Answer:
a. [tex]\frac{1}{2s} - \frac{2w}{2(s^{2} + 4w^{2} )}[/tex] b. F(s) exists when s ≥ w + wi√3
Step-by-step explanation:
a. Laplace Transform of f(t) = sin²(wt)
f(t) = sin²(wt)
Using the double angle identity
sin²(wt) = {1 - cos(2wt)}/2
f(t) = [1 - cos(2wt)]/2
Now Laplace of f(t) = F(s) = L{f(t)}
= L{[1 - cos(2wt)]/2}
= 1/2L{1} -1/2L{cos(2wt)}
= [tex]\frac{1}{2s} - \frac{2w}{2(s^{2} + 4(w)^{2} )}[/tex]
b. Range of values of s for which F(s) exists
The range of values of s for which F(s) exists is when F(s) ≥ 0
[tex]\frac{1}{2s} - \frac{2w}{2(s^{2} + 4w^{2} )} \geq 0\\\\\frac{1}{2s} \geq \frac{w}{2(s^{2} + 4w^{2} )}\\2(s^{2} + 4w^{2}) \geq 2sw \\s^{2} + 4w^{2} -2sw \geq 0 \\s^{2} -2sw + 4w^{2} \geq 0 \\[/tex]
Using the quadratic formula,
[tex]s = \frac{-(-2w) +/-\sqrt{(-2w)^{2} - 4 X 1 X 4w^{2} } }{2X1}\\=\frac{-(-2w) +/-\sqrt{4w^{2} - 16w^{2} } }{2}\\ = \frac{2w +/-\sqrt{ - 12w^{2} } }{2}\\= \frac{2w +/-2wi\sqrt{3} }{2}\\= w +/-wi\sqrt{3}[/tex]
So F(s) exists when s ≥ w + wi√3 or s ≥ w - wi√3
So F(s) exists when s ≥ w + wi√3
