What is the Fourier Transform of a constant signal?

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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.