Introduction

Is it possible to eliminate the need for synchronization of receiving station clocks while still maintaining the ability to perform the same task? Turns out that it is, and that will be the topic of this post. If you're not yet familiar with the basic form of TDoA with synchronized clocks, see the previous post in this series.

Why Synchronize Clocks?

Recall the TDoA equation, $M_U = T_{UA} - T_{UB}$. As stated in the previous post, the right hand side compares points-in-time from two different clocks, one at $R_A$ and one at $R_B$. For this to be valid, the two clocks must be synchronized. If the clocks were not synchronized we'd have this equation instead:

• $M_U = (T_{UA} - \delta_A) - (T_{UB} - \delta_B)$

Here we have extra $\delta_x$ terms which represent how far the clock at $R_x$ is ahead-of or behind some "true" time standard. TDoA requires that $\delta_x = 0$ for all $x$ and typically accomplishes this by synching all clocks to GPS time signals. But what if we don't want to require synchronization of clocks at receiving stations?

How to Deal with Unsynchronized Clocks

If we don't want to require synchronization then we'll need some other way to eliminate the unkwown $\delta_x$ terms from our math. One way to do that would be to have our receivers spot another sending station which we'll call $S_K$. We'd then have two simultaneous TDoA equations:

• $M_U = (T_{UA} - \delta_A) - (T_{UB} - \delta_B)$
• $M_K = (T_{KA} - \delta_A) - (T_{KB} - \delta_B)$

Subtracting the second from the first we get:

Hurray, we've eliminated the $\delta_x$ terms and what remains gives us a formula for the difference between the $M$ values of the branches associated with $S_U$ and $S_K$. All we need to do is measure two points-in-time at each of two receiving stations.

Scenario 1: Two Receivers, One Known Sender

Let's think of the sending stations this way:

• $S_U$ is a sending station at an (U)known location
• $S_K$ is a sending station at a (K)nown location

Since we know the location of $S_K$ we can replace $M_K$ with $\overline{S_K R_A} - \overline{S_K R_B}$ and rearange the equation to solve for $M_U$:

• $M_U$ $=$ $(\overline{S_K R_A} - \overline{S_K R_B})$ $+$ $(T_{UA} - T_{KA}) - (T_{UB} - T_{KB})$

And we're back to the situation where we can determine the hyperbolic branch that $S_U$ ought to be located on, simply by measuring the times at which stations first hear certain transmissions.

Let's work through the following example:

1. We know/trust the location of $S_K$ so we know $\overline{S_K R_A} = 29.7$ and $\overline{S_K R_B} = 13.7$
2. This puts $S_K$ on the branch with $M = {^+}16$.
3. Both $R_A$ and $R_B$ hear a message from $S_K$ and another from $S_U$ and note the arrival times.
4. We obtain the values from $R_A$ and $R_B$ and calculate $(T_{UA} - T_{KA}) - (T_{UB} - T_{KB}) = {^-}26$.
5. As a result, we know that $S_U$ should be on the branch with $M= {^+}16 + {^-}26 = {^-}10.$

Scenario 2: >2 Receivers, >1 Known Senders

For the most part, this scenario is the same as it was for the corresponding TDoA scenario with synchronized clocks, at least with respect to the addition of more receiving stations: Each new pair of receivers results in an additional branch which tells us more about the actual location of $S_U$. However, unlike the TDoA scenario, adding an additional DoTDoA receiver requires that the new receiver not only receives the $S_U$ message but also receives the $S_K$ message. Luckily, there is some flexibility in the choice of $S_K$ in any given scenario and there can, in fact, be several $S_{Kx}$ involved.

For example, the following is a valid DoTDoA scenario:

• $R_A$ receives $S_U$ and $S_{K1}$
• $R_B$ receives $S_U$ and $S_{K1}$ and $S_{K2}$
• $R_C$ receives $S_U$ and $S_{K2}$

... resulting in the following:

• Analysis of ($R_A$, $R_B$) will use $S_{K1}$ as a common known reference.
• Analysis of ($R_B$, $R_C$) will use $S_{K2}$ as a common known reference.
• ($R_C$, $R_A$) cannot be analyzed because they have no common known reference.

It's also possible for one receiver pair to have more than one $S_{Kx}$ in common as in the following:

• $R_A$ receives $S_U$ and $S_{K1}$ and $S_{K2}$ and $S_{K3}$
• $R_B$ receives $S_U$ and $S_{K1}$ and $S_{K2}$ and $S_{K3}$
• $R_C$ receives $S_U$ and $S_{K2}$

... in which case ($R_A$, $R_B$) use all three of $S_{K1}$ and $S_{K2}$ and $S_{K3}$ and then average the resulting $M_{U1}$, $M_{U2}$, and $M_{U3}$ values to arrive at one overall $M_U$ value with less error.

Conclusion

In the next post I'll discuss how WSJT-X can be used to do the "heavy lifting" in a DoTDoA system.

73s