Carol Gates
2021-01-27
Answered

Derive the transform of $f(t)=\mathrm{sin}kt$ by using the identity $\mathrm{sin}kt=\frac{1}{2}u({e}^{ikt}-{e}^{-ikt})$

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unessodopunsep

Answered 2021-01-28
Author has **105** answers

Step 1

Consider the given identity,

$\mathrm{sin}kt=\frac{1}{(2i)({e}^{ikt}-{e}^{-ikt}}$

Apply the Laplace transform on both sides,

$L\{\mathrm{sin}kt\}=L\left\{\frac{1}{(2i)({e}^{ikt}-{e}^{-ikt})}\right\}$

Step 2

Apply the linearity of Laplace transform,

$L\{\mathrm{sin}kt\}=\frac{1}{(2i)[L\left\{{e}^{ikt}\right\}-L\left\{{e}^{-ikt}\right\}]}$

$=\frac{1}{(2i)[\frac{1}{s-ik}-\frac{1}{s-(-ik)}]}\dots (L\left\{{e}^{at}\right\}=\frac{1}{(s-a)})$

$=\frac{1}{(2i)[\frac{1}{(s-ik)}-\frac{1}{(s+ik)}]}$

$=\frac{1}{(2i)\left[\frac{(s+ik)-(s-ik)}{(s-ik)(s+ik)}\right]}$

$=\frac{1}{(2i)\left[\frac{(2ik)}{({s}^{2}-{i}^{2}{k}^{2})}\right]}$

$=\frac{k}{({s}^{2}+{k}^{2})}$

Hence,$L\{\mathrm{sin}kt\}=\frac{k}{({s}^{2}+{k}^{2})}$

Consider the given identity,

Apply the Laplace transform on both sides,

Step 2

Apply the linearity of Laplace transform,

Hence,

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Find the inverse Laplace transform of $F\left(s\right)=\frac{1}{{(s+a)}^{2}(s+b)}\text{}\text{for}\text{}t\ge 0$

a)$\mathrm{sin}\left(at\right)+{e}^{-bt}$

b)$\frac{t{e}^{-bt}}{{(a-b)}^{2}}+\mathrm{sin}\left(at\right)$

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e)$\mathrm{sin}\left(at\right)+\frac{t{e}^{-bt}}{{(b-a)}^{2}}-\frac{t{e}^{-at}}{{(a-b)}^{2}}$

a)

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