Transforming Trigonometric Functions using Laplace Transformations

In summary, the conversation discusses the transformation of two functions, f(t) = cos((omega)t + tetha) and f(t) = sint*cost, and the inverse transform of F(s) = 8 / (s^2 + 4s). The suggested method is to take the definition of the Laplace transform and integrate by parts twice. However, there is some confusion about how to handle the limits, but it is clarified that the limits are not a problem and the Laplace transform of cos(t) is s/(1+s^2).
  • #1
kasse
384
1
I'm trying to transform some functions. These two I haven't succeeded transforming:

f(t) = cos((omega)t + tetha)

f(t) = sint*cost

Also, I need help to find the inverse transform of this function:

F(s) = 8 / (s^2 + 4s)
 
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  • #2
Show us what you tried. It looks to me like you can just take the definition of the laplace transform and integrate by parts twice for the first one.
 
  • #3
Hey Dick, I have the same question, laplace of cos(t). I did the integration by parts twice and got to the point where I need to take the limit. Now that's where I'm stuck. There is no limit of cosine and sine. They diverge to infinity. Rather they are bounded between 1 and -1. So what do I do?

Here is my solution.

http://i67.photobucket.com/albums/h304/john_ukranian/48maths.jpg

Too large to paste as an image. Its long but very very clearly defined what I did.

Thanks for the help in advance.

Edit: You can go ahead and write on my image to show mistakes. I noticed that I forgot about the 1/s when subtracting the integral from both sides. That should make a difference, but it still doesn't help with the limit as b goes to infinity.

I know I made mistakes, what are they exactly though.
 
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  • #4
The limits are not a problem sin(t) and cos(t) may oscillate but lim e^(-st) -> 0 at infinity. So evaluated between 0 and infinity sin(t)e^(-st) gives 0 and cos(t)e^(-st) gives -1. And the laplace transform of cos(t) is s/(1+s^2). That's what you are looking for. If A=laplace transform of cos(t), then the integration by parts should lead you to the conclusion A=1/s-A/s^2. Now solve for A.
 

1. What is a Laplace transformation?

A Laplace transformation is a mathematical tool used to convert a function from the time domain to the frequency domain. It is named after the French mathematician Pierre-Simon Laplace and is commonly used in engineering and physics to solve differential equations.

2. How is a Laplace transformation performed?

To perform a Laplace transformation, the function is multiplied by a decaying exponential term and then integrated over all time. The resulting integral is called the Laplace transform of the function. This process is typically done using tables or software, as it can be quite complex to perform by hand.

3. What are the benefits of using Laplace transformations?

Laplace transformations allow us to solve complicated differential equations in the frequency domain, which can be easier to manipulate than in the time domain. They also allow us to analyze the behavior of a system over time, such as its stability and response to different inputs.

4. What are some common applications of Laplace transformations?

Laplace transformations are widely used in fields such as engineering, physics, and economics. They are used to analyze systems in control theory, electrical circuits, heat transfer, and fluid mechanics, among others. They are also used in signal processing and image analysis.

5. Are there any limitations to Laplace transformations?

While Laplace transformations are a powerful tool, they do have some limitations. They can only be applied to functions that are defined for all positive values of time, and they may not always converge for certain functions. Additionally, they can be challenging to interpret for functions with complex behavior or multiple inputs.

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