Danelle Albright

2021-12-28

Find the inverse Laplace transform f(t) of the following transforms F(s). Match each of the following to the value of f(1.234) to two decimal places.

nghodlokl

Inverse Laplace ${L}^{t}\left\{\frac{9s+13}{{s}^{2}+2s+10}\right\}$
${L}^{t}\left\{\frac{9s+9+4}{{s}^{2}+2s+10}\right\}$
$={L}^{t}\left\{\frac{a\left(s+1\right)+4}{{\left(s+1\right)}^{2}+9}\right\}$
$=9{L}^{t}\left\{\frac{s+1}{{\left(s+1\right)}^{2}+9}\right\}+4{L}^{t}\left\{\frac{1}{{\left(s+1\right)}^{2}+9}\right\}$
$=9{e}^{-t}{L}^{t}\left\{\frac{s}{{s}^{2}+9}\right\}+4{e}^{-t}{L}^{t}\left\{\frac{1}{{s}^{2}+9}\right\}$
Using ${L}^{t}\left\{F\left(s-a\right)\right\}={e}^{at}t\left(t\right)$ (formula)
$=9{e}^{-t}\mathrm{cos}3t+\frac{4}{3}\mathrm{sin}3t$
Hence ${L}^{t}\left\{\frac{9s+13}{{s}^{2}+25+10}\right\}=9{e}^{-t}\mathrm{cos}3t+\frac{4}{3}{e}^{-t}\mathrm{sin}3t=f\left(t\right)$ (1)
Now $f\left(1.234\right)$ (2)
From (1) and (2)
$f\left(1.234\right)=9{e}^{-\left(1.234\right)}\mathrm{cos}\left(3\left(1.234\right)\right)+\frac{4}{3}{e}^{-\left(1.234\right)}\mathrm{sin}\left(3\left(1.234\right)\right)$
$=-2.42$

Mason Hall

a) $F\left(s\right)=\frac{9s+3}{{5}^{2}+28+10}$
$⇒{L}^{-1}\left\{\frac{9s+3}{{s}^{2}+28+10}\right\}=2$
$\frac{9s+3}{{s}^{2}+28+10}=\frac{9\left(s+1\right)}{{\left(s+1\right)}^{2}+9}=6\frac{1}{{\left(s+1\right)}^{2}+9}$
$⇒{L}^{-1}\left\{\frac{9s+3}{{s}^{2}+28+10}\right\}={L}^{-1}\left\{\frac{9\left(s+1\right)}{{\left(s+1\right)}^{2}+9}-\frac{6}{{\left(s+1\right)}^{2}+9}\right\}$
$=9{L}^{-1}\left\{\frac{s+1}{{\left(s+1\right)}^{2}+9}\right\}-6{L}^{-1}\left\{\frac{1}{{\left(s+1\right)}^{2}+9}\right\}$
$=9{e}^{-t}\mathrm{cos}\left(3t\right)-6{e}^{-t}×\frac{1}{3}\mathrm{sin}3t$
$=9{e}^{-t}\mathrm{cos}\left(3t\right)-2{e}^{-t}\mathrm{sin}3t$
$F\left(s\right)=\frac{{5}^{2}-28+1}{{5}^{4}-{45}^{3}+{105}^{2}-12+5}$
Take partial fraction of
$\frac{{5}^{2}-28+4}{{5}^{4}-{45}^{3}+{105}^{2}-125+5}=\frac{3}{4{\left(5-1\right)}^{2}}+\frac{1}{4\left({5}^{2}-28+5\right)}$

karton