# Problem with 1st order PLL update equation

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.

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

• Maybe this presentation can help: gnuradio.org/grcon/grcon17/presentations/… – Andy Walls Oct 6 '19 at 0:25
• The problem is that unlike standard TED, here I do not have an estimate of the error $\tau_k−\hat{\tau}_k$ but only $\tau_k$. – Elnaz Oct 6 '19 at 1:36
• 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. – Jason R Oct 6 '19 at 2:08
• 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 $\frac{1-\alpha z^{-1}}{stepsize(1-alpha)}$? – Elnaz Oct 7 '19 at 19:58