Fourier series of $f(t) = (t-|t|) ^2$

51 Views Asked by At

Apparently what I thought was a absolute value was a average down sign, I. E. if the value is 2.9 the sign will make it 2. Just got back from the professor.

In my last exam of wave physics that I didn't pass, this was one of the questions that I can't seem to get a grip on how he solved it by looking in the answer sheets, could you please help me to figure it out:

Give the fourier coefficients $a_0$,$b_1$,$a_1$,$b_2$ and $a_2$ from the signal $f(t) = (t-|t|)^2$.

Nothing about interval or anything else is mentioned and this is the solution that is given:

https://i.stack.imgur.com/pxXEa.jpg

And I don't know how he got this result which is why I'm asking here, to get help, guidance and tips of how he solved it.

1

There are 1 best solutions below

0
On BEST ANSWER

Apparently what I thought was a absolute value was a average down sign(truncation).

I. E. if the value is 0.9 the sign will make it 0 which in this case gives an interval of 0-1. In my opinion a horribel and not pedagogical way of writing a exam assignment but I hope I have it under control now.

An: https://i.stack.imgur.com/PdWeI.jpg

Bn: https://i.stack.imgur.com/JKoad.jpg