# Fourier series expansion. Find value of a term in expansion

1. Jan 28, 2017

### jaus tail

1. The problem statement, all variables and given/known data
Fourier series expansion of a signal f(t) is given as
f(t) = summation (n = -inf to n = +inf) [3/(4+(3n pi)2) ) * e j pi n t

A term in expansion is A0cos(6 pi )
find the value of A0

Repeat above question for A0 sin (6 pi t)

2. Relevant equations
Fourier expansion is summation n = -inf to +inf Cn ejwnt
where Cn is Integration over T0 x(t) e-jwnt dt

3. The attempt at a solution

In book they've said Cn is right.
But then they say an is 2 times real part of Cn
and have done an = 6/[4 + (18 pi)2

For third part they've put A0sin (6 pi t) = bn sin (n w t)
and for n = 6, this is zero.

I didn't understand this part. Why did they multiply it by 2 first and then how did it become zero in the second.

I understand that sin ( 6 pi) is zero but how can sin ( 6 pi t) be zero? Wouldn't this vary as 't' varies?

2. Jan 28, 2017

### pasmith

The coefficients of $\cos n\pi t$ and $\sin n\pi t$ come not just from $c_n$, but also $c_{-n}$: $$c_ne^{i n\pi t} + c_{-n}e^{-in\pi t} = (c_n + c_{-n})\cos n\pi t + i(c_n - c_{-n})\sin n\pi t$$

3. Jan 28, 2017

### jaus tail

Oh yeah. Thanks. I didn't think about C-n.