／var／log marcus chiu

❯

❯

Mathematical Analysis

❯

Harmonic Analysis

❯

Fourier Analysis

❯

(Signal-to-Frequency ／ Fourier-Transform) Algorithms

❯

Laplace Transform - Inverse Laplace Transform

Laplace Transform - Examples

Created on Dec 13, 2023 · Last Modified on May 23, 2026

Laplace Transforms - Examples

$L {1} = 1/ s$	Click here to expand... $L {1} = \int_{0}^{\infty} e^{- s t} 1 d t$ $L {1} = l i m_{A \to \infty} \int_{0}^{A} e^{- s t} d t$ $L {1} = l i m_{A \to \infty} [- \frac{1}{s} e^{- s t}]_{0}^{A}$ $L {1} = l i m_{A \to \infty} (- \frac{1}{s} e^{- s A} - - \frac{1}{s})$ $L {1} = l i m_{A \to \infty} (- \frac{1}{s} e^{- s A} + \frac{1}{s})$ $L {1} = - \frac{1}{s} e^{- s \infty} + \frac{1}{s}$ $L {1} = \frac{1}{s}$
$L {t^{n}} = n! / s^{n + 1}$	Click here to expand... $L {1} = 1/ s$ $L {f^{'} (t)} = s L {f (t)} - f (0)$ TODO Let’s consider:
$L {e^{a t}} = 1/ (s - a) for a < s$	Click here to expand... $L {e^{a t}} = \int_{0}^{\infty} e^{- s t} e^{a t} d t$ $L {e^{a t}} = \int_{0}^{\infty} e^{(a - s) t} d t$ $L {e^{a t}} = \frac{1}{a - s} [e^{(a - s) t}]_{0}^{\infty}$ $L {e^{a t}} = \frac{1}{a - s} [0 - 1] for a < s$ $L {e^{a t}} = - \frac{1}{a - s} for a < s$ $L {e^{a t}} = \frac{1}{s - a} for a < s$
$L {s in (a t)} = a / (s^{2} + a^{2})$	Click here to expand... $L {s in (a t)} = \int_{0}^{\infty} e^{- s t} s in (a t) d t$ Integration by parts states: $\int v \cdot d u = u \cdot v - \int u \cdot d v$ Let’s define: $d u = e^{- s t}$ $u = - \frac{1}{s} e^{- s t}$ $d v = a * cos (a t)$ $v = s in (a t)$ Thus: $\int_{0}^{\infty} e^{- s t} s in (a t) d t = - \frac{1}{s} e^{- s t} s in (a t) - \int_{0}^{\infty} - \frac{1}{s} e^{- s t} a * cos (a t) d t$ Substituting in: $L {s in (a t)} = \int_{0}^{\infty} e^{- s t} s in (a t) d t$ $L {s in (a t)} = - \frac{1}{s} e^{- s t} s in (a t) - \int_{0}^{\infty} - \frac{1}{s} e^{- s t} a * cos (a t) d t$ $L {s in (a t)} = - \frac{e ^{- s t}}{s} s in (a t) + \frac{a}{s} \int_{0}^{\infty} e^{- s t} * cos (a t) d t$ Let’s define: $d u = e^{- s t}$ $u = - \frac{1}{s} e^{- s t}$ $d v = - a * s in (a t)$ $v = cos (a t)$ Thus: $\int_{0}^{\infty} e^{- s t} * cos (a t) d t = - \frac{1}{s} e^{- s t} cos (a t) - \int_{0}^{\infty} - \frac{1}{s} e^{- s t} * (- a) * s in (a t)$ $\int_{0}^{\infty} e^{- s t} * cos (a t) d t = - \frac{1}{s} e^{- s t} cos (a t) - \int_{0}^{\infty} \frac{a}{s} e^{- s t} * s in (a t)$ Substituting in: $L {s in (a t)} = - \frac{e ^{- s t}}{s} s in (a t) + \frac{a}{s} \int_{0}^{\infty} e^{- s t} * cos (a t) d t$ $L {s in (a t)} = - \frac{e ^{- s t}}{s} s in (a t) + \frac{a}{s} [- \frac{1}{s} e^{- s t} cos (a t) - \int_{0}^{\infty} \frac{a}{s} e^{- s t} * s in (a t)]$ $L {s in (a t)} = - \frac{e ^{- s t}}{s} s in (a t) - \frac{a}{s ^{2}} e^{- s t} cos (a t) - \frac{a ^{2}}{s ^{2}} \int_{0}^{\infty} e^{- s t} * s in (a t)$ $L {s in (a t)} = - \frac{e ^{- s t}}{s} s in (a t) - \frac{a}{s ^{2}} e^{- s t} cos (a t) - \frac{a ^{2}}{s ^{2}} \int_{0}^{\infty} e^{- s t} * s in (a t)$ $L {s in (a t)} = - \frac{e ^{- s t}}{s} s in (a t) - \frac{a}{s ^{2}} e^{- s t} cos (a t) - \frac{a ^{2}}{s ^{2}} L {s in (a t)}$ $L {s in (a t)} + \frac{a ^{2}}{s ^{2}} L {s in (a t)} = - \frac{e ^{- s t}}{s} s in (a t) - \frac{a}{s ^{2}} e^{- s t} cos (a t)$ $L {s in (a t)} + \frac{a ^{2}}{s ^{2}} L {s in (a t)} = - e^{- s t} [\frac{1}{s} s in (a t) + \frac{a}{s ^{2}} cos (a t)]$ $L {s in (a t)} (1 + \frac{a ^{2}}{s ^{2}}) = - e^{- s t} [\frac{1}{s} s in (a t) + \frac{a}{s ^{2}} cos (a t)]$ $L {s in (a t)} \frac{s ^{2} + a ^{2}}{s ^{2}} = - e^{- s t} [\frac{1}{s} s in (a t) + \frac{a}{s ^{2}} cos (a t)]$ $L {s in (a t)} \frac{s ^{2} + a ^{2}}{s ^{2}} = [- e^{- s t} [\frac{1}{s} s in (a t) + \frac{a}{s ^{2}} cos (a t)]]_{0}^{\infty}$ $L {s in (a t)} \frac{s ^{2} + a ^{2}}{s ^{2}} = [0 + 1 (0 + \frac{a}{s ^{2}})]$ $L {s in (a t)} \frac{s ^{2} + a ^{2}}{s ^{2}} = \frac{a}{s ^{2}}$ $L {s in (a t)} = \frac{a}{s ^{2} + a ^{2}}$
$L {cos (a t)} = s / (s^{2} + a^{2})$	Click here to expand... $L {f^{'} (t)} = s L {f (t)} - f (0)$ $L {s in (a t)} = \frac{a}{s ^{2} + a ^{2}}$ Let’s define: $f^{'} (t) = cos (a t)$ $f (t) = \frac{1}{a} s in (a t)$ Thus: $L {cos (a t)} = s L {\frac{1}{a} s in (a t)} - \frac{1}{a} s in (a * 0)$ $L {cos (a t)} = s L {\frac{1}{a} s in (a t)} - 0$ $L {cos (a t)} = \frac{s}{a} L {s in (a t)}$ $L {cos (a t)} = \frac{s}{a} \frac{a}{s ^{2} + a ^{2}}$ $L {cos (a t)} = \frac{s}{s ^{2} + a ^{2}}$ Consider the following:
$L {f^{'} (t)} = s L {f (t)} - f (0)$	Click here to expand... $L {f^{'} (t)} = \int_{0}^{\infty} e^{- s t} f^{'} (t) d t$ Integrating by parts states: $\int u v^{'} = uv - \int u^{'} v$ Let’s define: $u = e^{- s t}$ $d u = - s * e^{- s t}$ $v = f (t)$ $d v = f^{'} (t)$ Thus, with integration-by-parts: $\int e^{- s t} f^{'} (t) d t = e^{- s t} f (t) - \int - s * e^{- s t} f (t) d t$ $\int_{0}^{\infty} e^{- s t} f^{'} (t) d t = [e^{- s t} f (t)]_{0}^{\infty} + s \int_{0}^{\infty} e^{- s t} f (t) d t$ Substituting in yields: $L {f^{'} (t)} = \int_{0}^{\infty} e^{- s t} f^{'} (t) d t$ $L {f^{'} (t)} = [e^{- s t} f (t)]_{0}^{\infty} + s \int_{0}^{\infty} e^{- s t} f (t) d t$ $L {f^{'} (t)} = [0 - f (0)] + s L {f (t)}$ $L {f^{'} (t)} = s L {f (t)} - f (0)$
$L {f^{''} (t)} = s L {f^{'} (t)} - f^{'} (0)$	Click here to expand... $L {f^{'} (t)}$ See explanation of:
Given: $L {f (t)} = F (s)$ Then: $L {e^{a t} f (t)} = F (s - a)$	Click here to expand... $L {e^{a t} f (t)} = \int_{0}^{\infty} e^{- s t} e^{a t} f (t) d t$ $L {e^{a t} f (t)} = \int_{0}^{\infty} e^{a t - s t} f (t) d t$ $L {e^{a t} f (t)} = \int_{0}^{\infty} e^{(a - s) t} f (t) d t$ $L {e^{a t} f (t)} = \int_{0}^{\infty} e^{- (s - a) t} f (t) d t$ Let’s define: $L {f (t)} = \int_{0}^{\infty} e^{- s t} f (t) d t = F (s)$ Thus: $L {e^{a t} f (t)} = F (s - a)$