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# Chapter 14 The Laplace Transform ## 14.1 Definition of the Laplace Transform ### Definition 14.1.1 Let $f(t)$ be defined for $t \ge 0$, then the integral $\qquad F(s) = \int_{0}^{\infty} e^{-st} f(t) dt$ is said to be the **Laplace Transform** of $f(t)$, provided that the integral converges....

# Chapter 14 The Laplace Transform ## 14.1 Definition of the Laplace Transform ### Definition 14.1.1 Let $f(t)$ be defined for $t \ge 0$, then the integral $\qquad F(s) = \int_{0}^{\infty} e^{-st} f(t) dt$ is said to be the **Laplace Transform** of $f(t)$, provided that the integral converges. **Notation:** $F(s) = L\{f(t)\}$ **Note:** 1. The Laplace Transform changes a function of $t$ into a function of $s$. 2. We are only concerned with $t \ge 0$. ### Example 14.1.1 Find $L\{1\}$ $L\{1\} = \int_{0}^{\infty} e^{-st} (1) dt = \lim_{b \to \infty} \int_{0}^{b} e^{-st} dt = \lim_{b \to \infty} [-\frac{e^{-st}}{s}]_{0}^{b}$ $= \lim_{b \to \infty} [-\frac{e^{-sb}}{s} + \frac{1}{s}] = 0 + \frac{1}{s} = \frac{1}{s}$ $\therefore L\{1\} = \frac{1}{s}, s > 0$ ### Example 14.1.2 Find $L\{e^{at}\}$ $L\{e^{at}\} = \int_{0}^{\infty} e^{-st} e^{at} dt = \int_{0}^{\infty} e^{-(s-a)t} dt = \lim_{b \to \infty} \int_{0}^{b} e^{-(s-a)t} dt$ $= \lim_{b \to \infty} [\frac{e^{-(s-a)t}}{-(s-a)}]_{0}^{b} = \lim_{b \to \infty} [\frac{e^{-(s-a)b}}{-(s-a)} + \frac{1}{s-a}] = 0 + \frac{1}{s-a}$ $\therefore L\{e^{at}\} = \frac{1}{s-a}, s > a$ ### Example 14.1.3 Find $L\{sin(at)\}$ $L\{sin(at)\} = \int_{0}^{\infty} e^{-st} sin(at) dt$ Use integration by parts twice: $\qquad u = sin(at) \qquad dv = e^{-st} dt$ $\qquad du = a cos(at) dt \qquad v = -\frac{e^{-st}}{s}$ $\int_{0}^{\infty} e^{-st} sin(at) dt = [-\frac{e^{-st}}{s}sin(at)]_{0}^{\infty} + \int_{0}^{\infty} \frac{e^{-st}}{s} a cos(at) dt$ $= 0 + \frac{a}{s} \int_{0}^{\infty} e^{-st} cos(at) dt$ Again: $\qquad u = cos(at) \qquad dv = e^{-st} dt$ $\qquad du = -a sin(at) dt \qquad v = -\frac{e^{-st}}{s}$ $\frac{a}{s} \int_{0}^{\infty} e^{-st} cos(at) dt = \frac{a}{s} ([-\frac{e^{-st}}{s} cos(at)]_{0}^{\infty} - \int_{0}^{\infty} \frac{e^{-st}}{s} a sin(at) dt)$ $= \frac{a}{s} (\frac{1}{s}) - \frac{a^{2}}{s^{2}} \int_{0}^{\infty} e^{-st} sin(at) dt$ So, $\int_{0}^{\infty} e^{-st} sin(at) dt = \frac{a}{s^{2}} - \frac{a^{2}}{s^{2}} \int_{0}^{\infty} e^{-st} sin(at) dt$ $\int_{0}^{\infty} e^{-st} sin(at) dt + \frac{a^{2}}{s^{2}} \int_{0}^{\infty} e^{-st} sin(at) dt = \frac{a}{s^{2}}$ $(1 + \frac{a^{2}}{s^{2}}) \int_{0}^{\infty} e^{-st} sin(at) dt = \frac{a}{s^{2}}$ $\frac{s^{2} + a^{2}}{s^{2}} \int_{0}^{\infty} e^{-st} sin(at) dt = \frac{a}{s^{2}}$ $\int_{0}^{\infty} e^{-st} sin(at) dt = \frac{a}{s^{2}} \frac{s^{2}}{s^{2} + a^{2}} = \frac{a}{s^{2} + a^{2}}$ $\therefore L\{sin(at)\} = \frac{a}{s^{2} + a^{2}}, s > 0$ ### Table of Laplace Transforms | $f(t)$ | $F(s)$ | | :------- | :------- | | 1 | $\frac{1}{s}, s > 0$ | | $e^{at}$ | $\frac{1}{s-a}, s > a$ | | $sin(at)$ | $\frac{a}{s^{2} + a^{2}}, s > 0$ |