# Fourier Transform

roldy

## Homework Statement

I need to determine the first three terms in the Fourier series pictured in the attachment.
Did I define the peace-wise functions correctly?

I'm re-posting this with the tex code instead of the attached document.

## Homework Equations

$$a_o=\frac{1}{2L}\int_{-L}^Lf(t)dt$$

$$a_n=\frac{1}{L}\int_{-L}^Lf(t)cos\left(\frac{n\pi{t}}{L}\right)dt$$

## The Attempt at a Solution

The plot of the series is symmetric, so therefore I am only going to find what $$a_o$$ and $$a_n$$.

$$a_o=\frac{2}{3T} \left[\int_\frac{-3T}{4}^\frac{-T}{4}(-1)dt+\int_\frac{-T}{4}^\frac{T}{4}(1)dt+\int_\frac{T}{4}^\frac{3T}{4}(-1)dt\right]$$
$$=\frac{2}{3T}\left[-\left(-T/4-(-3T/4)\right)+\left(T/4-(-T/4)\right)+\left(3T/4-T/4\right)\right] =\frac{2}{3T}\left(T/4-3T/4+T/4+T/4+3T/4-T/4\right)=1/3$$

$$a_n=\frac{4}{3T}\left[\int_\frac{-3T}{4}^\frac{-T}{4}(-1)cos\left(\frac{n\pi{t}}{3T/4}\right)dt+\int_\frac{-T}{4}^\frac{T}{4}(1)cos\left(\frac{n\pi{t}}{3T/4}\right)dt+\int_\frac{T}{4}^\frac{3T}{4}(-1)cos\left(\frac{n\pi{t}}{3T/4}\right)dt\right]$$

$$=\frac{4}{3T}\left[\left[-sin\left(\frac{n\pi{\left(-T/4\right)}}{3T/4}\right)+sin\left(\frac{n\pi{\left(-3T/4\right)}}{3T/4}\right)\right]\frac{3T}{4n\pi}+\left[sin\left(\frac{n\pi{\left(T/4\right)}}{3T/4}\right)-sin\left(\frac{n\pi{\left(-T/4\right)}}{3T/4}\right)\right]\frac{3T}{4n\pi}$$
$$-\left[sin\left(\frac{n\pi{\left(3T/4\right)}}{3T/4}\right)-sin\left(\frac{n\pi{\left(T/4\right)}}{3T/4}\right)\right]\frac{3T}{4n\pi}$$

$$\frac{1}{n\pi}\left[\left[-sin\left(\frac{-n\pi}{3}\right)+sin\left(-n\pi\right)\right]+\left[sin\left(\frac{n\pi}{3}\right)-sin\left(\frac{-n\pi}{3}\right)\right]-\left[sin\left(n\pi\right)-sin\left(\frac{n\pi}{3}\right)\right]\right]$$

Distributing the signs through and simplifying:

$$\frac{1}{n\pi}\left[4sin\left(\frac{n\pi}{3}\right)-2sin\left(n\pi\right)\right]$$

So for n=1,2,3 I get

$$a_1=\frac{2\sqrt{3}}{2\pi}$$

$$a_2=\frac{\sqrt{3}}{2\pi}$$

$$a_3=0$$

Homework Helper
What is f?

roldy
What is f?

It's not f. It's f(t), meaning f of t. If you look at the attached picture I came up with f(t) on three intervals. I'm not sure if they are correct. They look correct to me.

Homework Helper
yeah i get its f(t) but can't see any attached pic? pretty important part of the problem

roldy
Sorry about that. For some reason it didn't attach. Here's another try at it.

#### Attachments

• untitled.JPG
15.9 KB · Views: 283
Homework Helper
ok, so why integrate over more that a period?

3T/4 - (-3T/4) = 3T/2

roldy
I think that's where I might of made the mistake. I think I should of done it from -T/4 to 3T/4.

Homework Helper
That sounds better... Note that the integral over sin(tnpi/T) would be non zero over that interval as well, you could try your new interval on a sin as a check