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

AI Thread Summary
The Fourier series given is y(t) = 2 - (1/π)∑(1/n)sin(nπt/2), and the main discussion revolves around determining whether it is an odd or even function and identifying its period. The presence of the sine function suggests that the series is odd, as it meets the criteria for odd functions when evaluated at -t. The period is linked to the fundamental frequency, which is derived from the first sine term when n=1, indicating a period of 4 seconds. Participants emphasize the importance of checking the function's properties directly rather than solely relying on the multiple-choice options. The conversation highlights the need for clarity in understanding the Fourier series and its characteristics.
Northbysouth
Messages
241
Reaction score
2

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.
 

Attachments

  • ME Hwk 3 Question 1.png
    ME Hwk 3 Question 1.png
    5.5 KB · Views: 532
Physics news on Phys.org
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.
 
Back
Top