What's the meaning of the amount of frequency 0 in the Fourier Transform?

Question

In the Fourier series, I knew that the coefficient $a_0$ represents the DC value, shifting the signal up and down by the amount. Then, what's the actual meaning of the amount of frequency $0$ in the Fourier Transform? I saw many people refer to it as the DC component. But I thought the DC should be the average value and the $$ X(j0) = \int\limits_{-\infty}^\infty x(t)dt $$ It's not an averaged value. In addition, I saw the steps for obtaining the FT of the unit step function $u(t)$, however, $$ U(j0) = \int\limits_{-\infty}^\infty u(t)dt = \infty $$ What does it mean when the DC value is infinity? Therefore, I'm guessing that the Fourier Transform at frequency $0$ is not a DC value. So, what is it? Thanks!!!

Answer 1

The Continuous Fourier Transform is a metric of spectral density. Let's consider a simple DC signal $x(t) = 2$.

Since the signal constant and infinitely long, it's energy is infinite as well. All this energy is concentrated at a single frequency (0Hz) so the spectral density at 0Hz is infinite as well. To be precise the FT of $x(t) = 2$ is

$$X(\omega) = 2\delta(\omega)$$

You will find a similar effect for sine waves (complex or real) as well, i.e.

$$x(t) = e^{j\omega_0 t} \leftrightarrow X(\omega) = \delta(\omega-\omega_0)$$

$$x(t) = \cos(\omega_0 t) \leftrightarrow X(\omega) = \frac{1}{2}(\delta(\omega-\omega_0)+\delta(\omega+\omega_0))$$

One of tricky parts here is that a DC-only signal or cosine are "power signals", i.e. they have finite power but infinite energy. If this infinite energy is distributed over a finite number of frequencies the spectral density must be infinite.

In practice that's rarely a problem since power signals are physically impossible as they require infinite energy and infinite duration.