The 1090 Megahertz Riddle (second edition)

A Guide to Decoding Mode S and ADS-B Signals

Surface position

When an aircraft is on the ground, a different type of message is used to broadcast its position information. Unlike the airborne position message, the surface position message also includes the speed of the aircraft. Since no altitude information needs to be transmitted, this provides some extra bits for more information, such as speed and track angle.

The surface position message has Type Code from 5 to 8, which also represents the level of uncertainties in the position. And the structure of the surface position message ME field is:

+-------+--------+------+--------+------+------+-------------+-------------+
| TC, 5 | MOV, 7 | S, 1 | TRK, 7 | T, 1 | F, 1 | LAT-CPR, 17 | LON-CPR, 17 |
+-------+--------+------+--------+------+------+-------------+-------------+

We can see that the fields are defined similarly to the ones from airborne messages, except for the surveillance status and altitude bits which are replaced by movement and ground track. The details of the fields are shown in Table 1.1

Surface position message structure
FIELD MSG ME BITS
Type Code TC 33–37 1–5 5
Movement MOV 38–44 6–12 7
Status for ground track S 45 13 1
0: Invalid
1: Valid
Ground track TRK 46–52 14–20 7
Time T 53 21 1
CPR Format F 54 22 1
0: even frame
1: odd frame
Encoded latitude LAR-CPR 55–71 23–39 17
Encoded longitude LON-CPR 72–88 40–56 17

Movement

The movement field encodes the aircraft ground speed. The ground speed of the aircraft is encoded non-linearly and with different quantizations. This is to ensure that a lower speed can be encoded with a improved precision than a higher speed. Different levels of quantizations are defined in Table 1.2.

Surface position movement (ground speed) decoding
Encoded speed Ground speed range Quantization
0 Speed not available
1 Stopped (v $$<$$ 0.125 kt)
2–8 0.125 $$\leq$$ v $$<$$ 1 kt 0.125 kt steps
9–12 1 kt $$\leq$$ v $$<$$ 2 kt 0.25 kt steps
13–38 2 kt $$\leq$$ v $$<$$ 15 kt 0.5 kt steps
39–93 15 kt $$\leq$$ v $$<$$ 70 kt 1 kt steps
94–108 70 kt $$\leq$$ v $$<$$ 100 kt 2 kt steps
109–123 100 kt $$\leq$$ v $$<$$ 175 kt 5 kt steps
124 v $$\ge$$ 175 kt
125–127 Reserved

Ground track

When the status for the ground track field is set to 1, the ground track is encoded with a precision of (and rounded to) 360/128 degrees. Zero degrees represents an aircraft ground track that is aligned with the true north. Based on the encoded value ($$n$$), the ground track ($$\chi$$) is calculated as:

$\chi = \frac{360~n}{128} \quad \text{(degrees)}$

When the status for the ground track field is set to 0, the information contained in the ground track fields should be considered as invalid.

Position

Like coordinates from airborne messages, the surface latitude and longitude are also encoded in the Compact Position Reporting (CPR) format. CPR format bit indicates whether the message is an even or odd message. There are only a few slight differences in decoding the surface position positions.

First, the latitude zone size is four times smaller, which is defined as:

$\begin{split} \mathrm{dLat}_\mathrm{even} &= \frac{90^\circ}{4 N_Z} \\ \mathrm{dLat}_\mathrm{odd} &= \frac{90^\circ}{4 N_Z - 1} \end{split}$

Similarly, the longitude zone size is smaller too, which is:

$\begin{split} \mathrm{dLon}_\mathrm{even} &= \frac{90^\circ}{n_\mathrm{even}} \\ \mathrm{dLon}_\mathrm{odd} &= \frac{90^\circ}{n_\mathrm{odd}} \end{split}$

The algorithm used to encode the position is essentially the same as the one used for encoding airborne position. All the other equations from Chapter [chap:airborn-position] can be used to decode surface position.

The main difference, when compared to the airborne position, is that the original number of bits used for encoding surface coordinates is 19 instead of 17. However, in the message, only the lower-order 17 bits are included and transmitted in the surface message. Thus, there can be multiple solutions for a pair of odd and even messages, even when globally unambiguous decoding is used. To identify the correct solution, a reference position needs to be given. This can be the position of the receiver or the location of the airfield.

When employing locally unambiguous decoding, the reference position needs to be within 45 nautical miles of the actual location. Commonly, the positions of the airport is used.

With the equations from section [sec:cpr_airborne_global_lat], the solution refers to the latitude located at the northern hemisphere. We need to compute a southern hemisphere solution by subtracting this value with 90 degrees. The final correct latitude is the one that is closest to the reference position.

Similarly, with the equations from section [sec:cpr_airborne_global_lon], the solution refers only the longitude in the 0 to 90 degrees quadrant. We need to compute the other three possible solutions by adding 90, 180, and 270 degrees. The final correct longitude is also the one that is closest to the reference position.

Decoding example

Globally unambiguous decoding

First, we decode the location using globally unambiguous decoding. The following example contains an even/odd pair of surface position messages:

8C4841753AAB238733C8CD4020B1
8C4841753A8A35323FAEBDAC702D    (most recent)

Based on the ADS-B message structure, we can identify the ICAO address and message ME field as follows:

+----+--------+----------------+--------+
|    | ICAO   |       ME       |  PI    |
+----+--------+----------------+--------+
| 8C | 484175 | 3AAB238733C8CD | 4020B1 |
| 8C | 484175 | 3A8A35323FAEBD | AC702D |
+----+--------+----------------+--------+

We can covert the message data into binary format:

+-------+---------+---+---------+---+---+-------------------+-------------------+
| TC    | MOV     | S | TRK     | T | F | CPR-LAT           | CPR-LON           |
+-------+---------+-------------+---+---+-------------------+-------------------+
| 00111 | 0101010 | 1 | 0110010 | 0 | 0 | 11100001110011001 | 11100100011001101 |
| 00111 | 0101000 | 1 | 0100011 | 0 | 1 | 01001100100011111 | 11010111010111101 |
+-------+---------+---+---------+---+---+-------------------+-------------------+

reference latitude: 51.990
reference longitude: 4.375

it is possible to extract the encoded CPR latitude and longitude binary and convert them to decimal format. Then, they are divided by the $$2^{17}$$, representing the fractions of the positions within the latitude and longitude grids:

$\begin{split} \mathrm{lat}_\mathrm{cpr,even} &= \frac{115609}{2^{17}} \\ \mathrm{lon}_\mathrm{cpr,even} &= \frac{116941}{2^{17}} \\ \mathrm{lat}_\mathrm{cpr,odd} &= \frac{39199}{2^{17}} \\ \mathrm{lon}_\mathrm{cpr,odd} &= \frac{110269}{2^{17}} \end{split}$

We can calculate the latitude index, $$j$$, as:

$j = floor \left( 59 \times \frac{115609}{2^{17}} - 60 \times \frac{39199}{2^{17}} + \frac{1}{2} \right) = 34$

Then, we can decode the latitudes from both even and odd messages:

$\begin{split} \mathrm{lat}_\mathrm{even} &= \frac{90}{4 \times 15} \Big[ mod(34, 60) + \frac{115609}{2^{17}} \Big] = 52.323040008544920 \\ \mathrm{lat}_\mathrm{odd} &= \frac{90}{4 \times 15 - 1} \Big[ mod(34, 59) + \frac{39199}{2^{17}} \Big] = 52.320607072215964 \end{split}$

After validating the longitude zone of both messages are the same:

$\begin{split} \mathrm{NL}(\mathrm{lat}_\mathrm{even}) &= \mathrm{NL}(52.323040008544920) = 36 \\ \mathrm{NL}(\mathrm{lat}_\mathrm{odd}) &= \mathrm{NL}(52.320607072215964) = 36 \end{split}$

we can continue to calculate the global position. Since the even message is the most recent message, the northern hemisphere latitude solution is:

$\mathrm{lat}_N = \mathrm{lat}_\mathrm{odd} = 52.320607072215964$

The southern hemisphere latitude solution is:

$\begin{split} \mathrm{lat}_S &= 52.320607072215964 - 90 \\ &= -37.679392927784036 \end{split}$

By comparing the result with the reference latitude (51.990), the final latitude value is the one located in the northern hemisphere.

$\mathrm{lat} = \mathrm{lat}_N = 52.320607072215964$

The longitude solution in the first quadrant of 0 to 90 degrees is calculated based on the most recent (odd) message:

\begin{aligned} m &= floor \left( \frac{116941}{2^{17}} \times (36-1) - \frac{110269}{2^{17}} \times 36 + \frac{1}{2} \right) = 1\\ n &= max \Big( 36-1, 1 \Big) = 35\\ \mathrm{lon}_1 &= \frac{90}{35} \Big[ mod(1, 36-1) + \frac{110269}{2^{17}} \Big] = 4.734734671456474\end{aligned}

The three other possible solutions are:

$\begin{split} \mathrm{lon}_2 &= \mathrm{lon}_1 + 90 = 94.734734671456474 \\ \mathrm{lon}_3 &= \mathrm{lon}_1 + 180 = 184.734734671456474 \\ \mathrm{lon}_4 &= \mathrm{lon}_1 + 270 = 274.734734671456474 \\ \end{split}$

By comparing with the reference longitude (4.375), the final longitude is the one located in the first quadrant. Combining the result of previously obtained latitude, the final decoded coordinates (rounded up to the 6th decimal position) are:

latitude: 52.320607
longitude: 4.734735

Try it out Using pyModeS, we can perform the decoding of this globally unambiguous surface position as:

import pyModeS as pms

msg0 = "8C4841753AAB238733C8CD4020B1"
msg1 = "8C4841753A8A35323FAEBDAC702D"
t0 = 1457996410
t1 = 1457996412
lat_ref = 51.990
lon_ref = 4.375

pms.adsb.position(msg0, msg1, t0, t1, lat_ref, lon_ref)

Output:

(52.32061, 4.73473)

Note that the reference coordinates are needed for decoding the surface position. In this case the reference is the coordinates of the airfield.

Locally unambiguous decoding

Secondly, we decode the location using locally unambiguous decoding. In this case, we want to use the position obtained previously as the reference.

8C4841753A9A153237AEF0F275BE

The structure of the ADS-B message is:

+----+--------+----------------+--------+
|    | ICAO   |      Data      |  PI    |
+----+--------+----------------+--------+
| 8C | 484175 | 3A9A153237AEF0 | F275BE |
+----+--------+----------------+--------+

We can covert the message data into binary format:

+-------+---------+---+---------+---+---+-------------------+-------------------+
| TC    | MOV     | S | TRK     | T | F | CPR-LAT           | CPR-LON           |
+-------+---------+-------------+---+---+-------------------+-------------------+
| 00111 | 0101001 | 1 | 0100001 | 0 | 1 | 01001100100011011 | 11010111011110000 |
+-------+---------+---+---------+---+---+-------------------+-------------------+

\begin{aligned} \mathrm{dLat} &= \frac{90}{4 \times 15 - 1} = \frac{90}{59} \\ j &= floor \left( \frac{52.320607}{90/59} \right) + floor \left[ \frac{mod(52.320607, 90/59)}{90/59} - \frac{39195}{2^{17}} + \frac{1}{2} \right] = 34 \\ \mathrm{lat} &= \mathrm{lat}_\mathrm{odd} = \frac{90}{59} \times \left( 34 + \frac{39195}{2^{17}} \right) = 52.32056051997815\end{aligned}

Next, the longitude can be calculated as follows:

\begin{aligned} \mathrm{dLon} &= \frac{90}{max \Big( \mathrm{NL}(\mathrm{lat}_\mathrm{odd})-1, 1 \Big)} = \frac{90}{max(36-1, 1)} = \frac{90}{35} \\ m &= floor \left( \frac{4.734735}{90/35} \right) + floor \left( \frac{mod(4.734735, 90/35)}{90/35} - \frac{110320}{2^{17}} + \frac{1}{2} \right) = 1 \\ \mathrm{lon} &= \frac{90}{35} \times \left(1 + \frac{110320}{2^{17}} \right) = 4.735735212053571\end{aligned}

The final decoded coordinates (rounded up to 6 decimal positions) are:

latitude: 52.320561
longitude: 4.735735

Try it out Using pyModeS, decode this locally unambiguous surface position as:

import pyModeS as pms
msg = "8C4841753A9A153237AEF0F275BE"
lat_ref, lon_ref = 51.990, 4.375
pms.adsb.position_with_ref(msg, lat_ref, lon_ref)

Output:

(52.32061, 4.73473)

Movement and track

As a continuation of the previous example, we can also calculate the speed and track angle from the movement and track fields.

The decimal value of movement 0101001 is 41. Based on Table 1.2, the speed is encoded using 1 kt steps. Thus, the speed of aircraft is:

$15 + (42 - 39) \times 1 = 17 kt$

The decimal value of track angle 0100001 is 33. The track angle can be calculated as:

$\frac{360^\circ}{128} \times 33 = 92.8125 ^\circ$

Try it out Using pyModeS, decode aircraft speed and track angle as:

import pyModeS as pms
msg = "8C4841753A9A153237AEF0F275BE"
pms.adsb.velocity(msg)

Output (speed, track angle, vertical speed, tag):

17 92.8 0 GS

Note that the last two parameters returned by the previous function are vertical speed and a flag indicating whether the speed is ground speed (GS) or airspeed (IAS/TAS).