Laplace transform functions question

engineer_dave
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Homework Statement



Find the Laplace transform of the following functions

f(t)= sine squared t

Homework Equations



sin kt = k^2/(s^2 + k^2)


The Attempt at a Solution



I went like this;

sin squared t is sin t* sin t therefore it should be 1/(s^2 +1) (s^2 + 1) but unfortunately this was not the answer! any help would be very much appreciated. Thanks
 
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Well, the Laplace transform of a function is defined by an integral, isn't it? Could you write this integral, and try and solve it?
 
what do u mean?
 
u mean it would be the integral of e^-st multiplied by sine squared t??
 
Yes. Either that, or there may be a useful theorem you can use. I can't, however, remember whether the theorem I'm thinking of concerns the Laplace transform of products, or the product of Laplace transforms.
 
Hint use the trig indentity
sin^2(x)=\frac{1-cos(2x)}{2}
using this u get \frac{1}{2}*({\frac{1}{s}-\frac{s}{s^2+4})

which simplifies to the answer -> \frac{2}{s^3+4s}
 
cristo said:
Yes. Either that, or there may be a useful theorem you can use. I can't, however, remember whether the theorem I'm thinking of concerns the Laplace transform of products, or the product of Laplace transforms.

I think it was the theorem of the convolution operator, which is to be found at http://en.wikipedia.org/wiki/Convolution"

@engineer_dave; remember L{ f.g } is not equal to L{f} . L{g} ...However the convolution operator makes it so. Read the stuff at the link. Or the quickest way (since you're an engineer, it's the one you might be interested in) as real10 mentioned;

sin^2(t) = 1/2 - cos(2t)/2
=> L{sin^2(t)} = 1/2. L{1} - 1/2. L{cos2t} (=> due to the linearity of laplace operation, easily provable)
= 1/2 . 1/s - 1/2. 1/s^2 + 4
= 4 / s^3 + 4s indeed..
 
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MorrisonHotel said:
remember L{ f.g } is not equal to L{f} . L{g}

That's what I would say Because laplace is an integration transform so that it cannot be Distributed on 1 function times other.

And I want to take ur attention dave that ur Relevant equations is wrong and the right is :

L (sin kt )= k/(s^2 + k^2) not = k^2/(s^2 + k^2)

hope u get it
 
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