What is a Laplace transformation of sin^2(wt)?

Answer 1

There are several approaches to this problem, but the simplest would be to use trigonometric identities to reduce sin^2(wt) into a simpler form. Using the identity

sin²(wt) = (1/2)[1 - cos(2wt)],

we can then take the transform of each term:

(1/2)(L{1} - L{cos(2wt)}) = (1/2)(1/s - s/[s² +(2w)²]) = (1/2)[1/s - s/(s² + 4w²)]

Answer 2

The Laplace transform is a integral interpretation of a function using an exponential kernel as a mean to regulate the function to converge at infinity...

int_[lb=0]_[ub=infinity] f(t)e^-st dt

In this case the f(t) = sin²(ωt) = 1/2 - (1/2)*cos(2ωt) [ trigonometric identity ]

L{f(t)} = int_[lb=0]_[ub=infinity] [1/2 - (1/2)*cos(2ωt)] * e^-st dt = int_[lb=0]_[ub=infinity] (1/2)e^-st dt - int_[lb=0]_[ub=infinity] (1/2)cos(2ωt)e^-st dt

We will use integration-by-parts on the second part to help us solve this problem...

The first part yields int_[lb=0]_[ub=infinity] (1/2)e^-st dt = EVAL( -1/(2s)e^-st ) = 0 - [ -1/(2s) ] = 1/(2s)

Define Q = int_[lb=0]_[ub=infinity] cos(2ωt)e^-st dt

Using integration-by-parts gives...

u = cos(2ωt)

du = -2ωsin(2ωt)dt

dv = e^-st dt

v = (-s)^-1e^-st

Q = 1/2 * EVAL( [ 1/s * cos(2ωt)e^-st ] ) + 1/2 *(2ω/s) * int_[lb=0]_[ub=infinity] sin(2ωt)e^-st dt

= 1/2 * ( 0 - 1/s ) + ω/s * int_[lb=0]_[ub=infinity] sin(2ωt)e^-st dt

= 1/(2s) + ω/s * int_[lb=0]_[ub=infinity] sin(2ωt)e^-st dt

One more time, we see we return to the definition of the original integral (Q) in the integration by parts. When we solve for Q we get...

Q = s/2* 1/( s² + 4ω² )

Coupled with the first part we solved earlier we arrive at ...

L{f(t)} = 1/(2s) - s/2 * 1/( s² + 4ω² )

Answer 3

The Laplace Transform is (1/2s) -(1/2)(s/(s²+4w²))

This follows from using the half angle formula which transforms sin²(wt) to (1/2) - (1/2)cos(2wt) and then obtaining the Laplace Transforms from each.