Find Laplace transform of sin^32t
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hii dear here is ur answer
sin^3(2t) = (3*sin(2t)/4) - (sin(3*2t)/4)
Now, by applying laplace transform , we know that LT of
sin (wt) u(t) = w/(s^2+w^2),
our expression becomes,
(3/4)*(2/(s^2+4)) - (1/4)*(6/(s^2+36))
Now,
solve this expression to get
48/(s^4+40s^2+144)
hope it helps you.....
:-)
sin^3(2t) = (3*sin(2t)/4) - (sin(3*2t)/4)
Now, by applying laplace transform , we know that LT of
sin (wt) u(t) = w/(s^2+w^2),
our expression becomes,
(3/4)*(2/(s^2+4)) - (1/4)*(6/(s^2+36))
Now,
solve this expression to get
48/(s^4+40s^2+144)
hope it helps you.....
:-)
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