We accept that every sampled piece of data contains both the "true" signal and noise accumulated in the channel.
Let:
$d_i = f(t_i) + e_i$
Where,
| $d_i \equiv $ sampled data |
| $f(t_i) \equiv $ expected signal |
| $e_i \equiv$ noise at the given sample |
As with traditional methods, we let:
$f(t_i) = \sum\limits^{m}_{j=1} B_j G_j (t,\{w\})$
Where,
| $B_j \equiv $ a type of amplitude |
| $G_j \equiv$ a function of some kind |
| $t \equiv $ continuous time |
| $\{ \omega \} \equiv$ a set of parameters; could be frequencies, chirp ratios, decay rates, or some other quantity |
Recall the product rule:
$P(H|DI) = \frac{P(H|I)P(D|HI)}{P(D|I)}$
$H \equiv$ hypothesis
$D \equiv$ data
$I \equiv$ prior information
Let the hypothesis be the signal: $H = f(t)$
Accept all possible amplitudes by treating them as nuanced parameters:
$P(\omega |DI) = \int P(\omega B|DI) dB$
where $\omega \equiv \{ \omega_1 , \omega_2 , ... \omega_n \}$
Product rule now shows after direct substitution:
$P(H|DI) = \frac{P( \omega |I)P( D | \omega I)}{P(D|I)}$
By using Maximum Entropy method, we find a solution for the probability of the signal:
$P(\{\omega\} | D \sigma I) \propto exp(\frac{-N \bar{d^2} + m \bar{h^2} }{2 \sigma^2 })$
Where
| $h_j \equiv \sum\limits^{N}_{i=1} d_i H_j (t_i) $ and ($1 \leq j \leq m$) |
| $\bar{A^2} \equiv$ the arithmetic mean of $A^2$ for each value of $A \equiv \{A_1 , A_2 , A_3 , ... A_m\}$ More of a discussion what $H_j$ represents in a later revision |
| $\sigma \equiv $ the second moment; which is usually known prior to a calculation |
If $\sigma$ is not known, then
$P({\omega} | DI) \propto (1 - \frac{m \bar{h^2} }{N \bar{d^2}})^{\frac{m-N}{2}}$
$N \equiv$ total number of samples
If a sinusoidal model is expected (for example, a real and complex wave) then the model would be as follows:
$P(\omega | DI) \propto 1 - \left( \frac{ \frac{R(\omega)^2}{c} + \frac{I(\omega)^2}{s} }{N \bar{d^2}} \right) ^{\frac{2-N}{2}}$
Where
| $R(\omega) \equiv \sum\limits^{N}_{i=1} d_i \cos{\omega t_i}$ |
| $I(\omega) \equiv \sum\limits^{N}_{i=1} d_i \sin{\omega t_i}$ |
| $ c \equiv cos(\omega) \frac{N}{2} + \frac{sin(N \omega)}{2sin( \omega )}$ |
| $ s \equiv sin(\omega) \frac{N}{2} - \frac{sin(N \omega)}{2sin( \omega )}$ |