# Describe the Fourier Series

1. Feb 20, 2014

### Northbysouth

1. The problem statement, all variables and given/known data
For he following Fourier series, which of the answers correctly describes the following function

y(t) = 2 - $\stackrel{1}{π}$∑1inf1/nsin(n*πt/2)

a) odd function, period = 2 s
b) Even function, period = 2s
c) Odd function, period = 4s
d) Even functio, period = 4s

2. Relevant equations

3. The attempt at a solution

From the reading my professor assigned I'm pretty sure that it's odd because it has a sin function, though I still don't understand why.

I'm also confused as to how I find the period. I know that

T =1/f

w=2*π*f

So,

w = (n*π)/2

I'm not quite sure what to make of n.

Any help would be appreciated.

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2. Feb 20, 2014

### Simon Bridge

That would be: $$y(t)=2-\frac{1}{\pi}\sum_{n=1}^\infty \frac{1}{n\sin(n\pi t/2)}$$

Check y(-t), if it is the same as -y(t) then y(t) is odd, if it is the same as +y(t) then it is even.

y(t+T)=y(t), what is T?

Certainly the period of the sine function is something to do with 2 units.
But, since it is multi-choice, you don't actually have to find the period - just check to see which of the choices is correct.

3. Feb 21, 2014

### rude man

I think you're misreading what the OP meant, or maybe you're trying to point out that he /she miswrote it.

Your denominator is intended to be the terms in the series, not their inverse. Look at the OP's thumbnail.

4. Feb 21, 2014

### rude man

The series is a series of harmonics starting with a dc term ("2") and then a series of sine terms
So the first sine term must be the fundamental frequency where n = 1.
So compare sin(nπt/2) with sin(ωt). In other words, compare π/2 with ω since n = 1.

If you know ω, what is the period T?

BTW by "period" they mean the periodicity of the lowest frequency component, i.e. for which n=1.

5. Feb 21, 2014

### Simon Bridge

... a bit of both by the looks of things - neither thought completed. My excuse is that dinner was just about ready.

Yep - should have looked at the thumb.
$$y(t)=\frac{1}{\pi}\sum_{n=1}^\infty \frac{1}{n}\sin \left(\frac{n\pi t}{2}\right)$$

Which would make the RHS a Fourier expansion of something - a saw-tooth wave.
... advise still stands.

... or the period of y(t) - let OP work out if this is the same thing.

a pure sine wave would be: $y(t)=\sin(2\pi t/T)$ where T is the period ... comparing with any other sine wave would be how you find T.

But in this case there is a shortcut. Need only use the definition of the period to check to see if either of the options is the correct one.

Care needed though, if $y(t+T)=y(t)$ then $y(t+2T)=y(t)$ also.
But if $T$ is the period of $y(t)$ then $y(t+T/2)\neq y(t)$
So it is a matter of putting the numbers into the equation, and thinking about what "period" means.
Which, I suspect, is the point of the exercise.

6. Feb 21, 2014

### LCKurtz

Understandably so. None of the above, even without knowing what "s" is.

7. Feb 21, 2014

### Simon Bridge

@Northbysouth: any of this help?
You should check LCKurtz answer to see why... would y(t) be easier to handle without the "2 -" out the front?
(i.e. the way I wrote it in post #5?)

It is likely that there is a context missing from the problem statement you got.