## Frequency Domain and the Fourier Transform

## Sound Is Frequencies

Most sounds have high periodicity

Fourier's Theorem (FOO-ree-YAY or thereabouts) says that an infinitely repeating sound can be represented as a sum of sinusoids

The ear hears/decomposes a sum of sinusoids

Yet PCM is a sequence of samples over time: frequency is not represented

- The Nyquist Limit is hard to think of as a signal change over time thing

## Time and Frequency

We have: a continuous waveform, a function \(f(t)\) representing sound pressure

We want: a continuous

*spectrum*, showing the amplitude and phase of sine waves at every frequency \(\hat{f}(\omega)\)Wait, amplitude and phase from a single function? Yes, representing a frequency as a complex number with the usual geometric interpretation

$$ f(\omega) = a + b i $$

$$ |f(\omega)| = \sqrt{a^2 - b^2} $$

$$ \theta(f(\omega)) = tan^{-1}(a, b) $$

Note: you will see both

*i*and*j*for \(\sqrt{-1}\) in different contextsNote: we freely mix between

*angular frequency*\(\omega\) and "normal" frequency*f*(dammit — we'll be using*f*as a symbol for both frequency and a generic function) via$$ \omega = 2 \pi f $$

because once around the circle is one cycle

## The Euler Formula

Euler's Formula says complex exponential is a sinusoid:

$$ e^{i (\omega t + \theta)} = cos(\omega t + \theta) + i~sin(\omega t + \theta) $$

$$ = e^{i \omega t} e^{i \theta} $$

Starting point for "phasor analysis"

Now our sum of sinusoids can be represented as a sum of exponentials, making things easier (?)

## The Time Domain and the Frequency Domain

Reminder: Fourier claims that every \(f(t)\) can be represented as some \(\hat{f}(\omega)\) (more or less)

We think of the first kind of thing as "in the time domain", the second as "in the frequency domain"

Converting from a single frequency to its time domain representation is "easy":

$$ f(t) = e^{-i \omega t} $$

Even for a single sinusoid, converting the other way isn't immediately obvious

## Continuous Fourier Decomposition: The Fourier Transform

Let's just get the Fourier Transform out there:

$$ \hat{f}(\omega) = \int_{-\infty}^{\infty} f(t) e^{-i \omega t} dt $$

The minus sign in the exponent is easy to lose

The infinite integral is alarming

Still, we now can math up what we wanted

What about going the other way? Turns out that

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

So the transform is

*almost*self-inverse