Fourier Transform Homework: Determine First 3 Terms

[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

<br /> a_o=\frac{1}{2L}\int_{-L}^Lf(t)dt<br />

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

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.

<br /> 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]<br />
<br /> =\frac{2}{3T}\left[-\left(-T/4-(-3T/4)\right)+\left(T/4-(-T/4)\right)+\left(3T/4-T/4\right)\right]<br /> <br /> =\frac{2}{3T}\left(T/4-3T/4+T/4+T/4+3T/4-T/4\right)=1/3<br />

<br /> 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]<br />

<br /> =\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}<br />
<br /> -\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}<br />


<br /> \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]<br />

Distributing the signs through and simplifying:

<br /> \frac{1}{n\pi}\left[4sin\left(\frac{n\pi}{3}\right)-2sin\left(n\pi\right)\right]<br />

So for n=1,2,3 I get

<br /> a_1=\frac{2\sqrt{3}}{2\pi}<br />

<br /> a_2=\frac{\sqrt{3}}{2\pi}<br />

<br /> a_3=0<br />

 

[lanedance]

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.

 

[lanedance]

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.

 

[lanedance]

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.

 

[lanedance]

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