# 拉普拉斯变换计算示例及解决方案

## 拉普拉斯变换的定义

$$\displaystyle F(s) = \int_{0}^{+\infty} f(t) e^{-st} dt$$

$$\displaystyle F(s) = \int_{0}^{+\infty} e^{-st} dt$$

$$\displaystyle F(S) = \lim_{T \to +\infty} \left[ -\dfrac{1}{s} e^{-st} \right]_{0}^{T}$$

$$\quad \quad \displaystyle = \lim_{T \to +\infty} - \dfrac{e^{-sT} - e^{0}}{s}$$

$F(S) = \dfrac{1}{s}$

$$\displaystyle F(s) = \int_{0}^{+\infty} e^{at} e^{-st} dt$$

$$\displaystyle \quad \quad = \int_{0}^{+\infty} e^{(a-s)t} dt$$

$$F(S) = \lim_{T \to +\infty} \left[ \dfrac{1}{a - s} e^{(a-s)t} \right]_{0}^{T}$$

$$\displaystyle \quad \quad = \lim_{T \to +\infty} \dfrac{e^{(a-s)T} - e^{0}}{a-s}$$

$F(S) = \dfrac{1}{s - a}$

$$\displaystyle F(s) = \int_{0}^{+\infty} \sin(\omega t) e^{-st} dt$$

$$\sin(\omega t) = \dfrac{e^{j \omega t } - e^{ - j \omega t }}{2 j}$$

$$\displaystyle F(s) = \int_{0}^{+\infty} \dfrac{e^{j \omega t } - e^{ - j \omega t }}{2 j} e^{-st} dt$$

$$\quad \quad \displaystyle = \int_{0}^{+\infty} \dfrac{e^{j \omega t} e^{- s t}}{2 j} dt - \int_{0}^{+\infty} \dfrac{ e^{-j \omega t}e^{ - st}}{2 j} dt$$

$$\quad \quad \displaystyle = \int_{0}^{+\infty} \dfrac{e^{ (j \omega - s) t}}{2 j} dt - \int_{0}^{+\infty} \dfrac{ e^{ -(j \omega + s) t}}{2 j} dt$$

$$\quad \quad \displaystyle = \lim_{T \to +\infty} \left[ \dfrac{1}{2j( j \omega - s)} e^{(j\omega-s)t} \right]_{0}^{T} - \lim_{T \to +\infty} \left[ \dfrac{1}{-2j( j \omega + s)} e^{ - (j\omega+s)t} \right]_{0}^{T}$$

$$\quad \quad \displaystyle = \lim_{T \to +\infty} \dfrac{e^{(j\omega-s)T} - e^0}{2j( j \omega - s)} - \lim_{T \to +\infty} \dfrac{e^{ - (j\omega-s)T }- e^0}{-2j( j \omega + s)}$$

$$\displaystyle F(s) = - \dfrac {1}{2j( j \omega - s)} - \dfrac {1}{2j( j \omega + s)}$$

$\displaystyle F(s) = \dfrac{\omega}{\omega^2+s^2}$

$$\displaystyle F(s) = \int_{0}^{+\infty} \cosh(\omega t) e^{-st} dt$$

$$\cosh(\omega t) = \dfrac{e^{\omega t}+e^{-\omega t}}{2}$$

$$\displaystyle F(s) = \int_{0}^{+\infty} \dfrac{e^{\omega t } + e^{-\omega t }}{2 } e^{-st} dt$$

$$\quad \quad \displaystyle = \int_{0}^{+\infty} \dfrac{e^{ \omega t} e^{ - s t }}{2} dt + \int_{0}^{+\infty} \dfrac{ e^{ -\omega t}e^{ - st}}{2} dt$$

$$\quad \quad \displaystyle = \int_{0}^{+\infty} \dfrac{e^{ (\omega - s)t }}{2} dt + \int_{0}^{+\infty} \dfrac{ e^{ -(\omega + s)t}}{2} dt$$

$$\quad \quad \displaystyle = \lim_{T \to +\infty} \left[ \dfrac{1}{2( \omega - s)} e^{(\omega-s)t} \right]_{0}^{T} + \lim_{T \to +\infty} \left[ \dfrac{1}{ - 2( \omega + s)} e^{ - (\omega+s)t} \right]_{0}^{T}$$

$$\quad \quad \displaystyle = \lim_{T \to +\infty} \dfrac{e^{(\omega-s)T}-e^0}{2( \omega - s)} + \lim_{T \to +\infty} \dfrac{ e^{ - (\omega+s)T} - e^0 }{ - 2( \omega + s)}$$

$$\quad \quad \displaystyle F(s) = \dfrac{-1}{2(\omega - s)} + \dfrac{-1}{-2(\omega + s)}$$

$\displaystyle F(s) = \dfrac{s}{s^2 - \omega^2}$

$\displaystyle \int_{A}^{B} f(t) \delta(t - a) dt = \begin{cases} 1 & \text{当} A \lt a \lt B \\ 0 & \text{否则} \\ \end{cases}$
a)

$$\displaystyle \mathscr{L}\{\delta(t)\} = \int_{0^{-}}^{+\infty} \delta(t) e^{-st} dt$$

$$\displaystyle \mathscr{L}\{\delta(t)\} = \int_{0^{-}}^{+\infty} \delta(t) e^{-st} dt = e^0 = 1$$
b)

$$\displaystyle \mathscr{L}\{\delta(t - a)\} = \int_{0}^{+\infty} \delta(t - a) e^{-st} dt = e^{-as}$$

Dirac Delta 函数