Is This Fourier Series an Odd or Even Function with a Period of 4s?

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Discussion Overview

The discussion revolves around determining whether a given Fourier series represents an odd or even function and identifying its period. Participants explore the properties of the function defined by the series, which includes sine terms and a constant offset.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants suggest that the function is odd due to the presence of sine terms, but they express uncertainty about the reasoning behind this classification.
  • There is confusion regarding the calculation of the period, with references to the relationship between frequency and period (T = 1/f) and the angular frequency (ω = 2πf).
  • One participant points out a potential misreading of the original function, suggesting that the series should be interpreted correctly to understand its properties.
  • Another participant emphasizes that the period should be determined based on the lowest frequency component of the series, specifically when n = 1.
  • There is a suggestion that the function could represent a saw-tooth wave based on the Fourier expansion interpretation.
  • Some participants express doubt about the correctness of the multiple-choice options provided in the original question.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the function is odd or even, nor on the correct period. Multiple competing views remain regarding the interpretation of the function and its properties.

Contextual Notes

There are limitations in the discussion related to missing assumptions about the function's definition and the implications of the constant term in the series. Participants also note that the problem statement may lack necessary context for a complete understanding.

Northbysouth
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Homework Statement


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

Homework Equations





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

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.
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.

I'm also confused as to how I find the period.
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.
 
Simon Bridge said:
That would be: $$y(t)=2-\frac{1}{\pi}\sum_{n=1}^\infty \frac{1}{n\sin(n\pi t/2)}$$

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.
 
Northbysouth said:

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.


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.
 
I think you're misreading what the OP meant, or maybe you're trying to point out that he /she miswrote it.
... a bit of both by the looks of things - neither thought completed. My excuse is that dinner was just about ready.

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

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

Homework Statement


For he following Fourier series, which of the answers correctly describes the following function


y(t) = 2 - \stackrel{1}{π}∑1inf(1/n)sin(n*πt/2)

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

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.

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

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