What Is the DFT of a Constant Value?

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The discrete Fourier transform (DFT) of a constant value, such as 2, simplifies significantly due to the nature of the summation in the DFT formula. When applying the DFT formula, the constant can be factored out, leaving a sum of exponential terms. This results in the DFT yielding a delta function at k = 0, indicating that all the signal's energy is concentrated at this frequency. The discussion confirms that the DFT of a constant indeed produces a single impulse at the zero frequency, which is a crucial insight for understanding Fourier transforms. This clarification aids in solving related problems effectively.
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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