 + 1)^2}\mid_{s=0} = 0$$

$$Cs+D\mid_{s=i} = \frac{3s^2 + 4}{s^2}\mid_{s=i} = -1 \rightarrow Ci+D = -1 \rightarrow C = 0, D = -1$$

よって

$${\cal L}^{-1}\{\frac{3s^2 + 4}{s^4 + s^2}\} = {\cal L}^{-1}\{\frac{4}{s^2}\} - {\cal L}^{-1}\{\frac{1}{s^2 +1}\} = 4t - \sin{t} \ \ \framebox{終}$$

(g) 部分分数分解を行なうと

$${\cal L}^{-1}\{\frac{1+e^{-2s}}{(s^2 + 1)(s^2 + 4)}\} = {\cal L}^... ...1}{(s^2 + 1)(s^2 + 4)}\} + {\cal L}^{-1}\{\frac{e^{-2s}}{(s^2 + 1)(s^2 + 4)}\}$$

ここで

$$f(t) = {\cal L}^{-1}\{\frac{1}{(s^2 + 1)(s^2 + 4)}\} = {\cal L}^{-1}\{\frac{As+B}{s^2 + 1} + \frac{Cs+D}{s^2 + 4}\}$$

とおく

$$As+B\mid_{s=i} = (s^2 +1)(\frac{1}{(s^2 +1)(s^2 +4)})\mid_{s=i} = \frac{1}{3} \rightarrow Ai+B = \frac{1}{3} \rightarrow A = 0, B = \frac{1}{3}$$

$$Cs+D\mid_{s=2i} = \frac{1}{s^2 + 1}\mid_{s=2i} = -\frac{1}{3} \rightarrow Ci+D = -\frac{1}{3} \rightarrow C= 0, D = -\frac{1}{3}$$

よって

$$f(t) = {\cal L}^{-1}\{\frac{1}{(s^2 + 1)(s^2 + 4)}\} = {\cal L}^{... .../3}{s^2 + 1} + \frac{-1/3}{s^2 + 4}\} = \frac{\sin{t}}{3} - \frac{\sin{2t}}{6}$$

第2移動法則より

$${\cal L}^{-1}\{\frac{e^{-2s}}{(s^2 + 1)(s^2 + 4)}\} = u_{2}(t)f(t-2) = u_{2}(t)(\frac{\sin{(t-2)}}{3} - \frac{\sin{2(t-2)}}{6})$$

したがって

$${\cal L}^{-1}\{\frac{1+e^{-2s}}{(s^2 + 1)(s^2 + 4)}\} = \frac{\si... ...}{6} + u_{2}(t)(\frac{\sin{(t-2)}}{3} - \frac{\sin{2(t-2)}}{6}) \ \framebox{終}$$