Determine L^{-1}left[frac{(s-4)e^{-3s}}{s^2-4s+5}right]

Determine L^{-1}left[frac{(s-4)e^{-3s}}{s^2-4s+5}right]

Question
Laplace transform
asked 2021-03-18
Determine \(L^{-1}\left[\frac{(s-4)e^{-3s}}{s^2-4s+5}\right]\)

Answers (1)

2021-03-19
Concept used:
The inverse laplace transform rule.
If \(L^{-1}\left\{F(s)\right\}=f(t)\) then
1. \(L^{-1}\left\{e^{-as}F(s)\right\}=H(t-a)f(t-a)\), Here, H(t) is Heaviside step function.
2.\(L^{-1}\left\{F(s-a)\right\}=eatf(t)\)
3.\(L^{-1}\left\{sF(s)\right\}=f′(t)+f(0)\)
The linearity property of the inverse laplace transform is
\(L^{-1}\left\{a \cdot f(s)+b \cdot g(s)\right\}=a \cdot L^{-1}\left\{f(s)\right\}+b \cdot L^{-1}\left\{g(s)\right\}\)
From the Inverse Laplace transform table.
\(L^{-1}\left\{\frac{a}{s^2+a^2}\right\}=\sin(at)\)
Calculation
Apply the inverse transform rule "If \(L^{-1}\left\{F(s)\right\}=f(t)\)
then \(L^{-1}\left\{e^{-as}F(s)\right\}=H(t-a)f(t-a)\)".
For the function \(\frac{(s-4)e^{-3s}}{s^2-4s+5},\)
\(F(s)=\frac{s-4}{s^2-4s+5},   a=3\)
\(L^{-1}\left\{\frac{e^{-3s}s-4}{s^2-4s+5}\right\}=H(t-3)L^{-1}\left\{F(s-3)\right\} \dots (1)\)
\(\text{First find } L^{-1}\left\{F(s)\right\}=f(t) \text{ where } F(s)=\frac{s-4}{s^2-4s+5}\)
\(L^{-1}\left\{\frac{s-4}{s^2-4s+5}\right\}=L^{-1}\left\{\frac{s-2}{(s-2)^2+1}-2 \cdot \frac{1}{(s-2)^2+1}\right\}\)
Step 3
Now apply the linearity property of inverse Laplace transform.
\(L^{-1}\left\{\frac{s-4}{s^2-4s+5}\right\}=L^{-1}\left\{\frac{s-2}{(s-2)^2+1}-2 \cdot \frac{1}{(s-2)^2+1}\right\}\)
\(L^{-1}\left\{\frac{s-2}{(s-2)^2+1}\right\}-2L^{-1}\left\{\frac{1}{(s-2)^2+1}\right\}\)
\(\text{ Now apply the inverse transform rule }\\ L^{-1}\left\{F(s-a)\right\}=e^{at}L^{-1}(F(s))\)
\(L^{-1}\left\{\frac{s-2}{(s-2)^2+1}\right\}-2L^{-1}\left\{\frac{1}{(s-2)^2+1}\right\}=e^{2t}L^{-1}\left\{\frac{s}{s^2+1}\right\}-2e^{2t}L^{-1}\left\{\frac{1}{s^2+1}\right\} \dots (2)\)
Step 4
Now solve \(L^{-1}\left\{\frac{1}{s^2+1}\right\}\) by the use of formula  \(L^{-1}\left\{\frac{a}{s^2+a^2}\right\}=\sin(at)\).
Here, a is 1.
Therefore, \(L^{-1}\left\{\frac{1}{s^2+1}\right\}=\sin(t)\)
Step 5
Now solve \(L^{-1}\left\{\frac{1}{s^2+1}\right\}\) by the use of formula \(L^{-1}\left\{sF(s)\right\}=f′(t)+f(0)\)
The laplace \(L^{-1}\left\{\frac{s}{s^2+1}\right\}\) can be written as \(L^{-1}\left\{s\left\{\frac{1}{s^2+1}\right\}\right\}\)
Since, \(L^{-1}\left\{\left\{\frac{1}{s^2+1}\right\}\right\}= \sin(t), \therefore\)
\(L^{-1}\left\{\left\{\frac{s}{s^2+1}\right\}\right\}=L^{-1}\left\{s\left\{\frac{1}{s^2+1}\right\}\right\}\)
\(=(\sin t)'+\sin 0\)
\(=\cos t +0\)
\(=\cos t\)
Step 6
Now substitute \(L^{-1}\left\{\left\{1/(s^2+1)\right\}\right\}= \sin(t) \text{ and } L^{-1}\left\{\left\{\frac{s}{s^2+1}\right\}\right\}= \cos(t) \text{ in equation } (2)\)
\(L^{-1}\left\{\frac{s-2}{(s-2)^2+1}\right\}-2L^{-1}\left\{\frac{1}{(s-2)^2+1}\right\}=e^{2t}\cos t-2e^{2t}\sin t\)
\(=L^{-1}\left\{F(s)\right\}\)
\(\text{Now find } L^{-1}\left\{F(s-3)\right\} \text{ as },\)
\(L^{-1}\left\{F(s-3)\right\}=e^{2(t-3)}\cos(t-3)-2e^{2(t-3)}\sin(t-3)\)
\(\text{Now subtitute the value of } L^{-1}\left\{F(s-3)\right\} \text{ in equation } (1)\)
\(L^{-1}\left\{\frac{e^{-3s}s-4}{s^2-4s+5}\right\}=H(t-3)e^{2(t-3)}\cos(t-3)-2e^{2(t-3)}\sin(t-3)\)
Step 7
Answer: \(L^{-1}\left\{\frac{e^{-3s}s-4}{s^2-4s+5}\right\}=H(t-3)e^{2(t-3)}\cos(t-3)-2e^{2(t-3)}\sin(t-3)\)
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