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

In summary: Do you have a textbook? Can you post the actual question?In summary, the given Fourier series for function y(t) is odd and has a period of 4 seconds. The explanation for why it is odd and how to find the period is provided in the conversation.
  • #1
Northbysouth
249
2

Homework Statement


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


y(t) = 2 - [itex]\stackrel{1}{π}[/itex]∑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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
Northbysouth said:

Homework Statement


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


y(t) = 2 - [itex]\stackrel{1}{π}[/itex]∑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.
 
  • #7
@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.
 

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

What is a Fourier Series?

A Fourier Series is a mathematical representation of a periodic function as a sum of sinusoidal functions. It is named after French mathematician Joseph Fourier and is commonly used in signal processing, physics, and engineering.

How is a Fourier Series calculated?

To calculate a Fourier Series, the function is decomposed into a series of sine and cosine functions with different amplitudes and frequencies. The coefficients of these functions are then determined by integrating the original function over one period.

What is the significance of the Fourier Series?

The Fourier Series is significant because it allows complex periodic functions to be represented in a simpler form, making it easier to analyze and manipulate. It is also used in many real-world applications, such as in electrical engineering, image processing, and data compression.

What is the difference between a Fourier Series and a Fourier Transform?

A Fourier Series is used to represent a periodic function, while a Fourier Transform is used to represent a non-periodic function. Additionally, a Fourier Series uses discrete frequencies, while a Fourier Transform uses continuous frequencies.

Can any function be represented by a Fourier Series?

No, a function must meet certain criteria in order to be represented by a Fourier Series. It must be a periodic function with a finite number of discontinuities and the integral of the function must exist over one period. Otherwise, the Fourier Series will not converge to the original function.

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