# Fourier Transform and amplitude of waves

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 $\left|\hat{f}(\omega_0)\right|$.

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This is the definition of the Fourier transform. $\hat{f}(\omega)$ is the complex amplitude of the harmonic component of frequency $\omega$. –  Ron Gordon Aug 20 '13 at 19:55
You mean that is just a definition and is not the actual amplitude of the wave for the given frequency? –  Ambesh Aug 20 '13 at 19:59
I'd say that your intuition is right. Just like $f(t)$ tells you the contribution of time $t$ to the overall signal, $\hat{f}(\omega)$ tells you the contribution of frequency $\omega$ to the overall signal. And just like there is a continuum of time values, there is a continuum of frequency values. Also note that if you look at the definition of the FT and hold $\omega$ fixed, it is nothing more than the inner product of $f$ with a complex sinusoid at that frequency. –  AnonSubmitter85 Aug 20 '13 at 22:42

• 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: $P(a\le X\le b)=\int_{a}^b p(x)\,dx$
• 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 $\int_{a}^b |\widehat f(\xi)|\,d\xi$