Question

Find the Laplace transform F(s) = L{f(t)} of the function f(t) = sin^2 (wt), defined on the interval t >= 0. F(s) = L {sin^2 (wt)} = For what values of S does the Laplace transform exist

Answer 1

The laplace transform of f(t) is $F(s)=\frac{1}{2s}-\frac{2w}{2(s^2+4w^2)}$ and it's values exists for $s=w\pm\sqrt{3}wi$.

Given,

$f(t)=sin^2(wt)\\\\\therefore sin^2x=\frac{1-cos2x}{2}\\\\f(t)=\frac{1-cos2wt}{2}$

applying laplace transform on both sides,

$L[f(t)]=L[\frac{1-cos2wt}{2}]\\\\F(s)=\frac{1}{2s}-\frac{2w}{2(s^2+4w^2)}$

The range for values of s for which F(s) exists is when F(s)≥0

$\frac{1}{2s}-\frac{2w}{2(s^2+4w^2)} \geq0\\\\\frac{1}{2s} \geq\frac{2w}{2(s^2+4w^2)}\\\\2(s^2+4w^2) \geq 2sw\\\\s^2-2sw+4w^2 \geq0$

using formula to find roots,

$s=\frac{-(-2w)\pm\sqrt{(-2w^2)-4(4w^2)}}{2}\\\\s=\frac{-(-2w)\pm\sqrt{(12w^2}}{2}\\\\s=w\pm\sqrt{3}wi$

Thus,  the laplace transform of f(t) is $F(s)=\frac{1}{2s}-\frac{2w}{2(s^2+4w^2)}$ and it's values exists for $s=w\pm\sqrt{3}wi$.

To learn more about laplace transform refer here

https://brainly.com/question/13077895

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