Question

Tell me what you all think about the signal f(t) , I think it is cos(4*pi*t) not cos(2*pi*t)

Answer 1

cos 2pi t
I put it on demos graph, the period is 0.5

Answer 2

I think I got the details figured out...
f(x) looks like part of a cosine. the cosine has a total period of 1 radian.
Using
\[ f(x) = \cos \frac{2 \pi}{T} x \]
with T=1 , your function is
\[ f(x) = \cos 2 \pi x, \ -¼ \le x \le ¼ \]

The confusing part is f(x) is periodic with period ½ (it is a "half cosine", and repeats twice as fast as the cosine)

so f(x) has period ½
the "basis" functions cos(n x) will be multiples of this period:
\[ \cos \frac{2 \pi}{½} n x = \cos 4 \pi n x \]

\[ a_n = \frac{1}{\frac{1}{4}} \int_{-\frac{1}{4}}^{\frac{1}{4}} \cos(2 \pi x) \cos(4 \pi n x) \]
if we take advantage of the even property we can write this as
\[ a_n = 8 \int_0^{\frac{1}{4}} \cos(2 \pi x) \cos(4 \pi n x) \]

Answer 3

Thanks guys !