-
Team #12110
Page 1 of 9
Minimizing the
Number of repeaters
Introduction
Very high
frequency (VHF) is the radio
spectrum
，
whose frequency
band ranges from 30MHz to
300MHz. VHF
is always used for radio stations and television
broadcasts.
In addition, it is also
used
by
signal
transmission
of
sea
navigation
and
aviation.
Because
the
radio
spectrum
of
VHF
is
transmitted through
straight lines, a signal is easily influenced by
geographical factors easily.
Thus,
signals
become
weak
when
it
is
transmitted
and
some
low-
power
users
need
repeaters
to
amplify
them and increase
the transmission distance. We consider the
situation in which every two repeaters
are too close or the separate frequency
is not far enough apart which can interference
with each other.
In order to mitigate
the interference caused by the nearby repeaters,
this paper employs a continuous
tone-
coded squelch system (CTCSS). We associate to each
repeater a separate subaudible
tone
，
that
is,
the
subaudible
tone
(67Hz-250.3Hz)
is
added
to
VHF.
In
this
way,
repeaters
recognize
signals
attached to the same subaudible tones
just like secret keys.
In this system,
the nearby repeaters can
share the same
frequency pair. When users send the signals at one
frequency, different repeaters with
subaudible
tones
can
recognize
signals
from
the
users
the
same
subaudible
tone.
If
the
users
in
a
certain area
contact with each other, we should consider the
signal
’
s coverage area of
the users and
the repeaters. As long as
the users
’
signals are
accepted by repeaters, the signals could be
amplified
to
transmit
farther.
At
the
same
time,
the
repeaters
attached
with
the
subaudible
tones
could
only
recognize the users with the same
subaudible tones. Hence, we can consider repeaters
corresponding
to the number of the
users, which leads to the problem
of
frequency channel.
When the number of
users in this area increases, we can
add repeaters.
If two repeaters have
different subaudible tones,
they
would
not
communicate
with
each
other.
Thus,
we
should
consider
the
problem
of
how
the
repeaters
communicate
with
each
other
when
they
have
different
subaudible
tones.
In
the
mobile
communication
system
，
the spectrum is
influenced by many factors such as
reflex
，
diffraction and
dispersion.
Therefore, when
the radio spectrum transmits in the mountainous
area
，
we should still
consider the factors above.
Repeaters
[4]
Repeaters are a type of
equipment which can amplify
signals
，
make up the
deamplification signals
and support far
distance communication.
CTCSS
[5]
CTCSS(Continuous
Tone
Controlled
Squelch
System
)
is
short
for
subaudible
tones,
whose
frequency
ranges
from
67Hz
to
250.3Hz.
It
is
added
to
the
radio
spectrum
to
make
the
signal
carry
with
a
unique
secret key.
Assumption
The users in the area is uniform
distributed
The signal of the radio
spectrum in the
area can’t be effected
by environment
In a certain
period of time there are a small number of users
removing
All repeaters have the same
standard
Analysis and
solution of the model to the first problem
The problem is to find a
least number of repeaters in an area of radius 40
miles so that the users in this
area
can communicate with each other. Considering that
the given area is flat, we assume that the signal
of
Team #12110
Page 2 of 9
each repeater
covers a circular area and the repeater lies in
the center of the circle. The following Figure 1
shows the relationship of three
adjacent repeaters.
A
B
C
Figure 1.
The relationship
of three adjacent repeaters
For case B of Figure 1, if
three circles are tangent to each other, then we
find that the center area cannot
be
covered by the singles. In order to make the
signal cover the triangle area, we have to
consider adding a
circle, which means
that we need to add an extra repeater.
Undoubtedly, it is wasteful.
For case
C, if the
intersection of three circles
is not null, similar to case B, we also have to
add another repeater. Thus, it is
easy
to find that case A, comparing with cases B and C,
is optimal. Thus, we obtain the largest covering
area
and the smallest number of
repeaters.
When
more
circles
intersect
together,
we
get
some
closely
linked
hexagons,
as
shown
in
Figure
2.
Obviously, it looks like
a honeycomb structure. In fact, the honeycomb
pattern is one of the most efficient
arrangement for radio spectrum. It
transmits by the wireless medium of microwave,
satellites and radiation.
The
structure
has
a
feature
of
point-
to-point
transmission
or
multicast.
It
is
widely
used
in
UN
Urban
Network,
Campus Network and Enterprise Network.
Figure 2. some circles intersecting
together form the closely linked hexagons
Now we have a circle with
radius of 40 miles. Then we analyze the distances
of signals from users and
repeaters
covering in
the circle.
Because the differences
for
the users and repeaters in
energy
and height,
they have
different covering distances. We calculate the
distances with the theory of space loss. The
formula
[6]
is
L
M
?
88.1<
/p>
?
20
lg
F<
/p>
?
20
lg
h<
/p>
1
h
2
?
40
lg
d
L
M
F
,
the wireless loss
the communication working
frequency(MHz)
h
1
h
2
the
height of the repeater (m)
the height of the user(m)
d
Team #12110
the distance between the
user and repeater(km)
F
?
150
M
H
Z
,
h
1
?
1.5
m
Page 3
of 9
We assume that
and
h
2
?
3
0
m
,
under the condition of the cable loss and antenna
gain, we obtain the system gain
SG
?
P
t
< br>i
?
P
A
?
(
R
A
?
C
L
?
R
R
j
)
(
i
?
1,
2
j
?
1,
2)
.
The system gain is the allowed
decay maximum of the signal from the users to
repeaters. If the system
gain
value
is
higher
than
the
wireless
loss,
the
users
could
communicate
with
each
other.
Reversely,
the
users could not communicate. We make
the system gain value equals to the wireless loss,
thus, we get the
extremity distance
between the user and repeater. Then we have
SG
?
L
M
We choose a typical repeater and
the user facility. Thus, the
parameters
[6]
and data of
the repeaters are as
follows
The transmitting power
P
t
1
?
20
W
(43
dB
m
)
R
R
1
?
?
116
dB
m
The receiving
sensitivity
The antenna gain of the
repeaters
R
A
The
cable loss
C
L
?
2
dB
The
parameters of the interphone
The transmitting power
?
9.8
dB
P
t
2
?
4
W
(36
dB
m
)
The receiving sensitivity
R
R
2
?
?
116
dB
m
SG
1
?
144.2
dB
The antenna gain of the interphone
< br>P
A
?
0
dB
The
system
gain
of
the
system
from
users
to
repeaters
distance from the users to repeaters
system
from
the
repeaters
to
users
d
2
?
20.7
km
.
Thus,
we
get
the
sending
d
1
?
13.8
km
. Prove in the same way, we have the
system gain of the
,
the
sending
distance
from
the
repeaters
to
the
users
S
G
2
?
1
5
1.2
d
B
?
1
3.8
km
According
to
the
sending
distance
D
1
between
user
and
repeater
as
well
as
the
property
of <
/p>
D
2
?
23.0
9
km
regular hexagon, we
calculate the distance between two repeaters. We
obtain that
described in Figure 3.
Because
D
2
, which
is
is shorter than
D
2
'
,
users in this area cannot communicate with each
other.
Thus,
we
consider
the
sending
distance
D
2
'
between
two
repeaters
firstly.
Then
we
calculate
the
distance between the
user and the repeater again shown in Figure 4.
Finally, we get that
D
1
'
?
1
2.4
km
.
Team #12110
Page 4 of 9
D2=2
3.09km
D1=13.8km
Figure 3. the
calculation distance according to the sending
distance from users to repeaters.
D2'
=21.45km
D1'=12.4km
Figure 4. the calculation distance
according to the sending distance from repeaters
to the users.
According to the
calculated distance
D
1
'
?
12.4
km
D
2
'
?
21.45
km
, we know that
the given circle has
a
radius
of
40
miles.
We
firstly
consider
the
signals
’
covered
area
of
the
repeaters.
Thus,
we
get
the
distribution of the
repeater stations in this area showed in Figure 5.
The number of repeater stations is 37.
However, we need to decide the amount
of repeaters distributing in one station.