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pierce15
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Homework Statement
Find
$$ L^{-1} \left[ \frac{1}{ (p^2 + a^2)^2} \right] $$
Homework Equations
$$ L [ x \cos ax ] = \frac{p^2 - a^2} { (p^2 + a^2) ^2 } $$
The Attempt at a Solution
I have no idea. Any thoughts?
piercebeatz said:Homework Statement
Find
$$ L^{-1} \left[ \frac{1}{ (p^2 + a^2)^2} \right] $$
Homework Equations
$$ L [ x \cos ax ] = \frac{p^2 - a^2} { (p^2 + a^2) ^2 } $$
The Attempt at a Solution
I have no idea. Any thoughts?
piercebeatz said:I'm terribly sorry for that. Here's where I got (I'm supposed to use the above identity in the problem, and convolutions is in the next chapter):
$$ L^ {-1} \left[ \frac{1}{(p^2 + a^2)^2} \right] = L^{-1} \left[ \frac{1}{p^2 - a^2} \frac{p^2 - a^2}{(p^2 + a^2)^2} \right] $$
By the theorem
$$ L^{-1} [ F(p) G(p) ] = \int_0^x f(x-t)g(t) \, dt $$
(where F(p) = L(f(t)) etc.), this becomes
$$ L^{-1} \left[ \frac{1}{p^2 - a^2} \frac{p^2 - a^2}{(p^2 + a^2)^2} \right] = \int_0^x \frac{1}{a} \sinh (a (x-t)) \cos(at) \, dt $$
piercebeatz said:I don't know that theorem. Does anyone see any error in the above calculation? I can't find one
LCKurtz said:You don't know what theorem? Please quote the message to which you are replying.
You didn't take the inverse Laplace transform of the two pieces correctly. According to Mathematica, the inverse Laplace transform of ##\frac{p^2-a^2}{(p^2+a^2)^2}## is ##t \cos at##. You forgot the ##t## out front.piercebeatz said:I don't know that theorem. Does anyone see any error in the above calculation? I can't find one
A Laplace Transform is a mathematical tool used to convert a function of time into a function of complex frequency. It is often used in engineering and physics to solve differential equations and analyze systems in the frequency domain.
A Laplace Transform works by integrating a function of time with a complex exponential function. This converts the function from the time domain to the frequency domain. The resulting function is called the Laplace Transform of the original function.
Laplace Transforms are widely used in engineering and physics to solve differential equations, analyze systems in the frequency domain, and understand the behavior of complex systems. They are also used in signal processing, control systems, and circuit analysis.
A Laplace Transform and a Fourier Transform are similar, but the Laplace Transform includes a complex exponential function, while the Fourier Transform does not. This means that the Laplace Transform can handle functions that are not periodic, while the Fourier Transform is limited to periodic functions.
While Laplace Transforms are a powerful mathematical tool, they do have some limitations. They can only be applied to functions that have a Laplace Transform, and they may not always converge. Additionally, they do not always provide an intuitive understanding of the behavior of a system, as the frequency domain can be difficult to interpret visually.