Find Laplace transform for frac{s+8}{s^2+4s+5}

Find Laplace transform for frac{s+8}{s^2+4s+5}

Question
Laplace transform
asked 2021-01-31
Find Laplace transform for
\(\frac{s+8}{s^2+4s+5}\)

Answers (1)

2021-02-01

Step 1
Formula Used:
\(L\left\{e^{at}\sin bt\right\}=\frac{b}{(s-a)^{2}+b^{2}}\)
\(L\left\{e^{at}\cos bt\right\}=\frac{s-a}{(s-a)^{2}+b^{2}}\)
Step 2
\(\text{Consider }F(s)=\frac{s+8}{s^2+5s+5}\)
\(=\frac{s+8}{(s+2)^{2}+1}\)
\(=\frac{s+2+6}{(s+2)^{2}+1}\)
\(=\frac{(s+2)}{(s+2)^{2}+1}+\frac{6}{(s+2)^2+1}\)
Step 3
\(\text{Consider }F(s)=\frac{(s+2)}{(s+2)^{2}+1}+\frac{6}{(s+2)^2+1}\)
We now apply Inverse Laplace Transform, we get
\(L^{-1}\left\{F(s)\right\}=L^{-1}\left\{\frac{(s+2)}{(s+2)^2+1}\right\}+6L^{-1}\left\{\frac{1}{(s+2)^2+1}\right\}\)
\(\Rightarrow\ f(t)=e^{-2t}\cos t+6e^{-2t}\sin t\)
\(=e^{-2t}(\cos t+6\sin t)\)
Step 4
ANSWER
: The inverse Laplace Transform of F(s) is
\(f(t)=e^{-2t}(\cos t+6\sin t)\)

0

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