\(j\) は虚数単位である．
定義 #
フーリエ変換は収束条件が厳しい．そこで、工学的に意味のない \(t < 0\) では \(f = 0\) とし、さらに発散を防ぐために \(\def\e{\mathrm{e}} \e^{-\sigma t}\) を乗じて減衰させたものがラプラス変換である．
$$ \def\fourier{\mathcal{F}} \def\laplace{\mathcal{L}} \def\e{\mathrm{e}} \def\d{\mathrm{d}} \begin{align*} F(\omega) &= \fourier[f(t)] = \int_{-\infty}^\infty \e^{-j\omega t} \, \d t \\ F(s) &= \laplace[f(t)] = \int_0^\infty \e^{-\sigma t}\e^{-j\omega t}f(t) \, \d t = \int_0^\infty \e^{-st}f(t) \, \d t \end{align*} $$ラプラス変換
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} F(s) = \laplace[f(t)] = \int_0^\infty \e^{-st} f(t)\, \d t $$\(s=\sigma+j\omega\) を複素周波数といい、\(\sigma\) は包絡定数、\(\omega\) は周波数である．
\(f(t)\) を原関数、\(F(s)\) を像関数という．
逆ラプラス変換
$$ \def\laplace{\mathcal{L}} f(t) = \laplace^{-1}[F(s)] $$
ラプラス変換は積分が存在する \(s\) の定義域において用いられる．
性質 #
-
線形性
$$ c_1f_1(t) + c_2f_2(t) \xtofrom{} c_1F_1(s) + c_2F_2(s) $$ -
相似性
$$ f(at) \xtofrom{} \frac{1}{a} F\!\left( \frac{s}{a} \right) $$$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} \laplace[f(at)] &= \int_0^\infty \e^{-st}f(at)\,\d t \\ &= \frac{1}{a}\int_0^\infty \e^{-\frac{s}{a}\tau}f(\tau)\,\d\tau &&\tau\coloneqq at \\ &= \frac{1}{a}F\!\left(\frac{s}{a}\right) \end{align*} $$ -
移動法則
$$ \def\e{\mathrm{e}} \begin{align*} f(t-L) &\xtofrom{} \e^{- L s}F(s) \\ \e^{-\alpha t}f(t) &\xtofrom{} F(s+\alpha) \end{align*} $$-
原関数の移動法則
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} \laplace[f(t-L)] &= \int_L^\infty \e^{-st} f(t-L)\,\d t \\ &= \int_0^\infty \e^{-s(L+\tau)}f(\tau)\,\d\tau &&\tau\coloneqq t-L \\ &= \e^{-L s}\int_0^\infty \e^{-s\tau}f(\tau)\,\d\tau \\ &= \e^{-L s} F(s) \end{align*} $$ -
像関数の移動法則
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \laplace[\e^{-\alpha t}f(t)] = \int_0^\infty \e^{-st}\e^{-\alpha t}f(t)\,\d t = \int_0^\infty \e^{-(s+\alpha)t}f(t)\,\d t = F(s+\alpha) $$
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-
微分法則
$$ \def\d{\mathrm{d}} \begin{align*} \frac{\d}{\d t}f(t) &\xtofrom{} sF(s) - f(0) \\ -tf(t) &\xtofrom{} \frac{\d}{\d s}F(s) \end{align*} $$\(L\) をむだ時間という．
-
原関数の微分法則
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} \laplace\!\left[\frac{\d}{\d t}f(t)\right] &= \int_0^\infty \e^{-st} \frac{\d}{\d t}f(t) \,\d t \\ &= [\e^{-st}f(t)]_0^\infty -\int_0^\infty (-s)\e^{-st}f(t)\,\d t \\ &= 0-f(0)+s\int_0^\infty\e^{-st}f(t)\,\d t \\ &= sF(s)-f(0) \end{align*} $$ -
像関数の微分法則
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} \frac{\d}{\d s}F(s) &= \frac{\d}{\d s}\int_0^\infty \e^{-st}f(t)\,\d t\\ &= \int_0^\infty \frac{\partial}{\partial s}\e^{-st}f(t)\,\d t \\ &= \int_0^\infty (-t)\e^{-st}f(t)\,\d t \\ &= -\int_0^\infty \e^{-st}\,tf(t)\,\d t \\ &= -\laplace[tf(t)] \end{align*} $$
-
-
高次微分法則
$$ \begin{align*} f^{(n)}(t) &\xtofrom{} s^nF(s)-f(0)s^{n-1}-f^\prime(0)s^{n-2}-\cdots-f^{(n-1)}(0) \\ t^n f(t) &\xtofrom{} (-1)^n F^{(n)}(s) \end{align*} $$-
原関数の高次微分法則
$$ \def\laplace{\mathcal{L}}\def\d{\mathrm{d}} \begin{align*} \laplace[f^{(n)}(t)] &= s\laplace[f^{(n-1)}(t)]-f^{(n-1)}(0) \\ &= s\left(s\laplace[f^{(n-2)}(t)]-f^{(n-2)}(0)\right)-f^{(n-1)}(0) \\ &= s^2\laplace[f^{(n-2)}(t)]-f^{(n-2)}(0)s-f^{(n-1)}(0) \\ &= \cdots \\ &= s^n\laplace[f(t)]-f(0)s^{n-1}-f^\prime(0)s^{n-2}-\cdots-f^{(n-1)}(0) \\ \end{align*} $$ -
偶関数の高次微分法則
$$ \def\laplace{\mathcal{L}}\def\d{\mathrm{d}} \begin{align*} \laplace[t^n f(t)] &= \laplace[t\cdot t^{n-1}f(t)] \\ &= -\frac{\d}{\d s}\laplace[t^{n-1}f(t)] \\ &= -\frac{\d}{\d s}\laplace[t\cdot t^{n-2}f(t)] \\ &= (-1)^2 \frac{\d^2}{\d s^2}\laplace[t^{n-2}f(t)] \\ &= \cdots \\ &= (-1)^n\frac{\d^n}{\d s^n}\laplace[f(t)] \\ &= (-1)^n \frac{\d^n}{\d s^n} F(s) \end{align*} $$
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-
積分法則
$$ \def\d{\mathrm{d}} \begin{align*} \int_0^t f(t)\,\d t &\xtofrom{} \frac{1}{s}F(s) \\ \frac{1}{t}f(t) &\xtofrom{} \int_s^\infty F(s)\,\d s \end{align*} $$-
原関数の積分法則
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} \laplace\!\left[ \frac{\d}{\d t}\int_0^t f(\tau) \, \d\tau \right] &= s\laplace\!\left[ \int_0^t f(\tau) \, \d\tau \right] - \int_0^0f(\tau)\,\d\tau \\ \laplace[f(t)] &= s\laplace\!\left[ \int_0^t f(\tau) \, \d\tau \right] \\ F(s) &= s\laplace\!\left[ \int_0^t f(\tau) \, \d\tau \right] \end{align*} $$ -
像関数の積分法則
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} \int_s^\infty F(\sigma)\,\d \sigma &= \int_s^\infty \left(\int_0^\infty \e^{-\sigma t}f(t)\,\d t\right)\d\sigma \\ &= \int_0^\infty f(t)\left(\int_s^\infty \e^{-\sigma t}\,\d\sigma\right)\d t \\ &= \int_0^\infty f(t)\left[-\frac{1}{t}\e^{-\sigma t}\right]_{\sigma=s}^{\sigma=\infty} \d t \\ &= \int_0^\infty f(t)\frac{1}{t}\e^{-st}\,\d t \\ &= \int_0^\infty \e^{-st}\,\frac{1}{t}f(t)\,\d t \\ &= \laplace\!\left[\frac{1}{t}f(t)\right] \end{align*} $$
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-
最終値の定理
$$ f(\infty) = \lim_{t\to\infty} f(t) = \lim_{s\to\infty} sF(s) $$$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} \lim_{s\to0} \int_0^\infty \e^{-st} \frac{\d}{\d t} f(t) \, \d t &= \int_0^\infty \lim_{s\to0}\e^{-st} \frac{\d}{\d t} f(t) \, \d t \\ &= \int_0^\infty \frac{\d}{\d t} f(t) \, \d t \\ &= \lim_{\tau\to\infty}\int_0^\tau \frac{\d}{\d t} f(t) \, \d t \\ &= \lim_{\tau\to\infty}\{f(\tau) - f(0)\} \\ &= \lim_{\tau\to\infty}f(\tau) - f(0) \end{align*} $$一方で、微分のラプラス変換であることより
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \lim_{s\to0} \int_0^\infty \e^{-st} \frac{\d}{\d t} f(t) \, \d t = \lim_{s\to\infty}\{sF(s) - f(0)\} = \lim_{s\to0}sF(s) - f(0) $$以上より従う．
ステップとインパルス #
単位ステップ関数
$$ u(t) = \begin{cases} 0 &(t < 0) \\ 1 &(t \ge 0) \end{cases} $$
単位インパルス関数 (デルタ関数)
$$ \def\d{\mathrm{d}} \delta(t) = \begin{cases} \infty &(t = 0) \\ 0 &(t \ne 0) \end{cases}, \quad \int_{-\infty}^\infty \delta(t)\,\d t = 1 $$
\(u(t)\) を図のように \(\varepsilon \to 0\) としたものだと考えれば、\(\displaystyle \def\d{\mathrm{d}} \delta(t) = \frac{\d}{\d t}u(t)\) であると分かる．
主な関数のラプラス変換 #
-
単位ステップ関数
$$ u(t) \xtofrom{} \frac{1}{s} $$$$ \def\laplace{\mathcal{L}} \def\e{\mathrm{e}} \def\d{\mathrm{d}} \begin{align*} \laplace[u(t)] &= \int_0^\infty u(t) \e^{-st} \, \d t \\ &= \int_0^\infty \e^{-st} \cdot 1 \cdot \, \d t \\ &= \left[\frac{1}{-s}\e^{-st}\right]_0^\infty \\ &= \frac{1}{s} \end{align*} $$ -
デルタ関数
$$ \delta(t) \xtofrom{} 1 $$$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \newcommand\ntsize[1]{{\footnotesize #1}} \begin{align*} \laplace[\delta(t)] &= \int_0^\infty \e^{-st} \delta(t)\,\d t \\ &= \int_{-\infty}^\infty \e^{-st} \delta(t)\,\d t &&\delta(t) = 0 \ntsize{\textsf{ の限り積分区間を拡張}} \\ &= \e^{-s\cdot 0} \int_{-\infty}^\infty \delta(t) \, \d t \\ &= 1 \end{align*} $$または
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} \laplace[\delta(t)] &= \int_{0_-}^\infty \e^{-st}\delta(t) \, \d t \\ &= \int_{0_-}^\infty \e^{-st}\frac{\d}{\d t}u(t) \, \d t \\ &= s \laplace[u(t)] - u(0_-) \\ &= s \cdot \frac{1}{s} - 0 \\ &= 1 \end{align*} $$ -
\(t\)
$$ t \xtofrom{} \frac{1}{s^2} $$$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} \laplace[t] &= \int_0^\infty \e^{-st} t \, \d t \\ &= \left[\frac{1}{-s}\e^{-st}t\right]_0^\infty - \int_0^\infty \frac{1}{-s}\e^{-st} \cdot 1 \, \d t \\ &= 0 + \frac{1}{s} \lim_{t\to\infty}\frac{t}{\e^{st}} + \frac{1}{s}\left[\frac{1}{-s}\e^{-st}\right]_0^\infty \\ &= \frac{1}{s}\lim_{t\to\infty}\frac{1}{s\e^{st}} - \frac{1}{s^2}(0 - 1) \\ &= \frac{1}{s^2} \end{align*} $$ -
\(t^n\)
$$ t^n \xtofrom{} \frac{n!}{s^{n+1}} $$$$ \def\laplace{\mathcal{L}}\def\d{\mathrm{d}}\newcommand\ntsize[1]{{\footnotesize #1}} \begin{align*} \laplace[t^n] &= (-1)^n \frac{\d^n}{\d s^n}\laplace[1] &&\because\;\ntsize{\textsf{微分法則}} \\ &= (-1)^n \frac{\d^n}{\d s^n}\frac{1}{s} \\ &\hphantom{{}={}}\quad\begin{align*} F &= s^{-1} \\ F^\prime &= (-1)s^{-2} \\ F^{\prime\prime} &= (-2)(-1)s^{-3} \\ &\vdots \\ F^{(n)} &= (-1)^n\frac{n!}{s^{n+1}} \end{align*} \\ &= (-1)^{2n}\frac{n!}{s^{n+1}} \\ &= \frac{n!}{s^{n+1}} \end{align*} $$ -
三角関数
$$ \begin{align*} \sin\omega t &\xtofrom{} \frac{\omega}{s^2 + \omega^2} \\[6pt] \cos\omega t &\xtofrom{} \frac{s}{s^2 + \omega^2} \end{align*} $$$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} \laplace[\sin\omega t] &= \int_0^\infty \e^{-st}\sin\omega t \, \d t \\ &= \left[\frac{1}{-s}\e^{-st}\sin\omega t\right]_0^\infty - \int_0^\infty \frac{1}{-s}\e^{-st}\omega\cos\omega t \, \d t \\ &= \frac{\omega}{s}\left(\left[\frac{1}{-s}\e^{-st}\cos\omega t\right]_0^\infty - \int_0^\infty \frac{1}{-s}\e^{-st} (-\omega\sin\omega t)\,\d t\right) \\ &= 0 - \frac{\omega}{-s^2} - \frac{\omega^2}{s^2}\laplace[\sin\omega t] \\ \left(1 + \frac{\omega^2}{s^2}\right)\laplace[\sin\omega t] &= \frac{\omega}{s^2} \\ \laplace[\sin\omega t] &= \frac{\displaystyle\frac{\omega}{s^2}}{\displaystyle\frac{s^2 + \omega^2}{s^2}} = \frac{\omega}{s^2+\omega^2} \end{align*} $$または
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} \laplace[\sin\omega t] &= \laplace\!\left[ \frac{\e^{j\omega t} - \e^{-j\omega t}}{j2} \right] \\ &= \frac{1}{j2}\left(\int_0^\infty \e^{-(s-j\omega )t}\,\d t - \int_0^\infty \e^{-(s+j\omega)t}\,\d t\right) \\ &= \frac{1}{j2}\left(\frac{1}{s-j\omega} - \frac{1}{s+j\omega}\right) \\ &= \frac{1}{j2}\frac{(s+j\omega) - (s-j\omega)}{s^2+\omega^2} \\ &= \frac{\omega}{s^2+\omega^2} \end{align*} $$ -
双曲線関数
$$ \begin{align*} \sinh\omega t &\xtofrom{} \frac{\omega}{s^2 - \omega^2} \\[6pt] \cosh\omega t &\xtofrom{} \frac{s}{s^2 - \omega^2} \end{align*} $$$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} \laplace[\sinh\omega t] &= \laplace\!\left[\frac{\e^{\omega t} - \e^{-\omega t}}{2}\right] \\ &= \frac{1}{2}\left(\int_0^\infty \e^{-(s-\omega)t}\,\d t - \int_0^\infty \e^{-(s+\omega)t}\,\d t \right) \\ &= \frac{1}{2}\left(\frac{1}{s-\omega} - \frac{1}{s+\omega}\right) \\ &= \frac{1}{2}\frac{(s+\omega) - (s-\omega)}{s^2-\omega^2} \\ &= \frac{\omega}{s^2-\omega^2} \end{align*} $$
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}}\newcommand\ntsize[1]{{\footnotesize #1}} \begin{align*} \laplace[t \sin\omega t] &= -\frac{\d}{\d s}\laplace[\sin\omega t] &&\because\;\ntsize{\textsf{微分法則}}\\ &= -\frac{\d}{\d s}\frac{\omega}{s^2+\omega^2} \\ &= \frac{2\omega s}{(s^2+\omega^2)^2} \end{align*} $$\(\def\laplace{\mathcal{L}} \laplace[t \sin\omega t]\)
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}} \laplace^{-1}\left[ \frac{1}{(s-1)^3} \right] = \frac{1}{2!}\laplace^{-1}\left[\frac{2!}{(s-1)^3}\right] = \frac{1}{2}t^2 \e^t $$\(\def\laplace{\mathcal{L}} \laplace^{-1}\left[\dfrac{1}{(s-1)^3}\right]\)
$$ \def\laplace{\mathcal{L}}\def\d{\mathrm{d}} \laplace\left[\frac{\d^2x}{\d^2t}\right] = \laplace\left[\frac{\d}{\d t}\left(\frac{\d x}{\d t}\right)\right] = s\laplace\left[\frac{\d x}{\d t}\right] - \frac{\d x}{\d t}(0) = s\{s\laplace[x] - x(0)\} - \frac{\d x}{\d t}(0) $$次の微分方程式をラプラス変換を用いて解け．
$$ \ddot{x} + 3\dot{x} + 2x = 5 \quad (x(0)=-1, \, \dot{x}(0) = 2) $$
に注意して、与式の両辺をラプラス変換する．
$$ \newcommand\ntsize[1]{{\footnotesize #1}} \def\e{\mathrm{e}}\def\d{\mathrm{d}} \begin{align*} s^2X(s) - sx(0) - \dot{x}(0) + 3(sX(s) - x(0)) + 2X(s) &= \frac{5}{s} \\ s^2X(s) + s - 2 + 3(sX(s) + 1) + 2X(s) &= \frac{5}{s} \\ (s^2 + 3s + 2)X(s) +s+1 &= \frac{5}{s} \end{align*} \\ \begin{align*} X(s) &= \frac{\displaystyle -s-1+\frac{5}{s}}{s^2+3s+2} = \frac{-s^2-s+5}{s(s+1)(s+2)} \eqqcolon \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s+2} \\ &\hphantom{{}={}} \quad \begin{align*}\ntsize{\textsf{ここで }} A &= X(s)s|_{s=0} = \frac{5}{2}, \\ B &= X(s)(s+1)|_{s=-1} = \frac{5}{-1} = -5, \\ C &= X(s)(s+2)|_{s=-2} = \frac{3}{2} \end{align*} \\ &= \frac{5}{2}\frac{1}{s} -5\frac{1}{s+1} + \frac{3}{2}\frac{1}{s+2} \\ \therefore \quad x(t) &= \frac{5}{2} - 5\e^{-t} + \frac{3}{2}\e^{-2t} \end{align*} $$なお、部分分数分解の際は
$$ \newcommand\ntsize[1]{{\footnotesize #1}} \begin{align*} X(s) &= \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s+2} \\ X(s)s &= A + \frac{s}{s+1}B + \frac{s}{s+2}C &&(\times s) \\ X(s)s|_{s=0} &= A + 0 + 0 &&(s=0) \end{align*} $$となることを用いた (ヘビサイドの展開定理)．
畳み込み #
畳み込み
$$ \def\d{\mathrm{d}} f(t)*g(t) = \int_0^t f(\tau)g(t-\tau)\,\d\tau $$
次の性質がある．
-
可換性
$$ f(t)*g(t) = g(t)*f(t) $$ -
分配性
$$ f(t)*\{ g_1(t) + g_2(t) \} = f(t)*g_1(t) + f(t)*g_2(t) $$ -
\(t\) 領域での畳み込みは \(s\) 領域では積
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \laplace[f(t)*g(t)] = \laplace[f(t)]\laplace[g(t)] = F(s)G(s) $$$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}}\def\d{\mathrm{d}} \newcommand\ntsize[1]{{\footnotesize #1}} \begin{align*} \laplace[f(t)*g(t)] &= \int_0^\infty \e^{-st} \int_0^t f(\tau)g(t-\tau)\,\d\tau\,\d t \\ &= \int_0^\infty f(\tau)\int_\tau^\infty \e^{-st}g(t-\tau)\,\d t\,\d\tau &&\ntsize{\textsf{順序変更}} \\ &=\int_0^\infty f(\tau)\int_0^\infty \e^{-s(\tau+t^\prime)}g(t^\prime)\,\d t^\prime \, \d\tau &&t^\prime\coloneqq t-\tau \\ &= \int_0^\infty \e^{-s\tau}f(\tau)\,\d\tau\int_0^\infty \e^{-st^\prime}g(t^\prime)\,\d t^\prime \\ &= \laplace[f(t)]\laplace[g(t)] \end{align*} $$
$$ \def\laplace{\mathcal{L}}\def\e{\mathrm{e}} \laplace^{-1}\left[\frac{F(s)}{s-\alpha}\right] = \laplace^{-1}[F(s)]*\laplace^{-1}\left[\frac{1}{s-\alpha}\right] = f(t)*\e^{\alpha t} $$\(\def\laplace{\mathcal{L}}\displaystyle\laplace^{-1}\left[\frac{F(s)}{s-\alpha}\right]\)
$$ \def\laplace{\mathcal{L}} \begin{align*} x(t)*\sin t &= t^2 \\ \laplace[x(t)*\sin t] &= \laplace[t^2] \\ \laplace[x(t)]\laplace[\sin t] &= \laplace[t^2] \\ \laplace[x(t)] \cdot \frac{1}{s^2 + 1} &= \frac{2}{s^3} \\ \laplace[x(t)] &= \frac{2(s^2+1)}{s^3} = \frac{2}{s} + \frac{2}{s^3} \\ \therefore \quad x(t) &= 2 + t^2 \end{align*} $$次の積分方程式を満たす関数 \(x(t)\) を求めよ．
$$ \def\d{\mathrm{d}} \int_0^t x(\tau)\sin(t-\tau) \, \d\tau = t^2 $$
ラプラス変換公式表 #
| 原関数 | 像関数 |
|---|---|
| $$\delta(t)$$ | $$1$$ |
| $$u(t)$$ | $$\frac{1}{s}$$ |
| $$t$$ | $$\frac{1}{s^2}$$ |
| $$t^n$$ | $$\frac{n!}{s^{n+1}}$$ |
| $$\sin\omega t$$ | $$\frac{\omega}{s^2+\omega^2}$$ |
| $$\cos\omega t$$ | $$\frac{s}{s^2+\omega^2}$$ |
| $$\sinh\omega t$$ | $$\frac{\omega}{s^2-\omega^2}$$ |
| $$\cosh\omega t$$ | $$\frac{s}{s^2-\omega^2}$$ |
| $$\def\e{\mathrm{e}}\e^{-\alpha t}f(t)$$ | $$F(s+\alpha)$$ |
| $$\def\d{\mathrm{d}}\frac{\d}{\d t}f(t)$$ | $$sF(s)-f(0)$$ |
| $$-tf(t)$$ | $$\def\d{\mathrm{d}}\frac{\d}{\d s}F(s)$$ |
| $$\def\d{\mathrm{d}}\int_0^t f(t)\,\d t$$ | $$\frac{1}{s}F(s)$$ |
| $$f(t)*g(t)$$ | $$F(s)G(s)$$ |