拉普拉斯逆变换

Inverse Laplace Transform

反演积分公式

\begin{array}{r} f (t) = \frac{1}{2 π j} \int_{β - j \infty}^{β + j \infty} F (s) e^{s t} d s = L^{- 1} [F (s)] t > 0 \end{array}

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 $R e s = β$

\begin{array}{r} F (s) = L {f (t)} = \int_{0}^{+ \infty} f (t) e^{- s t} d t = \int_{- \infty}^{+ \infty} f (t) u (t) e^{- σ t} e^{j ω t} d t = F [f (t) u (t) e^{- σ t}] \end{array}

\begin{matrix} f (t) u (t) e^{- σ t} = \frac{1}{2 π} \int_{- \infty}^{+ \infty} F (s) e^{j ω t} d ω \\ f (t) u (t) = \frac{1}{2 π j} \int_{β - j \infty}^{β + j \infty} F (s) e^{s t} d s \\ f (t) = \frac{1}{2 π j} \int_{β - j \infty}^{β + j \infty} F (s) e^{s t} d s t > 0 \end{matrix}

先进行有理分式分解，使用留数法求极点（常见于控制系统的传递函数）

\begin{array}{r} F (s) = \frac{B (s)}{A (s)} = \frac{b_{0} s^{m} + b_{1} s^{m - 1} + \dots + b_{m}}{a_{0} s^{n} + a_{1} s^{n - 1} + \dots + a_{n}} (m \leq n) \end{array}

\begin{aligned} F (s) = \frac{c_{1}}{s - p_{1}} + \frac{c_{2}}{s - p_{2}} + \dots + \frac{c_{n}}{s - p_{n}} \\ L^{- 1} [\frac{1}{s - p_{i}}] = e^{p_{i} t} \\ f (t) = L^{- 1} [F (s)] = c_{1} e^{p_{1} t} + c_{2} e^{p_{2} t} + \dots + c_{n} e^{p_{n} t} \end{aligned}

\begin{aligned} F (s) = \frac{c_{1}}{s - p_{1}} + \frac{c_{1}^{*}}{s - p_{1}^{*}} + \frac{c_{3}}{s - p_{3}} + \dots + \frac{c_{n}}{s - p_{n}} \end{aligned}

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{aligned} p_{1} = σ + j ω \\ p_{1}^{*} = σ - j ω \\ [F (s) (s - p_{1})]_{s = p_{1}} & = c_{1} \\ [F (s) (s - p_{1}^{*})]_{s = p_{1}^{*}} & = c_{1}^{*} \\ L^{- 1} [\frac{c_{1}}{s - p_{1}} + \frac{c_{1}^{*}}{s - p_{1}^{*}}] & = c_{1} e^{p_{1} t} + c_{1}^{*} e^{p_{1}^{*} t} \\ = 2 R e (c_{1} e^{p_{1} t}) \\ = 2 R e {| c_{1} | e^{j a r g (c_{1})} e^{(σ + j w) t}} \\ = 2 | c_{1} | e^{σ t} \cos [a r g (c_{1}) + ω t] \end{aligned}

\begin{aligned} c_{1}, c_{1}^{*} p_{1}, p_{1}^{*} c_{1} e^{p_{1} t}, c_{1}^{*} e^{p_{1}^{*} t} \end{aligned}

\begin{aligned} F (s) = \frac{c_{r}}{(s - p_{1})^{r}} + \frac{c_{r - 1}}{(s - p_{1})^{r - 1}} + \dots + \frac{c_{1}}{(s - p_{1})} + \frac{c_{r + 1}}{(s - p_{r + 1})} + \dots + \frac{c_{n}}{(s - p_{n})} \end{aligned}

\begin{aligned} c_{r} = [F (s) (s - p_{1})^{r}]_{s = p_{1}} \\ c_{r - 1} = \frac{1}{1!} {\frac{d}{d s} [F (s) (s - p_{1})^{r}]}_{s = p_{1}} \\ \dots \dots \\ c_{r - j} = \frac{1}{j!} {\frac{d^{j}}{d s^{j}} [F (s) (s - p_{1})^{r}]}_{s = p_{1}} \end{aligned}

\begin{aligned} L^{- 1} [\frac{c_{r}}{(s - p_{1})^{r}} + \frac{c_{r - 1}}{(s - p_{1})^{r - 1}} + \dots + \frac{c_{1}}{(s - p_{1})}] \\ = \frac{c_{1}}{0!} e^{p_{1} t} + \frac{c_{2}}{1!} t e^{p_{1} t} + \frac{c_{3}}{2!} t^{2} e^{p_{1} t} + \dots + \frac{c_{r}}{(r - 1)!} t^{r - 1} e^{p_{1} t} \end{aligned}

有重根

$L [(- t)^{n} f (t)] = F^{(n)} (s)$

$L^{- 1} [\frac{1}{(s - p)^{r}}] = \frac{t^{r - 1}}{(r - 1)!} e^{p t}$

$L [\frac{1}{(s - 1)^{2}}] = t e^{t}$

\begin{aligned} L [t^{n}] & = \frac{n!}{s^{n + 1}} \\ L [f (t) e^{a t}] & = F (s - a) \\ L [\frac{1}{n!} t^{n} e^{p_{1} t}] & = \frac{1}{(s - p_{1})^{n + 1}} \\ L^{- 1} [\frac{c_{n}}{(s - p_{1})^{n}}] & = \frac{1}{(n - 1)!} t^{n - 1} c_{n} e^{p_{1} t} \end{aligned}

\begin{aligned} L [(- t)^{n} f (t)] & = F^{(n)} (s) \\ L [t e^{p_{1} t}] & = - \frac{d}{d s} [\frac{1}{s - p_{1}}] = \frac{1}{(s - p_{1})^{2}} \end{aligned}