Danelle Albright

Answered

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.

Answer & Explanation

nghodlokl

Expert

2021-12-29Added 33 answers

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\$

$={L}^{t}\left\{\frac{a(s+1)+4}{{(s+1)}^{2}+9}\right\}$

$=9{L}^{t}\left\{\frac{s+1}{{(s+1)}^{2}+9}\right\}+4{L}^{t}\left\{\frac{1}{{(s+1)}^{2}+9}\right\}$

$=9{e}^{-t}{L}^{t}\left\{\frac{s}{{s}^{2}+9}\right\}+4{e}^{-t}{L}^{t}\left\{\frac{1}{{s}^{2}+9}\right\}$

Using${L}^{t}\left\{F(s-a)\right\}={e}^{at}t\left(t\right)$ (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)

$f\left(1.234\right)=9{e}^{-\left(1.234\right)}\mathrm{cos}\left(3\left(1.234\right)\right)+\frac{4}{3}{e}^{-\left(1.234\right)}\mathrm{sin}\left(3\left(1.234\right)\right)$

$=-2.42$

Using

Hence

Now

From (1) and (2)

Mason Hall

Expert

2021-12-30Added 36 answers

a) $F\left(s\right)=\frac{9s+3}{{5}^{2}+28+10}$

$\Rightarrow {L}^{-1}\left\{\frac{9s+3}{{s}^{2}+28+10}\right\}=2$

$\frac{9s+3}{{s}^{2}+28+10}=\frac{9(s+1)}{{(s+1)}^{2}+9}=6\frac{1}{{(s+1)}^{2}+9}$

$\Rightarrow {L}^{-1}\left\{\frac{9s+3}{{s}^{2}+28+10}\right\}={L}^{-1}\{\frac{9(s+1)}{{(s+1)}^{2}+9}-\frac{6}{{(s+1)}^{2}+9}\}$

$=9{L}^{-1}\left\{\frac{s+1}{{(s+1)}^{2}+9}\right\}-6{L}^{-1}\left\{\frac{1}{{(s+1)}^{2}+9}\right\}$

$=9{e}^{-t}\mathrm{cos}\left(3t\right)-6{e}^{-t}\times \frac{1}{3}\mathrm{sin}3t$

$=9{e}^{-t}\mathrm{cos}\left(3t\right)-2{e}^{-t}\mathrm{sin}3t$

$F\left(s\right)=\frac{{5}^{2}-28+1}{{5}^{4}-{45}^{3}+{105}^{2}-12+5}$

Take partial fraction of

$\frac{{5}^{2}-28+4}{{5}^{4}-{45}^{3}+{105}^{2}-125+5}=\frac{3}{4{(5-1)}^{2}}+\frac{1}{4({5}^{2}-28+5)}$

$\Rightarrow {L}^{-1}\left\{\frac{{5}^{2}-25+4}{{5}^{4}-{45}^{3}+{105}^{2}-125+5}\right\}={L}^{t}\{\frac{3}{4{(5-1)}^{2}}+\frac{1}{4({5}^{2}-28+div>}$

Take partial fraction of

0

karton

Expert

2022-01-10Added 439 answers

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