Instructor: T. Kunz
1) Spectrum Allocations and Frequency Reuse (10 marks)
a) One of the major advantages
(I claimed) of the ISM band at 2.4 GHz is its global availability. Discuss
whether it is absolutely necessary to achieve spectrum harmonization (i.e.,
assign the same frequency bands everywhere) on a global scale. Even if spectrum
harmonization were not to be absolutely necessary, what are its advantages?
Answer (5 marks):
If we attempt to address the harmonization issue in
a pure technical sense, then a very simple answer presents itself due to the
fact that RF waves degrade in power as it travels. There will only be signal interference on the
boarders. This problem can be addressed
by mutual agreement on frequency usage on the borderline areas of the adjacent
countries. In this case the answer is
spectrum harmonization is not required.
However, to properly answer this question, we have
to consider all the issues associated with spectrum harmonization. This can only be achieved by looking at the
advantages and disadvantage of such harmonization.
There are a number of reasons where spectrum
harmonization would be advantageous:
1. It addresses the problem of
frequency interference at country boundaries, as radio waves transcend national
borders.
2. Standardization of radio
equipment (receivers an transmitters) per
application. Basically, the same
equipment may be used globally for the same purpose (cell phones).
3. Providing common services
using common equipment.
4. Facilitating global
satellite broadcasting.
5. Reduced research cost where
countries may share the cost of devising new technologies to address certain
global needs.
On the other hand, spectrum harmonization may be
disadvantageous due to the following reasons:
1. Increased complexity in
spectrum allocation (global agreements).
2. Increased cost of multiple
administrations.
3. Increased enforcement and
monitoring costs.
4. Conflicting interests in
various countries.
5. Varied geographic spectrum
requirement.
6. Sovereignty issues (being
forced to allocate frequencies by majority votes).
7. Equipment becomes a
commodity.
8. No distinction of offered
regional services (same everywhere).
Based on the information above it seems that spectrum harmonization is very
beneficial, but not absolutely necessary.
A clear case where total spectrum harmonization may not be necessary is
where a certain services may be essential in one region, these services are not
applicable to others and thus the spectrum may be put to better use for other
type of services.
On a final note, although spectrum harmonization is
not absolutely necessary; there is a clear need to harmonize a sub-set of the
spectrum where certain services are applicable globally,
this would include cellular phone systems and satellite systems.
b) Assume a regular cell layout
(i.e., cells are all identical in size and hexagonal in shape, as used in the
course notes). We discussed the idea of frequency reuse and the notion of reuse
patterns in class in the context of AMPS. The basic idea is that you would not
like to use the same set of frequencies in two adjacent cells, due to
interference that could cause. AMPS uses a reuse pattern of 7, I also mentioned
that another typical reuse pattern is 3 cells (i.e., each cell gets to use a
third of the spectrum). Would it make sense to base frequency reuse on groups
of 2, 4, 5, or 6 cells? Explain your answer.
Answer (5 marks):
The rule that correlates re-use distance and the
re-use pattern is D = R * sqrt ( 3 * N ), where D is
re-use distance, R is cell radius, and N is re-use pattern. Based on this it would seem possible to use
any re-use pattern number; however,
that is not the case. If we assume a
hexagonal cell shape, the re-use pattern must guarantee that we can lay down
the cell group and re-use the same cell group layout while maintaining the
approximate re-use distance for each cell.
Given that, and from the re-use pattern layouts detailed below, we can
see that:
(2) A re-use pattern of 2 is not valid
(frequency clash where the same frequencies would be used in adjacent cells).
(4) A re-use pattern of 4 is valid. There is no frequency overlap and the computed re-use distance is slightly
more than a re-use pattern of 3. Due to
the hexagonal shape, the actual
re-use distance varies and at a minimum is the same as that of 3.
(5) Re-use pattern of 5 is valid. There is no
frequency overlap and the computed
re-use distance is slightly more than a re-use pattern of 3. Due to the hexagonal cell shape and non-circular layout, the actual re-use distance is not the same
for every cell, and at a minimum, it is exactly the same as that of 3.
(6) Re-use patterns of 6 is
valid, however, the re-use distance and re-use pattern layout is only
slightly better than that of 3 due to the non
circular-like (and not fully
enclosed) shape produced.
In conclusion:
· A re-use pattern of 2 is not valid
·
Re-use patterns of 4, 5, and 6
are valid but offer no advantage
over patterns of 3 or 7.
2) Media Access Control (10 marks)
CDPD and
GPRS (among others) employ a random access strategy. In essence, a node that
wants to transmit data, competing for access to a shared channel with other
nodes. As a result, collusions can occur. For a cellular network, could you
envision other approaches for packet data transmission over a shared channel
that would
· Reduce/minimize/do away with collisions
· Allow an arbitrary (or at least large number)
of nodes in a cell to transmit some data
· Does not pre-allocate resources to each node
(i.e., nodes will only access the channel if and when they have data to send
Are there
any disadvantages to your proposal?
Answer (10 marks):
Essentially
there is a whole different group of MAC protocols that are based on a
reservation scheme. The nice thing about the cellular environment is that there
is a well-identified entity to manage the schedule: the base station. In
essence, this is what Bluetooth does in a more dynamic way in each PicoNet, where the Master controls the operation of all
slaves, or IEEE 802.11s PCF mode, where the Access Point polls the devices.
Since stations can only transmit when they are told (or have otherwise acquired
permission to access the channel) and other stations will NOT transmit at the
same time, collisions are avoided. As every computer network textbook will
show, reservation-based approaches are superior to random access approaches
under heavy load.
One of the
biggest issues with reservation-based approaches is how to let the base station
know that you have data to send and therefore would like to be included in the
schedule. If it is done in a way similar to IEEE 802.11 PCF or token ring
networks, each station that is known to be in the cell (part of the location
management) is given the right to transmit or at the least the base station
periodically inquires whether a given station has data to send. This can be
wasteful if not everyone has data to send all the time. If on the other hand
the station sends a message to the base station every time it wants to be
included in the schedule (i.e., it now has data to send), there is a
chicken-and-egg problem: how does the station do this without having been given
a reserved slot before? In essence, to start a new communication burst,
stations will have to use a separate random access control channel, similar to
GSM, to become part of the schedule. Once a station is part of the schedule,
changes to its schedules can probably be communicated together with the data
(for example, the flags in CDPD to indicate that there is more data to follow).
However, now there is a (smaller) collision problem on the random access
channel.
3) CDMA (20 marks)
a) You are to assign 4 DS-CDMA
senders a chip sequence. Below, three potential sets of 4 chip sequences are
given. Which of these four sets would you choose, and why?
(i) -1 +1 +1 -1 +1 +1 +1 -1 -1 -1 -1 -1 -1 +1 +1
+1
+1 +1 +1
+1 -1 -1 -1 -1 -1 +1 +1 -1 -1 +1 +1 -1
+1 +1 +1
-1 -1 +1 +1 +1 +1 +1 -1 -1 -1 +1 -1 +1
+1 -1 +1
-1 -1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1
(ii) +1 +1 +1 -1
+1 +1 -1 -1 -1 -1 -1 -1 +1 +1 +1
+1 +1 +1
+1 -1 -1 -1 -1 -1 +1 +1 -1 -1 +1 +1 -1
+1 +1 +1
-1 -1 +1 +1 +1 -1 -1 -1 -1 -1 -1
+1 -1 +1
-1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1
(iii) -1
+1 +1 -1 +1 +1 +1 -1 -1 -1 -1 -1 -1 +1 +1 +1
+1 +1 +1
+1 -1 -1 -1 -1 -1 +1 +1 -1 -1 +1 +1 -1
+1 +1 +1
-1 -1 +1 +1 +1 +1 +1 -1 -1 -1 -1 -1 -1
+1 -1 +1
-1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1 +1 -1
Answer (2 marks):
Chip sequences must be orthogonal (as many pairs are
the same as are different), that is the inner product of each pair of chip
sequences is 0. Based on this:
(I)
Not valid since the last two chip sequences are not orthogonal
(II)
Not valid since chip sequences are not of the same length
(III)
Valid, all pairwise inner products are 0.
b) If a 5th sender
were to join this group, could you assign it a chip sequence, based on your
answer to a)? If so, what would that chip sequence be? If not, what would you
have to do to accommodate the additional station?
Answer (3 marks):
The trick here is to find another chip sequence, Sb, whos inner product with the
rest of existing chip sequences is 0.
With 16 bits, there are many possibilities; however, it is still very
difficult to find one manually. A
program may be written to do this; however, by trial and error, the following
chip sequence was found to be a valid 5th sequence:
Let Sb be the new chip sequence, where
The results of the inner products with the existing
chip sequences are:
S1.Sb:
+1 +1 1 +1 1 1 +1 1 1 +1 1 1 +1 +1 +1 1 = 0 / 16 = 0
S2.Sb:
-1 +1 1 1 +1 +1 1 1 1 1 +1 1 +1 +1 +1 +1 = 0 / 16 = 0
S3.Sb:
-1 +1 1 +1 +1 1 +1 +1 +1 1 1 1 +1 1 1 +1 = 0 / 16 = 0
S4.Sb:
-1 1 1 +1 1 +1 +1 1 +1 +1 +1 1 1 1 +1 +1 = 0 / 16 = 0
c) Assume a receiver receives
the signal:
(+1 1 +1 3 +1 +1 +3 +1 +3 1 1 1 +1 3 1 1).
Which station send what
data bit?
Answer (5 marks):
Let Sc be
the received signal, then
S1.Sc: -1 1 +1 +3 +1 +1 +3 1
3 +1 +1 +1 1 3 1 1
= 0 / 16 = 0
S2.Sc: +1 1 +1 3 1 1 3 1
3 1 1 +1 1 3 1 +1
= -16 / 16 = -1
S3.Sc: +1 1 +1 +3 1 +1 +3 +1
+3 1 +1 +1 1 +3 +1 +1
= 16 / 16 = 1
S4.Sc: +1 +1 +1 +3 +1 1 +3 1
+3 +1 1 +1 +1 +3 1 +1
= 16 / 16 = 1
The results show that:
·
S1 sent nothing
·
S3 sent 1
·
S4 sent 1
d) Assume a receiver receives
the signal
(+1 +2 +2 0 0 1 0 2 2 0 0 1 2 +1 +2 0).
Which stations did send what
data bit?
Answer (5 marks):
Let Sd
be the received signal, then
S1.Sd:
-1 +2 +2 +0 +0 1 +0 +2 +2 +0 +0 +1 +2 +1 +2 +0 = 12 / 16
S2.Sd:
+1 +2 +2 +0 +0 +1 +0 +2 +2 +0 +0 +1 +2 +1 +2 +0
= 16 / 16
S3.Sd:
+1 +2 +2 +0 +0 1 +0 2 2 +0 +0 +1 +2 1 2 +0 = 0 / 16
S4.Sd:
+1 2 +2 +0 +0 +1 +0 +2 2 +0 +0 +1 2 1 +2 +0 = 2 / 16
The results are not conclusive! It would seem that there must have been some
interference (data loss/noise) in the signal.
There are two ways to handle this kind of situation:
1. Pessimistic approach: The
packet is invalid and must be re-sent
2. Optimistic approach: Adopt
error tolerance and rely on the levels above to determine if there really was
an error in the transmission.
Using the following assumptions regarding the inner product result:
·
Results of 16 /
·
Results of 6 /
· Results of +7 / 16 to +16 / 16 is considered as +1
We thus end up with the following esults:
· S1 sent 1
· S2 sent 1
· S3 sent nothing
· S4 sent nothing
e) In both frequency and time
division multiplexing, there is a need for guard space to make sure that
adjacent users do not interfere with each other. In frequency division
multiple access, this is done by not using small guard bands separating the
individual channels. In time division multiple access, guard bands exist in the
time domain (i.e., between one sender stopping transmission and the next sender
starting up, there is a brief pause). Does DS-CDMA require guards? If so, where
are they? If not, why not?
Answer (5 marks):
Indeed DS-CDMA requires guards, i.e., separating the
communication channels in the appropriate space. The guard bands in DS-CDMA are
the orthogonality requirement on the spreading codes.
To return to the language analogy (and to use a different
example from the textbook), if one pair of speakers were using German and the
second set the Swiss dialect of German, it would be difficult to keep them
apart. So the codes used by each speaker have to be sufficiently
different.
4) Location Management (10 marks)
Both CDPD
and GSM deal with mobile devices that move between cells. Briefly outline the
similarities and the differences between the two approaches? Do they really
solve the same problem?
Answer (10 marks):
The two approaches both deal with devices that move and have to be tracked within a cellular system. However, since CDPD provides connectionless packet data services, whereas GSM provides connection-oriented voice and data services, there are substantial differences between the two approaches.
In CDPD, a packet destined to a mobile device can
arrive at the network at any given moment. Therefore, a CDPD network keeps
close tabs on the location of each mobile, not just at the level of the cell,
but which shared channel in the cell a device is currently listening to.
In GSM, the device can either have an ongoing
connection or not. In the latter case, there is no need for GSM to keep tight
tabs on the devices location: before any data has to be delivered, the GSM
network will have to set up a connection. So in the absence of active
connections, GSM tracks devices only very loosely, at the level of a group of
cells, called location area. If a connection needs to be established, a
device is paged in multiple cells at the same time, once it replies to the
page, its exact location is also discovered.
However, if the device has ongoing connections, the
network has to be actively involved in handing over the device from cell to
cell, migrating these connections with the user
movement.