Fourier Transform and amplitude of waves

Question

Given this definition of the fourier transform:

$$f(t) \rightarrow \hat{f}(\omega)=\int\limits_{-\infty}^{+\infty}f(t)\,e^{-i\omega t}\,dt$$

and now

$$f(t)=\frac{1}{2\pi}\int\limits_{-\infty}^{+\infty}\hat{f}(\omega)e^{i\omega t}\,d\omega$$

It is supposed that for a particular frequency $\omega_0$ the amplitude is: $\left|\hat{f}(\omega)\right|$

I don't see why this is the amplitude.

What I'm trying to understand is an analogy with the fourier series. In the fourier series the coefficients are the amplitude for each of the particular waves that make our signal. I'm trying to see if in the fourier transform this means that we have infinite waves added and that each one has amplitude $\left|\hat{f}(\omega_0)\right|$.

Answer 1

An analogy with probability may be helpful.

• When we have a discrete random variable, its distribution is described by numbers that are actual probabilities of certain events, $P(X=a)$.
• For a continuous random variable, we use a probability density function. It relates to probabilities as follows: $P(a\le X\le b)=\int_{a}^b p(x)\,dx$

Same with Fourier series and integrals:

• Fourier series describes a periodic function by numbers (coefficients of Fourier series) that are actual amplitudes (and phases) associated with certain frequencies.
• Fourier transform is a "frequency density function". It relates to amplitudes as follows: if all frequencies in the range $a\le \xi\le b$ align perfectly at some moment, the resulting amplitude is $\int_{a}^b |\widehat f(\xi)|\,d\xi$