The output of a 2-input 2-output communication channel is given by:

$x^{(i)}(t) = \sum_{j=1}^{2}\sum_n {a^{(j)}_n}h_{ij}(t-nT-\tau^{(j)})$,

where $x^{(i)}(t)$ is the i-th output, $\{a^{(j)}_n\}$ are BPSK symbols of the j-th input, $h_{ij}(t)$ is the channel response from input j to output i, $T$ is the symbol period, and $\tau^{(j)}$ is the timing offset associated with input j.

I estimate the parameter $\tau^{(1)}$ using a linear combination of the past samples $x^{(i)}(t)$ according to:

1) $ \tau^{(1)} = x_k^{(1)}a_{k-1}^{(1)} - x_{k-1}^{(1)}a_{k}^{(1)}$

2) $ \tau^{(1)} = \sum_{i=1}^{2} (x_k^{(i)}a_{k-1}^{(1)} - x_{k-1}^{(i)}a_{k}^{(1)})$

where $x_k = x(kT)$. Both are unbiased estimators.

Method 2 is supposed to be an extension of method 1 where we utilize the secondary output, $x^{(2)}(t)$ in addition to $x^{(1)}(t)$ for a better estimation. However, my simulations shows method 2 has much higher estimation variance than method 1, if output 1 picks up more of input signal 1 than input signal 2, for example when:

$h_{11} (t) = h_{22} (t) = 0.8 \times sinc(t/T)$ and $h_{12} (t) = h_{21} (t) = 0.2 \times sinc(t/T)$.

Is there a better way to utilize both outputs without an increase in variance?

  • $\begingroup$ Since the channel is random and (slowly?) time-varying, I think you should focus on the variance on average; that is, the variance averaged over the channel. It is always possible to find a realization that is bad -- for example, the channel you give will also have a worse BER than the average. $\endgroup$ – MBaz Oct 16 '19 at 21:49
  • $\begingroup$ @MBaz I don't understand why you say the channel is random? I gave an example of an ideal ISI-free, Nyquist, channel. $\endgroup$ – Elnaz Oct 16 '19 at 22:13
  • $\begingroup$ In an actual radio system, you can't predict what the channel response for each pair of antennas is going to be like. In addition, the tx antennas, the rx antennas, and all reflectors may be moving, so the channel is time-varying. You almost never have a wireless channel that is fixed; that is why it is modeled as random. Usually, a particular set of channel responses, such as you have given, is called a channel realization -- but it is just one among an infinite number of possible realizations. Some realizations are 'bad" and some are "good", but a system is designed to be "good" on average. $\endgroup$ – MBaz Oct 16 '19 at 22:29
  • $\begingroup$ Sure, we don't know the actual channel responses before equalization. Here, I only focus on determining a MIMO extension of a 1D timing error detector. Here, I think adding the two estimates will inevitably raise the variance, regardless of the channel model chosen. I only chose the sinc functions to compare with the original 1976 M&M timing error detector. $\endgroup$ – Elnaz Oct 16 '19 at 22:45
  • $\begingroup$ A couple of thoughts: (1) is that also the case when output 1 picks up more of inputs signal 2 than 1 (i.e. the opposite of what you say in your question)? (2) Are you assuming some kind of coding, such as Alamouti? (3) I have extended Gardner's timing estimator to Alamouti: ieeexplore.ieee.org/document/8117181 -- even though you're working on M&M, and I didn't specifically look at the variance, my paper could maybe give you some ideas. $\endgroup$ – MBaz Oct 16 '19 at 22:51

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