Inverse Laplace Transform
反演积分公式
f(t)= \dfrac{1}{2\pi j} \int _{\beta-j\infty}^{\beta+j\infty}F(s) e^{ st }\, ds=\mathscr{L}^{-1}[F(s)] \quad t>0 \end{align}$$ 由像函数求像原函数的一般公式，积分路径是 $s$ 平面上的一条直线 $Re\; s=\beta$ ### 反演积分公式的推导 由[[拉普拉斯变换\|拉普拉斯变换]]知： $$\begin{align} F(s)=\mathscr{L}\{f(t)\} = \int_{0}^{+\infty} f(t)e^{-st} \, dt=\int _{-\infty}^{+\infty} f(t)u(t) e^{ -\sigma t }e^{ j\omega t }\, \mathrm{d}t=\mathscr{F}\left[ f(t)u(t)e^{ -\sigma t }\right] \end{align}$$ $$\begin{gathered} f(t)u(t)e^{ -\sigma t }=\dfrac{1}{2\pi} \int _{-\infty}^{+\infty} F(s)e^{ j\omega t } \, d\omega \\ f(t)u(t)= \dfrac{1}{2\pi j} \int _{\beta-j\infty}^{\beta+j\infty}F(s) e^{ st }\, ds \\ f(t)= \dfrac{1}{2\pi j} \int _{\beta-j\infty}^{\beta+j\infty}F(s) e^{ st }\, ds \quad t>0 \end{gathered}$$ ### 有理分式的拉普拉斯逆变换 先进行[[有理函数#有理分式的分解\|有理分式分解]]，使用[[留数\|留数]]法求极点（常见于控制系统的[[传递函数\|传递函数]]） $$\begin{align} F(s)=\frac{B(s)}{A(s)}=\frac{b_{0}s^{m}+b_{1}s^{m-1}+\cdots+b_{m}}{a_{0}s^{n}+a_{1}s^{n-1}+\cdots+a_{n}} \quad (m\leq n) \end{align}$$ - $B(s)=0$ 的解为**零点** - $A(s)=0$ 为 $F(s)$ 的特征方程，解为**极点** #### 1.互异的单实数极点 $$\begin{align} &F(s)=\frac{c_{1}}{s-p_{1}}+\frac{c_{2}}{s-p_{2}}+\cdots+\frac{c_{n}}{s-p_{n}} \\ &\mathscr{L}^{-1}\left[ \frac{1}{s-p_{i}} \right]=e^{ p_{i}t } \\ &f(t)=\mathscr{L}^{-1}[F(s)]=c_{1}e^{ p_{1}t }+c_{2}e^{ p_{2}t }+\cdots+c_{n}e^{ p_{n}t } \end{align}$$ #### 2.含有共轭复数极点 $$\begin{align} &F(s)=\frac{c_{1}}{s-p_{1}}+\frac{c_{1}^{*}}{s-p_{1}^{*}}+\frac{c_{3}}{s-p_{3}}+\cdots+\frac{c_{n}}{s-p_{n}} \\ \end{align}$$ 上式同乘 $(s-p_{1})$ 并取 $s=p_{1}$ 上式同乘 $(s-p_{1}^{*})$ 并取 $s=p_{1}^{*}$ $$\begin{align} p_{1}=\sigma+j\omega \\ p_{1}^{*}=\sigma-j\omega \\ \\ [F(s)(s-p_{1})]_{s=p_{1}}&=c_{1} \\ [F(s)(s-p_{1}^{*})]_{s=p_{1}^{*}}&=c_{1}^{*} \\ \\ \mathscr{L}^{-1}\left[ \frac{c_{1}}{s-p_{1}}+\frac{c_{1}^{*}}{s-p_{1}^{*}} \right] &=c_{1}e^{ p_{1}t }+c_{1}^{*}e^{ p_{1}^{*}t } \\ &=2Re(c_{1}e^{ p_{1}t }) \\ &=2Re\{|c_{1}|e^{j\,arg(c_{1})}e^{(\sigma+jw)t}\} \\ &=2|c_{1}|e^{ \sigma t }\cos [arg(c_{1})+\omega t] \end{align}$$ $$\begin{align} & c_{1} , c_{1}^{*} \quad p_{1} , p_{1}^{*} \quad c_{1}e^{ p_{1}t },c_{1}^{*}e^{p_{1}^{*}t } \end{align}$$ #### 3.有重根 $$\begin{align} & F(s)=\frac{c_{r}}{(s-p_{1})^{r}}+\frac{c_{r-1}}{(s-p_{1})^{r-1}}+\cdots+\frac{c_{1}}{(s-p_{1})}+\frac{c_{r+1}}{(s-p_{r+1})}+\cdots+\frac{c_{n}}{(s-p_{n})} \\ \end{align}$$ $$\begin{align} & c_{r}=[F(s)(s-p_{1})^{r}]_{s=p_{1}}\\ & c_{r-1}=\frac{1}{1!} \left\{\dfrac{\mathrm{d} }{\mathrm{d} s} [F(s)(s-p_{1})^{r}] \right\} _{s=p_{1}} \\ & \cdots \cdots\\ & c_{r-j}=\frac{1}{j!}\left \{\dfrac{\mathrm{d}^{j} }{\mathrm{d} s^{j}} [F(s)(s-p_{1})^{r}] \right\}_{s=p_{1}} \\ \end{align}$$ $$\begin{align} & \mathscr{L}^{-1}\left[ \frac{c_{r}}{(s-p_{1})^{r}}+\frac{c_{r-1}}{(s-p_{1})^{r-1}}+\cdots+\frac{c_{1}}{(s-p_{1})} \right] \\ & =\frac{c_{1}}{0!}e^{ p_{1}t}+\frac{c_{2}}{1!}te^{ p_{1}t }+\frac{c_{3}}{2!}t^{2}e^{ p_{1}t }+\cdots+\frac{c_{r}}{(r-1)!}t^{r-1}e^{ p_{1}t } \end{align}$$ >[!note] 有重根 > > $\mathscr{L}[(-t)^{n}f(t)]=F^{(n)}(s)$ > > $\mathscr{L}^{-1}[\dfrac{1} > {(s-p)^{r}}]=\dfrac{t^{r-1}}{(r-1)!}e^{ pt }$ $\mathscr{L}[\dfrac{1}{(s-1)^{2}}]=te^{ t }$ *** $$\begin{align} \mathscr{L}[t^{n}]&=\frac{n!}{s^{n+1}} \\ \mathscr{L}[f(t)e^{ at }]&=F(s-a) \\ \mathscr{L}\left[ \frac{1}{n!}t^{n}e^{ p_{1}t } \right]&=\frac{1}{(s-p_{1})^{n+1}} \\ \mathscr{L}^{-1}\left[ \frac{c_{n}}{(s-p_{1})^{n}} \right]&=\frac{1}{(n-1)!}t^{n-1}c_{n}e^{ p_{1}t } \end{align}$$ $$\begin{align} \mathscr{L}[(-t)^{n}f(t)]&=F^{(n)}(s)\\ \mathscr{L}[te^{ p_{1}t }]&=-\dfrac{\mathrm{d} }{\mathrm{d} s} \left[ \frac{1}{s-p_{1}} \right]=\frac{1}{(s-p_{1})^{2}}\\ \end{align}$$ ### 留数法计算反演积分