The output of a communication channel is given by:

$x(t) = \sum_n{a_n}h(t-nT)$,

where $\{a_n\}$ are BPSK symbols, $h(t)$ is the channel response, and $T$ is the symbol period. If there is an inherent unknown timing offset $\tau$ inside the channel response we will instead have:

$x(t) = \sum_n{a_n}h(t-nT-\tau)$.

Say we need to sample $x(t)$ at times $\{kT\}$ and we can detect the symbols using the shifted channel response $h(t-nT-\tau)$. In other words we do not wish to resample $x(t)$ to compensate for the timing offset but we can estimate $\tau$. An estimate of $\tau$ is given by $z_k = x_k \hat{a}_{k-1} - x_{k-1}\hat{a}_k$ such that $E\{z_k\} = \tau$, where $x_k = x(kT)$, and $\{\hat{a}_k\}$ are the decisions of a (Viterbi) detector. This is the $M \& M$ estimate.

The question is how to use this estimate to recursively update the estimates of $\tau$ in a 1-st order PLL sense since the following equation will diverge.

$\tau_{k+1} = \tau_k + stepsize \times z_k$

  • 2
    $\begingroup$ Maybe this presentation can help: gnuradio.org/grcon/grcon17/presentations/… $\endgroup$
    – Andy Walls
    Oct 6 '19 at 0:25
  • $\begingroup$ The problem is that unlike standard TED, here I do not have an estimate of the error $\tau_k−\hat{\tau}_k$ but only $\tau_k$. $\endgroup$
    – Elnaz
    Oct 6 '19 at 1:36
  • 1
    $\begingroup$ You might consider a leaky integrator to recursively average your $\tau_k$ estimate: $\tau_{k+1} = \alpha \tau_k + (1-\alpha) z_k$ for $\alpha \in [0, 1)$. The closer it is to $1$, the narrower the bandwidth of the filter. $\endgroup$
    – Jason R
    Oct 6 '19 at 2:08
  • $\begingroup$ Can it be $\tau_{k+1} = \alpha \tau_k + stepsize (1-\alpha) z_k$? Is this still a 1st order PLL with a transfer function of $\frac{1-\alpha z^{-1}}{stepsize(1-alpha)}$? $\endgroup$
    – Elnaz
    Oct 7 '19 at 19:58

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