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I am trying to figure out what the fourier transform of a constant signal is and for some reason i am coming to the conclusion that the answer is 1. Or better yet a step function.

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    $\begingroup$ the continuous Fourier Transform of a constant is not 1 (a constant), but is a dirac delta function. $$ \mathscr{F} \{ C \} = C \cdot \delta(f) $$ and that is not 1. $\endgroup$ – robert bristow-johnson May 4 '17 at 3:36
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    $\begingroup$ this SE thing actually imposes a length limitation (i forgot how many characters). that's one of the annoying things about SE. $\endgroup$ – robert bristow-johnson May 5 '17 at 0:46
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    $\begingroup$ Huh?${}{}{}{}{}$ $\endgroup$ – Dilip Sarwate May 5 '17 at 21:46
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    $\begingroup$ @robertbristow-johnson Comments need to have at least 15 characters in them, but as my Huh? above illustrates, one can still get around the requirement by surrounding pairs of {} with dollar signs. $\endgroup$ – Dilip Sarwate May 5 '17 at 21:49
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    $\begingroup$ @DilipSarwate, answers must be at least 30 characters. i s'pose my commment was longer than 30 characters. but it was still to trivial to deserve an answer. {} {} what really bothers me is, when i am on the meta site, that SE's bot copy edits my response and, due to terseness or lacking capitals, says that the quality of my answer is not good enough. that's when i would like to kick that bot's metal ass. $\endgroup$ – robert bristow-johnson May 6 '17 at 1:09
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I'll complete a bit the answer given in a comment above.

Intuitively first, to which frequency corresponds a signal constant in time, for exemple $x(t) = 1$ $\forall t$ ? Such a signal shows no variation in time and hence contains only a component with frequency 0 (this is a DC signal). This means that its Fourier transform must be 0 everywhere, except in $f=0$. Mathematically, $$X(f) = \delta(f).$$ Now, can we prove this? Yes, simply take the inverse Fourier transform of $\delta(f)$ and use the properties of the Dirac delta $\delta(f)$ $$x(t) = \int_{-\infty}^\infty \delta(f)e^{j2\pi ft} \mathrm{d}f = \int_{-\infty}^\infty \delta(f) \mathrm{d}f = 1.$$

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Fourier transforms (they are legion) somehow reflect the amplitude of (complex) sines in data. A flat signal "should" only have non-zero amplitudes on the $0$th frequency, and $0$ amplitude on the others. But what are we calling a flat signal? I will restrict to two common acceptions.

  1. In the continuous time, the signal spreads from $-\infty$ to $\infty$, and a continuous-time Fourier transform naturally transforms this infinite spread into an infinite amplitude at the $0$th frequency, theoretically turned into a distribution, denoted by the Dirac $\delta$ function, as answered by @anpar
  2. In a spatially bounded interval (like a constant-valued image), either continuous or discrete, assuming periodicity to maintain some flatness (using Fourier series or the discrete Fourier transform), you obtain a finite constant at $0$ frequency, and zero elsewhere.

This finite constant depends on how you normalize your Fourier transform.

Finally, on a single-sample signal, the DFT or FFT indeed gives you a constant "Fourier" transform:

fft(1)

ans = 1

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