Find the inverse Laplace transform of (any two) i) frac{(s^2+3)}{s(s^2+9)} ii) logleft(frac{(s+1)}{(s-1)}right)

Question
Laplace transform
asked 2021-02-20
Find the inverse Laplace transform of (any two)
i) \(\frac{(s^2+3)}{s(s^2+9)}\)
ii) \(\log\left(\frac{(s+1)}{(s-1)}\right)\)

Answers (1)

2021-02-21
(i) The given Laplace transformation is,\(\frac{(s^2+3)}{s(s^2+9)}\)
We can write it,
\(\frac{s^2+3}{s(s^2+9)}=\frac{s^2}{s(s^2+9)}+\frac{3}{s(s^2+9)}\)
\(=\frac{s}{(s^2+9)+\frac{s^2-(s^2+9)}{s(s^2+9))} \left(\frac{3}{-9}\right)}\)
\(=\frac{s}{(s^2+9)+\frac{1}{3} \cdot \frac{s}{s^2+9}+\frac{1}{3} \cdot \frac{1}{s}}\)
\(=\frac{2}{3} \cdot \frac{s}{s^2+9} + \frac{1}{3} \cdot \frac{1}{s}\)
\(=\frac{2}{3} \cdot \frac{s}{s^2+3^2}+ \frac{1}{3} \cdot \frac{1}{s}\)
Now taking inverse Laplace of both sides we have,
\(L^{-1}\left[\frac{s^2+3}{s(s^2+9)}\right]=\frac{2}{3}L^{-1}\left[\frac{s}{(s^2+3^2)}\right]+\frac{1}{3}L^{-1}\left[\frac{1}{s}\right]\)
\(=\frac{2}{3}\cos3t+\frac{1}{3}\)
ANSWER:\(L^{-1}\left[\frac{s^2+3}{s(s^2+9)}\right]=\frac{2}{3}\cos3t+\frac{1}{3}\)
Step 3
(ii)
The given Laplace transformation is,
\(\log\left(\frac{s+1}{s-1}\right)\)
We can write it,
\(\log\left(\frac{s+1}{s-1}\right)=\log(s+1)-\log(s-1)\)
Now,the property of Laplace transformation is,
\(tf(t) \leftrightarrow -\frac{d}{ds}[F(s)]\)
\(\text{Therefore, }\)
\(tf(t) \leftrightarrow -\frac{d}{ds}\left[\log(s+1)-\log(s-1)\right]\)
\(\Rightarrow tf(t) \leftrightarrow -\frac{1}{s+1}+\frac{1}{s-1}\)
\(\Rightarrow tf(t) \leftrightarrow -e^{-t}+e^t\)
\(\Rightarrow f(t) \leftrightarrow \frac{e^t-e^{-t}}{t}=2\left(\frac{e^t-e^{-t}} {2t}\right)=2\left(\frac{\sin ht}{t}\right)\)
Therefore,
\(L^{-1}\left[\log\frac{s+1}{s-1}\right]=2\left(\frac{\sin ht}{t}\right)\)
Answer: \(L^{-1}\left[\log\frac{s+1}{s-1}\right]=2\left(\frac{\sin ht}{t}\right)\)
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