Fourier Transform: Solve Homework Equations for fd

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Homework Help Overview

The discussion revolves around the Fourier Transform and its application to a function denoted as fd. Participants are exploring the relationship between the function fd and its representation in the context of Fourier analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the expression for fd and its dependence on variable x, questioning how the function is summed over a specified range. There are attempts to clarify the transformation process and the implications of the limits applied to the original function f(x).

Discussion Status

The discussion is active, with participants providing insights into the nature of the function fd and its Fourier transform. Some guidance has been offered regarding the relationship between fd and f(x), but there is no explicit consensus on the interpretation of the problem or the next steps to take.

Contextual Notes

There are indications of confusion regarding the function's dependence on variable x and the limits of integration. Participants are also addressing potential issues with the initial problem statement and its formulation.

Lengalicious
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Homework Statement


See Attachment


Homework Equations





The Attempt at a Solution



Ok so in a previous question I worked out fd = e-ipd*2*sinc(pa)/√(2∏), also worked out its Fourier transform if that helps.

Now I really am stuck on the question, any guidance would be appreciated, I don't understand how the function fd is summed over from -N to N, like I say any help to just send me in the right direction would be great.
 

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The problem suggests that the function fd should be a function of x, but I don't see an x in your expression.
 
MisterX said:
The problem suggests that the function fd should be a function of x, but I don't see an x in your expression.

Well the original function was f(x) which was a function of x that I then Fourier transformed with limits of x =-a to x=a, this replaced the x's with a's. fd(x) was then a slight variation of the original f(x) function, ultimately I ended up with said function where fd(x) = f(x - d) which transforms to e(-ipd)*f(p) where f(p) was the Fourier transform of f(x).
 
I made an issue with the question, not sure how to edit. Anyway f_d(x) = 1/a when |x-d|< a. What I said in the 1st post is actually the Fourier transform of this.
 

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