WIT2450
©
2000- 2005 Cirronet
™
Inc
17
M-2450-0000 Rev B
Take for example a network comprised of a base station and 10 remotes. A hop duration
of 10 ms is chosen. We decide that the base station needs to be able to send up to 32
bytes each hop (equivalent to a capacity for the base of ~ 32 kbps). Counting the 2.64ms
overhead for the base packet and making use of the fact that our RF rate is 460.8 kbps,
we determine that the base slot requires approximately:
Each remote time slot will be:
10 ms – 3.2 ms – (10)·0.417 ms
10
From our RF data rate of 460.8kbps we see that it takes 17.36
µ
s to send a byte of data,
so each remote will be able to send up to
= 15.1 bytes of data per hop.
However, the WIT2450 sends data in groups of 4 bytes. Thus, each remote will be able to
send 12 bytes of data. If the WIT2450 is using a protocol mode, the packet overhead is
stripped off (including start and stop bits) before transmission and does not need to be
considered when calculating RF capacity.
So in this example, the total capacity per remote would be:
If we figure a minimum margin of safety for lost packets and retransmissions of about
20%, we see that this would be more than sufficient to support 9.6 kbps of continuous
data per remote. It is also useful to remember that the asynchronous data input to the
WIT2450 is stripped of its start and stop bits during transmission by the radio, yielding a
"bonus" of 10/8 or 25% in additional capacity.
The above calculations are provided as a means of estimating the capacity of a multipoint
WIT2450 network. To determine the precise amount of capacity, you can actually set up
the radio system and then query the
maximum data length
from one of the remotes in
control mode to discover its exact setting. Divide this number by the hop duration as
above to get the remote's exact capacity.
= 0.263ms
0.263 ms
17.36
µ
s
32·8
460.8kbps
+ 2.64 ms = 3.20
12 bytes
10 ms
= 12 kbps