Understanding Fourier Coefficients for Step Functions: Homework Help

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Homework Help Overview

The discussion revolves around understanding Fourier coefficients for a step function defined piecewise over a specified interval. The original poster seeks clarification on the definitions of the Fourier coefficients, specifically a_n and b_n, and questions the normalization factor used in their calculations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the definitions of Fourier coefficients and their corresponding integrals. The original poster expresses uncertainty about the normalization factor and seeks a more precise definition beyond common sources.

Discussion Status

The conversation is ongoing, with participants prompting the original poster to provide their work and suggesting resources for reference. There is a focus on evaluating the integrals for the coefficients, but no consensus has been reached regarding the normalization factor or the definitions.

Contextual Notes

The original poster indicates a lack of reference material, which may be influencing their understanding of the definitions and calculations involved.

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Homework Statement



f(t) = 1 0<=t<T/2
-1 T/2 <=t<=T

ie. step function.frequency w_0 = 2pi/T

Homework Equations


The Attempt at a Solution



What's the definition for the Fourier coefficients a_n and b_n again? Not the one in wikipedia.
 
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Show us the work you have done on this. There are defining integrals for these coefficients, so set them up and try evaluating them.
 
[tex] <br /> a_n = \frac{1}{\pi} \int_0^{T/2} 1*cos(nt) dt +\frac{1}{\pi} \int_{T/2}^{T} (-1)*cos(nt) dt [/tex]

[tex] <br /> b_n = \frac{1}{\pi} \int_0^{T/2} 1*sin(nt) dt +\frac{1}{\pi} \int_{T/2}^{T} (-1)*sin(nt) dt [/tex]

I think the normalization factor is incorrect, which is why i asked for the definition, I think it was 2/L where L is the length of the interval - i can't remember.
 
If you can't remember, do you have a book?
 
No, that's why I asked.
 

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