Laplace transform functions question

In summary, the Laplace transform of sine squared t is equal to 1/2 divided by the Laplace transform of 1 plus 1/2 divided by the Laplace transform of cosine 2t. This simplifies to 4 divided by s cubed plus 4s. The convolution operator can also be used to find the Laplace transform of products. The relevant equation for the Laplace transform of sine kt is k divided by the quantity s squared plus k squared.
  • #1
engineer_dave
35
0

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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  • #2
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?
 
  • #3
what do u mean?
 
  • #4
u mean it would be the integral of e^-st multiplied by sine squared t??
 
  • #5
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.
 
  • #6
Hint use the trig indentity
[tex]sin^2(x)=\frac{1-cos(2x)}{2}[/tex]
using this u get [tex]\frac{1}{2}*({\frac{1}{s}-\frac{s}{s^2+4})[/tex]

which simplifies to the answer -> [tex]\frac{2}{s^3+4s}[/tex]
 
  • #7
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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  • #8
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
 

1. What is a Laplace transform function?

A Laplace transform function is a mathematical tool that allows us to convert a function from the time domain into the frequency domain. It is often used in engineering and physics to solve differential equations and analyze the behavior of systems.

2. How do I calculate a Laplace transform function?

To calculate a Laplace transform function, you need to use the Laplace transform integral, which involves integrating the function multiplied by an exponential term. The result of the integral is the Laplace transform of the function.

3. What is the purpose of using a Laplace transform function?

The purpose of using a Laplace transform function is to simplify the analysis of differential equations, which can be difficult to solve directly. By converting the function into the frequency domain, we can more easily analyze the behavior of the system and find solutions.

4. What types of functions can be transformed using the Laplace transform?

The Laplace transform can be applied to a wide range of functions, including exponential, trigonometric, and polynomial functions. However, the function must be piecewise continuous and have an exponential order for the transformation to be valid.

5. What are the advantages of using a Laplace transform function?

One of the main advantages of using a Laplace transform function is that it allows us to solve complicated differential equations and systems more easily. It also provides a way to analyze the stability and behavior of systems in the frequency domain, which can be useful in engineering and physics applications.

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