# What is the Fourier Transform of a constant signal?

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.

• 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. May 4, 2017 at 3:36
• this SE thing actually imposes a length limitation (i forgot how many characters). that's one of the annoying things about SE. May 5, 2017 at 0:46
• Huh?${}{}{}{}{}$ May 5, 2017 at 21:46
• @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. May 5, 2017 at 21:49
• @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. May 6, 2017 at 1:09

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 $\delta(f)$ $$x(t) = \int_{-\infty}^\infty \delta(f)e^{j2\pi ft} \mathrm{d}f = \int_{-\infty}^\infty \delta(f) \mathrm{d}f = 1.$$

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

When you take the Fourier Transform of a constant signal, the result will be a delta function centered at zero frequency. The delta function has a value of the amplitude of the constant signal.

To see why this happens, let's recall the formula for the Fourier Transform of a continuous-time signal $$x \left( t \right)$$:

$$X \left( f \right) = \int x \left( t \right) e^{-j 2 \pi f t } dt$$

where $$X \left( f \right)$$ is the Fourier Transform of $$x \left( t \right)$$, and $$f$$ is the frequency variable.

If $$x \left( t \right)$$ is a constant signal, then $$x \left( t \right) = C$$ for all $$t$$, where $$C$$ is a constant. Substituting this into the above formula, we get:

$$X \left( f \right) = \int C e^{-j 2 \pi f t} dt \implies X \left( f \right) = C \int e^{-j 2 \pi f t} dt$$

The integral in the above expression is equal to zero except when $$f = 0$$, in which case it is equal to the integration interval. Assuming that the integration interval is from $$- \infty$$ to $$\infty$$, we get:

$$X \left( f \right) = C \int e^{-j 2 \pi f t} dt \implies X \left( f \right) = C \int e^{0} dt \implies X \left( f \right) = C \left[ t \right]^{\infty}_{- \infty} \implies X \left( f \right) = C \left[ \infty - \left( - \infty \right) \right] \implies X \left( f \right) = C \cdot \infty$$

Therefore, the Fourier Transform of a constant signal $$x \left( t \right)$$ is a delta function centered at zero frequency with a value of $$C$$, the amplitude of the constant signal.

In summary, when you take the Fourier Transform of a constant signal, the frequency will be zero, and the result will be a delta function centered at zero frequency with a value equal to the amplitude of the constant signal.