Quadratic interpolation

The position of the peak is limited by the resolution of the DFT/FFT and its estimation can be improved using quadratic interpolation.

The general equation for a parabola is given by

$$y(x) \triangleq a(x-p)^2 + b,$$

where $p$ is the peak location and $b = y(p)$.

Considering the parabola at points $x = 0, 1, -1$ yields 3 equations,

$$y(0) = ap^2 + b = \beta$$

$$y(1) = a(1 + p^2 - 2p) + b = \gamma$$

$$y(-1) = a(1 + p^2 + 2p) + b = \alpha,$$

and 3 unknowns $a, p, b$.

Solve for $a$:

$$\alpha - \gamma = 4ap \longrightarrow a = \frac{\alpha - \gamma}{4p}.$$

Solve for $p$ using $a$:

$$\gamma - \beta = a-2ap$$

$$= \frac{\alpha - \gamma}{4p} - 2 \frac{\alpha - \gamma}{4p}p$$

$$4p(\gamma - \beta) = \alpha - \gamma -2(\alpha - \gamma)p$$

$$4p(\gamma - \beta) + 2p(\alpha - \gamma) = \alpha - \gamma$$

$$2p(\gamma - 2\beta + \alpha) = \alpha - \gamma$$

$$p = \frac{\alpha - \gamma}{2(\gamma - 2\beta + \alpha)}.$$

Finally, the height at peak $p$ is given by

$$y(p) = b$$

$$= \beta - ap^2$$

$$= \beta -\frac{\alpha - \gamma}{4p}p^2.$$

Another way (Dan Ellis)