The Fourier transform

Question

The Fourier transform is just an operation on certain functions. When the functions are signals, why is it that the variable chosen (w) has to correspond to the frequency of the signal, when it is just a variable?

Answer 1

and welcome! You're right that the Fourier transform is just a mathematical operation that you can perform on any function. You can perform a Fourier transform in space (take the Fourier transform of an image, for example), on voltages as a function of time, take the fourier transform of some differential equation, etc.

The reason why \$\omega\$ corresponds to the angular frequency when applied to a signal in the time domain comes from the details of the math. The Fourier transform for a signal \$f(t)\$ can be written as:

$$ F(\omega) = \int f(t) e^{-i\omega t}\, dt $$

so what does that mean? Well, you can write \$ e^{-i\omega t}= \cos(\omega t)-i\sin(\omega t)\$ (Euler's rule), so then you start to see clearly what \$\omega\$ means.

When \$\omega t = 2 \pi\$, then \$\cos\$ and \$ sin\$ would have gone one full cycle. So the amount of time for one full cycle (the period) is: $$ T_P = \frac{2 \pi}{\omega} $$

So that's the number of seconds per cycle. The number of cycles per second (the frequency in Hz) can then be found:

$$ f = \frac{1}{T_P}=\frac{\omega}{2\pi} $$

So that's why the \$\omega\$ of the Fourier transform of a signal is the angular frequency (\$\omega\$). Which is related to the frequency in Hz using the equation above.

Answer 2

Historically the Fourier analysis claimed that any function of a variable, whether continuous or discontinuous, can be written as a series of sines with multiple variables. I think that is where the term frequency comes in to picture. @KD9PDP has nicely explained the mathematical formulation.