Fourier Series Question: Express q(t) as a Fourier Expansion

[Nylex]

I got this question out of a book, but I can't get the book's answer. Since I can't draw, I'll just describe the graph given. Express q(t) as a Fourier series expansion.

The charge q(t) on the plates of a capacitor at time t is shown as a saw-tooth wave with period 2\pi and its peak is at t = \pi, where q(t) = Q.

At first, I tried to use the expressions for open boundary conditions (as q(t) is 0 at t = 0 and t = 2 pi). I got stuck, so I started again, with

q(t) = \frac{a_{0}}{\sqrt{L}} + \sum_{n=1}^\infty a_{n} \left(\frac{2}{L}\right)^\frac{1}{2} \cos \frac{2\pi nt}{L} + \sum_{n=1}^\infty b_{n} \left(\frac{2}{L}\right)^\frac{1}{2} \sin \frac{2\pi nt}{L}

where

a_{0} = \int_{0}^{L} \left(\frac{1}{L}\right)^\frac{1}{2} q(t) dt

a_{n} = \int_{0}^{L} \left(\frac{2}{L}\right)^\frac{1}{2} \cos \frac{2\pi nt}{L} q(t) dt

b_{n} = \int_{0}^{L} \left(\frac{2}{L}\right)^\frac{1}{2} \sin \frac{2\pi nt}{L} q(t) dt

Now, since the saw-tooth is symmetric, there was no need to calculate b_{n}, since they'd have been 0 anyway.

I represented q(t) using top hat functions:

q(t) = \left[\theta(t) - \theta(t - \frac{L}{2})\right]\frac{2tQ}{L} + \left[\theta(t - \frac{L}{2}) - \theta(t - L)\right]\frac{2(L - t)Q}{L}

Then I used that to calculate a_{0}[/tex] and a_{n}, using L = 2\pi<br /> <br /> After pages of working out, I got:<br /> <br /> a_{0} = Q\pi \left(\frac{1}{2\pi}\right)^\frac{1}{2}<br /> <br /> a_{n} = \left(\frac{1}{\pi}\right)^\frac{1}{2} \frac{Q}{\pi} \frac{2}{n^2} (\cos n\pi - 1)<br /> <br /> So,<br /> <br /> q(t) = Q\left[\frac{1}{2} + \frac{2}{\pi^2} \sum_{n=1}^\infty \frac{\cos nt \cos (n\pi - 1)}{n^2}\right]<br /> <br /> The book&#039;s answer is q(t) = Q\left[\frac{1}{2} - \frac{4}{\pi^2} \sum_{n=1}^\infty \frac{\cos (2n -1)t}{(2n - 1)^2}\right]<br /> <br /> I don&#039;t know where I&#039;ve gone wrong. I can&#039;t post all my working, as it&#039;s a lot and would take forever to LaTeX.

 

[Atrus]

So,

q(t) = Q\left[\frac{1}{2} + \frac{2}{\pi^2} \sum_{n=1}^\infty \frac{\cos nt \cos (n\pi-1)}{n^2}\right]

The book's answer is q(t) = Q\left[\frac{1}{2} - \frac{4}{\pi^2} \sum_{n=1}^\infty \frac{\cos (2n -1)t}{(2n - 1)^2}\right]

I don't know where I've gone wrong. I can't post all my working, as it's a lot and would take forever to LaTeX.

Well, your results are equivalent I'm poor at LaTeX, but I'll try to explain.

First, there is a typo in a final result (I suspect it is a typo), there should be

q(t) = Q\left[\frac{1}{2} + \frac{2}{\pi^2} \sum_{n=1}^\infty \frac{\cos nt (\cos (n\pi)-1)}{n^2}\right]

(notice the braces in second cos), but you got it right in a_n expression.

OK, now whenever n is even, expression under sumation is equal to 0 because (\cos (n\pi)-1) = 0 for even values of n.

So, we have to sum only if n is odd, otherwise written as n = 2k-1, for k going from 1 to inf.

For odd numbers (\cos (n\pi)-1) = -2

If you put it all together you can write
\sum_{k=1}^\infty \frac{-2\cos (2k-1)t}{(2k-1)^2}\right]

you can take the "-2" out of summation, and then you get the book's answer.

Hope my explanation was clear,
Cheers

 

[Nylex]

Well, your results are equivalent I'm poor at LaTeX, but I'll try to explain.

First, there is a typo in a final result (I suspect it is a typo), there should be

q(t) = Q\left[\frac{1}{2} + \frac{2}{\pi^2} \sum_{n=1}^\infty \frac{\cos nt (\cos (n\pi)-1)}{n^2}\right]

(notice the braces in second cos), but you got it right in a_n expression.

OK, now whenever n is even, expression under sumation is equal to 0 because (\cos (n\pi)-1) = 0 for even values of n.

So, we have to sum only if n is odd, otherwise written as n = 2k-1, for k going from 1 to inf.

For odd numbers (\cos (n\pi)-1) = -2

If you put it all together you can write
\sum_{k=1}^\infty \frac{-2\cos (2k-1)t}{(2k-1)^2}\right]

you can take the "-2" out of summation, and then you get the book's answer.

Hope my explanation was clear,
Cheers

Yay, thanks very much. Yes, I did make a typo with the brackets in my post.

Thanks again :D.