5.3.1 解答

5.3.1

1.

(a)

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

(b)

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

(c)

$${\cal L}\{\sin^{2}{t}\} = {\cal L}\{\frac{1 - \cos{2t}}{2}\} = \frac{1}{2}({\cal L}\{1\} - {\cal L}\{\cos{2t}\})$$

$$= \frac{1}{2}(\frac{1}{s} - \frac{s}{s^2 +4}) \ \ \ \framebox{終}$$

(d)公式より ${\cal L}\{\sin{t}\} = \frac{1}{s^2 + 1} = F(s)$．よって第一移動法則より

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

(e)公式より ${\cal L}\{t^{3}\} = \frac{3!}{s^4} = \frac{6}{s^4} = F(s)$．よって第一移動法則より

$${\cal L}\{t^{3}e^{3t}\} = F(s-3) = \frac{6}{(s-3)^{4}} \ \ \ \framebox{終}$$

(f)加法定理より

$${\cal L}\{\cos(t+a)\} = {\cal L}\{\cos{t}\cos{a} - \sin{t}\sin{a}\}$$

$$= \cos{a}{\cal L}\{\cos{t}\} - \sin{a}{\cal L}\{\sin{t}\}$$

$$= \frac{s\cos{a}}{s^2 + 1} - \frac{\sin{a}}{s^2 + 1} \ \ \framebox{終}$$

(g)$f(t)$を単位ステップ関数$u_{a}(t)$を用いて表すと，

$$f(t) = 2u_{0}(t) - 2u_{3}(t) + 2u_{6}(t)$$

よって

$${\cal L}\{f(t)\} = 2{\cal L}\{u_{0}(t)\} - 2{\cal L}\{u_{3}(t)\} ... ...{6}(t)\} = \frac{2}{s} - \frac{2e^{-3s}}{s} + \frac{2e^{-6s}}{s} \ \framebox{終}$$

(h) $f(t) = u_{\frac{\pi}{2}}(t)\sin{t}$より

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

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

(i) $f(t) = 2u_{0}(t) + (t^2 - 2)u_{2}(t) + (t - t^2)u_{4}(t) + (3 - t)u_{8}(t)$より第2移動法則を用いると

$${\cal L}\{f(t)\} = 2{\cal L}\{u_{0}(t)\} + {\cal L}\{(t^{2}-2)u_{2}(t)\} + {\cal L}\{(t-t^{2})u_{4}(t)\} + {\cal L}\{(3-t)u_{8}(t)\}$$

$$= \frac{2}{s} + e^{-2s}{\cal L}\{(t+2)^{2} - 2\} + e^{-4s}{\cal L}\{(t+4 - (t+4)^{2})\} + e^{-8s}{\cal L}\{3 - (t+8)\}$$

$$= \frac{2}{s} + e^{-2s}{\cal L}\{t^2 + 4t +2\} + e^{-4s}{\cal L}\{-(t^2 + 7t + 12)\} + e^{-8s}{\cal L}\{-t-5\}$$

$$= \frac{2}{s} +e^{-2s}(\frac{2}{s^3}+\frac{4}{s^2} + \frac{2}{s}) -... ...7}{s^2} + \frac{12}{s}) - e^{-8s}(\frac{1}{s^{2}} + \frac{5}{s}) \ \framebox{終}$$