where we assume s is a positive real number. For example, to find the Laplace transform of

Verify the following Laplace transforms, where u is a real number.

Tazmin Horton
2021-09-15
Answered

Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by

$F\left(s\right)={\int}_{0}^{\mathrm{\infty}}{e}^{-st}f\left(t\right)dt$

where we assume s is a positive real number. For example, to find the Laplace transform of$f\left(t\right)={e}^{-t}$ , the following improper integral is evaluated using integration by parts:

$F\left(s\right)={\int}_{0}^{\mathrm{\infty}}{e}^{-st}{e}^{-t}dt={\int}_{0}^{\mathrm{\infty}}{e}^{-(s+1)t}dt=\frac{1}{s+1}$

Verify the following Laplace transforms, where u is a real number.

$f\left(t\right)=\mathrm{cos}at\to F\left(s\right)=\frac{s}{{s}^{2}+{a}^{2}}$

where we assume s is a positive real number. For example, to find the Laplace transform of

Verify the following Laplace transforms, where u is a real number.

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FieniChoonin

Answered 2021-09-16
Author has **102** answers

Step 1

We have given the function.

$f\left(t\right)=\mathrm{cos}at$

Step 2

To prove this we plug the values in the formula.

$L\le ft\left\{\mathrm{cos}atright\right\}\left(s\right)={\int}_{0}^{\to +\mathrm{\infty}}{e}^{-st}\mathrm{cos}atdt$

$=\underset{L\to \mathrm{\infty}}{lim}{\int}_{0}^{L}{e}^{-st}\mathrm{cos}atdt$

$=\underset{L\to \mathrm{\infty}}{lim}{\left[\frac{{e}^{-st}(-s\mathrm{cos}at+a\mathrm{sin}at)}{{(-s)}^{2}+{a}^{2}}\right]}_{0}^{L}$

$=\underset{L\to \mathrm{\infty}}{lim}(\frac{{e}^{-sL}(-s\mathrm{cos}aL+a\mathrm{sin}aL)}{{s}^{2}+{a}^{2}}-\frac{{e}^{-s\times 0}(-s\mathrm{cos}(0\times a)+a\mathrm{sin}(0\times a))}{{s}^{2}+{a}^{2}})$

$=\underset{L\to \mathrm{\infty}}{lim}(\frac{s\mathrm{cos}(0\times a)-a\mathrm{sin}(0\times a)}{{s}^{2}+{a}^{2}}-\frac{{e}^{-sL}(-s\mathrm{cos}aL+a\mathrm{sin}aL)}{{s}^{2}+{a}^{2}})$

$=\frac{s\mathrm{cos}(0\times a)-a\mathrm{sin}(0\times a)}{{s}^{2}+{a}^{2}}-0$

$=\frac{s\mathrm{cos}0-a\mathrm{sin}0}{{s}^{2}+{a}^{2}}$

$=\frac{s}{{s}^{2}+{a}^{2}}$

We have given the function.

Step 2

To prove this we plug the values in the formula.

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