What Is the DFT of a Constant Value?

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SUMMARY

The discrete Fourier transform (DFT) of a constant value, specifically DFT{2}, simplifies to a delta function at k = 0. The DFT formula, X_{k} = ∑x[n] * e^{-2πkn/N} from n = 0 to N-1, allows for the constant value to be factored out, resulting in a sum of exponential terms. This leads to the conclusion that the DFT of a constant results in an impulse at the zero frequency component, confirming the expected behavior of the DFT for constant inputs.

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Mr.Tibbs
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The problem:

What is the discrete Fourier transform of a constant value?

Example DFT{2}

This is not my homework problem but will help me immensly in solving the actual problem.

DFT formula:

X_{k} =\sumx[n] * e^{\frac{-2\pi kn}{N}} from n = 0 to N-1

where N is the number of samples you can take in a 2\pi period.
 
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So x[n] = 2 for all n.

That means you can take x[n] outside the summation, and you're left with a sum of exponential terms. Do you know how to work that out?

Naively I would expect it to reduce to an impulse (delta function) at k = 0, or perhaps regularly repeating impulses, due to the discrete nature of the DFT. I've have to sit down and think about it some more.
 
Ah, that helps out immensely. As it turns out it does turn into a delta function at k = 0. Thank you so much for clearing that up for me.
 

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