Laplace transform of cosine squared function

In summary, the Laplace transform of the cosine squared function can be derived using the trigonometric identity that expresses cosine squared in terms of cosine of double angles. The transform is calculated by integrating the product of the cosine squared function and the exponential decay term \( e^{-st} \). The resulting expression involves simpler functions, leading to the final transform expression: \( \mathcal{L}\{\cos^2(at)\} = \frac{s}{s^2 + a^2} + \frac{1}{2s} \), providing a useful tool for solving differential equations and analyzing systems in engineering and physics.
  • #1
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Homework Statement
Please see below
Relevant Equations
##L[(\cos^2 (2t)] = L[\cos 2t] * L[\cos 2t]##
For part (b),
1713581521839.png

I have tried finding the Laplace transform of via the convolution property of Laplace transform.

My working is,

##L[\cos^2 (2t)] = L[\cos 2t] * L[\cos 2t]##
##L[\cos^2 (2t)] = \frac{s}{s^2 + 4} * \frac{s}{s^2 + 4}##
##\int_0^t \frac{s^2}{(s^2 + 4)^2} dt = \frac{ts^2}{(s^2 + 4)^2}##

However, I don't see how that is equivalent/equal to the expression they got for (b). Does some please know how or if I've made a mistake?

Thanks!
 
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  • #2
Your working is wrong. You are trying to make a convolution, but that is not how it is done. Please look up the actual expression for a convolution.

Nb: the easiest way to solve this is using trig identities as done in the solution.
 
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FAQ: Laplace transform of cosine squared function

What is the Laplace transform of the cosine squared function?

The Laplace transform of the cosine squared function, \( \cos^2(at) \), can be computed using the trigonometric identity \( \cos^2(at) = \frac{1 + \cos(2at)}{2} \). Thus, the Laplace transform \( \mathcal{L}\{\cos^2(at)\} \) is given by:

\( \mathcal{L}\{\cos^2(at)\} = \frac{1}{2} \mathcal{L}\{1\} + \frac{1}{2} \mathcal{L}\{\cos(2at)\} = \frac{1}{2s} + \frac{1}{2} \cdot \frac{s}{s^2 + (2a)^2} = \frac{1}{2s} + \frac{s}{2(s^2 + 4a^2)} \).

How do you derive the Laplace transform of the cosine squared function?

To derive the Laplace transform of \( \cos^2(at) \), start with the identity mentioned earlier: \( \cos^2(at) = \frac{1 + \cos(2at)}{2} \). Then, apply the linearity of the Laplace transform:

\( \mathcal{L}\{\cos^2(at)\} = \frac{1}{2} \mathcal{L}\{1\} + \frac{1}{2} \mathcal{L}\{\cos(2at)\} \). Compute each part separately: \( \mathcal{L}\{1\} = \frac{1}{s} \) and \( \mathcal{L}\{\cos(2at)\} = \frac{s}{s^2 + (2a)^2} \). Substituting these results gives the final expression for the Laplace transform.

What are the properties of the Laplace transform relevant to cosine squared?

Several properties of the Laplace transform are relevant when working with the cosine squared function, including linearity, time-shifting, and frequency-shifting. Linearity allows you to break down complex functions into simpler components. The time-shifting property can be useful when dealing with shifted cosine functions, while the frequency-shifting property helps in transforming functions involving \( \cos(at) \) or \( \sin(at) \) terms.

Can you compute the inverse Laplace transform of the result?

Yes, the inverse Laplace transform can be computed from the result obtained for \( \mathcal{L}\{\cos^2(at)\} \). The expression \( \frac{1

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