How to interpret the Fourier Transform 𝑋(𝜔) (Foundational Questions)

Question

The motivation behind the fourier transform is to somehow represent a non-periodic signal as a sum of sinusoids just as we do with the fourier series for periodic signals, correct?

With the Fourier series we have, $$x(t) = x(t+T) = \sum_{k = -\infty}^{\infty} X[k]e^{j\frac{2\pi}{T} t}$$

where $T$ is the period of our signal and $X[K]$ is the weight or coefficient of one specific sinusoid frequency, is that correct?

$$$$

On the other hand, we can't represent non-periodic the same way because each part of the signal is not (necessarily) the same. We attempt to remedy this issue by using Fourier transforms. Let's take the continuous time Fourier transform (the only one I've learned):

Let $x(t)$ be some non-periodic signal, let's say it looks like a signal that has a square wave first, then a triangle wave, then something random.

We take the Fourier transform:

$$X(\omega) = \int_{-\infty} ^{\infty} x(t)e^{-j\omega t}dt$$

where I believe $\omega = \frac{2\pi k}{T}$. In this case, $T$ here is some period of time we choose to repeat the signal to attempt to make a periodic signal... if that makes any sense to anyone out there

Now here's where my confusion really gets complicated.

Transferring the same logic as we did with the Fourier series $X(\omega)$ should be the the weight or coefficient of one particular sinusoid frequency that makes up the signal $x(t)$, right? Or is that just completely wrong? What's the interpretation of $X(\omega)$ here?

Also, we said our signal $x(t)$ was a square wave first, then a triangle wave, then something random. Shouldn't the $X(\omega)$ for a particular $\omega$ be different for the first part (square wave) than it is for the second part(triangle wave), and so on? How come we only get one value of $X(\omega)$?

Answer 1

If the Fourier transform

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{1}$$

of a function $x(t)$ exists, then the function $x(t)$ can be expressed in terms of its Fourier transform:

$$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega\tag{2}$$

Eq. $(2)$ is also called inversion formula. Note that $\omega$ is a real continuous variable. I.e., in general all frequencies are necessary to represent a non-periodic function $x(t)$, not just a discrete set of frequencies, as is the case for periodic functions.

The Fourier transform $X(\omega)$ is generally a complex-valued function. Its magnitude is often called the spectrum of $x(t)$. It represents the frequency content of $x(t)$. However, the phase of $X(\omega)$ is equally important for representing $x(t)$. The squared magnitude $|X(\omega)|^2$ is called the energy density of $x(t)$.

Even though mathematically unsatisfactory, it may help intuitively to think of $x(t)$ as a function whose "period" approaches infinity (which makes it non-periodic), and, consequently, the distance between the frequencies needed for its representation becomes smaller and smaller, until, in the limit, $\omega$ becomes a continuous variable, and the sum of the original Fourier series becomes and integral.