(2) 图2-2(b)
\[F(s) = \mathscr{L}[f(t)] = \int_0^{+\infty} f(t)e^{-st}\,dt = \int_0^{t_0} t e^{-st}\,dt = \frac{1-(1+t_0 s)e^{-t_0 s}}{s^2}\]
(3) 图2-2(c)
\[F(s) = \mathscr{L}[f(t)] = \int_0^{+\infty} f(t)e^{-st}\,dt = \int_0^{t_0} A\sin\left(\frac{\pi}{t_0}t\right)e^{-st}\,dt\]
\[= \frac{A\pi}{s t_0}\int_0^{t_0} e^{-st}\cos\left(\frac{\pi}{t_0}t\right)dt = \frac{A\pi}{s^2 t_0}\left(1+e^{-st_0} - \frac{\pi}{t_0}\int_0^{t_0} e^{-st}\sin\left(\frac{\pi}{t_0}t\right)dt\right)\]
即
\[F(s) = \frac{A\pi}{s^2 t_0}\left((1+e^{-st_0}) - \frac{\pi}{A t_0}F(s)\right)\]
故
\[F(s) = \frac{A\cdot \pi/t_0}{s^2+(\pi/t_0)^2}(1+e^{-st_0})\]
(4) 图2-2(d)
\[F(s) = \mathscr{L}[f(t)] = \int_0^{+\infty} f(t)e^{-st}\,dt\]
\[= \int_0^{T/2} e^{-st}\,dt - \int_{T/2}^{T} e^{-st}\,dt + \int_{T}^{3T/2} e^{-st}\,dt - \int_{3T/2}^{2T} e^{-st}\,dt + \cdots\]
\[= -\frac{1}{s}\left(e^{-s(1\cdot T/2)} - e^{-s(0\cdot T/2)} + e^{-s(3T/2)} - e^{-s(2\cdot T/2)} + \cdots\right)\]
\[+ \frac{1}{s}\left(e^{-s(2\cdot T/2)} - e^{-s(1\cdot T/2)} + e^{-s(4\cdot T/2)} - e^{-s(3\cdot T/2)} + \cdots\right)\]
\[= \frac{1}{s} + \frac{2}{s}\sum_{i=1}^{+\infty} e^{-isT} - \frac{2}{s}\sum_{i=1}^{+\infty} e^{-isT/2} = \frac{1}{s}\tanh\frac{sT}{4}\]
2-5 求下列函数的拉普拉斯反变换 \(f(t)\)。
(1) \(F(s)=\dfrac{s+1}{(s+2)(s+3)}\)； (2) \(F(s)=\dfrac{s}{(s+1)^2(s+2)}\)；
(3) \(F(s)=\dfrac{2s^2-5s+1}{s(s^2+1)}\)； (4) \(F(s)=\dfrac{s+2}{s(s+1)^2(s+3)}\)。
解 (1) \(f(t) = \mathscr{L}^{-1}[F(s)] = \mathscr{L}^{-1}\left[\dfrac{s+1}{(s+2)(s+3)}\right]\)
\[= \mathscr{L}^{-1}\left(\frac{2}{s+3}-\frac{1}{s+2}\right) = 2e^{-3t} - e^{-2t}\]
(2) 由于 \(\mathscr{L}^{-1}\left[\dfrac{s}{(s+a)(s+b)^2}\right]=\dfrac{[a-b(a-b)t]e^{-bt}-ae^{-at}}{(a-b)^2}\)，则
\[f(t) = \mathscr{L}^{-1}[F(s)] = \mathscr{L}^{-1}\left[\frac{s}{(s+1)^2(s+2)}\right]\]
\[= \frac{[2-(2-1)t]e^{-t}-2e^{-2t}}{(2-1)^2} = (2-t)e^{-t} - 2e^{-2t}\]
(3) \(f(t) = \mathscr{L}^{-1}[F(s)] = \mathscr{L}^{-1}\left[\dfrac{2s^2-5s+1}{s(s^2+1)}\right]\)
\[= \mathscr{L}^{-1}\left[\frac{1}{s}+\frac{s}{s^2+1}-\frac{5}{s^2+1}\right] = 1+\cos t - 5\sin t\]
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