Telecommunications
Random Transmitters
Random Transmitters
Context
Plextek Limited develop communications technology products and systems, such as
the electronics inside mobile telephones and telemetry units used in Formula One
motor racing. One project in which they have been involved was to develop electricity
meters that can be read remotely.
Cheap transmit-only radio units can be incorporated in electricity meters; this article
looks at how should they be set up to give the best data.
differentiation:
product rule
differentiation:
dx n
dx
chain rule
stationary points
optimisation
probability:
independent events
For a system like this to be feasible, many transmitters would need to share the same
radio channel. If signals were transmitted at regular intervals, all the transmitters
would need to be co-ordinated which is not practical. So the transmitters are set to
transmit at random; sometimes signals from different transmitters will interfere but
that does not matter as long as over a reasonable period a clear signal can be obtained.
Problem
Let there be N transmitters transmitting t-second messages randomly but on average
every T seconds. The probability that an individual message will be not overlap that
T  2t
from a second transmitter is
, so the probability that a message will not be
T
 T  2t 
overlapped any of N  1 others is 

 T 
N 1
. Since there are N transmitters, the
N 1
Nt  T  2t 

 . What is the maximum
T  T 
amount of data that can get through and how can this be achieved?
amount of data, D, getting through will be
Solution
D
Nt  T  2t 
.

T  T 
N 1
dD N  T  2t 
 .

dt T  T 
, so differentiating D with respect.to t gives
N 1

Nt  2  T  2t 
. 

T  T  T 
N 2

N (T  2t ) N 2
TN
(T  2 Nt ) .
Setting this equal to 0, to obtain values of t for which D is stationary, gives two
solutions: T  2t and T  2 Nt .
As many transmitters will share the same radio channel, T is much greater than 2t, so
it is only necessary to determine whether T  2 Nt gives a maximum.
dD
T
; T  2 Nt is positive if t 
and
dt
2N
dD
is positive, so, as t increases,
changes
dt
This can be done by considering the sign of
T
N (T  2t ) N 2
, and
2N
TN
T
from positive to negative where t 
, so this value of t maximises D.
2N
negative if t 
Substituting this value for t into the expression for D gives the maximum data rate as
1
1
1  
2 N 
N 1
.
To see how D behaves, for example to find out how sensitive it is to slight variations
in the value of t, it is useful to draw a graph. The one below shows Prob(t,T,N), the
probability than a message will get through as well as D or Data(t,T,N), the total
transmission time Nt/T x Prob(t,T,N), in the case where t = 1 and N = 1000.
© The Mathematical Association 2004
1
Random Transmitters
Telecommunications
0.2
0.2
0.19
0.18
0.17
0.16
0.15
0.14
0.13
0.12
Prob( t  T  N ) 0.11
0.1
Data( t  T  N )
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
0
0
0.2 0.4 0.6 0.8
1
1.2 1.4 1.6 1.8
0
n
 x
e = lim  1+ 
n  
n
or
 n xr 
e x = lim   

n  
 r =0 r! 
binomial expansion
x
2
2.2 2.4 2.6 2.8
3
3.2 3.4 3.6 3.8
4
T
N t
4
Extension
N 1
1
1
What happens to the maximum data rate, Dmax  1   , when a large number
2 N 
of transmitters share the same radio channel? This can be looked at in two ways.
1
1 
1   1 
Firstly, Dmax  .1   .1  
2  N 
N
1
1
so Dmax  .1.e1 
.
2
2e
N
1
and, for large N, 1   1 and
N
Alternatively, you can apply a binomial expansion to Dmax
N
x

1    e ,
N

1  1 
 1  
2
N
N 1
to give
2
3
1
( N  1)( N  2)( N  3)  1 
 1  ( N  1)( N  2)  1 
Dmax  1  ( N  1)   

 
  
2 
2
3!
N
N
N




2
3
1   N 1 
 N  1  N  2  ( 1)  N  1  N  2  N  3  ( 1)
 1  
(

1)











2   N 
 N  N  2
 N  N  N  3!
1 
1
1 
2  (1) 2 
1 
2 
3  ( 1)3

 1  1   (1)  1   1  
  1   1   1  

2  
N
N 
N 2
N 
N 
N  3!








1 1 1 1
1
 1
So, for large N, Dmax  1       e1 
.
2  1! 2! 3!
2e
 2
Source
The material in this article is based on a talk given by David Spreadbury of Plextek
Limited (http://www.plextek.co.uk) at Hills Road Sixth Form College, Cambridge on
8 July 1999. We are very grateful for his permission to use this material.
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© The Mathematical Association 2004