Basic Fourier Series question

Smythe

f(t) = [t]
(modulus of t)

-∏<t<∏
T=2∏
(period of 2∏)

What is a0?


The answer I have here is a0 = ∏

Is that correct? If so, why is a0 = ∏?

Thank you.

Michael Collins

You're close but not quite there with that answer - it should be

[latex] \displaystyle a_{0} = \frac{\pi}{2}. [/latex]

The [latex] a_{0} [/latex] term in a Fourier series is simply the average value of the function over one full period.

So [latex] f=a_{0} [/latex] is the constant valued function which gives you the same area over one full period as the function in question.

Smythe

Are you sure a0/2 doesn't equal pi/2 in this case?

In the notes I have it says;

a0/2 = pi/2 is the mean value of the signal

Michael Collins

Ah OK, yes you are correct. I forgot that some people only half the value of the [latex] a_{0} [/latex] term in the final formula, instead of calculating the actual average/mean term directly. (This is so you can use the [latex] a_{n} [/latex] formula for obtaining the [latex] a_{0} [/latex] term, instead of having to remember a separate formula for [latex] a_{0} [/latex].)

Smythe

Okay, no problem.

Also, I found essentially the same worked problem here;
http://www.exampleproblems.com/wiki/index.php/FS1

using an x instead of the t in my question.

It can be seen on that page what 'an' equals.

Would you have any idea why my notes state that 'an' equals
-4/[∏(k^2)]

Michael Collins

Provided in your notes there's a piece of text saying "for odd k only" or something similar, it actually works out the same whichever formula you use.

If you look at the [latex] a_{n} [/latex] term in those notes, they have

[latex] \displaystyle a_{n} = \frac{\left2((-1)^{n}-1)\right)}{\pi n^2}[/latex]

for all even values of n, this is zero, so we can just rewrite this as

[latex] \displaystyle \underbrace{a_{n}}_{n{\subscriptsize \hbox{ odd}}} = \frac{2(-2)}{\pi n^2} = \frac{-4}{\pi n^2}. [/latex]

since for all the odd values, the -1-1 add to give negative -2.

So it's the same thing at the end of the day.

Smythe

That's great. Thanks very much for that, I actually understand it now!

Michael Collins

Great. Glad to be of help.