# Fourier Series problem

Discussion in 'Homework Help' started by torx11, Mar 9, 2014.

1. ### torx11 Thread Starter New Member

Oct 6, 2011
13
0

Given that function, find the Fourier coefficients using the Fourier series method.

I am having a hard time figuring out how to find the a_k for this problem. The formula I am using to do so is:

I have tried to integrate it from 0 to the period T 2/3 and the answer always come out to be zero. I got an answer of zero also for the a_0.

Please could somebody help me with this problem.

Thank you.

2. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
We can't figure out where you are going wrong unless you show us your work.

3. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,504
512
Sine is an odd function. This means you don't need to do all the steps of finding Fourier series. You only need to do steps that apply to odd functions.

Here is one explanation: http://mathworld.wolfram.com/FourierSineSeries.html

If you want, I can scan relevant section from my textbook.

torx11 likes this.
4. ### torx11 Thread Starter New Member

Oct 6, 2011
13
0
Thank you for pointing me in the right direction. I will research the link you sent me.

5. ### blah2222 Well-Known Member

May 3, 2010
565
33
You can expand the function s(t) into a more usable form as show below:

$

s(t) = sin^3(3\pi t) = sin(3\pi t)sin^2(3\pi t) = \frac{1}{2}sin(3\pi t)(1 - cos(6\pi t)) = \frac{1}{2}(sin(3\pi t) - sin(3\pi t)cos(6\pi t))

= \frac{1}{2}(sin(3\pi t) - \frac{1}{2}(sin(9\pi t) - sin(3\pi t)))

= \frac{1}{2}sin(3\pi t) - \frac{1}{4}sin(9\pi t) + \frac{1}{4}sin(3\pi t)

= \frac{1}{4}(3sin(3\pi t) - sin(9\pi t))

= \frac{1}{4}(3sin(2\pi(\frac{3}{2})t) - sin(2\pi(\frac{9}{2})t))

$

By the looks of it you are using the complex exponential representation of the Fourier Series which is ultimately of the form:

$

s(t) = \sum\limits_{n=-\infty}^{\infty} \alpha_{n}e^{jn\omega t}

\alpha_{n} = \frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}} s(t)e^{-jn\omega t} dt

$

Euler's formula decomposes a sine function as follows:

$

sin(x) = \frac{1}{2j}(e^{jx} - e^{-jx})

$

Knowing the symmetry of a sine function on either side of the y-axis can also be a good clue for what will happen when you integrate over one period.

Putting that all together, you might just have a decent stab at this, but you really need to show your work for us to give more specific guidance.