4 Amplitude tracking

The analytic technique above relies on known values of A and B, but in a real system the amplitudes are not known, and furthermore, will slowly vary. A heuristic algorithm is used to track these variations. For the purposes of amplitude tracking one can assume that $\theta$ and $\phi$ are uniformly distributed in $[0,2\pi)$. Consequently, $(\theta-\phi) \mod 2\pi$ is also uniform in $[0,2\pi)$.

One simple heuristic for amplitude tracking is to compute the maximum and minimum values of $\Vert r[i]\Vert$ over $i=1, \ldots, n$, where $n$ is chosen such that $A[i]$ and $B[i]$ do not vary appreciably. Assuming without loss of generality that $A\ge B$, the minimum $\Vert r[i]\Vert$ will be close to $A-B$ and the maximum will be close to $A+B$, from which estimates of $A$ and $B$ can be determined. Unfortunately, it turns out that this heuristic does not perform well in the presence of noise.

A better performing heuristic is obtained by using the median $M$ of $\Vert r[1]\Vert^2, \ldots, \Vert r[n]\Vert^2$. Define

$$X \triangleq {1 \over n} \sum_{i=1}^n \Vert r[i]\Vert^2$$ (4)

$$Y \triangleq {2 \over n} \sum_{i: \Vert r[i]\Vert^2 > M} \Vert r[i]\Vert^2$$ (5)

$$Z \triangleq {2 \over n} \sum_{i: \Vert r[i]\Vert^2\le M} \Vert r[i]\Vert^2.$$ (6)

Since the expected value of $\Vert r\Vert^2$ is given by

$$E[\Vert r\Vert^2] = E[A^2 + 2AB\cos(\theta-\phi)+B^2] = A^2 + B^2,$$

it follows that $X \approx A^2 + B^2$ for reasonably large $n$. Conditioning on the event $\cos(\theta-\phi) > 0$, it follows that $(\theta-\phi) \mod 2\phi$ is uniform in $[0,\pi/2) \cup [3\pi/2,2\pi)$ Therefore,

$$\begin{split}& E[\Vert r\Vert^2 \, \vert \, \cos(\theta-\phi)... ...ha d\alpha\right] \\ & \quad = A^2 + B^2 + 4AB/\pi \end{split}$$

Similarly,

$$E[\Vert r\Vert^2 \, \vert \, \cos(\theta-\phi)<0] = A^2 + B^2 - 4AB/\pi.$$

Thus, $Y \approx A^2 + B^2 + 4AB/\pi$, and similarly $Z \approx A^2 + B^2 - 4AB/\pi$. If

$$\hat A \triangleq {1 \over 2} \left( \sqrt{X+{\pi\over 2}(Y-X)} + \sqrt{X+ {\pi\over 2}(Z-X)}\right)$$ (7)

$$\hat B \triangleq {1 \over 2} \left( \sqrt{X+{\pi\over 2}(Y-X)} - \sqrt{X+{\pi\over 2}(Z-X)}\right),$$ (8)

then $\hat A \approx A$ and $\hat B \approx B$. Using only the norms of $r$ over a set of samples, $A$ and $B$ can be estimated fairly accurately.