Fourier transform of rect(x/2)*comb(x) + sketch

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SUMMARY

The Fourier Transform of the function f(x) = rect(x/2) * comb(x) results in F(u) = 2sinc(2u) * comb(u). The sinc function is defined as sinc(u) = sin(pi u)/(pi u), which is crucial for accurate representation. The graph of this function consists of discrete points at integer values, specifically yielding {2, 0, 0, ...} for n ≥ 0, where the value at u=0 is conventionally defined as 2 despite the sinc function being undefined at that point.

PREREQUISITES
  • Understanding of Fourier Transform principles
  • Familiarity with the rectangle function (rect)
  • Knowledge of the Dirac comb function (comb)
  • Concept of the sinc function and its properties
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  • Study the properties of the Fourier Transform of convolution
  • Explore the applications of the sinc function in signal processing
  • Learn about the implications of sampling theory in Fourier analysis
  • Investigate graphical representations of Fourier Transforms for various functions
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Homework Statement



Take the Fourier Transform of f(x)=rect(x/2)*comb(x) where rect is the rectangle function and comb is the Dirac comb. Sketch the results.

Homework Equations


The FT of a convolution is the product of the individual FTs.


The Attempt at a Solution


Taking the FT is pretty simple, but it's the graph that I'm a little confused about. If we take the FT of f(x)<->F(u)

F(u)=2sinc(2u)comb(u), if we define sinc(u)=sin(pi u)/(pi u), rather than the other way which leaves out pi.

The graph of such a function is simply the sampling of points along 2sinc(2u) at the period of comb(u), which should just be 1. Therefore at u=0, you'd have a point of 2 (although 2sinc(2u) isn't defined at u=0, though I believe it's commonly considered to be defined as its limit, which in this case is 2). All of the other integer values of 2sinc(2u) are simply 0.

Am I correct in assuming that the graph of 2sinc(2u)comb(u) is just {2,0,0,...,0} for all n>=0?

This seems like a silly answer for a problem, but maybe that's the point. I just have a tendency to second guess myself when given a stupid problem with a simple answer.
 
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Sorry, this thread should be in Calculus & Beyond. I thought I was in that forum before I posted. If a mod could move it, I'd appreciate not having to re-post it.
 

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