 # g(t)={0, 0<t<11, 1<t<36, 3<t<54, 5<t g(t)={1,3}(t)+6prod_{3,5}(t)+4u(t-5) compute the Laplace transform g(t) coexpennan 2021-01-24 Answered

$g\left(t\right)=\left\{\begin{array}{lllll}0& 0

$g\left(t\right)=\prod _{1,3}\left(t\right)+6\prod _{3,5}\left(t\right)+4u\left(t-5\right)$
compute the Laplace transform of g(t)

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Step 1
According to the given information it is required to compute Laplace transform of g(t), where given that g(t) is expressed as,
$g\left(t\right)=\prod _{1,3}\left(t\right)+6\prod _{3,5}\left(t\right)+4u\left(t-5\right)$
Also from window , for $f\left(t\right)=\prod _{a,b}\left(t\right)$
$L\left(f\right)\left(s\right)=\frac{{e}^{-as}-{e}^{-bs}}{s},s>0$

$L\left(f\left(s\right)\right)=\frac{{e}^{-as}}{s},s>0$
Step 2
Laplace transform of g(t) can be expressed as:
$g\left(t\right)=L\left(g\right)\left(t\right)$
$=L\left[{g}_{1}\left(t\right)+6{g}_{2}\left(t\right)+3{g}_{3}\left(t\right)\right]$
$=L\left({g}_{1}\right)\left(t\right)+6L\left({g}_{2}\right)\left(t\right)+4L\left({g}_{3}\right)\left(t\right)$

${g}_{2}\left(t\right)=\prod _{3,5}\left(t\right)$

Step 3
Now, find

As,
$L\left({g}_{1}\right)\left(t\right)=\frac{{e}^{-1t}-{e}^{-3t}}{t},t>0$ (comparing by formula mentioned in previous step a=1 , b=3)
$L\left({g}_{2}\right)\left(t\right)=\frac{{e}^{-3t}-{e}^{-5t}}{t},t>0$ (similarly here a=3 , b=5)
$L\left({g}_{3}\right)\left(t\right)=\frac{{e}^{-5t}}{t},t>0$ (And here a=5)
Step 4
Substitute above values in formula for Laplace of g(t).
$g\left(t\right)=L\left\{\left(g\right)\left(t\right)\right\}$
$=L\left({g}_{1}\right)\left(t\right)+6L\left({g}_{2}\right)\left(t\right)+4L\left({g}_{3}\right)\left(t\right)$
$=\frac{{e}^{-t}-{e}^{-3t}}{t}+6\left(\frac{{e}^{-3t}-{e}^{-5t}}{t}\right)+4\left(\frac{{e}^{-5t}}{t}\right),t>0$
$=\frac{{e}^{-t}-{e}^{-3t}}{t}+\frac{6{e}^{-3t}-6{e}^{-5t}}{t}+\frac{4{e}^{-5t}}{t},t>0$
$=\frac{{e}^{-t}-{e}^{-3t}+6{e}^{-3t}-6{e}^{-5t}+4{e}^{-5t}}{t}$
$=\frac{{e}^{-t}+5{e}^{-3t}-2{e}^{-5t}}{t},t>0$
Thus , Laplace transform of g(t) is
$=\frac{{e}^{-t}+5{e}^{-3t}-2{e}^{-5t}}{t},t>0$