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 contexts
Note: 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