A Timing
Advance (TA) is used to compensate for the propagation delay as the
signal travels between the Mobile Station (MS) and Base Transceiver
Station (BTS). The Base Station System (BSS) assigns the TA to the MS
based on how far away it perceives the MS to be. Determination of the
TA is a normally a function of the Base Station Controller (BSC), bit
this function can be handled anywhere in the BSS, depending on the
manufacturer.
Time Division Multiple Access (TDMA) requires
precise timing of both the MS and BTS systems. When a MS wants to gain
access to the network, it sends an access burst on the RACH. The
further away the MS is from the BTS, the longer it will take the access
burst to arrive at the BTS, due to propagation delay. Eventually there
comes a certain point where the access burst would arrive so late that
it would occur outside its designated timeslot and would interfere with
the next time slot.
Access Burst
As you recall from the TDMA Tutorial, an access burst has 68.25 guard bits at the end of it.
This
guard time is to compensate for propagation delay due to the unknown
distance of the MS from the BTS. It allows an access burst to arrive up
to 68.25 bits later than it is supposed to without interfering with the
next time slot.
68.25
bits doesnt mean much to us in the sense of time, so we must convert
68.25 bits into a frame of time. To do this, it is necessary to
calculate the duration of a single bit, the duration is the amount of
time it would take to transmit a single bit.
Now, if
an access burst has a guard period of 68.25 bits this results in a
maximum delay time of approximately 252µs (3.69µs × 68.25 bits). This
means that a signal from the MS could arrive up to 252µs after it is
expected and it would not interfere with the next time slot.
The
next step is to calculate how far away a mobile station would have to
be for a radio wave to take 252µs to arrive at the BTS, this would be
the theoretical maximum distance that a MS could transmit and still
arrive within the correct time slot.
Using the speed of light,
we can calculate the distance that a radio wave would travel in a given
time frame. The speed of light (c) is 300,000 km/s.
Description
Formula
Result
Convert km to m
300,000km × 1000
300,000,000m
Convert m/s to m/µs
300,000,000 ÷ 1,000,000
300 m/µs
Calculate distance for 252µs
300 m/µs × 252µs
75600m
Convert m to km
75,600m ÷ 1000
75.6km
So,
we can determine that a MS could theoretically be up to 75.6km away
from a BTS when it transmits its access burst and still not interfere
with the next time slot.
However, we must take into account that
the MS synchronizes with the signal it receives from the BTS. We must
account for the time it takes for the synchronization signal to travel
from the BTS to the MS. When the MS receives the synchronization signal
from the BTS, it has no way of determining how far away it is from the
BTS. So, when the MS receives the syncronization signal on the SCH, it
synchronizes its time with the timing of the system. However, by the
time the signal arrives at the MS, the timing of the BTS has already
progressed some. Therefore, the timing of the MS will now be behind the
timing of the BTS for an amount of time equal to the travel time from
the BTS to the MS.
For example, if a MS were exactly 75.6km away
from the BTS, then it would take 252µs for the signal to travel from
the BTS to the MS.
The
MS would then synchronize with this timing and send its access burst on
the RACH. It would take 252µs for this signal to return to the BTS. The
total round trip time would be 504µs. So, by the time the signal from
the MS arrives at the BTS, it will be 504µs behind the timing of the
BTS. 504µs equals about 136.5 bits.
The
68.25 bits of guard time would absorb some of the delay of 136.5 bits,
but the access burst would still cut into the next time slot a whopping
68.25bits.
In
order to compensate for the two-way trip of the radio link, we must
divide the maximum delay distance in half. So, dividing 75.6km in half,
we get approximately 37.8 km. If a MS is further out than 37.8km and
transmits an access burst it will most likely interfere with the
following time slot. Any distance less than 37.8km and the access burst
should arrive within the guard time allowed for an access burst and it
will not interfere with the next time slot.
In GSM, the maximum
distance of a cell is standardized at 35km. This is due mainly to the
number of timing advances allowed in GSM, which is explained below.
In
order to determine the propagation delay between the MS and the BSS,
the BSS uses the synchronization sequence within an access burst. The
BSS examines the synchronization sequence and sees how long it arrived
after the time that it expected it to arrive. As we learned from above,
the duration of a single bit is approximately 3.69µs. So, if the BSS
sees that the synchronization is late by a single bit, then it knows
that the propagation delay is 3.69µs. This is how the BSS knows which
TA to send to the MS.
For
each 3.69µs of propagation delay, the TA will be incremented by 1. If
the delay is less than 3.69µs, no adjustment is used and this is known
as TA0. For every TA, the MS will start its transmission 3.69µs (or one
bit) early. Each TA really corresponds to a range of propagation delay.
Each TA is essentially equal to a 1-bit delay detected in the
synchronization sequence.
When
calculating the distances involved for each TA, we must remember that
the total propagation delay accounts for a two-way trip of the radio
wave. The first leg is the synchronization signal traveling from the
BTS to the MS, and the second leg is the access burst traveling from
the MS to the BTS. If we want to know the true distance of the MS from
the BTS, we must divide the total propagation delay in half.
For
example, if the BSS determines the total propagation delay to be
3.69µs, we can determine the distance of the MS from the BTS.
Description
Formula
Result
Determine one-way propagation time
3.69µs ÷ 2
1.845µs
Calculate distance (using speed of light.)
300 m/µs × 1.845µs
553.5m
We
determined earlier that for each propagation delay of 3.69µs the TA is
inceremented by one. We just learned that a propagation delay of 3.69µs
equals a one-way distance of 553.5 meters. So, we see that each TA is
equal to a distance of 553.5 meters from the tower. Starting from the
BTS (0 meters) a new TA will start every 553.5m.
TA Ring
Start
End
0
0
553.5m
1
553.5m
1107m
2
1107m
1660.5m
3
1660.5m
2214m
...
...
...
63
34.87km
35.42km
The
TA becomes very important when the MS switches over to using a normal
burst in order to transmit data. The normal burst does not have the
68.25 bits of guard time. The normal burst only has 8.25 bits of guard
time, so the MS must transmit with more precise timing. With a guard
time of 8.25 bits, the normal burst can only be received up to 30.44µs
late and not interfere with the next time slot. Because of the two-way
trip of the radio signal, if the MS transmits more than 15.22µs after
it is supposed to then it will interfere with the next time slot.
