Laplace transform of cosine squared function

member 731016
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!
 
Physics news on Phys.org
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.
 
  • Love
Likes member 731016
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top