Find the laplace inverse for the following 1) s2s2+2s+3 2) e−5ss

Name: Find the laplace inverse for the following 1) s2s2+2s+3 2) e−5ss
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find the laplace inverse for the following 1) s2s2+2s+3 2) e−5ss

s2s2+2s+3 here we use long division and completing the square by long division s2s2+2s+3=1+−2s−3s2+2s+3=1−2s+3s2+2s+3 complete the square of s2+2s+3 The coefficient of s is 2 . So we divide this by 2⇒2÷2=1 and add subtract thus term ⇒s2+2s+3=s2+2s+1−1+3=(s2+2s+1)+2=(s+1)2+2 ⇒s2s2+2s+3=1−2s+3(s+1)2+2 here we use L−1{6(s−a)2+b2}=eatsin⁡(bt) L−1{s−a(s−a)2+b2}=eatsin⁡(bt) ⇒L−1{s2s2+2s+3}=L−1{1}−L−1{2s−1+2(s+1)2+2} =d(t)−[L−1{2s−1(s+1)2+2}+22L−1{2(s+1)2+2}] =d(t)−12e−tsin⁡(2t)−2e−tcos⁡(2t) 2) e−5ss use L−1{e−ass}=u(t−a) , here a=5 ⇒L−1{e−5ss}=u(t−5)

Expert answers to "Find the laplace inverse for the following 1)..."

Burhan Hopper

Answered question

2021-09-06

Find the laplace inverse for the following
1)

\frac{s^{2}}{s^{2} + 2 s + 3}

\frac{e^{- 5 s}}{s}

Answer & Explanation

irwchh

Skilled2021-09-07Added 102 answers

$\frac{s^{2}}{s^{2} + 2 s + 3}$ here we use long division and completing the square
by long division $\frac{s^{2}}{s^{2} + 2 s + 3} = 1 + \frac{- 2 s - 3}{s^{2} + 2 s + 3} = 1 - \frac{2 s + 3}{s^{2} + 2 s + 3}$
complete the square of $s^{2} + 2 s + 3$
The coefficient of s is 2 . So we divide this by $2 \Rightarrow 2 \div 2 = 1$
and add subtract thus term
$\Rightarrow s^{2} + 2 s + 3 = s^{2} + 2 s + 1 - 1 + 3 = (s^{2} + 2 s + 1) + 2 = {(s + 1)}^{2} + 2$
$\Rightarrow \frac{s^{2}}{s^{2} + 2 s + 3} = 1 - \frac{2 s + 3}{{(s + 1)}^{2} + 2}$
here we use $L^{- 1} {\frac{6}{(s - a)^{2} + b^{2}}} = e^{a t} \sin (b t)$
$L^{- 1} {\frac{s - a}{(s - a)^{2} + b^{2}}} = e^{a t} \sin (b t)$
$\Rightarrow L^{- 1} {\frac{s^{2}}{s^{2} + 2 s + 3}} = L^{- 1} {1} - L^{- 1} {\frac{2 s - 1 + 2}{(s + 1)^{2} + 2}}$
$= d (t) - [L^{- 1} {\frac{2 s - 1}{(s + 1)^{2} + 2}} + \frac{2}{\sqrt{2}} L^{- 1} {\frac{\sqrt{2}}{(s + 1)^{2} + 2}}]$
$= d (t) - \frac{1}{\sqrt{2}} e^{- t} \sin (\sqrt{2} t) - 2 e^{- t} \cos (\sqrt{2} t)$
2) $\frac{e^{- 5 s}}{s} use L^{- 1} {\frac{e^{- a s}}{s}} = u (t - a)$ , here a=5
$\Rightarrow L^{- 1} {\frac{e^{- 5 s}}{s}} = u (t - 5)$

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